Asymptotic Distribution of Sample Mean Suppose I have a discrete random variable $X$ which follows a geometric distribution on $x=0,1,2,...$ and I take a random sample from this distribution of size $n$.  What is the asymptotic distribution of $\bar X$?
I already know that $E(X)=\frac{1-p}{p}$ and $V(X)=\frac{1-p}{p^2}$.
This seems like an application of central limit theorem, so I'm sure that $\bar X$ converges to a normal distribution.  However, the part that's tripping me up is calculating the mean and variance of the normal distribution that it's converging to.  How do you do this?
 A: Note that
$$
E\bar{X}=E\left(n^{-1}\sum_{i=1}^nX_{i}\right)
=n^{-1}\sum_1^nEX_i
=n^{-1}(nEX)=EX
$$
since the $X_i$ are identically distributed. Similarly,
$$
V(\bar{X})=V\left(n^{-1}\sum_{i=1}^nX_{i}\right)
=n^{-2}\sum_1^nV(X_i)
=n^{-2}(nV(X))
=n^{-1}V(X)
$$
since the $X_i$ are independent and identically distributed.
A: You have
$$
\operatorname E( \, \overline X \, ) = \operatorname E\left( \frac {X_1+\cdots+X_n} n \right) = \frac 1 n \left( \operatorname E(X_1) + \cdots + \operatorname E(X_n) \right) = \frac 1 n \cdot n \operatorname E(X) = \operatorname E(X) 
$$
and
\begin{align}
\operatorname{var}\left( \sqrt n \cdot \overline X \right) = \operatorname{var} \left( \frac{X_1 + \cdots + X_n}{\sqrt n} \right) = \frac 1 n \left(\operatorname{var}(X_1) + \cdots + \operatorname{var}(X_n) \right) = \frac 1n \cdot n \operatorname{var}(X)
\end{align}
So $\displaystyle \dfrac{\overline X - \operatorname E(X)}{\sqrt n}$ has expected value $0$ and variance $1,$ and approaches $N(0,1)$ as $n\to\infty.$
