According to Central Limit Theorem (CLT), the mean of any i.i.d. sample is Normal distributed (taking $n\rightarrow\infty$ samples).

Let $X_i\sim U(a,b)$.

Then $\bar{X}\sim N$ by CLT.

But as we know, $U$ has a.s. no support outside $(a,b)$ so its mean is never distributed Normal on $\mathbb{R}$? I expect it would be (a,b)truncated-Normal?


You need to look at a more precise statement of the central limit theorem. For i.i.d. random variables $X_i$ with mean $0$ it states that $\frac{X_1 + ... + X_n}{\sqrt{n}}$ (note the division by $\sqrt{n}$ rather than $n$) approaches a normal distribution with variance the variance of $X_i$. In particular, even if each $X_i$ is supported in $(a, b)$, this sum is supported in $(\sqrt{n} a, \sqrt{n} b)$.

  • $\begingroup$ To convince yourself that this is the right statement you should actually compute the variance of the sum above. This isn't hard but it's an important exercise in this context. $\endgroup$ Jun 29 '14 at 17:45
  • $\begingroup$ Could you maybe also discuss the case $a>0$, or nonnegative $X$ in general how this would fit CLT? $\endgroup$
    – emcor
    Jun 29 '14 at 17:58
  • 1
    $\begingroup$ @emcor: subtract the mean. CLT is a statement about how the sample mean fluctuates about its mean. That's why I specified mean $0$ above. $\endgroup$ Jun 29 '14 at 17:58

Central Limit Theorem says $\sqrt{n}(\bar X-\mu) \stackrel{d}{\rightarrow} N(0,\sigma)$, as $n\rightarrow\infty$. Or you can think of $\bar X$ approaches $N\big(\mu,\frac{\sigma}{\sqrt{n}}\big)$, so the normal distribution shrinks around the mean $\mu$ and the distribution approaches zero rapidly outside any fixed interval around $\mu$. Your statement lost the essential factor $\sqrt n$.

  • $\begingroup$ This isn't quite right; you get the wrong mean if you just multiply by $\sqrt{n}$. $\endgroup$ Jun 29 '14 at 17:40
  • $\begingroup$ @QiaochuYuan: You are right. My negligence. I have corrected it. $\endgroup$
    – Hans
    Jun 29 '14 at 17:42

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