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Suppose $X_1, X_2, ... X_n$ are independent and uniformly distributed (on $[0,1]$) random variables. Determine $\lim_{n\to\infty} \mathbb{P}(\sum_{i=1}^n X_i \leq \frac{n}{2})$

My thoughts were the following:

I suppose I can say that $$\lim_{n\to\infty} \mathbb{P}\left(\sum_{i=1}^n X_i \leq \frac{n}{2}\right)=\lim_{n\to\infty} \mathbb{P}\left(\frac{1}{n}\sum_{i=1}^n X_i \leq \frac{1}{2}\right)=\lim_{n\to\infty} \mathbb{P}\left(\overline X_n \leq \frac{1}{2}\right)$$ And isn't $\overline X_n$ also uniformly distributed? So the probability equals $\frac{1}{2}$?

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  • $\begingroup$ You can get the parentheses to adjust to their content by using \left and \right. $\endgroup$
    – joriki
    Commented Jun 13, 2016 at 8:56
  • $\begingroup$ @joriki: And also $\to$ with \to. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 13, 2016 at 8:58

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No, $\overline{X_n}$ isn't uniformly distributed; but it's distributed symmetrically about $\frac12$, so you can nevertheless conclude that the probability is $\frac12$ even without the limit.

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  • $\begingroup$ So what is the right way to go? I don't fully understand $\endgroup$
    – Di-lemma
    Commented Jun 13, 2016 at 9:08
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    $\begingroup$ If one is not comfortable thinking about this as intuitively as @joriki has described, we can also apply the Central Limit Theorem. In the limit, $\bar{X_n}$ will be distributed normally, and the mean of $\bar{X_n}$ will be $\mathbb{E}(X_i) = \frac{1}{2}.$ Thus, by the symmetry of the normal distribution, this probability would be $1/2.$ $\endgroup$
    – David
    Commented Jun 13, 2016 at 9:20
  • $\begingroup$ @Di-lemma: There's no more way to go. The statement is true by symmetry, even if you remove the limit. If you want to make the symmetry explicit, it's $X_i\to1-X_i$. $\endgroup$
    – joriki
    Commented Jun 13, 2016 at 9:21
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    $\begingroup$ @David: I think you meant "normally" instead of "uniformly"? But it would be quite a detour to prove something using the central limit theorem when it's actually true without the limit. $\endgroup$
    – joriki
    Commented Jun 13, 2016 at 9:22
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    $\begingroup$ @David: Then it's a bad exercise :-) $\endgroup$
    – joriki
    Commented Jun 13, 2016 at 9:35

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