Poisson Random Variable

I have written a much more detailed introduction to Poisson random variable here in my blog.

Definition We say a discrete r.v $\mathbb{X}$ is a Poisson r.v with parameters $\lambda$, denoted $\mathbb{X} \sim poisson(\lambda)$, if $$\mathbb{X} \sim p(n) = \begin{cases} \frac{\lambda^n}{n!}e^{-\lambda} & \text{if } n \in 0, 1, 2, 3... \\ 0 & \text{otherwise} \end{cases}$$

Property If $\mathbb{X} \sim posson(\lambda)$, then $\mu = \sigma^2 = \lambda$

Proof 1 By definition $\mathbb{E}[\mathbb{X}] = \sum_{n=0}^\infty n\frac{\lambda^n}{n!}e^{-\lambda} = \lambda \sum_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}e^{-\lambda} = \lambda$ $\mathbb{E}[\mathbb{X}^2] = \sum_{n=0}^\infty n^2\frac{\lambda^n}{n!}e^{-\lambda} = \lambda \sum_{n=1}^\infty n \frac{\lambda^{n-1}}{(n-1)!}e^{-\lambda} = \lambda \sum_{n=0}^\infty (n+1) \frac{\lambda^{n}}{n!}e^{-\lambda} = \lambda (\mathbb{E}[\mathbb{X}] + 1)$ so $var(\mathbb{X}) = \mathbb{E}[\mathbb{X}^2] - (\mathbb{E}[\mathbb{X}])^2 = \lambda$

Proof 2 By moment generating function

First lets check if it is eligible to apply moment generating function.

(a) $\frac{\mathrm{d}}{\mathrm{d}t} f(n)e^{nt} = \frac{\lambda^n}{(n-1)!}e^{-\lambda}e^{nt}$ $\sum_{n=1}^\infty \frac{\lambda^n}{(n-1)!}e^{-\lambda}e^{nt} = \lambda e^{t-\lambda} \sum_{n=0}^\infty \frac{(\lambda e^t)^n}{n!} = \lambda e^{\lambda e^t + t - \lambda}$

This convergence is definitely uniform for some closed interval containing $0$ (In fact, this convergence is uniform for all closed interval, see appendix for uniform convergence property on power series).

(b) $\sum f(n)e^{nt} = \sum_{n=0}^\infty e^{nt} \frac{\lambda^n}{n!}e^{-\lambda} = e^{-\lambda} \sum_{n=0}^\infty \frac{ (\lambda e^t)^n}{n!} = e^{\lambda (e^t - 1)}$ converges on all points

Since (a), (b) hold, $\mu = M_X'(0)$. $M_X(t) = e^{\lambda (e^t - 1)}$ as (b) already showed, so the mean $\mu = M_x'(0) = \lambda e^t e^{\lambda (e^t - 1)} |_{t=0} = \lambda$ and variance $\sigma^2 = \mathbb{E}[\mathbb{X}^2] - \mathbb{E}[\mathbb{X}]^2 = M_X''(0) - \mu^2 = \lambda + \lambda^2 - \lambda^2 = \lambda$