Prove that if $E(X\log X)<\infty$ then $E(\sup_n |S_n|/n)<\infty$.

Question

This is part 2 of a two part question. In the first part, we were asked to show that if you had a non-negative sub martingale $M_n$ then $$\sup_n E(\sup_{k\leq n} M_k)\leq \sup_n 2E(M_n \log M_n)+2$$

We need to use the above fact to Prove that if $E(X\log X)<\infty$ then $$E(\sup_n |S_n|/n)<\infty,$$ where $S_n=X_1+\dots+X_n$, with $X_i$ iid with distribution $X$.

I am unsure of how to introduce a sub-martingale into the problem since $S_n/n$ is not a martingale. Any help/hints would be appreciated.

Answer 1

Hints:

1. Show that $(S_n/n)_{n \in \mathbb{N}}$ is a backwards martingale.
2. Applying the first part to the (backwards) martingale gives a bound in terms of $$\mathbb{E}\left(\left| \frac{S_n}{n} \right| \cdot \left|\log\left( \frac{S_n}{n} \right) \right| \right).$$ In order to show that this expression is finite, use that $$[1,\infty) \ni x \mapsto x \log(x)$$ is a convex function and $$(0,1] \ni x \mapsto x \cdot \log(x)$$ is bounded.