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Math Help - Check my probability proof?

  1. #1
    Guy
    Guy is offline
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    Check my probability proof?

    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.
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  2. #2
    Super Member girdav's Avatar
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    Re: Check my probability proof?

    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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