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

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?

• you could have a look at continuous mapping theorem and delta rule. Dec 31 '18 at 23:23
• @JesperHybel Could you please be more specific about how to use delta rule? Dec 31 '18 at 23:24
• The non-zero probability of $\bar{X}=0$ (so taking $\log(0)$) is your main problem; this non-zero probability occurs at every finite sample size so you can't really argue for a sequence of random variables that converge to something. If you deal with this in some way (there are a couple of possible things you might do) then you could skirt this issue. Jan 1 '19 at 2:04
• @Glen_b, could you please suggest more details about "the couple of possible things"? Jan 1 '19 at 4:37
• e.g. One potential approach is to condition on being >0 and then construct a sequence of random variables from there.Presumably your simulation analysis would have excluded cases with $\bar{X}=0$ had you observed any. Jan 1 '19 at 5:14

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 $$\sigma^2/(n\mu^2)$$

• Don't you mean "so variance should be $\sigma^2/(n\mu^2)$" Jan 1 '19 at 0:41
• have edited to make more clear which varince I'm referring to when. But yes the variance you are interested in is $Var(\log \bar X_n) ) = \sigma^2/(n \mu^2)$. Jan 1 '19 at 0:49