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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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Here's a hint to get you started: $m_n$ is a quantity that depends on the distribution of $S_n$. What happens to the distribution of $S_n$ as $n \to \infty$? – Nate Eldredge Jul 2 '11 at 5:33
It is not a problem associated with a class that I am taking, so in that sense it isn't homework; it is an exercise in a text. Anyways, I had thought to use the Central Limit Theorem, but this problem appears before convergence in distribution is even discussed, so it must not be necessary. Is that some result that if $T_n$ converges in distribution to something then $T_n / {n^\epsilon}$ goes almost surely to $0$ for all $\epsilon > 0$? – guy Jul 2 '11 at 6:07
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. – guy Jul 2 '11 at 6:27
up vote 3 down vote accepted

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. – guy Jul 2 '11 at 15:52

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