# Determine $\lim_{n\to\infty} \mathbb{P}(\sum_{i=1}^n X_i \leq \frac{n}{2})$

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 $\frac{1}{2}$?

• You can get the parentheses to adjust to their content by using \left and \right. – joriki Jun 13 '16 at 8:56
• @joriki: And also $\to$ with \to. – Asaf Karagila Jun 13 '16 at 8:58

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.
• 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 $1/2.$ – David Jun 13 '16 at 9:20
• @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 $X_i\to1-X_i$. – joriki Jun 13 '16 at 9:21