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

The probability generating function of each $X_i$ is $\exp (z-1)$. Independence makes the generating function of $S_n$ equal to $\exp (n(z-1))$, so the addition of independent Poisson random variables gives $S_n\sim \mathrm{Pois}(n)$.

The Poisson distribution has mean and variance $n$, hence $\mathbb{E}{Y}_n=0$ and $\mathbb{E}[Y_n^2]=1$. Since $Y_n^{-}\le |Y_n|$, Chebyshev inequality gives

The Poisson central limit theorem, or the ordinary central limit theorem applied to the $X_i$, gives $Y_n\stackrel{d}{\to }Y$ for a standard normal random variable $Y$. The negative-part map $x\mapsto x^{-}:=\max (-x,0)$ is continuous, so the continuous mapping theorem gives

For $K>0$, the Cauchy-Schwarz inequality and part a giveThus $(Y_n^{-})$ is uniformly integrable. Combining this with the weak convergence of random variables from part b yields convergence of the first moments:By symmetry of the standard normal density,

For $S_n\sim \mathrm{Pois}(n)$,ConsequentlyPart c says that this tends to $1/\sqrt{2\pi }$. Rearranging gives the Stirling formula