A Strong Law using Medians 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. 
 A: 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.
