# Central Limit Theorem not valid?

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 $(\sqrt{n} a, \sqrt{n} b)$.

• 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. Jun 29 '14 at 17:45
• Could you maybe also discuss the case $a>0$, or nonnegative $X$ in general how this would fit CLT? Jun 29 '14 at 17:58
• @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. 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 $\sqrt n$.

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