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

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