Check my probability proof?

**Guy** · June 22nd 2011, 10:14 AM

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

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

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

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

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

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

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

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

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

and we are done.

**girdav** · November 24th 2012, 02:50 AM

What you did is correct. Just add the fact that $\left\{n^{-1}\lVert S_n\rVert_{L^1}\right\}$ is bounded.

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