Hello all. I'm not confident in my methodology for this, so confirmation that this is okay would be helpful.

Problem:Let $\displaystyle X_1, X_2, ...$ be iid random variables with $\displaystyle E(X_1) = 0$. Let $\displaystyle S_n = \sum_{i = 1} ^ n X_i$. Show that $\displaystyle E[|S_n|] = o(n)$.

Solution:We must show that $\displaystyle E\left|\frac{S_n}{n}\right| \to 0$, which is equivalent to $\displaystyle \frac{|S_n|}{n} \stackrel{L^1}{\to} 0$. By the weak law of large numbers, we know $\displaystyle \frac{S_n}{n} \stackrel{p}{\to} 0$, and hence $\displaystyle \frac{|S_n|}{n} \stackrel{L^1}{\to} 0$ holds if and only if $\displaystyle \frac{|S_n|}{n}$ is weakly uniformly integrable.

First we show $\displaystyle X_1, X_2, ...$ is uniformly integrable.

$\displaystyle \lim_{\alpha \to \infty} \sup \int_{|X_n| \ge \alpha} |X_n| \ dP = 0$

But because the rvs are identically distributed, this is equivalent to

$\displaystyle \lim_{\alpha \to \infty} \int_{|X_1| \ge \alpha} |X_1| \ dP = 0$

which is true. Next we show $\displaystyle \frac{|S_n|}{n}$ is weakly uniformly integrable. Fix $\displaystyle \epsilon > 0$ and using uniform integrability of the $\displaystyle X_n$ get $\displaystyle \delta$ such that if $\displaystyle P(A) < \delta$ then $\displaystyle \int_A |X_n| \ dP < \epsilon$. Fix $\displaystyle A$ with $\displaystyle P(A) < \delta$. Then

$\displaystyle \displaystyle \int_A \frac{|S_n|}{n} \ dP &\le n^{-1} \int_A \sum_{i = 1} ^ n |X_i| \ dP \\&= n^{-1} \sum_{i = 1} ^ n \int_A |X_i| \ dP \\&\le n^{-1} \sum_{i = 1} ^ n \epsilon = \epsilon $

and we are done.