I'm working on the following problem and I'm a little stuck

Suppose $X_1,X_2,\dots$ are iid.

(a) If $E|X_1|^\alpha$ is finite for some $\alpha>0$, show that $\max_{1\le k\le n} |X_k|/n^{1/\alpha}\to 0$ a.s.

(b) If $EX_1$ is finite and nonzero, show that $\max_{1\le k \le n}|X_k|/|S_n|\to 0$ a.s.

For part a) I'm thinking of using Borel-Cantelli by considering $\sum P(\max|X_k|^\alpha > \epsilon n)$ but I'm having trouble showing it's finite.

For part b) I think you would just use part a) and split up the product into $$\frac{\max|X_k|}{n^{1/\alpha}}\frac{n^{1/\alpha}}{|S_n|}$$ Then it's just a matter of showing $$\frac{n^{1/\alpha}}{|S_n|}$$ converges to something finite, which seems like an SLLN type problem, but not quite. Any advice would be appreciated.

source: Spring 1997

  • $\begingroup$ Markov's inequality for part (a)? $\endgroup$ – Brian Tung Aug 7 '18 at 21:29
  • 1
    $\begingroup$ You might be interested in my collection of solutions to (almost) all of the USC probability quals between 1994 and 2015: drive.google.com/file/d/11SxN17HnnaZjsSUJ7opqp_oa1UKlawvv/view $\endgroup$ – Mike Earnest Aug 7 '18 at 21:58
  • $\begingroup$ This is going to be extremely useful. Thanks a ton! $\endgroup$ – iYOA Aug 7 '18 at 22:07
  • $\begingroup$ Sure thing, please be on the lookout for errors though! I wrote those solutions pretty quickly. $\endgroup$ – Mike Earnest Aug 7 '18 at 22:08

I will use the following two Lemmas, try to prove them on your own:

  • If $X_i$ is an iid sequence with $\mathbb E|X_1|<\infty$, then $X_i/n\to 0$ a.s.

  • If $a_n$ is a sequence of real numbers with $a_n/n\to 0$, then $\frac1n \max_{1\le i\le n}a_i\to 0.$

Onto your problems:

  1. Since $E|X_1|^\alpha<\infty$, the first Lemma implies $|X_1|^\alpha/n\to 0$ a.s, then the second lemma implies $(\max_{1\le k\le n} |X_k|^\alpha)/n\to 0$ a.s. Conclude by raising each term to the $1/\alpha$.

  2. You should split it up like $$\frac{\max_{1\le k\le n}|X_k|}{n}\cdot \frac{n}{|S_n|}.$$The first fraction converges to $0$ by part (a). Since $n/S_n\to 1/EX_1$, you have $n/|S_n|\to 1/|EX_1|$.

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