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?

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    $\begingroup$ you could have a look at continuous mapping theorem and delta rule. $\endgroup$ Dec 31, 2018 at 23:23
  • $\begingroup$ @JesperHybel Could you please be more specific about how to use delta rule? $\endgroup$
    – zyxue
    Dec 31, 2018 at 23:24
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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jan 1, 2019 at 2:04
  • $\begingroup$ @Glen_b, could you please suggest more details about "the couple of possible things"? $\endgroup$
    – zyxue
    Jan 1, 2019 at 4:37
  • $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jan 1, 2019 at 5:14

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

  • $\begingroup$ Don't you mean "so variance should be $\sigma^2/(n\mu^2)$" $\endgroup$
    – zyxue
    Jan 1, 2019 at 0:41
  • $\begingroup$ have edited to make more clear which varince I'm referring to when. But yes the variance you are interested in is $Var(\log \bar X_n) ) = \sigma^2/(n \mu^2)$. $\endgroup$ Jan 1, 2019 at 0:49

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