Is the distribution of the logarithm of the mean of Bernoulli random variables ($\log \overline X$) still asymptotically normal?

Question

Let $\overline X$ be the mean of a Bernoulli random variable (r.v.)

$$\overline X = \frac{1}{n}\sum_{i=1}^{n} X_i$$

where $X_i \in \{0, 1\}$. So based on Central Limit Theoreom,

$$\overline X \sim \mathcal{N}\Big(p, \frac{p(1-p)}{n}\Big)$$

Obviously, $0 < \overline X < 1$ (let's ignore the boundary 0 and 1 edgecase for now).

I am interested in knowing if another r.v. $\log \overline X$ is asymptotically normal.

At first, I thought for $\log \overline X$ to be normal, then $\overline X$ should be log-normal, which isn't the case as described above.

However, I did some computer simulation, it seems to be the case that $\log \overline X$ is still normal, how to show it formally, please?

Answer 1

Have a look here delta method

So $\bar X_n$ is asymptotically normal with mean $\mu=p$ and variance $\sigma^2=p(1-p)$ in the sense that

$$ \sqrt n(\bar X_n - \mu ) \stackrel{d}{\rightarrow} \mathcal N(0,\sigma^2)$$

this is sometimes written as

$$\bar X_n \stackrel{a}{\sim} \mathcal N(\mu,\sigma^2/n)$$

so here $\sigma^2/n = p(1-p)/n = Var\left( \frac{1}{n} \sum_i^n X_i \right)$

Then with delta method which is based on among other thing the continuous mapping theorem

$\sqrt n(g(\bar X_n) - g(\mu)) \stackrel{d}{\rightarrow} \mathcal N(0,\sigma^2 g'(\mu)^2)$

where in the current case $g(t) = \log(t)$ and $g'(t) = 1/t$

so variance of $\sqrt n(g(\bar X_n) - g(\mu))$ should be $\sigma^2/\mu^2$ and hence

$\log \bar X_n \stackrel{a}{\sim} \mathcal N(g(\mu),\sigma^2/(n\mu^2))$.

so variance $\log \bar X_n$ is $\sigma^2/(n\mu^2)$