Let $X_1, X_2, ...$ be a sequence of random variables (finite almost surely), and let $0 < b_n \uparrow \infty$. As part of solving a problem I seem to have shown that $\frac{X_n}{b_n} \to 0$ almost surely implies $\frac{\max_{1 \le j \le n} |X_n|}{b_n} \to 0$ almost surely. I want to confirm that this is true since, as it pertains to the full problem I am solving, it doesn't use all of the hypotheses I am given. It seems simple enough, but I'm wondering if I am making a mistake somewhere.

Fix $n > k \ge 2$. Then $$\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \frac{\max_{1 \le j \le k} |X_j|}{b_n} + \frac{\max_{k \le j \le n} |X_j|}{b_n} \le \frac{\max_{1 \le j \le k} |X_j|}{b_n} + \max_{k \le j \le n} \frac{ |X_j|}{b_j}$$

the last inequlaity because $b_n \uparrow$. Letting $n \to \infty$ gives $$\limsup_n\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \sup_{j \ge k} \frac{|X_j|}{b_j}$$ almost surely and letting $k \to \infty$ gives $$\limsup_n\frac{\max_{1 \le j \le n} |X_j|}{b_n} \le \limsup_n \frac{|X_n|}{b_n}$$

and by hypothesis the term on the RHS is equal to $0$ almost surely.

Is this okay or are there holes?

  • $\begingroup$ Should the $s_n$ in the title be really $b_n$? Seems so. $\endgroup$ – Srivatsan Jul 27 '11 at 20:25
  • $\begingroup$ Yeah, sorry. That came from changing some notation. $\endgroup$ – guy Jul 27 '11 at 20:30
  • $\begingroup$ I went ahead and edited the title. I also added (sequences-and-series) tag since it seemed relevant. Hope that's ok. (My edit has to be approved before, so it might take some time to take effect.) $\endgroup$ – Srivatsan Jul 27 '11 at 20:33
  • 1
    $\begingroup$ Seems correct :). $\endgroup$ – fedja Jul 27 '11 at 22:57
  • $\begingroup$ The proof is correct and shows that the result has nothing to do with randomness but really concerns deterministic sequences. $\endgroup$ – Did Jul 30 '11 at 15:36

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