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2025 Paper 2 Q12
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.
- 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 \[\lambda - 1 < m < \lambda.\]
- Show that the minimum value of \(d_r\) occurs at \(r = k\), where \(k\) is such that \[k < \lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}} < k + 1.\]
- Show that the condition for the maximum value of \(d_r\) to occur at \(r = 1\) is \[1 < \lambda < 2 + \sqrt{2}.\]
- In the case \(\lambda = 3.36\), sketch a graph of \(p_r\) against \(r\) for \(r = 0, 1, 2, \ldots, 6, 7\).
Solution:
- 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 \(r = \lfloor \lambda \rfloor\).
- 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 \(k < \lambda + \frac{1}{2} + \sqrt{\lambda + \frac{1}{4}} < k + 1\). \(d_1 = e^{-\lambda}(\lambda - 1)\), \(d_k = e^{-\lambda} \frac{\lambda^{k-1}}{(k-1)!}(\frac{\lambda}{k}-1)\)
- 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 \(1 < \lambda < 2 + \sqrt{2}\)
1995 Paper 2 Q13
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 \(u_{k}.\)