UFM Statistics

Year 13 course of Further Statistics

Discrete random variables, including joint distributions and covariance

Showing 1-1 of 1 problems
1996 Paper 1 Q13
D: 1500.0 B: 1527.6

I have a Penny Black stamp which I want to sell to my friend Jim, but we cannot agree a price. So I put the stamp under one of two cups, jumble them up, and let Jim guess which one it is under. If he guesses correctly, I add a third cup, jumble them up, and let Jim guess correctly, adding another cup each time. The price he pays for the stamp is \(\pounds N,\) where \(N\) is the number of cups present when Jim fails to guess correctly. Find \(\mathrm{P}(N=k)\). Show that \(\mathrm{E}(N)=\mathrm{e}\) and calculate \(\mathrm{Var}(N).\)

Show Solution
\begin{align*} && \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\ &&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\ &&&= \frac{k-1}{k!} \\ &&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\ && \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\ && \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\ \Rightarrow && \textrm{Var}(N) &= 3e-e^2 \end{align*}

Showing 1-24 of 24 problems
2025 Paper 2 Q12
D: 1500.0 B: 1500.0

Let \(X\) be a Poisson random variable with mean \(\lambda\) and let \(p_r = P(X = r)\), for \(r = 0, 1, 2, \ldots\). Neither \(\lambda\) nor \(\lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}}\) is an integer.

  1. Show, by considering the sequence \(d_r \equiv p_r - p_{r-1}\) for \(r = 1, 2, \ldots\), that there is a unique integer \(m\) such that \(P(X = r) \leq P(X = m)\) for all \(r = 0, 1, 2, \ldots\), and that \[\lambda - 1 < m < \lambda.\]
  2. Show that the minimum value of \(d_r\) occurs at \(r = k\), where \(k\) is such that \[k < \lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}} < k + 1.\]
  3. Show that the condition for the maximum value of \(d_r\) to occur at \(r = 1\) is \[1 < \lambda < 2 + \sqrt{2}.\]
  4. In the case \(\lambda = 3.36\), sketch a graph of \(p_r\) against \(r\) for \(r = 0, 1, 2, \ldots, 6, 7\).

Show Solution
  1. Suppose \(d_r = p_r - p_{r-1}\) then \begin{align*} d_r &= p_r - p_{r-1} \\ &= \mathbb{P}(X = r) - \mathbb{P}(X = r-1) \\ &= e^{-\lambda} \left ( \frac{\lambda^r}{r!} - \frac{\lambda^{r-1}}{(r-1)!} \right) \\ &= e^{-\lambda} \frac{\lambda^{r-1}}{(r-1)!} \left ( \frac{\lambda}{r} - 1\right) \end{align*} Therefore \(d_r > 0 \Leftrightarrow \lambda > r\)ie, \(p_r\) is increasing while \(r < \lambda\) and reaches a (unique) maximum when \(r = \lfloor \lambda \rfloor\).
  2. Let \(dd_r = d_r - d_{r-1}\), so: \begin{align*} dd_r &= d_r - d_{r-1} \\ &= p_r - 2p_{r-1} + p_{r-2} \\ &= e^{-\lambda} \frac{\lambda^{r-2}}{r!} \left ( \lambda^2 - 2 \lambda r + r(r-1)\right ) \end{align*} Therefore \(dd_r < 0 \Leftrightarrow \lambda^2 - 2\lambda r +r(r-1) < 0 \Leftrightarrow r^2 -(1+2\lambda)r + \lambda^2 < 0\), but this has roots \(r = \frac{(1+2\lambda) \pm \sqrt{(1+2\lambda)^2-4\lambda^2}}{2} = \lambda + \frac12 \pm \sqrt{\lambda + \frac14}\). Therefore \(d_r\) is decreasing when \(r \in \left (\lambda + \frac12 -\sqrt{\lambda + \frac14},\lambda + \frac12 + \sqrt{\lambda + \frac14} \right)\), therefore the possible minimums are \(d_1\) and \(d_k\) where \(k < \lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}} < k + 1\). \(d_1 = e^{-\lambda}(\lambda - 1)\), \(d_k = e^{-\lambda} \frac{\lambda^{k-1}}{(k-1)!}(\frac{\lambda}{k}-1)\)
  3. If the maximum value of \(d_r\) is \(r = 1\) then \(d_r\) must be decreasing, ie considering \(dd_2\) we have \(\lambda^2 -4\lambda + 2< 0 \Leftrightarrow 2 - \sqrt{2} < \lambda < 2 + \sqrt{2}\). It must also be the case that it doesn't get beaten as \(\lambda \to \infty\). In this case \(d_r \to 0\), so we need \(d_1 > 0\), ie \(\lambda > 1\). Therefore \(1 < \lambda < 2 + \sqrt{2}\)
  4. TikZ diagram
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2019 Paper 3 Q11
D: 1500.0 B: 1500.0

The number of customers arriving at a builders' merchants each day follows a Poisson distribution with mean \(\lambda\). Each customer is offered some free sand. The probability of any given customer taking the free sand is \(p\).

  1. Show that the number of customers each day who take sand follows a Poisson distribution with mean \(p\lambda\).
  2. The merchant has a mass \(S\) of sand at the beginning of the day. Each customer who takes the free sand gets a proportion \(k\) of the remaining sand, where \(0 \leq k < 1\). Show that by the end of the day the expected mass of sand taken is $$\left(1 - e^{-kp\lambda}\right)S.$$
  3. At the beginning of the day, the merchant's bag of sand contains a large number of grains, exactly one of which is made from solid gold. At the end of the day, the merchant's assistant takes a proportion \(k\) of the remaining sand. Find the probability that the assistant takes the golden grain. Comment on the case \(k = 0\) and on the limit \(k \to 1\). In the case \(p\lambda > 1\) find the value of \(k\) which maximises the probability that the assistant takes the golden grain.

2017 Paper 2 Q12
D: 1600.0 B: 1563.6

Adam and Eve are catching fish. The number of fish, \(X\), that Adam catches in any time interval is Poisson distributed with parameter \(\lambda t\), where \(\lambda\) is a constant and \(t\) is the length of the time interval. The number of fish, \(Y\), that Eve catches in any time interval is Poisson distributed with parameter \(\mu t\), where \(\mu\) is a constant and \(t\) is the length of the time interval The two Poisson variables are independent. You may assume that that expected time between Adam catching a fish and Adam catching his next fish is \(\lambda^{-1}\), and similarly for Eve.

  1. By considering \(\P( X + Y = r)\), show that the total number of fish caught by Adam and Eve in time \(T\) also has a Poisson distribution.
  2. Given that Adam and Eve catch a total of \(k\) fish in time \(T\), where \(k\) is fixed, show that the number caught by Adam has a binomial distribution.
  3. Given that Adam and Eve start fishing at the same time, find the probability that the first fish is caught by Adam.
  4. Find the expected time from the moment Adam and Eve start fishing until they have each caught at least one fish.
\noindent [{\bf Note } This question has been redrafted to make the meaning clearer.]

2016 Paper 3 Q12
D: 1700.0 B: 1516.0

Let \(X\) be a random variable with mean \(\mu\) and standard deviation \(\sigma\). Chebyshev's inequality, which you may use without proof, is \[ \P\left(\vert X-\mu\vert > k\sigma\right) \le \frac 1 {k^2} \,, \] where \(k\) is any positive number.

  1. The probability of a biased coin landing heads up is \(0.2\). It is thrown \(100n\) times, where \(n\) is an integer greater than 1. Let \(\alpha \) be the probability that the coin lands heads up \(N\) times, where \(16n \le N \le 24n\). Use Chebyshev's inequality to show that \[ \alpha \ge 1-\frac 1n \,. \]
  2. Use Chebyshev's inequality to show that \[ 1+ n + \frac{n^2}{ 2!} + \cdots + \frac {n^{2n}}{(2n)!} \ge \left(1-\frac1n\right) \e^n \,. \]

Show Solution
  1. Let \(N\) be the number of times the coin lands heads up, ie \(N \sim Binomial(100n, 0.2)\), then \(\mathbb{E}(N) = \mu = 20n, \mathrm{Var}(N) = \sigma^2 = 100n \cdot 0.2 \cdot 0.8 = 16n \Rightarrow \sigma = 4\sqrt{n}\). \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - \mu| \leq k\sigma) &\leq \frac1{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - 20n| \leq \sqrt{n} \cdot 4\sqrt{n}) &\leq \frac1{{\sqrt{n}}^2} \\ \Rightarrow && 1 - \mathbb{P}(16n \leq N \leq 24n) &\leq \frac{1}{n} \\ \Rightarrow && 1 - \frac1n &\leq \alpha \end{align*}
  2. Suppose \(X \sim Pois(n)\), then \(\mathbb{E}(X) = n, \mathrm{Var}(X) = n\). Therefore \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1-\mathbb{P}(|X - n| \leq \sqrt{n} \cdot \sqrt{n}) &> \frac{1}{\sqrt{n}^2} \\ \Rightarrow && 1 - \sum_{i=0}^{2n} \mathbb{P}(X = i) & \leq \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} e^{-n} \frac{n^i}{i!} \geq 1 - \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} \frac{n^i}{i!} \geq \left ( 1 - \frac1n \right)e^n \end{align*}
2015 Paper 1 Q12
D: 1500.0 B: 1461.6

The number \(X\) of casualties arriving at a hospital each day follows a Poisson distribution with mean 8; that is, \[ \P(X=n) = \frac{ \e^{-8}8^n}{n!}\,, \ \ \ \ n=0, \ 1, \ 2, \ \ldots \ . \] Casualties require surgery with probability \(\frac14\). The number of casualties arriving on any given day is independent of the number arriving on any other day and the casualties require surgery independently of one another.

  1. What is the probability that, on a day when exactly \(n\) casualties arrive, exactly \(r\) of them require surgery?
  2. Prove (algebraically) that the number requiring surgery each day also follows a Poisson distribution, and state its mean.
  3. Given that in a particular randomly chosen week a total of 12 casualties require surgery on Monday and Tuesday, what is the probability that 8 casualties require surgery on Monday? You should give your answer as a fraction in its lowest terms.

2013 Paper 2 Q12
D: 1600.0 B: 1484.0

The random variable \(U\) has a Poisson distribution with parameter \(\lambda\). The random variables \(X\) and \(Y\) are defined as follows. \begin{align*} X&= \begin{cases} U & \text{ if \(U\) is 1, 3, 5, 7, \(\ldots\,\)} \\ 0 & \text{ otherwise} \end{cases} \\ Y&= \begin{cases} U & \text{ if \(U\) is 2, 4, 6, 8, \(\ldots\,\) } \\ 0 & \text{ otherwise} \end{cases} \end{align*}

  1. Find \(\E(X)\) and \(\E(Y)\) in terms of \(\lambda\), \(\alpha\) and \(\beta\), where \[ \alpha = 1+\frac{\lambda^2}{2!}+\frac{\lambda^4}{4!} +\cdots\, \text{ \ \ and \ \ } \beta = \frac{\lambda}{1!} + \frac{\lambda^3}{3!} + \frac{\lambda^5}{5!} +\cdots\,. \]
  2. Show that \[ \var(X) = \frac{\lambda\alpha+\lambda^2\beta}{\alpha+\beta} - \frac{\lambda^2\alpha^2}{(\alpha+\beta)^2} \] and obtain the corresponding expression for \(\var(Y)\). Are there any non-zero values of \(\lambda\) for which \( \var(X) + \var(Y) = \var(X+Y)\,\)?

Show Solution
  1. \begin{align*} \mathbb{E}(X) &= \sum_{r=1}^\infty r \mathbb{P}(X = r) \\ &= \sum_{j=1}^{\infty} (2j-1)\mathbb{P}(U=2j-1) \\ &= \sum_{j=1}^{\infty}(2j-1) \frac{e^{-\lambda} \lambda^{2j-1}}{(2j-1)!} \\ &= \sum_{j=1}^{\infty} e^{-\lambda} \frac{\lambda^{2j-1}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \sum_{j=1}^{\infty} \frac{\lambda^{2j-2}}{(2j-2)!} \\ &= \lambda e^{-\lambda} \alpha \end{align*} Since \(\mathbb{E}(X+Y) = \lambda, \mathbb{E}(Y) = \lambda(1-e^{-\lambda}\alpha) = \lambda(e^{-\lambda}(\alpha+\beta) - e^{-\lambda}\alpha) = \lambda e^{-\lambda} \beta\). Alternatively, as \(\beta + \alpha = e^{\lambda}\), \(\mathbb{E}(X) = \frac{\lambda \alpha}{\alpha+\beta}, \mathbb{E}(Y) = \frac{\lambda \beta}{\alpha+\beta}\)
  2. \begin{align*} \textrm{Var}(X) &= \mathbb{E}(X^2) - [\mathbb{E}(X) ]^2 \\ &= \sum_{odd} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(X) \right]^2 \\ &= \sum_{odd} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-2)!}+\sum_{odd} \frac{e^{-\lambda}\lambda^r}{(r-1)!} - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2 \beta + e^{-\lambda}\lambda \alpha - \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} \end{align*} Similarly, \begin{align*} \textrm{Var}(Y) &= \mathbb{E}(Y^2) - [\mathbb{E}(Y) ]^2 \\ &= \sum_{even} r^2 \mathbb{P}(U = r) - \left [ \mathbb{E}(Y) \right]^2 \\ &= \sum_{even} (r(r-1)+r)\frac{e^{-\lambda}\lambda^r}{r!} - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= e^{-\lambda}\lambda^2\alpha + e^{-\lambda}\lambda \beta - \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \end{align*} Since \(\textrm{Var}(X+Y) = \textrm{Var}(U) = \lambda\), we are interested in solving: \begin{align*} \lambda &= \frac{\lambda \alpha + \lambda^2 \beta}{\alpha+\beta}- \frac{\lambda^2 \alpha^2}{(\alpha+\beta)^2} + \frac{\lambda \beta + \lambda^2 \alpha}{\alpha+\beta}- \frac{\lambda^2 \beta^2}{(\alpha+\beta)^2} \\ &= \frac{\lambda(\alpha+\beta) + \lambda^2(\alpha+\beta)}{\alpha+\beta} - \frac{\lambda^2(\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{(\alpha+\beta)^2 - (\alpha^2+\beta^2)}{(\alpha+\beta)^2} \\ &= \lambda + \lambda^2 \frac{2\alpha\beta}{(\alpha+\beta)^2} \end{align*} which is clearly not possible if \(\lambda \neq 0\)
2012 Paper 2 Q13
D: 1600.0 B: 1516.0

In this question, you may assume that \(\displaystyle \int_0^\infty \!\!\! \e^{-x^2/2} \d x = \sqrt{\tfrac12 \pi}\,\). The number of supermarkets situated in any given region can be modelled by a Poisson random variable, where the mean is \(k\) times the area of the given region. Find the probability that there are no supermarkets within a circle of radius \(y\). The random variable \(Y\) denotes the distance between a randomly chosen point in the region and the nearest supermarket. Write down \(\P(Y < y)\) and hence show that the probability density function of \(Y\) is \(\displaystyle 2\pi y k \e^{-\pi k y^2}\) for \(y\ge0\). Find \(\E(Y)\) and show that \(\var(Y) = \dfrac{4-\pi}{4\pi k}\).

Show Solution
A circle radius \(y\) has a number of supermarkets \(X\) where \(X \sim Po(k \pi y^2)\). \[ \mathbb{P}(X = 0) = e^{-k\pi y^2} \frac{1}{0!} = e^{-k\pi y^2} \] The probability \(\mathbb{P}(Y < y) = 1-\mathbb{P}(Y \geq y) = 1-e^{-k\pi y^2}\), and in particular \(f_Y(y) = 2k\pi y e^{-k\pi y^2}\) (by differentiating). \begin{align*} && \mathbb{E}(Y) &= \int_0^\infty yf_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^2 k e^{-\pi k y^2} \d y \\ \sigma^2 = \frac{1}{2k\pi}:&&&= \pi k \sqrt{2 \pi}\sigma \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi} \sigma }y^2 e^{-\frac12 \cdot 2\pi k y^2} \d y \\ &&&=\pi k \sqrt{2 \pi}\sigma \mathbb{E}\left (N(0, \sigma^2)^2 \right) \\ &&&= \pi k \sqrt{2 \pi}\sigma\sigma^2 \\ &&&= \pi k \sqrt{2 \pi} \frac{1}{(2k\pi)^{3/2}} \\ &&&= \frac{1}{2\sqrt{k}} \end{align*} \begin{align*} && \mathbb{E}(Y^2) &= \int_0^\infty y^2f_Y(y) \d y \\ &&&= \int_0^\infty 2\pi y^3 k e^{-\pi k y^2} \d y \\ &&&= \int_0^{\infty}y^2 2y \pi k e^{-\pi k y^2} \d y \\ \\ &&&= \left [-y^2 e^{-\pi k y^2}\right]_0^{\infty}+\int_0^\infty 2ye^{-\pi k y^2} \d y \\ &&&= \left [-\frac{1}{\pi k}e^{-\pi k y^2} \right]_0^{\infty} \\ &&&= \frac{1}{\pi k} \\ \Rightarrow && \textrm{Var}(Y) &= \mathbb{E}(Y^2) - \left [ \mathbb{E}(Y)\right]^2 \\ &&&= \frac{1}{\pi k} - \frac{1}{4k} \\ &&&= \frac{4 - \pi}{4\pi k} \end{align*}
2010 Paper 1 Q13
D: 1484.0 B: 1516.0

The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean \(\lambda\) texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is \(p\), show that \[ p\e^{2\lambda}-\e^{\lambda}+1=0. \] Given that \(4p<1\), show that there are two positive values of \(\lambda\) that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means \(\lambda_{1}\) and \(\lambda_{2}\) texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also \(p\), find an expression for \(\lambda_{1}+\lambda_{2}\) in terms of~\(p\). Find the probability, in terms of \(p\), that she waits between 1 and 2 hours in the morning to receive her first text.

2007 Paper 1 Q14
D: 1500.0 B: 1484.0

The discrete random variable \(X\) has a Poisson distribution with mean \(\lambda\).

  1. Sketch the graph \(y=\l x+1 \r \e^{-x}\), stating the coordinates of the turning point and the points of intersection with the axes. It is known that \(\P(X \ge 2) = 1-p\), where \(p\) is a given number in the range \(0 < p <1\). Show that this information determines a unique value (which you should not attempt to find) of \(\lambda\).
  2. It is known (instead) that \(\P \l X = 1 \r = q\), where \(q\) is a given number in the range \(0 < q <1\). Show that this information determines a unique value of \(\lambda\) (which you should find) for exactly one value of \(q\) (which you should also find).
  3. It is known (instead) that \(\P \l X = 1 \, \vert \, X \le 2 \r = r\), where \(r\) is a given number in the range \(0

2006 Paper 1 Q13
D: 1484.0 B: 1468.0

A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.

  1. I go to the shop and roll a fair six-sided die once. I decide that if I roll a score of \(N\), I will buy \(100N\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6} }{ 1 - \e^{-0.1}} \r \] Show also that the expected number of diamonds I find is 0.35.
  2. Instead, I decide to roll a fair six-sided die repeatedly until I score a 6. If I roll my first 6 on my \(T\)th throw, I will buy \(100T\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}} \] Calculate also the expected number of diamonds that I find. (You may find it useful to consider the the binomial expansion of \(\l 1 - x \r^{-2}\).)

2005 Paper 2 Q13
D: 1600.0 B: 1500.0

The number of printing errors on any page of a large book of \(N\) pages is modelled by a Poisson variate with parameter \(\lambda\) and is statistically independent of the number of printing errors on any other page. The number of pages in a random sample of \(n\) pages (where \(n\) is much smaller than \(N\) and \(n\ge2\)) which contain fewer than two errors is denoted by \(Y\). Show that \(\P(Y=k) = \binom n k p^kq^{n-k}\) where \(p=(1+\lambda)e^{-\lambda}\) and \(q=1-p\,\). Show also that, if \(\lambda\) is sufficiently small,

  1. \(q\approx \frac12 \lambda^2\,\);
  2. the largest value of \(n\) for which \(\P(Y=n)\ge 1-\lambda\) is approximately \(2/\lambda\,\);
  3. $ \P(Y>1 \;\vert\; Y>0) \approx 1-n(\lambda^2/2)^{n-1}\;.$

2003 Paper 3 Q12
D: 1700.0 B: 1470.9

Brief interruptions to my work occur on average every ten minutes and the number of interruptions in any given time period has a Poisson distribution. Given that an interruption has just occurred, find the probability that I will have less than \(t\) minutes to work before the next interruption. If the random variable \(T\) is the time I have to work before the next interruption, find the probability density function of \(T\,\). I need an uninterrupted half hour to finish an important paper. Show that the expected number of interruptions before my first uninterrupted period of half an hour or more is \(\e^3-1\). Find also the expected length of time between interruptions that are less than half an hour apart. Hence write down the expected wait before my first uninterrupted period of half an hour or more.

2003 Paper 1 Q14
D: 1500.0 B: 1475.2

Jane goes out with any of her friends who call, except that she never goes out with more than two friends in a day. The number of her friends who call on a given day follows a Poisson distribution with parameter \(2\). Show that the average number of friends she sees in a day is~\(2-4\e^{-2}\,\). Now Jane has a new friend who calls on any given day with probability \(p\). Her old friends call as before, independently of the new friend. She never goes out with more than two friends in a day. Find the average number of friends she now sees in a day.

2001 Paper 1 Q13
D: 1500.0 B: 1500.0

Four students, one of whom is a mathematician, take turns at washing up over a long period of time. The number of plates broken by any student in this time obeys a Poisson distribution, the probability of any given student breaking \(n\) plates being \(\e^{-\lambda} \lambda^n/n!\) for some fixed constant \(\lambda\), independent of the number of breakages by other students. Given that five plates are broken, find the probability that three or more were broken by the mathematician.

Show Solution
Let \(X\) be the number of plates broken by the mathematician and \(Y\) by the other student. Then \(X \sim Po(\lambda), Y \sim Po(3\lambda)\) and \(X+Y \sim Po(4\lambda)\) \begin{align*} && \mathbb{P}(X = k | X+Y = n) &= \frac{\mathbb{P}(X = k, Y = n-k)}{\mathbb{P}(X+Y=n)} \\ &&&= \frac{e^{-\lambda} \lambda^k/k! \cdot e^{-3\lambda} (4\lambda)^{n-k}/(n-k)!}{e^{-4\lambda}(4\lambda)^n/n!} \\ &&&= \binom{n}{k} \left ( \frac{1}{4} \right)^k \left ( \frac{3}{5} \right)^{n-k} \end{align*} Therefore \(X | X+Y = n \sim Binomial(n, \tfrac14)\) \begin{align*} \mathbb{P}(X \geq 3 | X + Y = n) &= \binom{5}{3} \frac{3^2}{4^5} + \binom{5}{4} \frac{3}{4^5} + \binom{5}{5} \frac{1}{4^5} \\ &= \frac{1}{4^5} \left ( 90+ 15 + 1 \right) \\ &= \frac{106}{4^5} = \frac{53}{512} \approx \frac1{10} \end{align*}
2000 Paper 3 Q12
D: 1700.0 B: 1553.7

In a lottery, any one of \(N\) numbers, where \(N\) is large, is chosen at random and independently for each player by machine. Each week there are \(2N\) players and one winning number is drawn. Write down an exact expression for the probability that there are three or fewer winners in a week, given that you hold a winning ticket that week. Using the fact that $$ {\biggl( 1 - {a \over n} \biggr) ^n \approx \e^{-a}}$$ for \(n\) much larger than \(a\), or otherwise, show that this probability is approximately \({2 \over 3}\) . Discuss briefly whether this probability would increase or decrease if the numbers were chosen by the players. Show that the expected number of winners in a week, given that you hold a winning ticket that week, is \( 3-N^{-1}\).

1999 Paper 3 Q14
D: 1700.0 B: 1487.9

In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$

1998 Paper 2 Q14
D: 1600.0 B: 1500.0

The staff of Catastrophe College are paid a salary of \(A\) pounds per year. With a Teaching Assessment Exercise impending it is decided to try to lower the student failure rate by offering each lecturer an alternative salary of \(B/(1+X)\) pounds, where \(X\) is the number of his or her students who fail the end of year examination. Dr Doom has \(N\) students, each with independent probability \(p\) of failure. Show that she should accept the new salary scheme if $$A(N+1)p < B(1-(1-p)^{N+1}).$$ Under what circumstances could \(X\), for Dr Doom, be modelled by a Poisson random variable? What would Dr Doom's expected salary be under this model?

1997 Paper 2 Q14
D: 1600.0 B: 1469.6

Traffic enters a tunnel which is 9600 metres long, and in which overtaking is impossible. The number of vehicles which enter in any given time is governed by the Poisson distribution with mean 6 cars per minute. All vehicles travel at a constant speed until forced to slow down on catching up with a slower vehicle ahead. I enter the tunnel travelling at 30 m\(\,\)s\(^{-1}\) and all the other traffic is travelling at 32 m\(\,\)s\(^{-1}\). What is the expected number of vehicles in the queue behind me when I leave the tunnel? Assuming again that I travel at 30 m\(\,\)s\(^{-1}\), but that all the other vehicles are independently equally likely to be travelling at 30 m\(\,\)s\(^{-1}\) or 32 m\(\,\)s\(^{-1}\), find the probability that exactly two vehicles enter the tunnel within 20 seconds of my doing so and catch me up before I leave it. Find also the probability that there are exactly two vehicles queuing behind me when I leave the tunnel. \noindent [Ignore the lengths of the vehicles.]

1996 Paper 3 Q14
D: 1700.0 B: 1484.0

Whenever I go cycling I start with my bike in good working order. However if all is well at time \(t\), the probability that I get a puncture in the small interval \((t,t+\delta t)\) is \(\alpha\,\delta t.\) How many punctures can I expect to get on a journey during which my total cycling time is \(T\)? When I get a puncture I stop immediately to repair it and the probability that, if I am repairing it at time \(t\), the repair will be completed in time \((t,t+\delta t)\) is \(\beta\,\delta t.\) If \(p(t)\) is the probability that I am repairing a puncture at time \(t\), write down an equation relating \(p(t)\) to \(p(t+\delta t)\), and derive from this a differential equation relating \(p'(t)\) and \(p(t).\) Show that \[ p(t)=\frac{\alpha}{\alpha+\beta}(1-\mathrm{e}^{-(\alpha+\beta)t}) \] satisfies this differential equation with the appropriate initial condition. Find an expression, involving \(\alpha,\beta\) and \(T\), for the time expected to be spent mending punctures during a journey of total time \(T\). Hence, or otherwise, show that, the fraction of the journey expected to be spent mending punctures is given approximately by \[ \quad\frac{\alpha T}{2}\quad\ \mbox{ if }(\alpha+\beta)T\text{ is small, } \] and by \[ \frac{\alpha}{\alpha+\beta}\quad\mbox{ if }(\alpha+\beta)T\text{ is large.} \]

1995 Paper 2 Q13
D: 1600.0 B: 1484.0

Fly By Night Airlines run jumbo jets which seat \(N\) passengers. From long experience they know that a very small proportion \(\epsilon\) of their passengers fail to turn up. They decide to sell \(N+k\) tickets for each flight. If \(k\) is very small compared with \(N\) explain why they might expect \[ \mathrm{P}(r\mbox{ passengers fail to turn up})=\frac{\lambda^{r}}{r!}\mathrm{e}^{-\lambda} \] approximately, with \(\lambda=N\epsilon.\) For the rest of the question you may assume that the formula holds exactly. Each ticket sold represents \(\pounds A\) profit, but the airline must pay each passenger that it cannot fly \(\pounds B\) where \(B>A>0.\) Explain why, if \(r\) passengers fail to turn up, its profit, in pounds, is \[ A(N+k)-B\max(0,k-r), \] where \(\max(0,k-r)\) is the larger of \(0\) and \(k-r.\) Write down the expected profit \(u_{k}\) when \(k=0,1,2\) and \(3.\) Find \(v_{k}=u_{k+1}-u_{k}\) for general \(k\) and show that \(v_{k}>v_{k+1}.\) Show also that \[ v_{k}\rightarrow A-B \] as \(k\rightarrow\infty.\) Advise Fly By Night on how to choose \(k\) to maximise its expected profit \(u_{k}.\)

1995 Paper 1 Q13
D: 1500.0 B: 1484.0

A scientist is checking a sequence of microscope slides for cancerous cells, marking each cancerous cell that she detects with a red dye. The number of cancerous cells on a slide is random and has a Poisson distribution with mean \(\mu.\) The probability that the scientist spots any one cancerous cell is \(p\), and is independent of the probability that she spots any other one.

  1. Show that the number of cancerous cells which she marks on a single slide has a Poisson distribution of mean \(p\mu.\)
  2. Show that the probability \(Q\) that the second cancerous cell which she marks is on the \(k\)th slide is given by \[ Q=\mathrm{e}^{-\mu p(k-1)}\left\{ (1+k\mu p)(1-\mathrm{e}^{-\mu p})-\mu p\right\} . \]

1993 Paper 2 Q16
D: 1600.0 B: 1500.0

At the terminus of a bus route, passengers arrive at an average rate of 4 per minute according to a Poisson process. Each minute, on the minute, one bus arrives with probability \(\frac{1}{4},\) independently of the arrival of passengers or previous buses. Just after eight o'clock there is no-one at the bus stop.

  1. What is the probability that the first bus arrives at \(n\) minutes past 8?
  2. If the first bus arrives at 8:05, what is the probability that there are \(m\) people waiting for it?
  3. Each bus can take 25 people and, since it is the terminus, the bus arrive empty. Explain carefully how you would calculate, to two significant figures, the probability that when the first bus arrives it is unable to pick up all the passengers. Your method should need the use of a calculator and standard tables only. There is no need to carry out the calculation.

1992 Paper 3 Q16
D: 1700.0 B: 1484.0

The probability that there are exactly \(n\) misprints in an issue of a newspaper is \(\mathrm{e}^{-\lambda}\lambda^{n}/n!\) where \(\lambda\) is a positive constant. The probability that I spot a particular misprint is \(p\), independent of what happens for other misprints, and \(0 < p < 1.\)

  1. If there are exactly \(m+n\) misprints, what is the probability that I spot exactly \(m\) of them?
  2. Show that, if I spot exactly \(m\) misprints, the probability that I have failed to spot exactly \(n\) misprints is \[ \frac{(1-p)^{n}\lambda^{n}}{n!}\mathrm{e}^{-(1-p)\lambda}. \]

Show Solution
  1. \(\binom{m+n}{m} p^m (1-p)^n\)
  2. \(\,\) \begin{align*} \mathbb{P}(\text{failed to spot }n\text{ misprints}|\text{spotted }m\text{ misprints}) &= \frac{\mathbb{P}(\text{failed to spot }n\text{ misprints and spotted }m\text{ misprints}) }{\mathbb{P}(\text{spotted }m\text{ misprints})} \\ &= \frac{\binom{m+n}{n}p^m(1-p)^n e^{-\lambda} \lambda^{m+n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}p^m(1-p)^k e^{-\lambda} \lambda^{m+k}/(n+k)!} \\ &= \frac{\binom{m+n}{n}(1-p)^n \lambda^{n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}(1-p)^k \lambda^{k}/(n+k)!} \\ &= \frac{(1-p)^n \lambda^{n}/n!}{\sum_{k=0}^{\infty} (1-p)^k \lambda^{k}/k!} \\ &= \frac{(1-p)^n\lambda^n}{n!} e^{-(1-p)\lambda} \end{align*} Alternatively, given the missed misprints and spotted misprints are independent, we can view them as both following \(Po(p\lambda)\) and \(Po((1-p)\lambda)\) and so we obtain exactly this result, without calculation.
1989 Paper 1 Q15
D: 1500.0 B: 1516.0

I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5\(\,\)km, and I am held up at a level crossing for \(A\) minutes, where \(A\) is a continuous random variable uniformly distributed between \(0\) and 10. Via Bend Boulevard the distance is 4\(\,\)km, and I am delayed, by talking to each of \(B\) friends for 3\(\,\)minutes, for a total of \(3B\) minutes, where \(B\) is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2\(\,\)km, but in addition, due to never-ending road works, there are five places at each of which, with probability \(\frac{4}{5},\) I have to make a detour that increases the distance by 1\(\,\)km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12\(\,\)km\(\,\)h\(^{-1}\). For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard? Which route should I use all the time: \begin{questionparts} \item if I wish my average journey time to be as small as possible; \item if I wish my journey time to be less than 32 minutes as often as possible? \end{questionpart} Justify your answers.

Show Solution
\(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}
  1. \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
  2. \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).

Showing 1-1 of 1 problems
2003 Paper 2 Q13
D: 1600.0 B: 1469.5

The random variable \(X\) takes the values \(k=1\), \(2\), \(3\), \(\dotsc\), and has probability distribution $$ \P(X=k)= A{{{\lambda}^k\e^{-{\lambda}}} \over {k!}}\,, $$ where \(\lambda \) is a positive constant. Show that \(A = (1-\e^{-\lambda})^{-1}\,\). Find the mean \({\mu}\) in terms of \({\lambda}\) and show that $$ \var(X) = {\mu}(1-{\mu}+{\lambda})\;. $$ Deduce that \({\lambda} < {\mu} < 1+{\lambda}\,\). Use a normal approximation to find the value of \(P(X={\lambda})\) in the case where \({\lambda}=100\,\), giving your answer to 2 decimal places.

Showing 1-2 of 2 problems
2016 Paper 2 Q13
D: 1600.0 B: 1516.0

  1. The random variable \(X\) has a binomial distribution with parameters \(n\) and \(p\), where \(n=16\) and \(p=\frac12\). Show, using an approximation in terms of the standard normal density function $\displaystyle \tfrac{1}{\sqrt{2\pi}} \, \e ^{-\frac12 x^2} $, that \[ \P(X=8) \approx \frac 1{2\sqrt{2\pi}} \,. \]
  2. By considering a binomial distribution with parameters \(2n\) and \(\frac12\), show that \[ (2n)! \approx \frac {2^{2n} (n!)^2}{\sqrt{n\pi}} \,. \]
  3. By considering a Poisson distribution with parameter \(n\), show that \[ n! \approx \sqrt{2\pi n\, } \, \e^{-n} \, n^n \,. \]

Show Solution
  1. \(X \sim B(16, \tfrac12)\), then \(X \approx N(8, 2^2)\), in particular \begin{align*} && \mathbb{P}(X = 8) &\approx \mathbb{P} \left ( 8 - \frac12 \leq 2Z + 8 \leq 8 + \frac12 \right) \\ &&&= \mathbb{P} \left (-\frac14 \leq Z \leq \frac14 \right) \\ &&&= \int_{-\frac14}^{\frac14} \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2\pi}} \int_{-\frac14}^{\frac14} 1\d x\\ &&&= \frac{1}{2 \sqrt{2\pi}} \end{align*}
  2. Suppose \(X \sim B(2n, \frac12)\) then \(X \approx N(n, \frac{n}{2})\), and \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left ( n - \frac12 \leq \sqrt{\frac{n}{2}} Z + n \leq n + \frac12 \right) \\ &&&= \mathbb{P} \left ( - \frac1{\sqrt{2n}} \leq Z \leq \frac1{\sqrt{2n}}\right) \\ &&&= \int_{-\frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2 \pi}} e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{n\pi}}\\ \Rightarrow && \binom{2n}{n}\frac1{2^n} \frac{1}{2^n} & \approx \frac{1}{\sqrt{n \pi}} \\ \Rightarrow && (2n)! &\approx \frac{2^{2n}(n!)^2}{\sqrt{n\pi}} \end{align*}
  3. \(X \sim Po(n)\), then \(X \approx N(n, (\sqrt{n})^2)\), therefore \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left (-\frac12 \leq \sqrt{n} Z \leq \frac12 \right) \\ &&&= \int_{-\frac{1}{2 \sqrt{n}}}^{\frac{1}{2 \sqrt{n}}} \frac{1}{\sqrt{2\pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && e^{-n} \frac{n^n}{n!} & \approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && n! &\approx \sqrt{2 \pi n} e^{-n}n^n \end{align*}
2001 Paper 2 Q12
D: 1600.0 B: 1484.0

The national lottery of Ruritania is based on the positive integers from \(1\) to \(N\), where \(N\) is very large and fixed. Tickets cost \(\pounds1\) each. For each ticket purchased, the punter (i.e. the purchaser) chooses a number from \(1\) to \(N\). The winning number is chosen at random, and the jackpot is shared equally amongst those punters who chose the winning number. A syndicate decides to buy \(N\) tickets, choosing every number once to be sure of winning a share of the jackpot. The total number of tickets purchased in this draw is \(3.8N\) and the jackpot is \(\pounds W\). Assuming that the non-syndicate punters choose their numbers independently and at random, find the most probable number of winning tickets and show that the expected net loss of the syndicate is approximately \[ N\; - \; %\textstyle{ \frac{5 \big(1- e^{-2.8}\big)}{14} \;W\;. \]

Showing 1-8 of 8 problems
2018 Paper 3 Q13
D: 1700.0 B: 1484.0

The random variable \(X\) takes only non-negative integer values and has probability generating function \(\G(t)\). Show that \[ \P(X = 0 \text{ or } 2 \text{ or } 4 \text { or } 6 \ \ldots ) = \frac{1}{2}\big(\G\left(1\right)+\G\left(-1\right)\big). \] You are now given that \(X\) has a Poisson distribution with mean \(\lambda\). Show that \[ \G(t) = \e^{-\lambda(1-t)} \,. \]

  1. The random variable \(Y\) is defined by \[ \P(Y=r)= \begin{cases} k\P(X=r) & \text{if \(r=0, \ 2, \ 4, \ 6, \ \ldots\) \ }, \\[2mm] 0& \text{otherwise}, \end{cases} \] where \(k\) is an appropriate constant. Show that the probability generating function of \(Y\) is \(\dfrac{\cosh\lambda t}{\cosh\lambda}\,\). Deduce that \(\E(Y) < \lambda\) for \(\lambda > 0\,\).
  2. The random variable \(Z\) is defined by \[\P(Z=r)= \begin{cases} c \P(X=r) & \text{if \(r = 0, \ 4, \ 8, \ 12, \ \ldots \ \)}, \\[2mm] 0& \text{otherwise,} \end{cases} \] where \(c\) is an appropriate constant. Is \(\E(Z) < \lambda\) for all positive values of \(\lambda\,\)?

Show Solution
\begin{align*} &&G_X(t) &= \mathbb{E}(t^N) \\ &&&= \sum_{k=0}^{\infty} \mathbb{P}(X = k) t^k \\ \Rightarrow && G_X(1) &= \sum_{k=0}^{\infty} \mathbb{P}(X = k) \\ \Rightarrow && G_X(-1) &= \sum_{k=0}^{\infty} (-1)^k\mathbb{P}(X = k) \\ \Rightarrow && \frac12 (G_X(1) + G_X(-1) &= \sum_{k=0}^{\infty} \frac12 (1 + (-1)^k) \mathbb{P}(X = k) \\ &&&= \sum_{k=0}^{\infty} \mathbb{P}(X =2k) \end{align*}
  1. \begin{align*} 1 &= \sum_r \mathbb{P}(Y = r) \\ &= \sum_{k=0}^\infty k \cdot \mathbb{P}(X = 2k) \\ &= k \cdot \frac12 \l e^{-\lambda(1-1) } + e^{-\lambda(1+1) }\r \\ &= \frac{k}{2}(1+e^{-2\lambda}) \end{align*} Therefore \(k = \frac{2}{1+e^{-2\lambda}} = e^{\lambda} \frac{1}{\cosh \lambda}\) \begin{align*} && G_X(t) + G_X(-t) &= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\ &&&= \sum_{k=0}^\infty \mathbb{P}(X = k)t^k(1^k + (-1)^k) \\ &&&= 2\sum_{k=0}^\infty \mathbb{P}(X = 2k)t^{2k} \\ &&&= 2\sum_{k=0}^\infty \frac{1}{k}\mathbb{P}(Y = 2k)t^{2k} \\ &&&= \frac{2}{k}G_Y(t) \\ \Rightarrow && G_Y(t) &= k \cdot \frac{G_X(t) + G_X(-t)}{2} \\ &&&= k\frac{e^{-\lambda(1-t)} + e^{-\lambda(1+t)}}{2} \\ &&&= \frac{e^\lambda}{\cosh \lambda} \frac{e^{-\lambda} (e^{\lambda t}+e^{-\lambda t}) }{2} \\ &&&= \frac{\cosh \lambda t}{\cosh \lambda} \end{align*} Since \(\mathbb{E}(Y) = G_Y'(1)\) and \begin{align*} && G_Y'(t) &= \frac{\lambda \sinh \lambda t}{\cosh \lambda t} \\ \Rightarrow && G_Y'(1) &= \lambda \tanh \lambda \\ &&&< \lambda \end{align*} since \(\tanh x < 1\)
  2. \begin{align*} && \frac14 \l G_X(t) + G_X(it) +G_X(-t) + G_X(-it) \r &= \sum_{k=0}^\infty \mathbb{P}(X=k)t^k (1 + i^k + (-1)^k + (-i)^k) \\ &&&= \sum_{k=0}^\infty \mathbb{P}(X = 4k)t^{4k} \\ &&&= \frac{G_Z(t)}{c} \end{align*} Since \(G_Z(1) = 1\) we must have \(c = \frac1{\frac14 \l G_X(1) + G_X(i) +G_X(-1) + G_X(-i) \r}\) \begin{align*} && c &= \frac{4e^{\lambda}}{e^{\lambda} + e^{-\lambda} + e^{i\lambda} + e^{-i\lambda}} \\ &&&= \frac{2e^{\lambda}}{\cosh \lambda + \cos \lambda} \\ && G_Z(t) &= c \cdot \frac14 \l e^{-\lambda(1-t)}+e^{-\lambda(1-it)}+e^{-\lambda(1+t)}+e^{-\lambda(1+it)} \r \\ &&&= \frac{ce^{-\lambda t}}{4} \l 2\cosh \lambda t + 2 \cos \lambda t\r \\ &&&= \frac{\cosh \lambda t + \cos \lambda t}{\cosh \lambda + \cos \lambda} \end{align*} We are interested in \(G_Z'(1)\) so: \begin{align*} && G_Z'(t) &= \frac{\lambda (\sinh \lambda t - \sin \lambda t)}{\cosh \lambda + \cos \lambda } \end{align*} Considering various values of \(\lambda\), it makes sense to look at \(\lambda = \pi\) (since \(\cos \lambda = -1\) and the denominator will be small). From this we can see: \begin{align*} G'_Z(1) &= \frac{\pi (\sinh \pi-0)}{\cosh \pi-1} \\ &= \frac{\pi}{\tanh \frac{\pi}{2}} > \pi \end{align*} So \(\mathbb{E}(Z)\) is larger than \(\lambda\) for \(\lambda = \pi\) (and probably many others)
2015 Paper 3 Q12
D: 1700.0 B: 1500.0

A 6-sided fair die has the numbers 1, 2, 3, 4, 5, 6 on its faces. The die is thrown \(n\) times, the outcome (the number on the top face) of each throw being independent of the outcome of any other throw. The random variable \(S_n\) is the sum of the outcomes.

  1. The random variable~\(R_n\) is the remainder when \(S_n\) is divided by 6. Write down the probability generating function, \(\G(x)\), of \(R_1\) and show that the probability generating function of \(R_2\) is also \(\G(x)\). Use a generating function to find the probability that \(S_n\) is divisible by 6.
  2. The random variable \(T_n\) is the remainder when \(S_n\) is divided by 5. Write down the probability generating function, \(\G_1(x)\), of \(T_1\) and show that \(\G_2(x)\), the probability generating function of \(T_2\), is given by \[ {\rm G}_2(x) = \tfrac 1 {36} (x^2 +7y) \] where \(y= 1+x+x^2+x^3+x^4\,\). Obtain the probability generating function of \(T_n\) and hence show that the probability that \(S_n\) is divisible by \(5\) is \[ \frac15\left(1- \frac1 {6^n}\right) \] if \(n\) is not divisible by 5. What is the corresponding probability if \(n\) is divisible by 5?

Show Solution
  1. \(G(x) = \frac{1}{6} (1 + x + x^2 + x^3 + x^4 + x^5)\) The pgf for \(R_2\) is: \begin{align*} \frac1{36}x^2 + \frac{2}{36}x^3 + \frac{3}{36}x^4 + \frac{4}{36}x^5 + \frac{5}{36} +\\ \quad \quad + \frac{6}{36}x^1 + \frac{5}{36}x^2 + \frac4{36}x^3 + \frac3{36}x^4 + \frac{2}{36}x^5 + \frac{1}{36} \\ = \frac{1}{6}(1 + x + x^2 + x^3 + x^4 + x^5) = G(x) \end{align*} Since rolling the dice twice is the same as rolling the dice once, rolling the dice \(n\) times will be the same as rolling it once, ie the pgf for \(R_n\) will be \(G(x)\) and the probability \(S_n\) is divisible by \(6\) is \(\frac16\)
  2. \(G_1(x) = \frac{1}{6} + \frac{1}{3}x^1 + \frac{1}{6}x^2 + \frac16x^3+ \frac16x^4 = \frac16(1 + 2x+x^2+x^3+x^4)\). If \(G_n\) is the probability generating function for \(T_n\) then we can obtain \(G_n\) by multiplying \(G_{n-1}\) by \(G(x)\) and replacing any terms of order higher than \(5\) with their remainder on division by \(5\). (Or equivalently, working over \(\mathbb{R}[x]/(x^5-1)\). If \(y = 1 + x + x^2 + x^3 + x^4\) then: \begin{align*} xy &= x + x^2 + x^3 + x^4 +x^5 \\ &= x + x^2 + x^3 + x^4 + 1 \\ &= y \\ \\ y^2 &= (1 + x+x^2 + x^3+x^4)^2 \\ &= 1 + 2x + 3x^2 + 4x^3+5x^4+4x^5+3x^6 + 2x^7 + x^8 \\ &= (1+4) + (2+3)x+(3+2)x^2 + (4+1)x^3 + 5x^4 \\ &= 5y \end{align*} \begin{align*} \frac{1}{36}(y+x)(y+x) &= \frac1{36}(y^2 + 2xy + x^2) \\ &= \frac1{36}(5y + 2y + x^2 ) \\ &= \frac1{36}(7y + x^2) \end{align*} Similarly, \begin{align*} G_n(x) &= \l\frac{1}{6}(x+y) \r^n \\ &= \frac1{6^n} \l \sum_{i=0}^n \binom{n}{i} y^ix^{n-i} \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} y^i + x^n \r \\ &= \frac1{6^n} \l \sum_{i=1}^n \binom{n}{i} 5^{i-1}y + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y((5+1)^n-1) + x^n \r \\ &= \frac1{6^n} \l \frac{1}{5}y(6^n-1) + x^n \r \\ \end{align*} Therefore if \(n \not \equiv 0 \pmod{5}\), we can find the probability of \(T_n = 0\) by looking at the constant coefficient, ie plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) \r = \frac{1}{5} \l 1- \frac{1}{6^n} \r \] When \(n \equiv 0 \pmod{5}\) we can also find the constant coefficient by plugging in \(x = 0\), which is: \[\frac1{6^n} \l \frac{1}{5}(6^n-1) + 1 \r = \frac{1}{5} \l 1+ \frac{4}{6^n} \r \]
Note: this whole question can be considered a "roots-of-unity" filter in disguise. Our computations in \(\mathbb{R}[x]/(x^5 - 1)\) are the same as computations using \(\omega\), in fact \(\mathbb{R}[x]/(x^5 - 1) \cong \mathbb{R}[\omega]\) where \(\omega\) is a primitive \(5\)th root of unity
2014 Paper 3 Q13
D: 1700.0 B: 1500.0

I play a game which has repeated rounds. Before the first round, my score is \(0\). Each round can have three outcomes: \begin{compactenum}[1.] \item my score is unchanged and the game ends; \item my score is unchanged and I continue to the next round; \item my score is increased by one and I continue to the next round. \end{compactenum} The probabilities of these outcomes are \(a\), \(b\) and~\(c\), respectively (the same in each round), where \(a+b+c=1\) and \(abc\ne0\). The random variable \(N\) represents my score at the end of a randomly chosen game. Let \(\.G(t)\) be the probability generating function of \(N\).

  1. Suppose in the first round, the game ends. Show that the probability generating function conditional on this happening is 1.
  2. Suppose in the first round, the game continues to the next round with no change in score. Show that the probability generating function conditional on this happening is~\(\.G(t)\).
  3. By comparing the coefficients of \(t^n\), show that $ \.G(t) = a + b\.G(t) + ct\.G(t)\,. $ Deduce that, for \(n\ge0\), \[ P(N=n) = \frac{ac^n}{(1-b)^{n+1}}\,. \]
  4. Show further that, for \(n\ge0\), \[ P(N=n) = \frac{\mu^n}{(1+\mu)^{n+1}}\,, \] where \(\mu=\.E(N)\).

2011 Paper 3 Q12
D: 1700.0 B: 1516.0

The random variable \(N\) takes positive integer values and has pgf (probability generating function) \(\G(t)\). The random variables \(X_i\), where \(i=1\), \(2\), \(3\), \(\ldots,\) are independently and identically distributed, each with pgf \({\H}(t)\). The random variables \(X_i\) are also independent of \(N\). The random variable \(Y\) is defined by \[ Y= \sum_{i=1}^N X_i \;. \] Given that the pgf of \(Y\) is \(\G(\H(t))\), show that \[ \E(Y) = \E(N)\E(X_i) \text{ \ \ \ \ and \ \ \ \ } \var(Y) = \var(N)\big(\E(X_i)\big)^2 + \E(N) \var(X_i) \,.\] A fair coin is tossed until a head occurs. The total number of tosses is \(N\). The coin is then tossed a further \(N\) times and the total number of heads in these \(N\) tosses is \(Y\). Find in this particular case the pgf of \(Y\), \(\E(Y)\), \(\var(Y)\) and \(\P(Y=r)\).

2009 Paper 3 Q12
D: 1700.0 B: 1516.0

  1. Albert tosses a fair coin \(k\) times, where \(k\) is a given positive integer. The number of heads he gets is \(X_1\). He then tosses the coin \(X_1\) times, getting \(X_2\) heads. He then tosses the coin \(X_2\) times, getting \(X_3\) heads. The random variables \(X_4\), \(X_5\), \(\ldots\) are defined similarly. Write down \(\E(X_1)\). By considering \(\E(X_2 \; \big\vert \; X_1 = x_1)\), or otherwise, show that \(\E(X_2) = \frac14 k\). Find \(\displaystyle \sum_{i=1}^\infty \E(X_i)\).
  2. Bertha has \(k\) fair coins. She tosses the first coin until she gets a tail. The number of heads she gets before the first tail is \(Y_1\). She then tosses the second coin until she gets a tail and the number of heads she gets with this coin before the first tail is \(Y_2\). The random variables \(Y_3\), \(Y_4\), \(\ldots\;\), \(Y_k\) are defined similarly, and \(Y= \sum\limits_{i=1}^k Y_i\,\). Obtain the probability generating function of \(Y\), and use it to find \(\E(Y)\), \(\var(Y)\) and \(\P(Y=r)\).

2003 Paper 3 Q14
D: 1700.0 B: 1599.8

Write down the probability generating function for the score on a standard, fair six-faced die whose faces are labelled \(1, 2, 3, 4, 5, 6\). Hence show that the probability generating function for the sum of the scores on two standard, fair six-faced dice, rolled independently, can be written as \[ \frac1{36} t^2 \l 1 + t \r^2 \l 1 - t + t^2 \r^2 \l 1 + t + t^2 \r^2 \;. \] Write down, in factorised form, the probability generating functions for the scores on two fair six-faced dice whose faces are labelled with the numbers \(1, 2, 2, 3, 3, 4\) and \(1, 3, 4, 5, 6, 8,\) and hence show that when these dice are rolled independently, the probability of any given sum of the scores is the same as for the two standard fair six-faced dice. Standard, fair four-faced dice are tetrahedra whose faces are labelled \(1, 2, 3, 4,\) the score being taken from the face which is not visible after throwing, and each score being equally likely. Find all the ways in which two fair four-faced dice can have their faces labelled with positive integers if the probability of any given sum of the scores is to be the same as for the two standard fair four-faced dice.

2000 Paper 3 Q13
D: 1700.0 B: 1516.0

A set of \(n\) dice is rolled repeatedly. For each die the probability of showing a six is \(p\). Show that the probability that the first of the dice to show a six does so on the \(r\)th roll is $$q^{n r } ( q^{-n} - 1 )$$ where \(q = 1 - p\). Determine, and simplify, an expression for the probability generating function for this distribution, in terms of \(q\) and \(n\). The first of the dice to show a six does so on the \(R\)th roll. Find the expected value of \(R\) and show that, in the case \(n = 2\), \(p=1/6\), this value is \(36/11\). Show that the probability that the last of the dice to show a six does so on the \(r\)th roll is \[ \big(1-q^r\big)^n-\big(1-q^{r-1}\big)^n. \] Find, for the case \(n = 2\), the probability generating function. The last of the dice to show a six does so on the \(S\)th roll. Find the expected value of \(S\) and evaluate this when \(p=1/6\).

1997 Paper 3 Q12
D: 1700.0 B: 1500.0

  1. I toss a biased coin which has a probability \(p\) of landing heads and a probability \(q=1-p\) of landing tails. Let \(K\) be the number of tosses required to obtain the first head and let \[ \mathrm{G}(s)=\sum_{k=1}^{\infty}\mathrm{P}(K=k)s^{k}. \] Show that \[ \mathrm{G}(s)=\frac{ps}{1-qs} \] and hence find the expectation and variance of \(K\).
  2. I sample cards at random with replacement from a normal pack of \(52\). Let \(N\) be the total number of draws I make in order to sample every card at least once. By expressing \(N\) as a sum \(N=N_{1}+N_{2}+\cdots+N_{52}\) of random variables, or otherwise, find the expectation of \(N\). Estimate the numerical value of this expectation, using the approximations \(\mathrm{e}\approx2.7\) and \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\approx0.5+\ln n\) if \(n\) is large.

Showing 1-14 of 14 problems
2025 Paper 3 Q11
D: 1500.0 B: 1500.0

  1. Let \(\lambda > 0\). The independent random variables \(X_1, X_2, \ldots, X_n\) all have probability density function $$f(t) = \begin{cases} \lambda e^{-\lambda t} & t \geq 0 \\ 0 & t < 0 \end{cases}$$ and cumulative distribution function \(F(x)\). The value of random variable \(Y\) is the largest of the values \(X_1, X_2, \ldots, X_n\). Show that the cumulative distribution function of \(Y\) is given, for \(y \geq 0\), by $$G(y) = (1 - e^{-\lambda y})^n$$
  2. The values \(L(\alpha)\) and \(U(\alpha)\), where \(0 < \alpha \leq \frac{1}{2}\), are such that $$P(Y < L(\alpha)) = \alpha \text{ and } P(Y > U(\alpha)) = \alpha$$ Show that $$L(\alpha) = -\frac{1}{\lambda}\ln(1 - \alpha^{1/n})$$ and write down a similar expression for \(U(\alpha)\).
  3. Use the approximation \(e^t \approx 1 + t\), for \(|t|\) small, to show that, for sufficiently large \(n\), $$\lambda L(\alpha) \approx \ln(n) - \ln\left(\ln\left(\frac{1}{\alpha}\right)\right)$$
  4. Hence show that the median of \(Y\) tends to infinity as \(n\) increases, but that the width of the interval \(U(\alpha) - L(\alpha)\) tends to a value which is independent of \(n\).
  5. You are given that, for \(|t|\) small, \(\ln(1 + t) \approx t\) and that \(e^3 \approx 20\). Show that, for sufficiently large \(n\), there is an interval of width approximately \(4\lambda^{-1}\) in which \(Y\) lies with probability \(0.9\).

Show Solution
  1. Note that \(\displaystyle F(y) = \mathbb{P}(X_i < y) = \int_0^y \lambda e^{-\lambda t} \d t = 1-e^{-\lambda y}\). Notice also that \begin{align*} G(y) &= \mathbb{P}(Y < y) \\ &= \mathbb{P}(\max_i(X_i) < y) \\ &= \mathbb{P}(X_i < y \text{ for all }i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i < y) \\ &= \prod_{i=1}^n (1-e^{-\lambda y})\\ &= (1-e^{-\lambda y})^n \end{align*} as required.
  2. \begin{align*} && \mathbb{P}(Y < L(\alpha)) &= \alpha \\ \Rightarrow && (1-e^{-\lambda L(\alpha)})^n &= \alpha \\ \Rightarrow && 1-e^{-\lambda L(\alpha)} &= \alpha^{\tfrac1n} \\ \Rightarrow && L(\alpha) &= -\frac{1}{\lambda}\ln \left (1-\alpha^{\tfrac1n} \right) \end{align*} Notice also: \begin{align*} && \mathbb{P}(Y > U(\alpha)) &= \alpha \\ \Rightarrow && 1 - (1-e^{-\lambda U(\alpha)})^n &= \alpha \\ \Rightarrow && U(\alpha) &= -\frac{1}{\lambda}\ln \left ( 1-(1-\alpha)^{\tfrac1n} \right) \end{align*}
  3. \begin{align*} \lambda L(\alpha) &= -\ln \left (1-\alpha^{\tfrac1n} \right) \\ &= -\ln \left (1-e^{\tfrac1n \ln \alpha} \right) \\ &\approx - \ln \left ( 1 - 1 - \frac1n \ln \alpha\right) \tag{\(e^t \approx 1 + t\)} \\ &= -\ln \left ( \frac{1}{n} \ln \frac{1}\alpha \right) \\ &= - \ln \frac{1}{n} - \ln \left ( \ln \frac{1}{\alpha} \right )\\ &= \ln n - \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \end{align*} since if \(n\) is large, \(\frac{\ln \alpha}{n}\) is small.
  4. The median is the value where \(\mathbb{P}(Y < M) = \frac12\), or in other words \(L(\frac12)\), but this is \(\approx \frac{\ln n - \ln (\ln 2)}{\lambda} \to \infty\). \begin{align*} && \lambda U(\alpha) &\approx \ln n - \ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right) \\ \Rightarrow && \lambda(U(\alpha) - L(\alpha)) &\approx -\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right)+ \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right) \\ \Rightarrow && U(\alpha) - L(\alpha) &\to \frac{1}{\lambda} \left ( \ln \left ( \ln \left ( \frac{1}{\alpha} \right ) \right)-\ln \left ( \ln \left ( \frac{1}{1-\alpha} \right ) \right ) \right) \end{align*} which doesn't depend on \(n\).
  5. Suppose \(\alpha = \frac{1}{20}\) then \begin{align*} U(\alpha) - L(\alpha) &\approx \frac{1}{\lambda} \left (\ln \ln 20 - \ln \ln \frac{20}{19} \right) \\ &= \lambda^{-1} \left (\ln \ln 20 - \ln \ln (1 + \frac{1}{19}) \right) \\ &\approx \lambda^{-1} \left (\ln 3 - \ln \frac{1}{19} \right) \tag{\(\ln(1+t) \approx t\)} \\ &\approx \lambda^{-1} \ln 3 \cdot 19 \\ &\approx \lambda^{-1} (1 + 3) \\ &\approx 4\lambda^{-1} \end{align*} [Note that \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]
2014 Paper 3 Q12
D: 1700.0 B: 1500.0

The random variable \(X\) has probability density function \(\f(x)\) (which you may assume is differentiable) and cumulative distribution function \(\F(x)\) where $-\infty < x < \infty \(. The random variable \)Y\( is defined by \)Y= \e^X$. You may assume throughout this question that \(X\) and \(Y\) have unique modes.

  1. Find the median value \(y_m\) of \(Y\) in terms of the median value \(x_m\) of \(X\).
  2. Show that the probability density function of \(Y\) is $\f(\ln y)/y\(, and deduce that the mode~\)\lambda\( of \)Y$ satisfies $\f'(\ln \lambda) = \f(\ln \lambda)\(.
  3. Suppose now that \)X \sim {\rm N} (\mu,\sigma^2)$, so that \[ \f(x) = \frac{1}{\sigma \sqrt{2\pi}\,} \e^{-(x-\mu)^2/(2\sigma^2)} \,. \] Explain why \[ \frac{1}{\sigma \sqrt{2\pi}\,} \int_{-\infty}^{\infty}\e^{-(x-\mu-\sigma^2)^2/(2\sigma^2)} \d x = 1 \] and hence show that \( \E(Y) = \e ^{\mu+\frac12\sigma^2}\).
  4. Show that, when \(X \sim {\rm N} (\mu,\sigma^2)\), \[ \lambda < y_m < \E(Y)\,. \]

2014 Paper 2 Q12
D: 1600.0 B: 1484.8

The lifetime of a fly (measured in hours) is given by the continuous random variable~\(T\) with probability density function \(\.f(t)\) and cumulative distribution function \(\.F(t)\). The \emph{hazard function}, \(\.h(t)\), is defined, for \(\.F(t)<1\), by \[ \.h(t) = \frac{\.f(t)}{1-\.F(t)}\,. \]

  1. Given that the fly lives to at least time \(t\), show that the probability of its dying within the following \(\delta t\) is approximately \(\.h (t) \, \delta t\) for small values of \(\delta t\).
  2. Find the hazard function in the case \(\.F(t) = t/a\) for \(0< t < a\). Sketch \(\.f(t)\) and \(\.h(t)\) in this case.
  3. The random variable \(T\) is distributed on the interval $t> a\(, where \)a>0\(, and its hazard function is \)t^{-1}$. Determine the probability density function for \(T\).
  4. Show that \(\.h(t)\) is constant for \(t>b\) and zero otherwise if and only if \(\.f(t) =k\.e^{-k(t-b)}\) for \(t>b\), where \(k\)~is a positive constant.
  5. The random variable \(T\) is distributed on the interval \(t> 0\) and its hazard function is given by \[ \.h(t) = \left(\frac{\lambda}{\theta^\lambda}\right)t^{\lambda-1}\,, \] where \(\lambda\) and \(\theta\) are positive constants. Find the probability density function for \(T\).

2013 Paper 3 Q13
D: 1700.0 B: 1484.0

  1. The continuous random variable \(X\) satisfies \(0\le X\le 1\), and has probability density function \(\f(x)\) and cumulative distribution function \(\F(x)\). The greatest value of \(\f(x)\) is~\(M\), so that \(0\le \f(x) \le M\).
    1. The continuous random variable \(Y\) satisfies \(0\le Y\le 1\), and has probability density function \(k \F(y) \f(y)\), where \(\f\) and \(\F\) are as above.
      1. [\bf (a)] Determine the value of the constant \(k\).
      2. [\bf (b)] Show that \[ 1+ \frac{nM}{n+1}\mu_{n+1} - \frac{nM}{n+1} \le \E(Y^n) \le 2M\mu_{n+1}\,, \] where \(\mu_{n+1} = \E(X^{n+1})\) and \(n\ge0\).
      3. [\bf (c)] Hence show that, for \(n\ge 1\), \[ \mu _n \ge \frac{n}{(n+1)M} -\frac{n-1}{n+1} \,.\]

2011 Paper 2 Q13
D: 1600.0 B: 1500.0

What property of a distribution is measured by its {\em skewness}?

  1. One measure of skewness, \(\gamma\), is given by \[ \displaystyle \gamma= \frac{ \E\big((X-\mu)^3\big)}{\sigma^3}\,, \] where \(\mu\) and \(\sigma^2\) are the mean and variance of the random variable \(X\). Show that \[ \gamma = \frac{ \E(X^3) -3\mu \sigma^2 - \mu^3}{\sigma^3}\,. \] The continuous random variable \(X\) has probability density function \(\f\) where \[ \f(x) = \begin{cases} 2x & \text{for } 0\le x\le 1\,, \\[2mm] 0 & \text{otherwise}\,. \end{cases} \] Show that for this distribution \(\gamma= -\dfrac{2\sqrt2}{5}\).
  2. The {\em decile skewness}, \(D\), of a distribution is defined by \[D= \frac { {\rm F}^{-1}(\frac9{10}) - 2{\rm F} ^{-1}(\frac12) + {\rm F}^{-1} (\frac1{10}) } {{\rm F}^{-1}(\frac9{10}) - {\rm F} ^{-1} (\frac1{10})}\,, \] where \({\rm F}^{-1}\) is the inverse of the cumulative distribution function. Show that, for the above distribution, $ D= 2 -\sqrt5\,. $ The {\em Pearson skewness},~\(P\), of a distribution is defined by \[ P = \frac{3(\mu-M)}{\sigma} \,,\] where \(M\) is the median. Find \(P\) for the above distribution and show that \(D>P>\gamma\,\).

2010 Paper 2 Q12
D: 1600.0 B: 1486.8

The continuous random variable \(X\) has probability density function \(\f(x)\), where \[ \f(x) = \begin{cases} a & \text {for } 0\le x < k \\ b & \text{for } k \le x \le 1\\ 0 & \text{otherwise}, \end{cases} \] where \(a > b > 0\) and \(0 < k < 1\). Show that \(a > 1\) and \(b < 1\).

  1. Show that \[ \E(X) = \frac{1-2b+ab}{2(a-b)}\,. \]
  2. Show that the median, \(M\), of \(X\) is given by \(\displaystyle M=\frac 1 {2a}\) if \(a+b\ge 2ab\) and obtain an expression for the median if \(a+b\le 2ab\).
  3. Show that \(M < \E(X)\,\).

2007 Paper 2 Q14
D: 1600.0 B: 1484.0

The random variable \(X\) has a continuous probability density function \(\f(x)\) given by \begin{equation*} \f(x) = \begin{cases} 0 & \text{for } x \le 1 \\ \ln x & \text{for } 1\le x \le k\\ \ln k & \text{for } k\le x \le 2k\\ a-bx & \text{for } 2k \le x \le 4k \\ 0 & \text{for } x\ge 4k \end{cases} \end{equation*} where \(k\), \(a\) and \(b\) are constants.

  1. Sketch the graph of \(y=\f(x)\).
  2. Determine \(a\) and \(b\) in terms of \(k\) and find the numerical values of \(k\), \(a\) and \(b\).
  3. Find the median value of \(X\).

Show Solution
  1. TikZ diagram
  2. Since \(f(x)\) is continuous, \(a -bx\) joins \((2k, \ln k)\) and \((4k ,0)\). ie it has a gradient of \(\frac{-\ln k}{2k}\) and is zero at \(4k\), hence \(\displaystyle b = -\frac{\ln k}{2k}, a = 2\ln k\). The \(3\) sections have areas \(\int_1^k \ln x \d x = k \ln k -k +1\), \(k \ln k, k \ln k\). Therefore \begin{align*} &&1&= 3k\ln k - k +1 \\ \Rightarrow &&0 &= k(3\ln k - 1) \\ \Rightarrow &&\ln k &= \frac13 \\ \Rightarrow &&k &= e^{1/3} \\ && a &= \frac23 \\ && b&= -\frac16e^{-1/3} \end{align*}
  3. Clearly \(1 > k \ln k > \frac{1}{3}\), therefore the median must lie between \(k\) and \(2k\). So we need, \(\frac12\) to be the area of the rectangle + the triangle, ie: \begin{align*} && \frac12 &= k \ln k + (2k-M) \ln k \\ &&&= \frac13 k + \frac13 (2k - M) \\ \Rightarrow && M &= 3k - \frac32 \\ \Rightarrow && M &= 3e^{1/3} - \frac32 \end{align*}
2005 Paper 3 Q14
D: 1700.0 B: 1484.0

In this question, you may use the result \[ \displaystyle \int_0^\infty \frac{t^m}{(t+k)^{n+2}} \; \mathrm{d}t =\frac{m!\, (n-m)!}{(n+1)! \, k^{n-m+1}}\;, \] where \(m\) and \(n\) are positive integers with \(n\ge m\,\), and where \(k>0\,\). The random variable \(V\) has density function \[ \f(x) = \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \quad \quad (0 \le x < \infty) \;, \] where \(a\) is a positive integer. Show that \(\displaystyle C = \frac{(2a+1)!}{a! \, a!}\;\). Show, by means of a suitable substitution, that \[ \int_0^v \frac{x^a}{(x+k)^{2a+2}} \; \mathrm{d}x = \int_{\frac{k^2}{v}}^\infty \frac{u^a}{(u+k)^{2a+2}} \; \mathrm{d}u \] and deduce that the median value of \(V\) is \(k\). Find the expected value of \(V\). The random variable \(V\) represents the speed of a randomly chosen gas molecule. The time taken for such a particle to travel a fixed distance \(s\) is given by the random variable \(\ds T=\frac{s}{V}\). Show that \begin{equation} \mathbb{P}( T < t) = \ds \int_{\frac{s}{t}}^\infty \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}}\; \mathrm{d}x \tag{\( *\)} \end{equation} and hence find the density function of \(T\). You may find it helpful to make the substitution \(\ds u = \frac{s}{x}\) in the integral \((*)\). Hence show that the product of the median time and the median speed is equal to the distance \(s\), but that the product of the expected time and the expected speed is greater than \(s\).

Show Solution
\begin{align*} && f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\ \Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\ &&&= \int_0^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ &&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\ &&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\ &&&= C \frac{a!a!}{(2a+1)!} \\ \Rightarrow && C &= \frac{(2a+1)!}{a!a!} \end{align*} \begin{align*} && I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\ u = k^2/x, \d x = -k^2u^{-2} \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\ \end{align*} At the median we want a value \(M\) such that \(M = k^2/M\) ie \(M = k\) \begin{align*} && \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\ &&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\ &&&= \frac{k(a+1)}{a} = \frac{a+1}a k \end{align*} \begin{align*} && \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\ &&&= \mathbb{P}(V > \frac{s}{t}) \\ &&&= \int_{s/t}^{\infty} f(x) \d x \\ &&&= \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ \\ \Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\ &&&= \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\ &&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\ &&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\ &&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\ &&&= \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}} \end{align*} Therefore \(T\) follows the same distribution, but with parameter \(s/k\) rather than \(k\). In particular it has median \(s/k\) (and the product of the medians is \(s\)). However, the product of the expected time and expected speed is \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)
2004 Paper 2 Q12
D: 1600.0 B: 1516.0

Sketch the graph, for \(x \ge 0\,\), of $$ y = kx\e^{-ax^2} \;, $$ where \(a\) and \(k\) are positive constants. The random variable \(X\) has probability density function \(\f(x)\) given by \begin{equation*} \f(x)= \begin{cases} kx\e^{-ax^2} & \text{for \(0 \le x \le 1\)}\\[3pt] 0 & \text{otherwise}. \end{cases} \end{equation*} Show that \(\ds k=\frac{2a}{1-\e^{-a}}\) and find the mode \(m\) in terms of \(a\,\), distinguishing between the cases \(a < \frac12\) and \(a > \frac12\,\). Find the median \(h\) in terms of \(a\)\, and show that \(h > m\) if \(a > -\ln\left(2\e^{-1/2} - 1\right).\) Show that, \(-\ln\left(2\e^{-1/2}-1\right)> \frac12 \,\). Show also that, if \(a > -\ln\left(2\e^{-1/2} - 1\right) \,\), then $$ P(X > m \;\vert\; X < h) = {{2\e^{-1/2}-\e^{-a}-1} \over 1-\e^{-a}}\;. $$

2002 Paper 2 Q13
D: 1600.0 B: 1484.0

Let \(\F(x)\) be the cumulative distribution function of a random variable \(X\), which satisfies \(\F(a)=0\) and \(\F(b)=1\), where \(a>0\). Let \[ \G(y) = \frac{\F(y)}{2-\F(y)}\;. \] Show that \(\G(a)=0\,\), \(\G(b)=1\,\) and that \(\G'(y)\ge0\,\). Show also that \[ \frac12 \le \frac2{(2-\F(y))^2} \le 2\;. \] The random variable \(Y\) has cumulative distribution function \(\G(y)\,\). Show that \[ {\ts \frac12} \,\E(X) \le \E(Y) \le 2 \E(X) \;, \] and that \[ \var(Y) \le 2\var(X) +{\ts \frac 74} \big(\E(X)\big)^2\;. \]

1996 Paper 2 Q12
D: 1600.0 B: 1500.0

  1. Let \(X_{1}\), \(X_{2}\), \dots, \(X_{n}\) be independent random variables each of which is uniformly distributed on \([0,1]\). Let \(Y\) be the largest of \(X_{1}\), \(X_{2}\), \dots, \(X_{n}\). By using the fact that \(Y<\lambda\) if and only if \(X_{j}<\lambda\) for \(1\leqslant j\leqslant n\), find the probability density function of \(Y\). Show that the variance of \(Y\) is \[\frac{n}{(n+2)(n+1)^{2}}.\]
  2. The probability that a neon light switched on at time \(0\) will have failed by a time \(t>0\) is \(1-\mathrm{e}^{-t/\lambda}\) where \(\lambda>0\). I switch on \(n\) independent neon lights at time zero. Show that the expected time until the first failure is \(\lambda/n\).

1992 Paper 3 Q15
D: 1700.0 B: 1500.0

A goat \(G\) lies in a square field \(OABC\) of side \(a\). It wanders randomly round its field, so that at any time the probability of its being in any given region is proportional to the area of this region. Write down the probability that its distance, \(R\), from \(O\) is less than \(r\) if \(0 < r\leqslant a,\) and show that if \(r\geqslant a\) the probability is \[ \left(\frac{r^{2}}{a^{2}}-1\right)^{\frac{1}{2}}+\frac{\pi r^{2}}{4a^{2}}-\frac{r^{2}}{a^{2}}\cos^{-1}\left(\frac{a}{r}\right). \] Find the median of \(R\) and probability density function of \(R\). The goat is then tethered to the corner \(O\) by a chain of length \(a\). Find the conditional probability that its distance from the fence \(OC\) is more than \(a/2\).

1990 Paper 2 Q15
D: 1600.0 B: 1500.0

A target consists of a disc of unit radius and centre \(O\). A certain marksman never misses the target, and the probability of any given shot hitting the target within a distance \(t\) from \(O\) it \(t^{2}\), where \(0\leqslant t\leqslant1\). The marksman fires \(n\) shots independently. The random variable \(Y\) is the radius of the smallest circle, with centre \(O\), which encloses all the shots. Show that the probability density function of \(Y\) is \(2ny^{2n-1}\) and find the expected area of the circle. The shot which is furthest from \(O\) is rejected. Show that the expected area of the smallest circle, with centre \(O\), which encloses the remaining \((n-1)\) shots is \[ \left(\frac{n-1}{n+1}\right)\pi. \]

Show Solution
Another way to describe \(Y\) is the maximum distance of any shot from \(O\). Let \(X_i\), \(1 \leq i \leq n\) be the \(n\) shots then, \begin{align*} F_Y(y) &= \mathbb{P}(Y \leq y) \\ &= \mathbb{P}(X_i \leq y \text{ for all } i) \\ &= \prod_{i=1}^n \mathbb{P}(X_i \leq y) \tag{each shot independent}\\ &= \prod_{i=1}^n y^2\\ &= y^{2n} \end{align*} Therefore \(f_Y(y) = \frac{\d}{\d y} (y^{2n}) = 2n y^{2n-1}\). \begin{align*} \mathbb{E}(\pi Y^2) &= \int_0^1\pi y^2 \f_Y(y) \d y \\ &=\pi \int_0^1 2n y^{2n+1} \d y \\ &=\left ( \frac{n}{n+1} \right )\pi \end{align*}. Let \(Z\) be the distance of the second furthest shot, then: \begin{align*} && F_Z(z) &= \mathbb{P}(Z \leq z) \\ &&&= \mathbb{P}(X_i \leq z \text{ for at least } n - 1\text{ different } i) \\ &&&= n\mathbb{P}(X_i \leq z \text{ for all but 1}) + \mathbb{P}(X_i \leq z \text{ for all } i) \\ &&&= n \left ( \prod_{i=1}^{n-1} \mathbb{P}(X_i \leq z) \right) \mathbb{P}(X_n > z) + z^{2n} \\ &&&= nz^{2n-2}(1-z^2) + z^{2n} \\ &&&= nz^{2n-2} -(n-1)z^{2n} \\ \Rightarrow && f_Z(z) &= n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \\ \Rightarrow && \mathbb{E}(\pi Z^2) &= \int_0^1 \pi z^2 \left (n(2n-2)z^{2n-3}-2n(n-1)z^{2n-1} \right) \d z \\ &&&= \pi \left ( \frac{n(2n-2)}{2n} - \frac{2n(n-1)}{2n+2}\right) \\ &&&= \left ( \frac{n-1}{n+1} \right) \pi \end{align*}
1989 Paper 3 Q15
D: 1700.0 B: 1503.8

The continuous random variable \(X\) is uniformly distributed over the interval \([-c,c].\) Write down expressions for the probabilities that:

  1. \(n\) independently selected values of \(X\) are all greater than \(k\),
  2. \(n\) independently selected values of \(X\) are all less than \(k\),
where \(k\) lies in \([-c,c]\). A sample of \(2n+1\) values of \(X\) is selected at random and \(Z\) is the median of the sample. Show that \(Z\) is distributed over \([-c,c]\) with probability density function \[ \frac{(2n+1)!}{(n!)^{2}(2c)^{2n+1}}(c^{2}-z^{2})^{n}. \] Deduce the value of \({\displaystyle \int_{-c}^{c}(c^{2}-z^{2})^{n}\,\mathrm{d}z.}\) Evaluate \(\mathrm{E}(Z)\) and \(\mathrm{var}(Z).\)

Show Solution
  1. \begin{align*} \mathbb{P}(n\text{ independent values of }X > k) &= \prod_{i=1}^n \mathbb{P}(X > k) \\ &= \left ( \frac{c-k}{2c}\right)^n \end{align*}
  2. \begin{align*} \mathbb{P}(n\text{ independent values of }X < k) &= \prod_{i=1}^n \mathbb{P}(X < k) \\ &= \left ( \frac{k+c}{2c}\right)^n \end{align*}
\begin{align*} &&\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta) &= \mathbb{P}(n\text{ values } < z - \delta \text{ and } n \text{ values} > z + \delta) \\ &&&= \binom{2n+1}{n,n,1} \left ( \frac{c-(z+\delta)}{2c}\right)^n\left ( \frac{(z-\delta)+c}{2c}\right)^n \frac{2 \delta}{2 c} \\ &&&= \frac{(2n+1)!}{n! n!} \frac{((c-(z+\delta))(c+(z-\delta)))^n 2\delta}{2^n c^n} \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-(z+\delta))(c+(z-\delta)))^n 2\delta \\ \Rightarrow && \lim_{\delta \to 0} \frac{\mathbb{P}(\text{median} < z+\delta \text{ and median} > z - \delta)}{2 \delta} &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}((c-z)(c+z))^n \\ &&&= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2) \\ \end{align*} \begin{align*} && 1 &= \int_{-c}^c \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ \Rightarrow && \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!} &= \int_{-c}^c (c^2-z^2)^n \d z \end{align*} \begin{align*} \mathbb{E}(Z) &= \int_{-c}^c z \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z (c^2-z^2)^n \d z \\ &= 0 \end{align*} \begin{align*} \mathrm{Var}(Z) &= \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 \\ &= \mathbb{E}(Z^2) \\ &= \int_{-c}^c z^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}}(c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-c}^c z^2 (c^2-z^2)^n \d z \\ &=\frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \left ( \left [ -\frac{1}{2(n+1)}z(c^2-z^2)^{n+1} \right]_{-c}^c + \frac{1}{2(n+1)}\int_{-c}^c (c^2-z^2)^{n+1} \d z \right) \\ &= \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \frac{1}{2(n+1)} \frac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} \\ &= \frac{(n+1)^2(2c)^2}{(n+1)(2n+2)(2n+3)} \\ &= \frac{2c^2}{2n+3} \end{align*}

Showing 1-7 of 7 problems
2016 Paper 1 Q13
D: 1500.0 B: 1500.0

An internet tester sends \(n\) e-mails simultaneously at time \(t=0\). Their arrival times at their destinations are independent random variables each having probability density function \(\lambda \e^{-\lambda t}\) (\(0\le t<\infty\), \( \lambda >0\)).

  1. The random variable \(T\) is the time of arrival of the e-mail that arrives first at its destination. Show that the probability density function of \(T\) is \[ n \lambda \e^{-n\lambda t} \,, \] and find the expected value of \(T\).
  2. Write down the probability that the second e-mail to arrive at its destination arrives later than time \(t\) and hence derive the density function for the time of arrival of the second e-mail. Show that the expected time of arrival of the second e-mail is \[ \frac{1}{\lambda} \left( \frac1{n-1} + \frac 1 n \right) . \]

2015 Paper 2 Q13
D: 1600.0 B: 1516.0

The maximum height \(X\) of flood water each year on a certain river is a random variable with probability density function \(\f\) given by \[ \f(x) = \begin{cases} \lambda \e^{-\lambda x} & \text{for \(x\ge0\)}\,, \\ 0 & \text{otherwise,} \end{cases} \] where \(\lambda\) is a positive constant. It costs \(ky\) pounds each year to prepare for flood water of height \(y\) or less, where \(k\) is a positive constant and \(y\ge0\). If \(X \le y\) no further costs are incurred but if \(X> y\) the additional cost of flood damage is \(a(X - y )\) pounds where \(a\) is a positive constant.

  1. Let \(C\) be the total cost of dealing with the floods in the year. Show that the expectation of \(C\) is given by \[\mathrm{E}(C)=ky+\frac{a}{\lambda}\mathrm{e}^{-\lambda y} \, . \] How should \(y\) be chosen in order to minimise \(\mathrm{E}(C)\), in the different cases that arise according to the value of \(a/k\)?
  2. Find the variance of \(C\), and show that the more that is spent on preparing for flood water in advance the smaller this variance.

Show Solution
  1. \(\,\) \begin{align*} && \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\ &&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\ &&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\ &&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\ &&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\ \\ && \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\ \Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right) \end{align*} Since \(\mathbb{E}(C)\) is clearly increasing when \(y\) is very large, the optimal value will be \(\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)\), if \(\frac{a}{k} > 1\), otherwise you should spend nothing on flood defenses.
  2. \begin{align*} && \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\ &&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\ &&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\ && \textrm{Var}(C) &= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\ &&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\ &&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\ \\ && \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\ &&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0 \end{align*} so \(\textrm{Var}(C)\) is decreasing in \(y\).
2002 Paper 3 Q13
D: 1700.0 B: 1516.0

A continuous random variable is said to have an exponential distribution with parameter \(\lambda\) if its density function is \(\f(t) = \lambda \e ^{- \lambda t} \; \l 0 \le t < \infty \r\,\). If \(X_1\) and \(X_2\), which are independent random variables, have exponential distributions with parameters \(\lambda_1\) and \(\lambda_2\) respectively, find an expression for the probability that either \(X_1\) or \(X_2\) (or both) is less than \(x\). Prove that if \(X\) is the random variable whose value is the lesser of the values of \(X_1\) and \(X_2\), then \(X\) also has an exponential distribution. Route A and Route B buses run from my house to my college. The time between buses on each route has an exponential distribution and the mean time between buses is 15 minutes for Route A and 30 minutes for Route B. The timings of the buses on the two routes are independent. If I emerge from my house one day to see a Route A bus and a Route B bus just leaving the stop, show that the median wait for the next bus to my college will be approximately 7 minutes.

2002 Paper 1 Q14
D: 1500.0 B: 1516.0

In order to get money from a cash dispenser I have to punch in an identification number. I have forgotten my identification number, but I do know that it is equally likely to be any one of the integers \(1\), \(2\), \ldots , \(n\). I plan to punch in integers in order until I get the right one. I can do this at the rate of \(r\) integers per minute. As soon as I punch in the first wrong number, the police will be alerted. The probability that they will arrive within a time \(t\) minutes is \(1-\e^{-\lambda t}\), where \(\lambda\) is a positive constant. If I follow my plan, show that the probability of the police arriving before I get my money is \[ \sum_{k=1}^n \frac{1-\e^{-\lambda(k-1)/r}}n\;. \] Simplify the sum. On past experience, I know that I will be so flustered that I will just punch in possible integers at random, without noticing which I have already tried. Show that the probability of the police arriving before I get my money is \[ 1-\frac1{n-(n-1)\e^{-\lambda/r}} \;. \]

1998 Paper 2 Q13
D: 1600.0 B: 1516.0

A random variable \(X\) has the probability density function \[ \mathrm{f}(x)=\begin{cases} \lambda\mathrm{e}^{-\lambda x} & x\geqslant0,\\ 0 & x<0. \end{cases} \] Show that $${\rm P}(X>s+t\,\vert X>t) = {\rm P}(X>s).$$ The time it takes an assistant to serve a customer in a certain shop is a random variable with the above distribution and the times for different customers are independent. If, when I enter the shop, the only two assistants are serving one customer each, what is the probability that these customers are both still being served at time \(t\) after I arrive? One of the assistants finishes serving his customer and immediately starts serving me. What is the probability that I am still being served when the other customer has finished being served?

1997 Paper 1 Q14
D: 1484.0 B: 1484.0

The maximum height \(X\) of flood water each year on a certain river is a random variable with density function \begin{equation*} {\mathrm f}(x)= \begin{cases} \exp(-x)&\text{if \(x\geqslant 0\),}\\ 0&\text{otherwise}. \end{cases} \end{equation*} It costs \(y\) megadollars each year to prepare for flood water of height \(y\) or less. If \(X\leqslant y\) no further costs are incurred but if \(X\geqslant y\) the cost of flood damage is \(r+s(X-y)\) megadollars where \(r,s>0\). The total cost \(T\) megadollars is thus given by \begin{equation*} T= \begin{cases} y&\text{if \(X\leqslant y\)},\\ y+r+s(X-y)&\text{if \(X>y\)}. \end{cases} \end{equation*} Show that we can minimise the expected total cost by taking \[y=\ln(r+s).\]

1994 Paper 3 Q13
D: 1700.0 B: 1484.0

During his performance a trapeze artist is supported by two identical ropes, either of which can bear his weight. Each rope is such that the time, in hours of performance, before it fails is exponentially distributed, independently of the other, with probability density function \(\lambda\exp(-\lambda t)\) for \(t\geqslant0\) (and 0 for \(t<0\)), for some \(\lambda>0.\) A particular rope has already been in use for \(t_{0}\) hours of performance. Find the distribution for the length of time the artist can continue to use it before it fails. Interpret and comment upon your result. Before going on tour the artist insists that the management purchase two new ropes of the above type. Show that the probability density function of the time until both ropes fail is \[ \mathrm{f}(t)=\begin{cases} 2\lambda\mathrm{e}^{-\lambda t}(1-\mathrm{e}^{-\lambda t}) & \text{ if }t\geqslant0,\\ 0 & \text{ otherwise.} \end{cases} \] If each performance lasts for \(h\) hours, find the probability that both ropes fail during the \(n\)th performance. Show that the probability that both ropes fail during the same performance is \(\tanh(\lambda h/2)\).

Showing 1-16 of 16 problems
2018 Paper 3 Q12
D: 1700.0 B: 1516.0

A random process generates, independently, \(n\) numbers each of which is drawn from a uniform (rectangular) distribution on the interval 0 to 1. The random variable \(Y_k\) is defined to be the \(k\)th smallest number (so there are \(k-1\) smaller numbers).

  1. Show that, for \(0\le y\le1\,\), \[ {\rm P}\big(Y_k\le y) =\sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} . \tag{\(*\)} \]
  2. Show that \[ m\binom n m = n \binom {n-1}{m-1} \] and obtain a similar expression for \(\displaystyle (n-m) \, \binom n m\,\). Starting from \((*)\), show that the probability density function of \(Y_k\) is \[ n\binom{ n-1}{k-1} y^{k-1}\left(1-y\right)^{ n-k} \,.\] Deduce an expression for \(\displaystyle \int_0^1 y^{k-1}(1-y)^{n-k} \, \d y \,\).
  3. Find \(\E(Y_k) \) in terms of \(n\) and \(k\).

Show Solution
  1. \begin{align*} && \mathbb{P}(Y_k \leq y) &= \sum_{j=k}^n\mathbb{P}(\text{exactly }j \text{ values less than }y) \\ &&&= \sum_{j=k}^m \binom{m}{j} y^j(1-y)^{n-j} \end{align*}
  2. This is the number of ways to choose a committee of \(m\) people with the chair from those \(m\) people. This can be done in two ways. First: choose the committee in \(\binom{n}{m}\) ways and choose the chair in \(m\) ways so \(m \binom{n}{m}\). Alternatively, choose the chain in \(n\) ways and choose the remaining \(m-1\) committee members in \(\binom{n-1}{m-1}\) ways. Therefore \(m \binom{n}{m} = n \binom{n-1}{m-1}\) \begin{align*} (n-m) \binom{n}{m} &= (n-m) \binom{n}{n-m} \\ &= n \binom{n-1}{n-m-1} \\ &= n \binom{n-1}{m} \end{align*} \begin{align*} f_{Y_k}(y) &= \frac{\d }{\d y} \l \sum^{n}_{m=k}\binom{n}{m}y^{m}\left(1-y\right)^{n-m} \r \\ &= \sum^{n}_{m=k} \l \binom{n}{m}my^{m-1}\left(1-y\right)^{n-m} -\binom{n}{m}(n-m)y^{m}\left(1-y\right)^{n-m-1} \r \\ &= \sum^{n}_{m=k} \l n \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n \binom{n-1}{m} y^{m}\left(1-y\right)^{n-m-1} \r \\ &= n\sum^{n}_{m=k} \binom{n-1}{m-1}y^{m-1}\left(1-y\right)^{n-m} -n\sum^{n+1}_{m=k+1} \binom{n-1}{m-1} y^{m-1}\left(1-y\right)^{n-m} \\ &= n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \end{align*} \begin{align*} &&1 &= \int_0^1 f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k-1}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1} \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \Rightarrow && \frac{1}{n \binom{n-1}{k-1}} &= \int_0^1 y^{k-1}(1-y)^{n-k} \d y \\ \end{align*}
  3. \begin{align*} && \mathbb{E}(Y_k) &= \int_0^1 y f_{Y_k}(y) \d y \\ &&&= \int_0^1 n \binom{n-1}{k-1} y^{k}(1-y)^{n-k} \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k}(1-y)^{n-k} \d y \\ &&&= n \binom{n-1}{k-1}\int_0^1 y^{k+1-1}(1-y)^{n+1-(k+1)} \d y \\ &&&= n \binom{n-1}{k-1} \frac{1}{(n+1) \binom{n}{k}}\\ &&&= \frac{n}{n+1} \cdot \frac{k}{n} \\ &&&= \frac{k}{n+1} \end{align*}
2017 Paper 3 Q12
D: 1700.0 B: 1500.2

The discrete random variables \(X\) and \(Y\) can each take the values \(1\), \(\ldots\,\), \(n\) (where \(n\ge2\)). Their joint probability distribution is given by \[ \P(X=x, \ Y=y) = k(x+y) \,, \] where \(k\) is a constant.

  1. Show that \[ \P(X=x) = \dfrac{n+1+2x}{2n(n+1)}\,. \] Hence determine whether \(X\) and \(Y\) are independent.
  2. Show that the covariance of \(X\) and \(Y\) is negative.

2016 Paper 3 Q13
D: 1700.0 B: 1500.0

Given a random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), we define the kurtosis, \(\kappa\), of \(X\) by \[ \kappa = \frac{ \E\big((X-\mu)^4\big)}{\sigma^4} -3 \,. \] Show that the random variable \(X-a\), where \(a\) is a constant, has the same kurtosis as \(X\).

  1. Show by integration that a random variable which is Normally distributed with mean 0 has kurtosis 0.
  2. Let \(Y_1, Y_2, \ldots, Y_n\) be \(n\) independent, identically distributed, random variables with mean 0, and let \(T = \sum\limits_{r=1}^n Y_r\). Show that \[ \E(T^4) = \sum_{r=1}^n \E(Y_r^4) + 6 \sum_{r=1}^{n-1} \sum_{s=r+1}^{n} \E(Y^2_s) \E(Y^2_r) \,. \]
  3. Let \(X_1\), \(X_2\), \(\ldots\)\,, \(X_n\) be \(n\) independent, identically distributed, random variables each with kurtosis \(\kappa\). Show that the kurtosis of their sum is \(\dfrac\kappa n\,\).

Show Solution
\begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}
  1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
  2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
  3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}
2015 Paper 3 Q13
D: 1700.0 B: 1500.0

Each of the two independent random variables \(X\) and \(Y\) is uniformly distributed on the interval~\([0,1]\).

  1. By considering the lines \(x+y =\) \(\mathrm{constant}\) in the \(x\)-\(y\) plane, find the cumulative distribution function of \(X+Y\).
  2. Hence show that the probability density function \(f\) of \((X+Y)^{-1}\) is given by \[ \f(t) = \begin{cases} 2t^{-2} -t^{-3} & \text{for \( \tfrac12 \le t \le 1\)} \\ t^{-3} & \text{for \(1\le t <\infty\)}\\ 0 & \text{otherwise}. \end{cases} \] Evaluate \(\E\Big(\dfrac1{X+Y}\Big)\,\).
  3. Find the cumulative distribution function of \(Y/X\) and use this result to find the probability density function of \(\dfrac X {X+Y}\). Write down \(\E\Big( \dfrac X {X+Y}\Big)\) and verify your result by integration.

Show Solution
  1. \(\mathbb{P}(X + Y \leq c) \) is the area between the \(x\)-axis, \(y\)-axis and the line \(x + y = c\). There are two cases for this: \[\mathbb{P}(X + Y \leq c) = \begin{cases} 0 & \text{ if } c \leq 0 \\ \frac{c^2}{2} & \text{ if } c \leq 1 \\ 1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\ 1 & \text{ otherwise} \end{cases}\]
  2. \begin{align*} && \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\ \Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases} 0 & \text{ if } \frac1{t} \leq 0 \\ \frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\ \frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ (2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ 2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \end{align*} Therefore, \begin{align*} \E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\ &= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\ &= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\ &= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\ &= 1 + 2 \ln 2 -2 + 1 \\ &= 2 \ln 2 \end{align*}
  3. \begin{align*} &&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\ &&&= \begin{cases} 0 & \text{if } c \leq 0 \\ \frac{c}{2} & \text{if } 0 \leq c \leq 1 \\ 1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\ \\ \Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\ &&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\ &&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\ &&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\ &&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\ 1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\ && f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t^2} & \text{if } t \geq 1 \\ \frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\ \Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\ &&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\ &&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ \\ && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\ &&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\ &&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\ &&&= \frac{1}{2} \end{align*} We can also notice that \(1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r\) so it's clearly true as long as we can show that the integral converges.
2013 Paper 3 Q12
D: 1700.0 B: 1500.0

A list consists only of letters \(A\) and \(B\) arranged in a row. In the list, there are \(a\) letter \(A\)s and \(b\) letter \(B\)s, where \(a\ge2\) and \(b\ge2\), and \(a+b=n\). Each possible ordering of the letters is equally probable. The random variable \(X_1\) is defined by \[ X_1 = \begin{cases} 1 & \text{if the first letter in the row is \(A\)};\\ 0 & \text{otherwise.} \end{cases} \] The random variables \(X_k\) (\(2 \le k \le n\)) are defined by \[ X_k = \begin{cases} 1 & \text{if the \((k-1)\)th letter is \(B\) and the \(k\)th is \(A\)};\\ 0 & \text{otherwise.} \end{cases} \] The random variable \(S\) is defined by \(S = \sum\limits_ {i=1}^n X_i\,\).

  1. Find expressions for \(\E(X_i)\), distinguishing between the cases \(i=1\) and \(i\ne1\), and show that \(\E(S)= \dfrac{a(b+1)}n\,\).
  2. Show that:
    1. [ for \(j\ge3\), \(\E(X_1X_j) = \dfrac{a(a-1)b}{n(n-1)(n-2)}\,\);
    2. $\ds \sum\limits_{i=2}^{n-2} \bigg( \sum\limits_{j=i+2}^n \E(X_iX_j)\bigg) = \dfrac{a(a-1)b(b-1)}{2n(n-1)}\,\(;
    3. \)\var(S) = \dfrac {a(a-1)b(b+1)}{n^2(n-1)}\,$.

2010 Paper 3 Q13
D: 1700.0 B: 1516.0

In this question, \({\rm Corr}(U,V)\) denotes the product moment correlation coefficient between the random variables \(U\) and \(V\), defined by \[ \mathrm{Corr}(U,V) \equiv \frac{\mathrm{Cov}(U,V)}{\sqrt{\var(U)\var(V)}}\,. \] The independent random variables \(Z_1\), \(Z_2\) and \(Z_3\) each have expectation 0 and variance 1. What is the value of \(\mathrm{Corr} (Z_1,Z_2)\)? Let \(Y_1 = Z_1\) and let \[ Y_2 = \rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2\,, \] where \(\rho_{12}\) is a given constant with $-1<\rho _{12}<1$. Find \(\E(Y_2)\), \(\var(Y_2)\) and \(\mathrm{Corr}(Y_1, Y_2)\). Now let \(Y_3 = aZ_1 + bZ_2 + cZ_3\), where \(a\), \(b\) and \(c\) are real constants and \(c\ge0\). Given that \(\E(Y_3) = 0\), \(\var(Y_3) = 1\), \( \mathrm{Corr}(Y_1, Y_3) =\rho^{{2}}_{13} \) and \( \mathrm{Corr}(Y_2, Y_3)= \rho^{{2}} _{23}\), express \(a\), \(b\) and \(c\) in terms of \(\rho^{2} _{23}\), \(\rho^{2}_{13}\) and \(\rho^{2} _{12}\). Given constants \(\mu_i\) and \(\sigma_i\), for \(i=1\), \(2\) and \(3\), give expressions in terms of the \(Y_i\) for random variables \(X_i\) such that \(\E(X_i) = \mu_i\), \(\var(X_i) = \sigma_ i^2\) and \(\mathrm{Corr}(X_i,X_j) = \rho_{ij}\).

Show Solution
\begin{align*} \mathrm{Corr} (Z_1,Z_2) &= \frac{\mathrm{Cov}(Z_1,Z_2)}{\sqrt{\var(Z_1)\var(Z_2)}} \\ &= \frac{\mathbb{E}(Z_1 Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{\mathbb{E}(Z_1)\mathbb{E}(Z_2)}{\sqrt{1 \cdot 1}} \\ &= \frac{0}{1} \\ &= 0 \end{align*} \begin{align*} && \mathbb{E}(Y_2) &= \mathbb{E}(\rho_{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \mathbb{E}(\rho_{12} Z_1) + \mathbb{E}( (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}\mathbb{E}( Z_1) + (1 - {\rho_{12}^2})^{ \frac12}\mathbb{E}( Z_ 2) \\ &&&= 0\\ \\ && \textrm{Var}(Y_2) &= \textrm{Var}(\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \textrm{Var}(\rho_{12} Z_1)+\textrm{Cov}(\rho_{12} Z_1,(1 - {\rho_{12}^2})^{ \frac12} Z_ 2 ) + \textrm{Var}((1 - {\rho_{12}^2})^{ \frac12} Z_ 2) \\ &&&= \rho_{12}^2\textrm{Var}( Z_1)+\rho_{12} (1 - {\rho_{12}^2})^{ \frac12} \textrm{Cov}(Z_1, Z_ 2 ) + (1 - {\rho_{12}^2})\textrm{Var}(Z_ 2) \\ &&&= \rho_{12}^2 + (1-\rho_{12}^2) = 1 \\ \\ && \textrm{Cov}(Y_1, Y_2) &= \mathbb{E}((Y_1-0)(Y_2-0)) \\ &&&= \mathbb{E}(Z_1 \cdot (\rho _{12} Z_1 + (1 - {\rho_{12}^2})^{ \frac12} Z_ 2)) \\ &&&= \rho_{12} \mathbb{E}(Z_1^2) + (1-\rho_{12}^2)^{\frac12}\mathbb{E}(Z_1, Z_2) \\ &&&= \rho_{12} \\ \Rightarrow && \textrm{Corr}(Y_1, Y_2) &= \frac{\textrm{Cov}(Y_1, Y_2)}{\sqrt{\textrm{Var}(Y_1)\textrm{Var}(Y_2)}} \\ &&&= \frac{\rho_{12}}{1 \cdot 1} = \rho_{12} \end{align*} Suppose \(Y_3 =aZ_1 +bZ_2+cZ_3\) with \(\mathbb{E}(Y_3) = 0\) (must be true), \(\textrm{Var}(Y_3) = 1 = a^2+b^2+c^2\) and \(\textrm{Corr}(Y_1, Y_3) = \rho_{13}, \textrm{Corr}(Y_2, Y_3) = \rho_{23}\). \begin{align*} && \textrm{Corr}(Y_1,Y_3) &= \textrm{Cov}(Y_1, Y_3) \\ &&&= \textrm{Cov}(Z_1, aZ_1 +bZ_2+cZ_3) \\ &&&= a \\ \Rightarrow && a &= \rho_{13} \\ \\ && \textrm{Corr}(Y_2,Y_3) &= \textrm{Cov}(Y_2, Y_3) \\ &&&= \textrm{Cov}(\rho_{12}Z_1+(1-\rho_{12}^2)^\frac12Z_2, \rho_{13}Z_1 +bZ_2+cZ_3) \\ &&&= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && \rho_{23} &= \rho_{12}\rho_{13}+(1-\rho_{12}^2)^\frac12b \\ \Rightarrow && b &= \frac{\rho_{23}-\rho_{12}\rho_{13}}{(1-\rho_{12}^2)^\frac12} \\ && c &= \sqrt{1-\rho_{13}^2-\frac{(\rho_{23}-\rho_{12}\rho_{13})^2}{(1-\rho_{12}^2)}} \end{align*} Finally, let \(X_i = \mu_i + \sigma_i Y_i\)
2007 Paper 3 Q12
D: 1700.0 B: 1487.4

I choose a number from the integers \(1, 2, \ldots\), \((2n-1)\) and the outcome is the random variable~\(N\). Calculate \( \E(N)\) and \(\E(N^2)\). I then repeat a certain experiment \(N\) times, the outcome of the \(i\)th experiment being the random variable \(X_i\) ($1\le i \le N\(). For each \)i$, the random variable \(X_i\) has mean \(\mu\) and variance~\(\sigma^2\), and \(X_i\) is independent of \(X_j\) for \(i\ne j\) and also independent of \(N\). The random variable \(Y\) is defined by \(Y= \sum\limits_{i=1}^NX_i\). Show that \(\E(Y)=n\mu\) and that \(\mathrm{Cov}(Y,N) = \frac13n(n-1)\mu\). Find \(\var(Y) \) in terms of \(n\), \(\sigma^2\) and \(\mu\).

2006 Paper 3 Q14
D: 1700.0 B: 1516.0

For any random variables \(X_1\) and \(X_2\), state the relationship between \(\E(aX_1+bX_2)\) and \(\E(X_1)\) and \(\E(X_2)\), where \(a\) and \(b\) are constants. If \(X_1\) and \(X_2\) are independent, state the relationship between \(\E(X_1X_2)\) and \(\E(X_1)\) and \(\E(X_2)\). An industrial process produces rectangular plates. The length and the breadth of the plates are modelled by independent random variables \(X_1\) and \(X_2\) with non-zero means \(\mu_1\) and \(\mu_2\) and non-zero standard deviations \(\sigma_1\) and \(\sigma_2\), respectively. Using the results in the paragraph above, and without quoting a formula for \(\var(aX_1+bX_2)\), find the means and standard deviations of the perimeter \(P\) and area \(A\) of the plates. Show that \(P\) and \(A\) are not independent. The random variable \(Z\) is defined by \(Z=P-\alpha A\), where \(\alpha \) is a constant. Show that \(Z\) and \(A\) are not independent if \[ \alpha \ne \dfrac{2(\mu_1^{\vphantom2} \sigma_2^2 +\mu_2^{\vphantom2}\sigma_1^2)} { \mu_1^2 \sigma_2^2 +\mu_2^2\sigma_1^2 + \sigma_1^2\sigma_2^2 } \;. \] Given that \(X_1\) and \(X_2\) can each take values 1 and 3 only, and that they each take these values with probability \(\frac 12\), show that \(Z\) and \(A\) are not independent for any value of \(\alpha\).

Show Solution
\(\E(aX_1+bX_2) = a \E(X_1) + b\E(X_2)\) for any \(X_1, X_2\) \(\E(X_1X_2)=\E(X_1)\E(X_2)\). if \(X_1, X_2\) are independent. \begin{align*} && \E(P) &= \E(2(X_1+X_2)) = 2(\E[X_1]+\E[X_2]) \\ &&&= 2(\mu_1 + \mu_2) \\ && \var(P) &= \E[\left ( 2(X_1+X_2) \right)^2] - \E[2(X_1+X_2)]^2 \\ &&&= 4\E[X_1^2+2X_1X_2+X_2^2] -4(\mu_1 + \mu_2)^2 \\ &&&= 4(\mu_1^2 + \sigma_1^2 + 2\mu_1\mu_2 + \mu_2^2 + \sigma_2^2) - 4(\mu_1 + \mu_2)^2 \\ &&&= 4(\sigma_1^2+\sigma_2^2) \\ && \textrm{SD}(P) &= 2 \sqrt{\sigma_1^2+\sigma_2^2}\\ \\ && \E(A) &= \E[X_1X_2] = \E[X_1]\E[X_2] \\ &&&= \mu_1\mu_2 \\ && \var(A) &= \E[(X_1X_2)^2] - (\mu_1\mu_2)^2 \\ &&&= (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) - (\mu_1\mu_2)^2\\ &&&= \mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2\\ && \textrm{SD}(A) &= \sqrt{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2} \end{align*} \begin{align*} \E[PA] &= \E[2(X_1+X_2)X_1X_2] \\ &= 2\E[X_1^2X_2] + 2\E[X_1X_2^2]\\ &= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2)\\ &\neq 2(\mu_1 + \mu_2)\mu_1\mu_2 \\ &= \E[P]\E[A] \end{align*} \begin{align*} && \E[Z] &= \E[P] - \alpha \E[A] \\ &&&= 2(\mu_1+\mu_2) - \alpha \mu_1 \mu_2 \\ \\ && \E[ZA] &= \E[PA - \alpha A^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha \E[A^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha \E[X_1^2]\E[X_2^2] \\ &&&= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \text{if ind.} && \E[Z]\E[A] &= \E[ZA]\\ && (2(\mu_1+\mu_2) - \alpha \mu_1 \mu_2) \mu_1\mu_2 &= 2(\mu_1^2 + \sigma_1^2)\mu_2 + 2\mu_1 (\mu_2^2+\sigma_2^2) - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \Rightarrow && 2(\mu_1^2\mu_2+\mu_1\mu_2^2) - \alpha \mu_1^2\mu_2^2 &= 2(\mu_1^2\mu_2+\mu_1\mu_2^2) + 2\sigma_1^2\mu_2 + 2\sigma_2^2\mu_1 - \alpha (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) \\ \Rightarrow && \alpha ((\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) - \mu_1^2\mu_2^2) &= 2(\sigma_1^2\mu_2 + \sigma_2^2\mu_1) \\ \Rightarrow && \alpha &= \frac{ 2(\sigma_1^2\mu_2 + \sigma_2^2\mu_1) }{\mu_1^2 \sigma_2^2 + \mu_2^2 \sigma_1^2 + \sigma_1^2 \sigma_2^2} \end{align*} Therefore if they are not independent if \(\alpha \neq \) the expression. \begin{array}{c|c|c|c|c|c} & X_1 & X_2 & A & P & Z \\ \hline 0.25 & 1 & 1 & 1 & 4 & 4-\alpha \\ 0.25 & 1 & 3 & 3 & 8 & 8-3\alpha \\ 0.25 & 3 & 1 & 3 & 8 & 8-3\alpha \\ 0.25 & 3 & 3 & 9 & 12 & 12-9\alpha \\ \end{array} If \(\mathbb{P}(A = 1, Z = 4-\alpha) = \mathbb{P}(A = 1)\mathbb{P}(Z = 4-\alpha)\) then \(\mathbb{P}(Z = 4-\alpha) = 1\), but that mean \(4-\alpha = 8-3\alpha = 12-9\alpha\) which is not a consistent set of equations as the first two are solved by \(\alpha = 2\) and the second by \(\alpha = \frac23\)
2005 Paper 3 Q12
D: 1700.0 B: 1516.0

Five independent timers time a runner as she runs four laps of a track. Four of the timers measure the individual lap times, the results of the measurements being the random variables \(T_1\) to \(T_4\), each of which has variance \(\sigma^2\) and expectation equal to the true time for the lap. The fifth timer measures the total time for the race, the result of the measurement being the random variable \(T\) which has variance \(\sigma^2\) and expectation equal to the true race time (which is equal to the sum of the four true lap times). Find a random variable \(X\) of the form \(aT+b(T_1+T_2+T_3+T_4)\), where \(a\) and \(b\) are constants independent of the true lap times, with the two properties: (1) \ whatever the true lap times, the expectation of \(X\) is equal to the true race time; (2) \ the variance of \(X\) is as small as possible. Find also a random variable \(Y\) of the form \(cT+d(T_1+T_2+T_3+T_4)\), where \(c\) and \(d\) are constants independent of the true lap times, with the property that, whatever the true lap times, the expectation of \(Y^2\) is equal to \(\sigma^2\). In one particular race, \(T\) takes the value 220 seconds and \((T_1 + T_2 + T_3 + T_4)\) takes the value \(220.5\) seconds. Use the random variables \(X\) and \(Y\) to estimate an interval in which the true race time lies.

2004 Paper 3 Q12
D: 1700.0 B: 1500.0

A team of \(m\) players, numbered from \(1\) to \(m\), puts on a set of a \(m\) shirts, similarly numbered from \(1\) to \(m\). The players change in a hurry, so that the shirts are assigned to them randomly, one to each player. Let \(C_i\) be the random variable that takes the value \(1\) if player \(i\) is wearing shirt \(i\), and 0 otherwise. Show that \(\mathrm{E}\left(C_1\right)={1 \over m}\) and find \(\var \left(C_1\right)\) and \(\mathrm{Cov}\left(C_1 \, , \; C_2 \right) \,\). Let \(\, N = C_1 + C_2 + \cdots + C_m \,\) be the random variable whose value is the number of players who are wearing the correct shirt. Show that \(\mathrm{E}\left(N\right)= \var \left(N\right) = 1 \,\). Explain why a Normal approximation to \(N\) is not likely to be appropriate for any \(m\), but that a Poisson approximation might be reasonable. In the case \(m = 4\), find, by listing equally likely possibilities or otherwise, the probability that no player is wearing the correct shirt and verify that an appropriate Poisson approximation to \(N\) gives this probability with a relative error of about \(2\%\). [Use \(\e \approx 2\frac{72}{100} \,\).]

Show Solution
There are \(m!\) different ways of assigning the shirts, and in \((m-1)!\) of them player \(1\) gets their own shirt, ie \(\mathbb{E}(C_1) = \mathbb{P}(\text{player }1\text{ gets own shirt}) = \frac{(m-1)!}{m!} = \frac{1}{m}\). \(\var(C_1) = \mathbb{E}(C_1^2) - [\mathbb{E}(C_1)]^2 = \frac{1}{m} - \frac{1}{m^2} = \frac{m-1}{m^2}\). If we have two players, there are \((m-2)!\) ways they both get their own shirts, therefore \(\textrm{Cov}(C_1,C_2) = \mathbb{E}(C_1C_2) - \mathbb{E}(C_1)\mathbb{E}(C_2) = \frac{(m-2)!}{m!} - \frac{1}{m^2} = \frac{1}{m(m-1)} - \frac{1}{m^2} = \frac{m-m+1}{m^2(m-1)} = \frac{1}{m^2(m-1)}\). \begin{align*} \mathbb{E}(N) &= \mathbb{E}(C_1 + C_2 + \cdots + C_m) \\ &= \mathbb{E}(C_1) + \mathbb{E}(C_2) + \cdots + \mathbb{E}(C_m) \\ &= \frac{1}{m} + \frac{1}{m} +\cdots+ \frac1m \\ &= 1 \\ \\ \var(N) &= \sum_{r=1}^m \var(C_r) + 2\sum_{r=1}^{m-1} \sum_{s=2}^{m} \textrm{Cov}(C_r,C_s) \\ &= m \frac{m-1}{m^2} + 2 \frac{m(m-1)}{2}\frac{1}{m^2(m-1)} \\ &=\frac{m-1}{m} + \frac{1}{m} \\ &= 1 \end{align*} If we were to take a normal approximation, we would want to take \(N(1,1)\), but this would say things like \(-1\) is as likely as \(3\) shirts being correct, which is clearly a bad model. A Poisson is much more likely to be a sensible model as they have the same mean and variance as the parameter, and if \(m\) is large, the covariance between shirts is going to be very small, so it will appear similar to random events occurring. We can have \begin{align*} BADC \\ BCDA \\ BDAC \\ CADB \\ CDAB\\ CDBA \\ DABC\\ DCAB \\ DCBA \end{align*} Ie \(\frac{9}{24}\) ways to have no player wearing their own shirt with \(4\) players. \(Po(1)\) would say this probability is \(e^{-1}\), giving a relative error of: \begin{align*} \frac{e^{-1}-\frac{9}{24}}{\frac9{24}} &\approx \frac{\frac{100}{272} - \frac{9}{24}}{\frac9{24}} \\ &= -\frac{1}{51} \\ &\approx -2\% \end{align*}
2002 Paper 3 Q14
D: 1700.0 B: 1500.0

Prove that, for any two discrete random variables \(X\) and \(Y\), \[ \mathrm{Var} \left(X + Y \right) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2 \, \mathrm{Cov}(X,Y), \] where \(\mathrm{Var}(X)\) is the variance of \(X\) and \(\mathrm{Cov}(X,Y)\) is the covariance of \(X\) and \(Y\). When a Grandmaster plays a sequence of \(m\) games of chess, she is, independently, equally likely to win, lose or draw each game. If the values of the random variables \(W\), \(L\) and \(D\) are the numbers of her wins, losses and draws respectively, justify briefly the following claims:

  1. \(W + L + D\) has variance \(0\,\);
  2. \(W + L\) has a binomial distribution.
Find the value of %$\displaystyle \rho \left[ W , \, L \right] = \(\displaystyle {\mathrm{Cov}(W,L) \over \sqrt{\mathrm{Var}(W) \mathrm{Var}(L)}}\;\).

2000 Paper 3 Q14
D: 1700.0 B: 1500.0

The random variable \(X\) takes only the values \(x_1\) and \(x_2\) (where \( x_1 \not= x_2 \)), and the random variable \(Y\) takes only the values \(y_1\) and \(y_2\) (where \(y_1 \not= y_2\)). Their joint distribution is given by $$ \P ( X = x_1 , Y = y_1 ) = a \ ; \ \ \P ( X = x_1 , Y = y_2 ) = q - a \ ; \ \ \P ( X = x_2 , Y = y_1 ) = p - a \ . $$ Show that if \(\E(X Y) = \E(X)\E(Y)\) then $$ (a - p q ) ( x_1 - x_2 ) ( y_1 - y_2 ) = 0 . $$ Hence show that two random variables each taking only two distinct values are independent if \(\E(X Y) = \E(X) \E(Y)\). Give a joint distribution for two random variables \(A\) and \(B\), each taking the three values \(- 1\), \(0\) and \(1\) with probability \({1 \over 3}\), which have \(\E(A B) = \E( A)\E (B)\), but which are not independent.

Show Solution
\begin{align*} \mathbb{P}(X = x_1) &= a + q - a = q \\ \mathbb{P}(X = x_2) &= 1 - q \\ \mathbb{P}(Y = y_1) & = a + p - a = p \\ \mathbb{P}(Y = y_2) & = 1 - p \end{align*} \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &= \l qx_1 + (1-q)x_2 \r \l p y_1 + (1-p)y_2\r \\ &= qpx_1y_1 + q(1-p)x_1y_2 + (1-q)px_2y_1 + (1-q)(1-p)x_2y_2 \\ \mathbb{E}(XY) &= ax_1y_1 + (q-a)x_1y_2 + (p-a)x_2y_1 + (1 + a - p - q)x_2y_2 &= \end{align*} Therefore \(\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)\) is a degree 2 polynomial in the \(x_i, y_i\). If \(x_1 = x_2\) then we have: \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &=x_1 \l p y_1 + (1-p)y_2\r \\ \mathbb{E}(XY) &= x_1(ay_1 + (q-a)y_2 + (p-a)y_1 + (1 + a - p - q)y_2) \\ &= x_1 (py_1 + (1-p)y_2) \end{align*} Therefore \(x_1 - x_2\) is a root and by symmetry \(y_1 - y_2\) is a root. Therefore it remains to check the coefficient of \(x_1y_1\) which is \(a - pq\) to complete the factorisation. For any two random variables taking two distinct values, we can find \(a, q, p\) satisfying the relations above. We also note that \(X\) and \(Y\) are independent if \(\mathbb{P}(X = x_i, Y = y_i) = \mathbb{P}(X = x_i)\mathbb{P}(Y = y_i)\). Since \(x_1 \neq x_2\) and \(y_1 \neq y_2\) and \(\E(A B) = \E( A)\E (B) \Rightarrow a = pq\). But if \(a = pq\), we have \(\mathbb{P}(X = x_1, Y = y_1) = \mathbb{P}(X = x_1)\mathbb{P}(Y = y_1)\) and all the other relations drop out similarly. Consider \begin{align*} \mathbb{P}(A = -1, B = 1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = 0, B = 0) &= \frac{1}{3} \\ \mathbb{P}(A = 1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \end{align*}
1997 Paper 3 Q14
D: 1700.0 B: 1516.0

An industrial process produces rectangular plates of mean length \(\mu_{1}\) and mean breadth \(\mu_{2}\). The length and breadth vary independently with non-zero standard deviations \(\sigma_{1}\) and \(\sigma_{2}\) respectively. Find the means and standard deviations of the perimeter and of the area of the plates. Show that the perimeter and area are not independent.

1995 Paper 3 Q12
D: 1700.0 B: 1484.0

The random variables \(X\) and \(Y\) are independently normally distributed with means 0 and variances 1. Show that the joint probability density function for \((X,Y)\) is \[ \mathrm{f}(x,y)=\frac{1}{2\pi}\mathrm{e}^{-\frac{1}{2}(x^{2}+y^{2})}\qquad-\infty < x < \infty,-\infty < y < \infty. \] If \((x,y)\) are the coordinates, referred to rectangular axes, of a point in the plane, explain what is meant by saying that this density is radially symmetrical. The random variables \(U\) and \(V\) have a joint probability density function which is radially symmetrical (in the above sense). By considering the straight line with equation \(U=kV,\) or otherwise, show that \[ \mathrm{P}\left(\frac{U}{V} < k\right)=2\mathrm{P}(U < kV,V > 0). \] Hence, or otherwise, show that the probability density function of \(U/V\) is \[ \mathrm{g}(k)=\frac{1}{\pi(1+k^{2})}\qquad-\infty < k < \infty. \]

1991 Paper 3 Q16
D: 1700.0 B: 1504.3

The random variables \(X\) and \(Y\) take integer values \(x\) and \(y\) respectively which are restricted by \(x\geqslant1,\) \(y\geqslant1\) and \(2x+y\leqslant2a\) where \(a\) is an integer greater than 1. The joint probability is given by \[ \mathrm{P}(X=x,Y=y)=c(2x+y), \] where \(c\) is a positive constant, within this region and zero elsewhere. Obtain, in terms of \(x,c\) and \(a,\) the marginal probability \(\mathrm{P}(X=x)\) and show that \[ c=\frac{6}{a(a-1)(8a+5)}. \] Show that when \(y\) is an even number the marginal probability \(\mathrm{P}(Y=y)\) is \[ \frac{3(2a-y)(2a+2+y)}{2a(a-1)(8a+5)} \] and find the corresponding expression when \(y\) is off. Evaluate \(\mathrm{E}(Y)\) in terms of \(a\).

1990 Paper 3 Q16
D: 1700.0 B: 1484.0

  1. A rod of unit length is cut into pieces of length \(X\) and \(1-X\); the latter is then cut in half. The random variable \(X\) is uniformly distributed over \([0,1].\) For some values of \(X\) a triangle can be formed from the three pieces of the rod. Show that the conditional probability that, if a triangle can be formed, it will be obtuse-angled is \(3-2\sqrt{2.}\)
  2. The bivariate distribution of the random variables \(X\) and \(Y\) is uniform over the triangle with vertices \((1,0),(1,1)\) and \((0,1).\) A pair of values \(x,y\) is chosen at random from this distribution and a (perhaps degenerate) triangle \(ABC\) is constructed with \(BC=x\) and \(CA=y\) and \(AB=2-x-y.\) Show that the construction is always possible and that \(\angle ABC\) is obtuse if and only if \[ y>\frac{x^{2}-2x+2}{2-x}. \] Deduce that the probability that \(\angle ABC\) is obtuse is \(3-4\ln2.\)

Show Solution
  1. TikZ diagram
    The construction is possible if \(x + y > 2-x-y \Rightarrow x+y > 1\) (which is as the triangle is above the diagonal line), and \(x + (2-x-y) > y \Rightarrow 1 > y\) (true as the triangle is below the horizontal line) and \(y + (2-x-y) > x \Rightarrow 1 > x\) (true as the triangle is left of the vertical arrow). By the cosine rule: \begin{align*} && y^2 &= x^2 + (2-x-y)^2 - 2 x (2-x-y) \cos \angle ABC \\ \Rightarrow && \cos \angle ABC &= \frac{x^2+(2-x-y)^2 - y^2}{2x(2-x-y)} \\ &&&= \frac{4+2x^2-4x-4y+2xy}{2x(2-x-y)} \\ \underbrace{\Rightarrow}_{\cos \angle ABC < 0} && 0 &> 4+2x^2-4x-4y+2xy \\ \Rightarrow && 0 &> 2x^2-4x+4 - 2(x-2)y \\ \Rightarrow && y &> \frac{x^2-2x+2}{2-x} \\ &&&= -x + \frac{2}{2-x} \end{align*}
    TikZ diagram
    Therefore the area we want is: \begin{align*} A &= 1 - \int_0^1 \left ( -x + \frac{2}{2-x} \right)\d x \\ &= 1 - \left [-\frac12 x^2 - 2 \ln(2-x) \right]_0^1 \\ &= 1 + \frac12 -2 \ln 2 \\ &= \frac32 - 2 \ln 2 \end{align*} Therefore the relative area is: \(\frac{\frac32 - 2 \ln 2}{1/2} = 3 - 4 \ln 2\)

No problems in this section yet.

Showing 1-4 of 4 problems
2008 Paper 3 Q12
D: 1700.0 B: 1516.0

Let \(X\) be a random variable with a Laplace distribution, so that its probability density function is given by \[ \f(x) = \frac12 \e^{-\vert x \vert }\;, \text{ \ \ \ \(-\infty < x < \infty \)\ }. \tag{\(*\)} \] Sketch \(\f(x)\). Show that its moment generating function \({\rm M}_X(\theta)\) is given by ${\rm M}_X(\theta) = (1-\theta^2)^{-1}\( and hence find the variance of \)X$. A frog is jumping up and down, attempting to land on the same spot each time. In fact, in each of \(n\) successive jumps he always lands on a fixed straight line but when he lands from the \(i\)th jump (\(i=1\,,2\,,\ldots\,,n\)) his displacement from the point from which he jumped is \(X_i\,\)cm, where \(X_i\) has the distribution \((*)\). His displacement from his starting point after \(n\)~jumps is \(Y\,\)cm (so that \(Y=\sum\limits_{i=1}^n X_i\)). Each jump is independent of the others. Obtain the moment generating function for \(Y/ \sqrt {2n}\) and, by considering its logarithm, show that this moment generating function tends to \(\exp(\frac12\theta^2)\) as \(n\to\infty\). Given that \(\exp(\frac12\theta^2)\) is the moment generating function of the standard Normal random variable, estimate the least number of jumps such that there is a \(5\%\) chance that the frog lands 25~cm or more from his starting point.

1999 Paper 3 Q12
D: 1700.0 B: 1500.0

In the game of endless cricket the scores \(X\) and \(Y\) of the two sides are such that \[ \P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\] for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).

  1. Find \(\P(X+Y=n)\) for each \(n>0\).
  2. Show that \(2\lambda \e^{2\lambda-1}=1\).
  3. Show that \(2x \e^{2x-1}\) is an increasing function of \(x\) for \(x>0\) and deduce that the equation in (ii) has at most one solution and hence determine \(\lambda\).
  4. Calculate the expectation \(\E(2^{X+Y})\).

Show Solution
  1. \begin{align*} && \mathbb{P}(X+Y = n) &= \sum_{i = 0}^n \mathbb{P}(X = i, Y = n-i) \\ &&&= \sum_{i = 0}^n e^{-1} \frac{n \lambda^n}{i! (n-i)!} \\ &&&=e^{-1} n \lambda^n \sum_{i = 0}^n\frac{1}{i! (n-i)!} \\ &&&=\frac{e^{-1} n}{n!} \lambda^n \sum_{i = 0}^n\frac{n!}{i! (n-i)!} \\ &&&= \frac{n\lambda^n}{e n!} 2^n \\ &&&= \frac{n (2 \lambda)^n}{e \cdot n!} \end{align*}
  2. \begin{align*} && 1 &= \sum_{n = 0}^{\infty} \mathbb{P}(X+Y =n ) \\ &&&= \sum_{n = 0}^{\infty}\frac{n (2 \lambda)^n}{e \cdot n!} \\ &&&= \sum_{n = 1}^\infty \frac{ (2 \lambda)^n}{e \cdot (n-1)!} \\ &&&= \frac{2 \lambda}{e}\sum_{n = 0}^\infty \frac{ (2 \lambda)^n}{n!} \\ &&&= \frac{2 \lambda}{e} e^{2\lambda} \\ &&&= 2 \lambda e^{2\lambda - 1} \end{align*} \\
  3. Consider \(f(x) = 2xe^{2x-1}\), then \begin{align*} && f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\ &&&= e^{2x-1} (2 + 4x) > 0 \end{align*} Therefore \(f(x)\) is an increasing function of \(x\), which means \(f(x) = 1\) has at most one solution for \(\lambda\). Therefore \(\lambda = \frac12\)
  4. \begin{align*} \mathbb{E}(2^{X+Y}) &= \sum_{n = 0}^\infty \mathbb{P}(X+Y = n) 2^n \\ &= \sum_{n = 1}^\infty \frac{1}{e(n-1)!} 2^{n} \\ &= \frac{2}{e} \sum_{n=0}^\infty \frac{2^n}{n!} \\ &= \frac{2}{e} e^2 \\ &= 2e \end{align*}
1997 Paper 3 Q13
D: 1700.0 B: 1500.0

Let \(X\) and \(Y\) be independent standard normal random variables: the probability density function, \(\f\), of each is therefore given by \[ \f(x)=\left(2\pi\right)^{-\frac{1}{2}}\e^{-\frac{1}{2}x^{2}}. \]

  1. Find the moment generating function \(\mathrm{E}(\e^{\theta X})\) of \(X\).
  2. Find the moment generating function of \(aX+bY\) and hence obtain the condition on \(a\) and \(b\) which ensures that \(aX+bY\) has the same distribution as \(X\) and \(Y\).
  3. Let \(Z=\e^{\mu+\sigma X}\). Show that \[ \mathrm{E}(Z^{\theta})=\e^{\mu\theta+\frac{1}{2}\sigma^{2}\theta^{2}}, \] and hence find the expectation and variance of \(Z\).

1995 Paper 3 Q14
D: 1700.0 B: 1516.0

A candidate finishes examination questions in time \(T\), where \(T\) has probability density function \[ \mathrm{f}(t)=t\mathrm{e}^{-t}\qquad t\geqslant0, \] the probabilities for the various questions being independent. Find the moment generating function of \(T\) and hence find the moment generating function for the total time \(U\) taken to finish two such questions. Show that the probability density function for \(U\) is \[ \mathrm{g}(u)=\frac{1}{6}u^{3}\mathrm{e}^{-u}\qquad u\geqslant0. \] Find the probability density function for the total time taken to answer \(n\) such questions.

Show Solution
\begin{align*} && M_T(x) &= \mathbb{E}[e^{xT}] \\ &&&= \int_0^{\infty} e^{xt}te^{-t} \d t \\ &&&= \int_0^{\infty}te^{(x-1)t} \d t \\ &&&= \left [ \frac{t}{x-1} e^{(x-1)t} \right]_0^{\infty} - \int_0^\infty \frac{e^{(x-1)t}}{x-1} \d t \\ &&&= \left [ \frac{e^{(x-1)t}}{(x-1)^2} \right]_0^{\infty} \\ &&&= \frac{1}{(x-1)^2} \\ \\ && M_U(x) &= M_{T_1+T_2}(x) \\ &&&= \frac1{(x-1)^4} \\ \\ && I_n &= \int_0^{\infty} t^ne^{(x-1)t} \d t \\ &&&= \left[ \frac{1}{(x-1)}t^ne^{(x-1)t} \right]_0^{\infty} - \frac{n}{(x-1)} \int_0^{\infty}t^{n-1}e^{(x-1)t} \d t \\ &&&= -\frac{n}{(x-1)}I_{n-1} \\ \Rightarrow && I_n &= \frac{n!}{(1-x)^{n+1}} \\ \\ \Rightarrow && \int_0^{\infty} e^{xt} \frac16u^3e^{-u} \d u &= \int_0^{\infty} \frac16u^3e^{(x-1)u} \d u \\ &&&= \frac{1}{(1-x)^4} \\ \Rightarrow && f_U(u) &= \frac16u^3e^{-u} \\ \\ && M_{X_1+\cdots+X_n}(x) &= \frac{1}{(x-1)^{2n}} \\ \Rightarrow && f_{X_1+\cdots+X_n}(t) &= \frac1{(2n-1)!} t^{2n-1}e^{-t} \end{align*} (NB: This is the gamma distribution)

No problems in this section yet.

Showing 1-9 of 9 problems
2009 Paper 3 Q13
D: 1700.0 B: 1488.4

  1. The point \(P\) lies on the circumference of a circle of unit radius and centre \(O\). The angle,~\(\theta\), between \(OP\) and the positive \(x\)-axis is a random variable, uniformly distributed on the interval \(0\le\theta<2\pi\). The cartesian coordinates of \(P\) with respect to \(O\) are \((X, Y)\). Find the probability density function for \(X\), and calculate \(\var (X)\). Show that \(X\) and \(Y\) are uncorrelated and discuss briefly whether they are independent.
  2. The points \(P_i\) (\(i=1\), \(2\), \(\ldots\) , \(n\)) are chosen independently on the circumference of the circle, as in part (i), and have cartesian coordinates \((X_i, Y_i)\). The point \(\overline P\) has coordinates \((\overline X, \overline Y)\), where \(\overline X =\dfrac1n \sum\limits _{i=1}^n X_i\) and \(\overline Y =\dfrac1n \sum\limits _{i=1}^n Y_i\). Show that \(\overline X\) and \(\overline Y\) are uncorrelated. Show that, for large \(n\), $\displaystyle \P\left(\vert \overline X \vert \le \sqrt{\frac2n}\right) \approx 0.95\,$.

2001 Paper 3 Q14
D: 1700.0 B: 1484.0

A random variable \(X\) is distributed uniformly on \([\, 0\, , \, a\,]\). Show that the variance of \(X\) is\phantom{ }\({1 \over 12} a^2\). A sample, \(X_1\) and \(X_2\), of two independent values of the random variable is drawn, and the variance \(V\) of the sample is determined. Show that \(V = {1 \over 4} \l X_1 -X_2 \r ^2\), and hence prove that \(2 V\) is an unbiased estimator of the variance of X. Find an exact expression for the probability that the value of \(V\) is less than \({1 \over 12} a^2\) and estimate the value of this probability correct to one significant figure.

2001 Paper 2 Q14
D: 1600.0 B: 1500.0

Two coins \(A\) and \(B\) are tossed together. \(A\) has probability \(p\) of showing a head, and \(B\) has probability \(2p\), independent of \(A\), of showing a head, where \(0 < p < \frac12\). The random variable \(X\) takes the value 1 if \(A\) shows a head and it takes the value \(0\) if \(A\) shows a tail. The random variable \(Y\) takes the value 1 if \(B\) shows a head and it takes the value \(0\) if \(B\) shows a tail. The random variable \(T\) is defined by \[ T= \lambda X + {\textstyle\frac12} (1-\lambda)Y. \] Show that \(\E(T)=p\) and find an expression for \(\var(T)\) in terms of \(p\) and \(\lambda\). Show that as \(\lambda\) varies, the minimum of \(\var(T)\) occurs when \[ \lambda =\frac{1-2p}{3-4p}\;. \] The two coins are tossed \(n\) times, where \(n>30\), and \(\overline{T}\) is the mean value of \(T\). Let \(b\) be a fixed positive number. Show that the maximum value of \(\P\big(\vert \overline{T}-p\vert < b\big)\) as \(\lambda\) varies is approximately \(2\Phi(b/s)-1\), where \(\Phi\) is the cumulative distribution function of a standard normal variate and \[ s^2= \frac{p(1-p)(1-2p)}{(3-4p)n}\;. \]

2000 Paper 2 Q14
D: 1600.0 B: 1484.0

The random variables \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\) are independently and uniformly distributed on the interval \(0 \le x \le 1\). The random variable \(Y\) is defined to be the median of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). Given that the probability density function of \(Y\) is \(\g(y)\), where \[ \mathrm{g}(y)=\begin{cases} ky^{n}(1-y)^{n} & \mbox{ if }0\leqslant y\leqslant1\\ 0 & \mbox{ otherwise} \end{cases} \] use the result $$ \int_0^1 {y^{r}}{{(1-y)}^{s}}\,\d y = \frac{r!s!}{(r+s+1)!} $$ to show that \(k={(2n+1)!}/{{(n!)}^2}\), and evaluate \(\E(Y)\) and \({\rm Var}\,(Y)\). Hence show that, for any given positive number \(d\), the inequality $$ {\P\left({\vert {Y - 1/2} \vert} < {d/{\sqrt {n}}} \right)} < {\P\left({\vert {{\bar X} - 1/2} \vert} < {d/{\sqrt {n}}} \right)} $$ holds provided \(n\) is large enough, where \({\bar X}\) is the mean of \(X_1\), \(X_2\), \(\ldots\) , \(X_{2n+1}\). [You may assume that \(Y\) and \(\bar X\) are normally distributed for large \(n\).]

1998 Paper 3 Q14
D: 1700.0 B: 1500.0

A hostile naval power possesses a large, unknown number \(N\) of submarines. Interception of radio signals yields a small number \(n\) of their identification numbers \(X_i\) (\(i=1,2,...,n\)), which are taken to be independent and uniformly distributed over the continuous range from \(0\) to \(N\). Show that \(Z_1\) and \(Z_2\), defined by $$ Z_1 = {n+1\over n} {\max}\{X_1,X_2,...,X_n\} \hspace{0.3in} {\rm and} \hspace{0.3in} Z_2 = {2\over n} \sum_{i=1}^n X_i \;, $$ both have means equal to \(N\). Calculate the variance of \(Z_1\) and of \(Z_2\). Which estimator do you prefer, and why?

1997 Paper 1 Q12
D: 1500.0 B: 1500.0

An experiment produces a random number \(T\) uniformly distributed on \([0,1]\). Let \(X\) be the larger root of the equation \[x^{2}+2x+T=0.\] What is the probability that \(X>-1/3\)? Find \(\mathrm{E}(X)\) and show that \(\mathrm{Var}(X)=1/18\). The experiment is repeated independently 800 times generating the larger roots \(X_{1}\), \(X_{2}\), \dots, \(X_{800}\). If \[Y=X_{1}+X_{2}+\dots+X_{800}.\] find an approximate value for \(K\) such that \[\mathrm{P}(Y\leqslant K)=0.08.\]

1996 Paper 2 Q14
D: 1600.0 B: 1500.0

The random variable \(X\) is uniformly distributed on \([0,1]\). A new random variable \(Y\) is defined by the rule \[ Y=\begin{cases} 1/4 & \mbox{ if }X\leqslant1/4,\\ X & \mbox{ if }1/4\leqslant X\leqslant3/4\\ 3/4 & \mbox{ if }X\geqslant3/4. \end{cases} \] Find \({\mathrm E}(Y^{n})\) for all integers \(n\geqslant 1\). Show that \({\mathrm E}(Y)={\mathrm E}(X)\) and that \[{\mathrm E}(X^{2})-{\mathrm E}(Y^{2})=\frac{1}{24}.\] By using the fact that \(4^{n}=(3+1)^{n}\), or otherwise, show that \({\mathrm E}(X^{n}) > {\mathrm E}(Y^{n})\) for \(n\geqslant 2\). Suppose that \(Y_{1}\), \(Y_{2}\), \dots are independent random variables each having the same distribution as \(Y\). Find, to a good approximation, \(K\) such that \[{\rm P}(Y_{1}+Y_{2}+\cdots+Y_{240000} < K)=3/4.\]

1992 Paper 1 Q14
D: 1500.0 B: 1484.8

The average number of pedestrians killed annually in road accidents in Poldavia during the period 1974-1989 was 1080 and the average number killed annually in commercial flight accidents during the same period was 180. Discuss the following newspaper headlines which appeared in 1991. (The percentage figures in square brackets give a rough indication of the weight of marks attached to each discussion.)

  1. {[}10\%{]} Six Times Safer To Fly Than To Walk. 1974-1989 Figures Prove It.
  2. {[}10\%{]} Our Skies Are Safer. Only 125 People Killed In Air Accidents In 1990.
  3. {[}30\%{]} Road Carnage Increasing. 7 People Killed On Tuesday.
  4. {[}50\%{]} Alarming Rise In Pedestrian Casualties. 1350 Pedestrians Killed In Road Accidents During 1990.

1991 Paper 1 Q15
D: 1516.0 B: 1484.0

A fair coin is thrown \(n\) times. On each throw, 1 point is scored for a head and 1 point is lost for a tail. Let \(S_{n}\) be the points total for the series of \(n\) throws, i.e. \(S_{n}=X_{1}+X_{2}+\cdots+X_{n},\) where \[ X_{j}=\begin{cases} 1 & \text{ if the }j \text{ th throw is a head}\\ -1 & \text{ if the }j\text{ th throw is a tail.} \end{cases} \]

  1. If \(n=10\,000,\) find an approximate value for the probability that \(S_{n}>100.\)
  2. Find an approximate value for the least \(n\) for which \(\mathrm{P}(S_{n}>0.01n)<0,01.\)
Suppose that instead no points are scored for the first throw, but that on each successive throw, 2 points are scored if both it and the first throw are heads, two points are deducted if both are tails, and no points are scored or lost if the throws differ. Let \(Y_{k}\) be the score on the \(k\)th throw, where \(2\leqslant k\leqslant n.\) Show that \(Y_{k}=X_{1}+X_{k}.\) Calculate the mean and variance of each \(Y_{k}\) and determine whether it is true that \[ \mathrm{P}(Y_{2}+Y_{3}+\cdots+Y_{n}>0.01(n-1))\rightarrow0\quad\mbox{ as }n\rightarrow\infty. \]

Show Solution
Notice that \(\mathbb{E}(X_i) = 0, \mathbb{E}(X_i^2) = 1\) and so \(\mathbb{E}(S_n) =0, \textrm{Var}(S_n) = n\).
  1. Then by the central limit theorem (or alternatively the normal approximation to the binomial), \begin{align*} && \mathbb{P}(S_n > 100) &\underbrace{\approx}_{\text{CLT}} \mathbb{P} \left (Z > \frac{100}{\sqrt{10\, 000}} \right) \\ &&&= \mathbb{P}(Z > 1) \\ &&&= 1-\Phi(1) \\ &&&\approx 15.9\% \end{align*}
  2. \begin{align*} &&\mathbb{P}(S_n > 0.01n) &\approx \mathbb{P} \left (Z > \frac{0.01n}{\sqrt{n}} \right) \\ &&&= \mathbb{P}(Z > 0.01 \sqrt{n}) \\ &&&= 1-\Phi(0.01\sqrt{n}) \\ &&&< 0.01 \\ && \Phi^{-1}(0.01) &= -2.3263\ldots \\ \Rightarrow && 0.01 \sqrt{n} &= 2.3263\ldots \\ \Rightarrow && n &\approx 233^2 \end{align*}
\begin{array}{cc|cc} 1\text{st throw}& k\text{th throw} & X_1 + X_k & Y_k \\ \hline \text{head} & \text{head} & 1 + 1 & 2 \\ \text{head} & \text{tail} & 1 - 1 & 0 \\ \text{tail} & \text{head} & -1 + 1 & 0 \\ \text{tail} & \text{tail} & -1- 1 & -2 \\ \end{array} Across all possible cases \(Y_k = X_1 + X_k\) so therefore these random variables are equal. \begin{align*} \mathbb{E}(Y_k) &= \mathbb{E}(X_1) + \mathbb{E}(Y_k) \\ &= 0 + 0 = 0 \\ \\ \textrm{Var}(Y_k) &= \textrm{Var}(X_1)+\textrm{Var}(X_k) \\ &= 2 \\ \\ \mathbb{E}\left (\sum_{k=2}^n Y_k \right) &= 0 \\ \textrm{Var}\left (\sum_{k=2}^n Y_k \right) &= 2(n-1) \end{align*} Therefore approximately \(\displaystyle \sum_{k=2}^n Y_k \approx N(0, 2(n-1))\) \begin{align*} \mathbb{P} \left (\sum_{k=2}^n Y_k > 0.01(n-1) \right) &\approx \mathbb{P} \left (Z > \frac{0.01(n-1)}{\sqrt{2(n-1)}} \right) \\ &= \mathbb{P} \left (Z > c \sqrt{n-1} \right) \\ &\to 0 \text{ as } n \to \infty \end{align*}

Showing 1-1 of 1 problems
2004 Paper 3 Q14
D: 1700.0 B: 1488.4

\textit{In this question, \(\Phi(z)\) is the cumulative distribution function of a standard normal random variable.} A random variable is known to have a Normal distribution with mean \(\mu\) and standard deviation either \(\sigma_0\) or \(\sigma_1\), where \(\sigma_0 < \sigma_1\,\). The mean, \(\overline{X}\), of a random sample of \(n\) values of \(X\) is to be used to test the hypothesis \(\mathrm{H}_0: \sigma = \sigma_0\) against the alternative \(\mathrm{H}_1: \sigma = \sigma_1\,\). Explain carefully why it is appropriate to use a two sided test of the form: accept \(\mathrm{H}_0\) if \phantom{} \(\mu - c < \overline{X} < \mu+c\,\), otherwise accept \(\mathrm{H}_1\). Given that the probability of accepting \(\mathrm{H}_1\) when \(\mathrm{H}_0\) is true is \(\alpha\), determine \(c\) in terms of \(n\), \(\sigma_0\) and \(z_{\alpha}\), where \(z_\alpha \) is defined by \(\ds\Phi(z_{\alpha}) = 1 - \tfrac{1}{2}\alpha\). The probability of accepting \(\mathrm{H}_0\) when \(\mathrm{H}_1\) is true is denoted by \(\beta\). Show that \(\beta\) is independent of \(n\). Given that \(\Phi(1.960)\approx 0.975\) and that \(\Phi(0.063) \approx 0.525\,\), determine, approximately, the minimum value of \(\ds \frac{\sigma_1}{\sigma_0}\) if \(\alpha\) and \(\beta\) are both to be less than \(0.05\,\).

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