My question is relatively simple: what is the limiting distribution of the sample mean? But there are some technicalities I want to discuss.

context: I was asked this problem in an exam, and I feel my answer is correct. Professor doesn't think so and gives me a roundabout explanation that doesn't really address the issue.

By the Weak Law of Large Numbers, we know that $\bar{X}$ converges to $\mu$ in distribution, where $\bar{X}$ denotes the sample mean and $\mu$ denotes the true mean.

I was asked to find the limiting distribution of $\bar{X}$. I used the idea that convergence in probability to a constant $\mu$ implies convergence in distribution to that constant. So, I specified the limiting distribution of $\bar{X}$ as:

$F_{\bar{X}}(\bar{X} \leq x) = 1$ if $x \geq \mu$ and 0 otherwise.

The answer I was expected to give was:

$\sqrt{n}(\bar{X}-\mu) \to N(0, \sigma^2)$ in distribution.

My problem is that this isn't the limiting distribution of $\bar{X}$ itself -- it's the limiting distribution of a function of $\bar{X}$. Am I correct in stating that $\bar{X}$ converges to $\mu$ in distribution, or did I miss the point of what a limiting distribution is?

Thanks in advance.

  • $\begingroup$ I do not think it is incorrect to say that for very large, infinite $n$, $\bar{X}$ converges to a degenerate distribution on $\mu$. You also need to bear in mind though that since for large but not infinite $n$ by the CLT we know that , $\sqrt{n} \frac{\left( \bar{X}-\mu \right) }{\sigma} \rightarrow^D N \left(0,1 \right)$, it follows that $\bar{X} \rightarrow^D N \left(\mu, \frac{\sigma^2}{n} \right)$. $\endgroup$
    – JohnK
    Nov 24, 2014 at 18:05
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    $\begingroup$ In the latter case, I don't think $\bar{X}$ converges in distribution to $N(\mu, \frac{\sigma^2}{n})$ because as $n \to \infty$, the variance goes to 0, and that just shows $\bar{X} \to \mu$. $\endgroup$ Nov 24, 2014 at 18:08
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    $\begingroup$ Your professor could be wrong for two reasons: one you have given and the other is that the CLT does not apply unless additional assumptions are made about the underlying distribution. However, the WLLN also requires additional assumptions, so neither of the positions you portray here is correct. I would presume, though, that (1) the question stipulated more assumptions than you have revealed here and (2) your professor may have defined "limiting distribution" in a more general sense than "limit in distribution." Everything hinges on these details! $\endgroup$
    – whuber
    Nov 24, 2014 at 18:34
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    $\begingroup$ A usual "understood" shortcut in statistical communication, is that the expression "the limiting distribution of $X_n$" means more generally "a limiting distribution of some function of $X_n$". To newcomers in statistics, this indeed can cause confusion. $\endgroup$ Nov 24, 2014 at 20:07

1 Answer 1


You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d $ a constant. A different way to frame a similar question is to say, what is an approximating distribution of $\bar{X}_n$ ($n$ being the sample size in question). Then it would be right to say $\bar{X}_n \dot{\sim} \mathcal{N} \left( \mu, \sigma^2/n \right)$

I think it's sloppy notation and the professor should have been clearer. In fact, in my theory classes, our professor had the deepest ire for what he considered a serious deficiency of understanding if students found limiting distributions that were functions of the $n$.


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