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?
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 $\mu=\sigma^2=\lambda$.