$\lim_{n\rightarrow \infty}P\left(\sum_{i=1}^n X_i \leq \frac{n}{2}+ n^{3/4} \right)$?

Question

Let $X_1,.....,X_n$ be i.i.d random variables with $U(0, 1)$. Then

$$\lim_{n\rightarrow \infty}P\left(\sum_{i=1}^n X_i \leq \frac{n}{2}+ n^{3/4} \right)$$

I tried to use CLT but somehow I cant solve the problem. Please help. Thanks

Answer 1

$\newcommand{\Var}{\operatorname{Var}}\newcommand{\P}{\mathbb{P}}$No need for CLT. Let $Y_n =\sum_{i=1}^n X_i$. Now notice the following: \begin{align} \P\left(Y_n> \frac{n}{2}+n^{3/4}\right)=\P\left(Y_n -\frac{n}{2}>n^{3/4}\right)\leq \P\left(\bigg|Y_n-\frac{n}{2}\bigg|>n^{3/4}\right)\leq \frac{\Var(Y_n)}{(n^{3/4})^2} \end{align} where the last inequality is obtained via Chebyshev's inequality and the fact that $\mathbb{E}[Y_n]=\frac{n}{2}$. We have $\Var(Y_n)=\frac{n}{12}$ hence: \begin{align} \P\left(Y_n> \frac{n}{2}+n^{3/4}\right)\leq \frac{1}{12n^{1/2}}\to 0 \end{align} as $n\to\infty$. So we conclude: \begin{align} \lim_{n\to\infty}\P\left(\sum_{i=1}^nX_i\leq \frac{n}{2}+n^{3/4}\right)=\lim_{n\to\infty}1-\P\left(Y_n> \frac{n}{2}+n^{3/4}\right)=1-0=1 \end{align}