# Is the sample mean of a Poisson Distribution Normal? [closed]

I am trying to figure out why we are allowed to use the sample mean in a normal distribution.

I tried using moment generating function for $\bar{x}$ to prove that the distribution was normal but didn't have any luck.

Are we able to use the sample mean for the poisson distribution with a normal distribution because of CLT?

If the CLT is the only appropriate solution, this is only for larger samples. Cannot we not use this on smaller samples?

## closed as off-topic by Xander Henderson, Did, user223391, Saad, NChMar 17 '18 at 0:25

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The sample mean is not normally distributed for any finite sample size $n$. Recall that if $X_1$, $X_2$ are independent $\mathrm{Pois}(\lambda)$ random variables, then for $k\geqslant0$ \begin{align} \mathbb P(X_1+X_2=k) &= \sum_{j=0}^k\mathbb P(X_1=j,X_2=k-j)\\ &= \sum_{j=0}^k\mathbb P(X_1=j)\mathbb P(X_2=k-j)\\ &= \sum_{j=0}^ke^{-\lambda} \left(\frac{\lambda^j}{j!} \right)e^{-\lambda}\left(\frac{\lambda^{k-j}}{(k-j)!}\right)\\ &= e^{-2\lambda}\lambda^k\sum_{j=0}^k \frac{1}{j!(k-j)!}\\ &= e^{-2\lambda}\frac{\lambda^k}{k!}\sum_{j=0}^k \binom kj\\ &= e^{-2\lambda}\frac{\lambda^k}{k!}, \end{align} so by induction, $\sum_{j=1}^n X_j$ has $\mathrm{Pois}(n\lambda)$ distribution. Hence for $\bar X_n := \frac1n\sum_{j=1}^n X_j$ we have $$\mathbb P\left(\bar X_n = \frac kn\right) = e^{-n\lambda}\frac{(n\lambda)^k}{k!}.$$ Instead, we look at the asymptotic distribution of $\bar X_n$. Note that \begin{align} \mathbb E\left[ \bar X_n\right] &= \mathbb E\left[\frac1n\sum_{j=1}^n X_j \right]\\ &= \frac1n\sum_{j=1}^n \mathbb E[X_j]\\ &= \frac1n\cdot n\lambda = \lambda, \end{align} and \begin{align} \mathbb E[X_1(X_1-1)] &= \sum_{k=0}^\infty k(k-1)e^{-\lambda}\frac{\lambda^k}{k!}\\ &= e^{-\lambda}\sum_{k=0}^\infty k\frac{\lambda^k}{(k-2)!}\\ &= \lambda ^2 e^{-\lambda}\sum_{k=0}^\infty k\frac{\lambda^k}{k!}\\ &= \lambda^2, \end{align} hence \begin{align} \mathrm{Var}\left(X_1\right) &= \mathbb E\left[X_1^2\right] - \mathbb E\left[ X_1\right]^2\\ &= \mathbb E[X_1(X_1-1)] + \mathbb E[X_1] - \mathbb E[X_1]^2\\ &= \lambda^2 + \lambda - \lambda^2 = \lambda. \end{align} It follows that \begin{align} \mathrm{Var}\left(\bar X_n\right) &= \mathrm{Var}\left(\frac1n \sum_{j=1}^n X_j\right)\\ &= \frac1{n^2}\sum_{j=1}^n \mathrm{Var}(X_j)\\ &=\frac{\lambda}n<\infty. \end{align} By the Lindeberg–Lévy Central Limit Theorem, we conclude that $$\sqrt n\left(\bar X_n - \mu\right) \stackrel d\longrightarrow N(0,\sigma^2),$$ where $\mu=\sigma^2=\lambda$.