\(X_{1},X_{2},\ldots,X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1].\) The random variables \(A\) and \(B\) are defined by
\[
A=\min(X_{1},\ldots,X_{n}),\qquad B=\max(X_{1},\ldots,X_{n}).
\]
For any fixed \(k\), such that \(0< k< \frac{1}{2},\) let
\[
p_{n}=\mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k).
\]
What happens to \(p_{n}\) as \(n\rightarrow\infty\)? Comment briefly on this result.
Lord Copper, the celebrated and imperious newspaper proprietor, has decided to run a lottery in which each of the \(4,000,000\) readers of his newspaper will have an equal probability \(p\) of winning \(\pounds 1,000,000\) and their changes of winning will be independent. He has fixed all the details leaving to you, his subordinate, only the task of choosing \(p\). If nobody wins \(\pounds 1,000,000\), you will be sacked, and if more than two readers win \(\pounds 1,000,000,\) you will also be sacked. Explaining your reasoning, show that however you choose \(p,\) you will have less than a 60\% change of keeping your job.
Solution:
\begin{align*}
&& p_n &= \mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k) \\
&&&= \mathrm{P}(A\leqslant k) +\P(B\geqslant1-k) - \mathrm{P}(A\leqslant k\mbox{ or }B\geqslant1-k)\\
&&&= 1-\mathrm{P}(A\geq k) +1-\P(B \leq 1-k) - \l 1- \mathrm{P}(A\geq k\mbox{ and }B\leq 1-k)\r\\
&&&= 1 - \P(X_i \geq k) - \P(X_i \leq 1-k) + \P(k \leq X_i \leq 1-k) \\
&&&= 1 - k^n - (1-k)^n + (1-2k)^n
\end{align*}
Therefore as \(n \to \infty\) \(p_n \to 1\), since \(k, (1-k), (1-2k)\) are all between \(0\) and \(1\) and so their powers will tend to \(0\).
Let \(N = 4\,000\,000\). The probability exactly one person wins is \(Np(1-p)^{N-1}\). The probability exactly two people win is \(\binom{N}{2} p^2 (1-p)^{N-2}\). We wish to maximise the sum of these probabilities. To find this maximum, differentiate wrt \(p\).
\begin{align*}
\frac{\d}{\d p} : && \small N(1-p)^{N-1}-N(N-1)p(1-p)^{N-2} + N(N-1)p(1-p)^{N-2} - \frac12 N(N-1)(N-2)p^2(1-p)^{N-3} \\
&&= N(1-p)^{N-3} \l (1-p)^2 - \frac12(N-1)(N-2)p^2\r \\
\Rightarrow && \frac{(1-p)}{p} = \sqrt{\frac{(N-1)(N-2)}{2}} \\
\Rightarrow && p = \frac{1}{1+ \sqrt{\frac{(N-1)(N-2)}{2}}}
\end{align*}
This will be a maximum, since this is an increasing function at \(p=0\) and decreasing at \(p=1\) and there's only one stationary point.
Note that \(p > \frac{\sqrt{2}}{(N-2)}\) and \(p < \frac{\sqrt{2}}{N-1+\sqrt{2}} < \frac{\sqrt{2}}{N}\) and so:
\begin{align*}
Np(1-p)^{N-1} &< \sqrt{2}(1-\frac{\sqrt{2}}{N-2})^{N-1} \\
&\approx \sqrt{2} e^{-\sqrt{2}}
\end{align*}
\begin{align*}
\frac{N(N-1)}{2}p^2(1-p)^{N-2} &<(1-\frac{\sqrt{2}}{N-2})^{N-1} \\
&\approx e^{-\sqrt{2}}
\end{align*}
Alternatively, we can use a Poisson approximation. The number of winners is \(B(N, p)\) where we are hoping \(np\) is small but not zero. Therefore it's reasonable to approximation \(B(N,p)\) by \(Po(Np)\). (Call this value \(\lambda\)). Then we wish to maximise:
\begin{align*}
&& p &= e^{-\lambda} \l \lambda + \frac{\lambda^2}{2} \r \\
&&&= e^{-\lambda} \lambda \l 1+ \frac{\lambda}{2} \r \\
\Rightarrow && \ln p &= -\lambda + \ln \lambda + \ln(1+\frac12 \lambda) \\
\frac{\d}{\d \lambda}: && \frac{p'}{p} &= -1 + \frac{1}{\lambda} + \frac{1}{2+\lambda} \\
&&&= \frac{-(2+\lambda)\lambda+2+2\lambda}{\lambda(2+\lambda)} \\
&&&= \frac{2-\lambda^2}{\lambda(2+\lambda)} \\
\Rightarrow && \lambda &= \sqrt{2}
\end{align*}
\begin{align*}
\frac{\sqrt{2}+1}{e^{\sqrt{2}}} &< \frac{\sqrt{2}+1}{1+\sqrt{2}+1+\frac{1}{3}\sqrt{2}+\frac{1}{6}} \\
&= \frac{30\sqrt{2}-18}{41}
\end{align*}
Either way, we find we want to estimate \(e^{-\sqrt{2}}(1+\sqrt{2})\)