-2
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Did, Community, Saad, NCh
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.