How to prove that if $X\sim P(\lambda) \Rightarrow Var(X)=\lambda$?

Question

How to prove that if $X\sim P(\lambda) \Rightarrow Var(X)=\lambda$?
$P(X)$ means: Poisson distribution.

Thank you!

Answer 1

As $X\sim P(\lambda)$, we know that $\mathbb{P}\left(X=k\right)=\frac{\lambda^k}{k!}\mathrm{e}^{-\lambda}$. Then as $\mathrm{Var}\left(X\right)=\mathbb{E}\left[X^2\right]-\left(\mathbb{E}\left[X\right]\right)^2$, we can calculate $\mathrm{Var}\left(X\right)$.

First we will find $\mathbb{E}\left[X\right]$ \begin{align*} \mathbb{E}\left[X\right] &= \sum_{k=0}^{\infty}k\cdot\mathbb{P}\left(X=k\right) \\ &= \sum_{k=0}^{\infty}\frac{k\lambda^k}{k!}\mathrm{e}^{-\lambda} \\ &= \lambda\mathrm{e}^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{\left(k-1\right)!} \\ &= \lambda\mathrm{e}^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!} \\ &= \lambda\mathrm{e}^{-\lambda}\mathrm{e}^{\lambda} \\ &= \lambda \end{align*}

Now we find $\mathbb{E}\left[X^2\right]$ \begin{align*} \mathbb{E}\left[X^2\right] &= \sum_{k=0}^{\infty}k^2\mathrm{e}^{-\lambda}\frac{\lambda^k}{k!} \\ &=\sum_{k=1}^{\infty}k^2\mathrm{e}^{-\lambda}\frac{\lambda^k}{k!} \\ &=\sum_{y=0}^{\infty}(y+1)^2\mathrm{e}^{-\lambda}\frac{\lambda^{(y+1)}}{(y+1)!} \end{align*} by substituting $y= k - 1 $. Since $ (y+1)! = (y+1)y!$, we have \begin{align*} \sum_{y=0}^{\infty}(y+1)^2\mathrm{e}^{-\lambda}\frac{\lambda^y}{(y+1)y!}\lambda &=\lambda\sum_{y=0}^{\infty}(y+1)\mathrm{e}^{-\lambda}\frac{1}{y!}\lambda^{y}\\ &=\lambda\sum_{y=0}^{\infty}(y+1)\left(\frac{\lambda^y}{y!}\mathrm{e}^{-\lambda}\right) \\ &= \lambda \left[ \sum_{y=0}^{\infty}y \left(\frac{\lambda^y}{y!}\mathrm{e}^{-\lambda}\right) + \sum_{y=0}^{\infty}\frac{\lambda^y}{y!}\mathrm{e}^{-\lambda} \right]\\ &=\lambda \left( \lambda + 1 \right) \\ &=\lambda^2+\lambda \end{align*}

So $\mathrm{Var}\left(X\right)=\lambda^2+\lambda-\lambda^2=\lambda$