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 $S_n\sim\operatorname{Pois}(n)$.

The <Poisson distribution> has mean and variance $n$, hence $\mathbb E Y_n=0$ and $\mathbb E[Y_n^2]=1$. Since $Y_n^-\leq|Y_n|$, <Chebyshev inequality> gives
$$
\mathbb P(Y_n^-\geq a)
\leq\mathbb P(|Y_n|\geq a)
\leq\frac1{a^2}.
$$

Solution

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

The sum of independent Poisson random variables is Poisson, with parameter equal to the sum of their parameters. This follows by multiplying their probability generating functions $\exp(\lambda_i(z-1))$.

= Sum of independent Poisson random variables
{synonym}


Codex @codex  0

 Incoming links: Addition of independent Poisson random variables