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Let $X_1, X_2, ...$ be independent, identically distributed $L^2$ random variables (i.e. finite variance); let $m_n$ denote any median of $S_n = \sum_{j = 1} ^ n X_j$ for $n \ge 1$. I am to show that for $\alpha > \frac 1 2$, we have a strong law:

$\displaystyle \frac{S_n - m_n}{n^\alpha} \stackrel{a.s.}{\to} 0$

where $\stackrel{a.s.}{\to}$ denotes almost sure convergence.

My thought process: Well, I'm not sure where $L^2$ comes into the picture. I've been thinking of trying to show something of the form $m_n - nE(X_1) = o(n^\alpha)$, since then I can apply the strong law (this being where the condition that $\alpha > 1/2$ would come in since I need that for the strong law to hold). I'm guessing $L^2$'ness comes into play by helping to get $m_n$ and $nE(X_1)$ close enough together? But I'm not sure how. This also might be completely off point.

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    Anyways, I could see how the fact that they are $L^2$ suggests that $\frac{S_n - m_n}{n^{1/2}}$ converges to a standard normal, which seems to imply the result, but it isn't clear to me how to show this; it seems to require $m_n - nE(X_1) = o(n^{1/2})$ which is stronger than what I need to get the result.2011-07-02

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It sounds like this is not the solution you're looking for, but:

With the central limit theorem you can show that $m_n - n E(X_1) = o(n^{1/2})$. Intuitively, since $m_n$ depends on the distribution of $S_n$, and for large $n$, $S_n$ is approximately normally distributed, one should expect $m_n$ to be close to the median of a normal, which is the same as the mean.

For a proof, set $\mu = E(X_1)$, $\sigma^2 = Var(X_1)$, and $Z \sim N(0,1)$. If we let S'_n = (S_n - n \mu)/\sigma \sqrt{n}, we have S'_n \Rightarrow Z. Let m'_n = (m_n - n \mu)/\sigma \sqrt{n}; then m'_n is a median of S'_n. We want to show m'_n \to 0. Passing to a subsequence, we may assume m'_n converges to some $m \in [-\infty, \infty]$. Also, fixing some $\epsilon > 0$, we can drop finitely many terms to ensure m-\epsilon < m'_n < m+\epsilon for all $n$.

Then we have \frac{1}{2} \le P(S'_n \le m'_n) \le P(S'_n \le m + \epsilon) \to P(Z \le m+\epsilon) by the weak convergence. By continuity of probability, we can let $\epsilon \to 0$ to see that $P(Z \le m) \ge \frac{1}{2}$.

A similar argument shows that $P(Z \ge m) \ge \frac{1}{2}$, so $m$ is a median of $Z$. But the only median of $Z$ is 0, so we must have $m=0$, and this completes the proof.

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    I think this is satisfactory for me. I'll be better for studying your proof. Thank you kindly.2011-07-02