# Determine the asymptotic distribution of $\bar X_n$, properly centered and $\sqrt n$ scaled

Let $X_1, X_2,...X_n$ be i.i.d. with $P(X_i =1)=1-P(X_i =0)=p,p \in (0,1)$. (a) Show that $\bar X_n$ is the MLE of p. (b) Find the mean $\mu_n$ and variance $\sigma^{2}_n$ of $\bar X_n$ and invoke the CLT to determine the asymptotic distribution of $\bar X_n$, properly centered and $\sqrt n$ scaled.

Thoughts: I know how to solve part (a) and also know the mean and variance of part (b). The question that bothers me is the second part of part (b). I don't know to apply CLT here with $\sqrt n$ scaled. Any explicit explanation of this part would be much appreciated.

## 1 Answer

Here is a review of some facts I suppose you already know.

$S_n = \sum_{i=1}^n X_i \sim Binom(n, p).$ Also, $E(X_i) = p\;$ and $V(X_i) = p(1-p).$ Thus, $E(S_n) = np\;$ and $SD(S_n) = \sqrt{np(1-p)}.$

If we define $Z_n = \frac{S_n - np}{\sqrt{np(1-p)}},$ then $Z_n$ is approximately standard normal. According the the CLT its distribution gets closer to standard normal as $n$ increases. I assume this is what is meant by 'centered and $\sqrt{n}$-scaled'.

Given that $\bar X = S_n/n,$ you ought to be able to re-express the fraction for $Z_n$ in the required format.

• thanks!!! BTW, what is V? Is that the variance? If so, I think it should be $p(1-p)$. What does SD mean here? – hil316 Jan 3 '16 at 3:41
• Right. My mistake. Fixing it. Thanks. SD=standard deviation. – BruceET Jan 3 '16 at 7:49