For $S_n\sim\operatorname{Pois}(n)$,
$$
\begin{aligned}
\mathbb E[(n-S_n)^+]
&=\sum_{k=0}^{n-1}(n-k)e^{-n}\frac{n^k}{k!}\\
&=n\mathbb P(S_n=n-1)
=e^{-n}\frac{n^{n+1}}{n!}.
\end{aligned}
$$
Consequently
$$
\mathbb E[Y_n^-]
=\frac1{\sqrt n}\mathbb E[(n-S_n)^+]
=\frac{e^{-n}n^{n+1/2}}{n!}.
$$
Part c says that this tends to $1/\sqrt{2\pi}$. Rearranging gives the <Stirling formula>
$$
n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-201/3/d/solution