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2025 Paper 3 Q11
D: 1500.0 B: 1500.0
cumulative distribution function exponential distribution order statistics probability density function logarithmic approximation median confidence interval independent random variables

  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 \(0.9\).

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 \(\ln \ln 20 - \ln \ln \frac{20}{19} = 4.0673\ldots\)]

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2018 Paper 2 Q12
D: 1600.0 B: 1500.0
probability geometric distribution expected value optimisation coin tossing logarithmic approximation consecutive events

In a game, I toss a coin repeatedly. The probability, \(p\), that the coin shows Heads on any given toss is given by \[ p= \frac N{N+1} \,, \] where \(N\) is a positive integer. The outcomes of any two tosses are independent. The game has two versions. In each version, I can choose to stop playing after any number of tosses, in which case I win £\(H\), where \(H\) is the number of Heads I have tossed. However, the game may end before that, in which case I win nothing.

  1. In version 1, the game ends when the coin first shows Tails (if I haven't stopped playing before that). I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Find, in terms of \(h\) and \(p\), an expression for my expected winnings and show that I can maximise my expected winnings by choosing \(h=N\).
  2. In version 2, the game ends when the coin shows Tails on two consecutive tosses (if I haven't stopped playing before that). I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Show that my expected winnings are \[ \frac{ hN^h (N+2)^h}{(N+1)^{2h}} \,.\] In the case \(N=2\,\), use the approximation \(\log_3 2 \approx 0.63\) to show that the maximum value of my expected winnings is approximately £3.

Solution:

  1. Since we either win \(h\) or \(0\), to calculate the expected winnings we just need to calculate the probability that we get \(h\) consecutive heads, therefore: \begin{align*} && \mathbb{E}(\text{winnings}) &= E_h \\ &&&= h \cdot \left ( \frac{N}{N+1} \right)^h \\ && \frac{E_{h+1}}{E_h} &= \frac{h+1}{h }\left ( \frac{N}{N+1} \right) \end{align*} Therefore \(E_h\) is increasing if \(h \leq N\), so we can maximise our winnings by taking \(h = N\). (In fact, we could take \(h = N\) or \(h = N+1\), but arguably \(h = N\) is better as we have the same expected value but lower variance).
  2. We can have up to \(h\) tails appearing (if we imagine slots for tails of the form \(\underbrace{\_H\_H\_H\_\cdots\_H}_{h\text{ spaces and }h\, H}\) so, we have \begin{align*} && \mathbb{P}(\text{wins}) &= \sum_{t = 0}^h \mathbb{P}(\text{wins and } t\text{ tails}) \\ &&&= \sum_{t = 0}^h\binom{h}{t} \left ( \frac{N}{N+1} \right)^h\left ( \frac{1}{N+1} \right)^t \\ &&&= \left ( \frac{N}{N+1} \right)^h \sum_{t = 0}^h\binom{h}{t}\left ( \frac{1}{N+1} \right)^t \cdot 1^{h-t} \\ &&&= \left ( \frac{N}{N+1} \right)^h \left ( 1 + \left ( \frac{1}{N+1} \right) \right)^h \\ &&&= \left ( \frac{N}{N+1} \right)^h \left ( \frac{N+2}{N+1}\right)^h \\ &&&= \frac{N^h(N+2)^h}{(N+1)^{2h}} \\ \Rightarrow && \E(\text{winnings}) &= h \cdot \frac{N^h(N+2)^h}{(N+1)^{2h}} \end{align*} If \(N = 2\), we have \begin{align*} && \E(\text{winnings}) &= E_h \\ &&&= h \cdot \frac{2^h\cdot2^{2h}}{3^{2h}}\\ &&&= h \cdot \frac{2^{3h}}{3^{2h}} \\ \Rightarrow && \frac{E_{h+1}}{E_h} &= \frac{h+1}{h} \frac{8}{9} \\ \end{align*} Therefore to maximise the winnings we should take \(h = 8\), and the expected winnings will be: \begin{align*} && E_8 &= 8 \cdot \frac{2^{24}}{3^{16}} \\ \Rightarrow && \log_3 E_8 &= 27 \log_3 2 - 16 \\ &&&\approx 24 \cdot 0.63 - 16 \\ &&&\approx 17 - 16 \\ &&&\approx 1 \\ \Rightarrow && E_8 &\approx 3 \end{align*}

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2011 Paper 1 Q7
D: 1500.0 B: 1500.0
differential equation first order separable partial fractions logarithmic approximation flow rate modelling integration

In this question, you may assume that \(\ln (1+x) \approx x -\frac12 x^2\) when \(\vert x \vert \) is small. The height of the water in a tank at time \(t\) is \(h\). The initial height of the water is \(H\) and water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.

  1. Suppose that water leaks out at a rate proportional to the height of the water in the tank, and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that \[ \frac {\d h}{\d t } = k( \alpha^2 H -h)\,, \] for some positive constant \(k\). Deduce that the time \(T\) taken for the water to reach height \(\alpha H\) is given by \[ kT = \ln \left(1+\frac1\alpha\right)\,, \] and that \(kT\approx \alpha^{-1}\) for large values of \(\alpha\).
  2. Suppose that the rate at which water leaks out of the tank is proportional to \(\sqrt h\) (instead of \(h\)), and that when the height reaches \(\alpha^2 H\), where \(\alpha\) is a constant greater than 1, the height remains constant. Show that the time \(T'\) taken for the water to reach height \(\alpha H\) is given by \[ cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln \left(1+\frac1 {\sqrt\alpha} \right)\right)\, \] for some positive constant \(c\), and that \(cT'\approx \sqrt H\) for large values of \(\alpha\).

Solution:

  1. \begin{align*} \frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}} \end{align*}. We also know that when \(h = \alpha^2 H\), \(\frac{\d h}{\d t} = 0\), ie \(c - k \alpha^2 H = 0\) therefore: \[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \] \begin{align*} && \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\ && \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\ && - \ln |\alpha^2H -h| &= kt + C \\ t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\ \Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\ && kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\ &&&= \ln \frac{1+\alpha}{\alpha} \\ &&&= \ln \left (1 + \frac1{\alpha} \right) \\ &&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\ &&&\approx \alpha^{-1} \end{align*}
  2. \begin{align*} && \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\ \Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\ u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\ &&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\ &&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)} \right)\right)\\ &&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}} \right)\right)\\ &&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}} \right)\right)}\\ &&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha} \right)\right) \\ &&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\ &&&= \sqrt{H} \end{align*} as required.

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2009 Paper 2 Q11
D: 1600.0 B: 1516.0
work-energy-power Newton's second law power resistance train logarithmic approximation work-energy theorem kinematics integration optimisation

A train consists of an engine and \(n\) trucks. It is travelling along a straight horizontal section of track. The mass of the engine and of each truck is \(M\). The resistance to motion of the engine and of each truck is \(R\), which is constant. The maximum power at which the engine can work is \(P\). Obtain an expression for the acceleration of the train when its speed is \(v\) and the engine is working at maximum power. The train starts from rest with the engine working at maximum power. Obtain an expression for the time \(T\) taken to reach a given speed \(V\), and show that this speed is only achievable if \[ P>(n+1)RV\,. \]

  1. In the case when \((n+1) RV/P\) is small, use the approximation \(\ln (1-x) \approx -x -\frac12 x^2\) (valid for small \( x \)) to obtain the approximation \[ PT\approx \tfrac 12 (n+1) MV^2\, \] and interpret this result.
  2. In the general case, the distance moved from rest in time \(T\) is \(X\). {\em Write down}, with explanation, an equation relating \(P\), \(T\), \(X\), \(M\), \(V\), \(R\) and \(n\) and hence show that \[ X= \frac{2PT - (n+1)MV^2}{2(n+1)R} \,. \]

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