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I am struggling with an exercise… the first part of the exercise was ok, but the second part is a lot less trivial.

$X_n$ are non negative iid random variables.

First part: check that if $E(X_i)$ finite then $\lim X_n/n=0$ almost surely. I showed this using events $A_{k,\epsilon}=\{X_n>n\epsilon\}$, tail formula for $E(X_i)$ and 1st Borel Cantelli lemma.

Second part: on the other hand show that if $E(X_i)=\infty$ then $\operatorname*{p-lim} X_n/n =0 $ (that is $X_n/n \to 0$ in probability) but $P(\limsup X_n/n=\infty)=1$.

I get stuck on this one as it seems to contradict what I learnt in class. I try to use the 2nd Borel Cantelli lemma but can’t get to the conclusion.

Brian Moehring
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JenniferK
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    See https://math.stackexchange.com/help/notation for tips on formatting your post with MathJax to make it more readable. See also How to ask a good question. As it stands your question contains not much more than just the problem statement, which is likely to get closed. Include more about what you have tried. – Nate Eldredge Feb 11 '24 at 18:25

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For the first half: let $X$ be a fixed random variable with the same distribution as the $X_n$ (you can take $X=X_1$ if you like). Then $P(X_n/n \ge \epsilon) = P(X/n \ge \epsilon) = P(X \ge \epsilon n)$. Show that this converges to $0$ as $n \to \infty$. Hint: what can you say about $\bigcap_{n =1}^{\infty} \{X \ge \epsilon n\}$?

For the second half, start by showing the following lemma: if $X$ is any nonnegative random variable, then $$E[X]-1 \le \sum_{n=1}^\infty P(X \ge n) \le E[X]$$ Hint: show $(X-1) \le \sum_{n=1}^\infty 1_{X \ge n} \le X$ for each $\omega$, by considering what happens for each side when $k-1 \le X(\omega) < k$ for a given integer $k$.

Use this with the second Borel-Cantelli lemma to show that $P(X_n \ge n \text{ i.o.}) = 1$, which says that $\limsup X_n/n \ge 1$ almost surely. Now repeat, replacing $X_n$ with $X_n/m$ for arbitrary $m$, to get $\limsup X_n/n \ge m$ a.s. Finally take a countable intersection to get $\limsup X_n/n = \infty$ a.s.

Nate Eldredge
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  • Can you please clarify the last sentence? To get from $limsupX_n/n>=m$ a.s. to $limsup = infinity$ a.s.? – JenniferK Feb 11 '24 at 20:35
  • @JenniferK: Let $A_m = { \limsup X_n/n \ge m}$. We showed $P(A_m) = 1$. Let $A = \bigcap_{m=1}^\infty A_m$; then $P(A)=1$ and on $A$ we have $\limsup X_n/n = \infty$. – Nate Eldredge Feb 11 '24 at 20:48
  • Thank you! Wouldn’t it be enough to say that $P(A_m)=1 $ for any m including very large m and thus $ P(limsupX_n/n= infinity)=1$ ? – JenniferK Feb 11 '24 at 20:52
  • @JenniferK: That's the basic idea, yes. I just wanted to be clear about the details. We're using the countable additivity of $P$ (the claim is not true for finitely additive measures) and so it's important that we only need to know $\limsup X_n/n \ge m$ for a countable set of values for $m$. (Formatting hint: use \limsup and \infty.) – Nate Eldredge Feb 11 '24 at 20:59
  • Thanks! And thanks for the formatting tips! – JenniferK Feb 11 '24 at 21:02