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2013 Paper 3 Q13
D: 1700.0 B: 1484.0
cumulative distribution function probability density function integration by parts inequality expectation moments continuous random variable bounding

  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. Show that \(0\le \F(x) \le Mx\) for \(0\le x\le1\).
    2. For any function \(\g(x)\), show that \[ \int_0^1 2 \g(x) \F(x) \f(x) \d x = \g(1) - \int_0^1 \g'(x) \big( \F(x)\big)^2 \d x \,. \]
  2. 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. Determine the value of the constant \(k\).
    2. 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. Hence show that, for \(n\ge 1\), \[ \mu _n \ge \frac{n}{(n+1)M} -\frac{n-1}{n+1} \,.\]

Solution:

    1. \(\,\) \begin{align*} && 0 &\leq f(t) &\leq M \\ \Rightarrow && \int_0^x 0 \d t &\leq \int_0^x f(t) \d t & \leq \int_0^x M \d x \\ \Rightarrow && 0 &\leq F(x) &\leq Mx \end{align*}
    2. \(\,\) \begin{align*} && \int_0^1 2g(x)F(x)f(x) \d x &= \left [ g(x) F(x)^2 \right] - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \\ &&&= g(1) - \int_0^1 g'(x) \left ( F(x)\right)^2 \d x \end{align*}
    1. \(\,\) \begin{align*} && 1 &= \int_0^1 kF(y)f(y) \d y \\ &&&= k\left [ \frac12 F(y)^2\right]_0^1 \\ &&&= \frac{k}{2} \\ \Rightarrow && k &= 2 \end{align*}
    2. \(\,\) \begin{align*} \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &\geq \int_0^1 y^n 2My f(y) \d y \\ &= 2M\int_0^1 y^{n+1} f(y) \d y \\ &= 2M \E[X^{n+1}] = 2M\mu_{n+1} \\ \\ \E[Y^n] &= \int_0^1 y^n 2F(y)f(y) \d y \\ &= 1 - \int_0^1 ny^{n-1} F(y)^2 \d y \\ &\geq 1 - \int_0^1 ny^{n-1}My F(y) \d y \\ &= 1 - M\int_0^1 ny^n F(y) \d y \\ &= 1 - M[\frac{n}{n+1}y^{n+1} F(y)]_0^1 + M\int_0^1\frac{n}{n+1} y^{n+1} f(y) \d y \\ &= 1 - \frac{nM}{n+1} + \frac{nM}{n+1} \mu_{n+1} \end{align*}
    3. Since \(\E[Y^{n-1}] \geq 0\) we must have \begin{align*} && 2M\mu_n \geq 1 + \frac{(n-1)M}{n}\mu_n - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \left (2M + \frac{(n-1)M}{n} \right) \geq 1 - \frac{(n-1)M}{n} \\ \Rightarrow && \mu_n \frac{3Mn-M}{n} & \geq \frac{n-(n-1)M}{n} \\ \Rightarrow && \mu_n & \geq \frac{n-(n-1)M}{3Mn-M} \end{align*}

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2012 Paper 2 Q13
D: 1600.0 B: 1516.0
Poisson distribution probability density function continuous random variable expected value variance Gaussian integral CDF to PDF spatial Poisson process nearest neighbour distance

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 \(\var(Y) = \dfrac{4-\pi}{4\pi k}\).

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*}

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2012 Paper 3 Q12
D: 1700.0 B: 1469.4
geometric probability equilateral triangle tetrahedron area decomposition probability density function expected value minimum of random variables continuous random variable

  1. A point \(P\) lies in an equilateral triangle \(ABC\) of height 1. The perpendicular distances from \(P\) to the sides \(AB\), \(BC\) and \(CA\) are \(x_1\), \(x_2\) and \(x_3\), respectively. By considering the areas of triangles with one vertex at \(P\), show that \(x_1+x_2+x_3=1\). Suppose now that \(P\) is placed at random in the equilateral triangle (so that the probability of it lying in any given region of the triangle is proportional to the area of that region). The perpendicular distances from \(P\) to the sides \(AB\), \(BC\) and \(CA\) are random variables \(X_1\), \(X_2\) and \(X_3\), respectively. In the case \(X_1= \min(X_1,X_2,X_3)\), give a sketch showing the region of the triangle in which \(P\) lies. Let \(X= \min(X_1,X_2,X_3)\). Show that the probability density function for \(X\) is given by \[ \f(x) = \begin{cases} 6(1-3x) & 0 \le x \le \frac13\,, \\ 0 & \text{otherwise}\,. \end{cases} \] Find the expected value of \(X\).
  2. A point is chosen at random in a regular tetrahedron of height 1. Find the expected value of the distance from the point to the closest face. \newline [The volume of a tetrahedron is \(\frac13 \times \text{area of base}\times\text{height}\) and its centroid is a distance \(\frac14\times \text{height}\) from the base.]

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2011 Paper 2 Q13
D: 1600.0 B: 1500.0
skewness cumulative distribution functions probability density function moments expectation median quantiles inverse CDF continuous random variable inequality

What property of a distribution is measured by its 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 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 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 \(D > P > \gamma\,\).

Solution: Skewness is a measure of the symmetry (specifically the lack-thereof) in the distribution. How much mass is there on one side rather than another.

  1. \(\,\) \begin{align*} && \gamma &= \frac{\E \left [ (X - \mu)^3 \right ]}{\sigma^3} \\ &&&= \frac{\E \left [ X^3 - 3\mu X^2 + 3\mu^2 X - \mu^3 \right ]}{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \E[X^2] + 3\mu^2 \E[X] - \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu (\mu^2 + \sigma^2) + 3\mu^2\cdot \mu- \mu^3 }{\sigma^3} \\ &&&= \frac{\E [ X^3 ]- 3\mu \sigma^2 - \mu^3 }{\sigma^3} \\ \end{align*} \begin{align*} && f(x) &= \begin{cases} 2x & \text{for } 0\le x\le 1\,, \\[2mm] 0 & \text{otherwise}\,. \end{cases} \\ && \E[X] &= \int_0^1 2x^2 \d x \\ &&&= \frac23 \\ && \E[X^2] &= \int_0^1 2x^3 \d x \\ &&&= \frac12 \\ && \E[X^3] &= \int_0^1 2x^4 \d x \\ &&&= \frac25 \\ \\ && \mu &= \frac23 \\ && \sigma^2 &= \frac12 - \frac49 = \frac{1}{18} \\ && \gamma &= \frac{\frac25 - 3 \cdot \frac23 \cdot \frac1{18} - \frac8{27}}{\frac{1}{54\sqrt2}} \\ &&&= -\frac{2\sqrt2}{5} \end{align*}
  2. First note that \(\displaystyle F(x) = \int_0^x 2t \d t = x^2\) for \(x \in [0,1]\). In particular, \(F^{-1}(x) = \sqrt{x}\), so \begin{align*} && 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})} \\ &&&= \frac{\sqrt{\frac9{10}} - 2 \sqrt{\frac5{10}} + \sqrt{\frac1{10}}}{\sqrt{\frac9{10}}-\sqrt{\frac1{10}}} \\ &&&= \frac{3-2\sqrt5+1}{3 - 1} \\ &&&= \frac{4-2\sqrt5}{2} = 2-\sqrt5 \end{align*} \begin{align*} && P &= \frac{3(\mu - M)}{\sigma} \\ &&&= \frac{3(\frac23 - \sqrt{\frac12})}{\frac{1}{3\sqrt2}} \\ &&&= 6 \sqrt2 - 9 \end{align*} First we compare \(P\) and \(D\), \(6\sqrt2-9\) and \(2-\sqrt5\) \begin{align*} && D & > P \\ \Leftrightarrow && 2-\sqrt5 &> 6\sqrt2 - 9 \\ \Leftrightarrow && 11 -6\sqrt2 &> \sqrt 5 \\ \Leftrightarrow && (121 + 72 - 132\sqrt2) & > 5 \\ \Leftrightarrow && 188 & > 132\sqrt2 \\ \Leftrightarrow && 47 & > 33 \sqrt 2\\ \Leftrightarrow && 2209 & > 2178 \end{align*} also \begin{align*} && P &> \gamma \\ \Leftrightarrow && 6\sqrt2 - 9 &> -\frac{2\sqrt2}{5} \\ \Leftrightarrow && 30\sqrt2 - 45 & > -2\sqrt2 \\ \Leftrightarrow && 32 \sqrt 2 &> 45 \\ \Leftrightarrow && 2048 &> 2025 \end{align*}

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2010 Paper 2 Q12
D: 1600.0 B: 1486.8
probability cumulative distribution functions piecewise PDF expectation median inequality continuous random variable

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 \(M < \E(X)\,\).

Solution: \begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= ak + b(1-k) \\ &&&= b + (a-b)k \\ \Rightarrow && k &= \frac{1-b}{a-b} \\ \Rightarrow && b & < 1 \tag{\(0 < k, \,a > b\)} \\ && k &> 1 \\ \Rightarrow && a-b & > 1-b \\ \Rightarrow && a > 1 \end{align*}

  1. \(\,\) \begin{align*} && \E[X] &= \int_0^1 x \cdot f(x) \d x \\ &&&= \int_0^k ax \d x + \int_k^1 b x \d x \\ &&&= a \frac{k^2}{2} + b \frac{1-k^2}{2} \\ &&&= \frac12b + (a-b) \frac{(1-b)^2}{2(a-b)^2} \\ &&&= \frac{(1-b)^2+b(a-b)}{2(a-b)} \\ &&&= \frac{1-2b+ab}{2(a-b)} \end{align*}
  2. \(\,\) The median \(M\) satisfies \[\frac12 = \int_0^M f(x) \d x \] If \(ka = \frac{a-ab}{a-b} \leq \frac12 \Leftrightarrow 2a-2ab \leq a-b \Leftrightarrow a+b \leq 2ab\) then \(M > k\) otherwise \(M < k\). In the latter case: \begin{align*} && \frac12 &= Ma \\ \Rightarrow && M &= \frac{1}{2M} \end{align*} In the former case \begin{align*} && \frac12 &= ka + (M-k)b \\ &&&= k(a-b) + Mb \\ &&&= 1-b + M b \\ \Rightarrow && M &= 1-\frac1{2b} \end{align*}

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2002 Paper 3 Q13
D: 1700.0 B: 1516.0
exponential distribution probability minimum of random variables CDF independence median memoryless property continuous random variable

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.

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1999 Paper 2 Q13
D: 1600.0 B: 1484.0
probability continuous random variable uniform distribution probability density function mean and variance ratio geometric probability

A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, \(R\), of the length of the shorter piece to the length of the longer piece is less than \(r\). Find the probability density function for \(R\), and calculate the mean and variance of \(R\).

Solution: Let \(X \sim U[0, \tfrac12]\) be the shorter piece, so \(R = \frac{X}{1-X}\), and \begin{align*} && \mathbb{P}(R \leq r) &= \mathbb{P}(\tfrac{X}{1-X} \leq r) \\ &&&= \mathbb{P}(X \leq r - rX) \\ &&&= \mathbb{P}((1+r)X \leq r) \\ &&&= \mathbb{P}(X \leq \tfrac{r}{1+r} ) \\ &&&= \begin{cases} 0 & r < 0 \\ \frac{2r}{1+r} & 0 \leq r \leq 1 \\ 1 & r > 1 \end{cases} \\ \\ && f_R(r) &= \begin{cases} \frac{2}{(1+r)^2} & 0 \leq r \leq 1 \\ 0 & \text{otherwise} \end{cases} \end{align*} Let \(Y \sim U[\tfrac12, 1]\) be the longer piece, then \(R = \frac{1-Y}{Y} = Y^{-1} - 1\) and \begin{align*} \E[R] &= \int_{\frac12}^1 (y^{-1}-1) 2 \d y \\ &= 2\left [\ln y - y \right]_{\frac12}^1 \\ &= -2 + 2\ln2 +2\frac12 \\ &= 2\ln2 -1 \\ \\ \E[R^2] &= \int_{\frac12}^1 (y^{-1}-1)^2 2 \d y\\ &= 2\left [-y^{-1} -2\ln y + 1 \right]_{\frac12}^1 \\ &= 2 \left ( 2 - 2\ln 2+\frac12\right) \\ &= 3-4\ln 2 \\ \var[R] &= 3 - 4 \ln 2 -(2\ln 2-1)^2 \\ &= 2 - 4(\ln 2)^2 \end{align*}

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1999 Paper 3 Q13
D: 1700.0 B: 1484.0
probability conditional Bayes' theorem continuous random variable density function binomial distribution expected value geometric probability

The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion \(X\) of a cake where \(X\) is a random variable with density function \[{\mathrm f}(x)=Ax\] for \(0\leqslant x\leqslant 1\) where \(A\) is a constant.

  1. What is the expected number of currants in my portion?
  2. If I find all four currants in my portion, what is the probability that I took more than half the cake?

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1987 Paper 1 Q15
D: 1500.0 B: 1516.7
geometric probability circle distance mean and variance trigonometric uniform distribution integration continuous random variable

A point \(P\) is chosen at random (with uniform distribution) on the circle \(x^{2}+y^{2}=1\). The random variable \(X\) denotes the distance of \(P\) from \((1,0)\). Find the mean and variance of \(X\). Find also the probability that \(X\) is greater than its mean.

Solution: Consider the angle from the origin, then \(P = (\cos \theta, \sin \theta)\) where \(\theta \sim U(0, 2\pi)\), and \(X = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta}\) \begin{align*} \mathbb{E}[X] &= \int_0^{2\pi} \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \frac1{2\pi} \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} \sqrt{2 - 2\cos \theta} \d \theta \\ &= \frac{1}{2\pi}\int_0^{2\pi} \sqrt{4\sin^2 \frac{\theta}{2}} \d \theta \\ &= \frac{1}{\pi}\int_0^{2\pi} \left |\sin \frac{\theta}{2} \right| \d \theta \\ &= \frac{1}{\pi} \left [ -2\cos \frac{\theta}{2} \right]_0^{2\pi} \\ &= \frac1{\pi} \l 2 + 2\r \\ &= \frac{4}{\pi} \end{align*} \begin{align*} \mathbb{E}(X^2) &= \frac1{2\pi}\int_0^{2\pi} (\cos \theta - 1)^2 + \sin^2 \theta \d \theta \\ &= \frac1{2\pi}\int_0^{2\pi} 2 - 2 \cos \theta \d \theta \\ &= \frac{4\pi}{2\pi} \\ &= 2 \\ \end{align*} \(\Rightarrow\) \(\mathrm{Var}(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2 = 2 - \frac{16}{\pi^2} = \frac{2\pi^2 - 16}{\pi^2}\).

TikZ diagram
Where the line makes a length longer than \(\frac{4}{\pi}\) it will make an angle at the origin of \(2\sin^{-1} \frac{2}{\pi}\). Therefore the probability of being larger than this is \(\frac{2\pi - 2 \times 2\sin^{-1} \frac{2}{\pi}}{2 \pi} = 1 - \frac{2}{\pi} \sin^{-1} \frac{2}{\pi} \approx 0.560\)

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