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Let $X_1,\dots,X_n \overset{iid}\sim exp(1)$.

But then $\bar{X}$ is supposed to approach a normal distribution?

I agree that the skewness will slow such convergence, but that isn't my issue. The exponential distribution does not have support on the negative numbers, so $P(\bar{X} <0) = 0 $. It's not that the probability too small for software to calculate. The probability is zero.

Ditto if $X_1,\dots,X_n \overset{iid}\sim U(0,1)$ or $\chi^2_2$.

So what's going on?

What I can reason is that the distribution for finite $n$ never has support on all of $\mathbb{R}$ like a normal distribution does. However, when we "get" to $\infty$, we get support on all of $\mathbb{R}$. This reminds me of the Cauchy sequence of rational numbers 3, 3.1, 3.14, 3.141,... converging to the irrational $\pi$.

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  • $\begingroup$ Two approaches to answer: (1) The definition of convergence dist'n is satisfied, and it if doesn't match with your intuitive views about support, then your intuitive views don't matter. (2) Focus on the shape of the distribution over regions where most of the probability is. The dist'n of the mean of $n$ iid exponentials is gamma with shape parameter $n$. If you plot the gamma dist'n for $\bar X_{100}$ and superimpose the normal dist'n with same $\mu$ and $\sigma,$ you won't be able to distinguish between them. And 0 is so many SD's from the mean that the probability below 0 is essentially 0. $\endgroup$
    – BruceET
    Commented Aug 20, 2019 at 0:27
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    $\begingroup$ The variable $\bar{X}$ does indeed not become negative when every $X_i>0$. But, it's not the variable $\bar{X}$ that's approaching the normal distribution, and instead, it is some transformed version of it that approaches the normal distribution. $\endgroup$
    – Sextus Empiricus
    Commented Aug 20, 2019 at 0:27
  • $\begingroup$ My answer at stats.stackexchange.com/a/3904/919 makes a point of explaining and illustrating this with a more extreme example: that of a Bernoulli distribution. You need to remember that the CLT concerns the standardized distribution of the mean (or sum). BTW, there is no such thing as "getting to" infinity. $\endgroup$
    – whuber ♦
    Commented Aug 20, 2019 at 0:44
  • $\begingroup$ @Ben Your answer in the linked thread nailed it. $\endgroup$
    – Dave
    Commented Aug 20, 2019 at 0:45

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