I hope this question is within the scope of this site. Please note that I have solved this Exercise, I do have doubts about my presentation though and about how to rigorously empathize on the arguments.

Problem: Let $(X_n)_{n \geq 1}$ be a sequence of i.i.d. RV with exponential distribution (standard, parameter 1). Show that $$\limsup \frac{X_n}{\ln(n)} =1 \text{ a.s.} $$

"Proof": As before mentioned I do understand the idea behind this proof, my problem lies in the presentation. I will highlight the important steps that I do not understand.

Certainly since $X_n \sim X \sim $Exponential(1) we have for all $t>0$ that $$P(X_n \geq t) = \int_t^\infty e^{-x}dx = e^{-t} $$

Now let $\epsilon \in (0,1)$ be fixed but arbitrary, we then get $$P(X_n/\ln(n) \geq 1- \epsilon ) =\frac{1}{n^{1-\epsilon}} \implies \sum P(X_n/\ln(n) \geq 1- \epsilon) = \infty $$ Since our sequence of RV is i.i.d. we can use Borel Cantelli 2 to conclude that $$P( \limsup \lbrace X_n/ \ln(n) \geq 1-\epsilon \rbrace ) = 1 $$ Please note the braces $\lbrace. \rbrace$ here, they will later on contribute to my confusion. Since the above statement is true for all $\epsilon \in (0,1)$ I can let $\epsilon = \frac{1}{k}$ for $k > 1$ and then define $$A_k:= \lbrace \limsup X_n/\ln(n) \geq 1 - 1/k \rbrace $$

Question 1: I have seen the above being done before. Namely the $\limsup$ which was 'outside' of the event (namely outside the brackets) is now inside the event. Is this step okay? How can I be certain that pulling the $\limsup$ inside the event will still give me an event that has probability $1$?

Now I know that $P(A_k)=1$ for all $k>1$, therefore $P(\cap A_k$)=1 and therefore I have $$\omega \in \cap A_k \iff \limsup (X_n(\omega) / \ln(n)) \geq 1 - 1/k \ , \forall k >1 \\ \implies \lim\sup X_n/\ln(n) \geq 1 \ \text{ almost surely}$$


Now for other direction I want to go a similar route. Here I run into trouble though. Obviously for $\epsilon \in (0,1)$ I have $$ P ( X_n/\ln (n) \geq 1+\epsilon)= \frac{1}{n^{\epsilon +1}} $$ Similar as in the above steps I will this time end up with a finite sum, hence by Borel Cantelli 1st. Version I obtain that $$P (\limsup \lbrace X_n/ \ln(n) \geq 1+\epsilon \rbrace) = 0 $$ Here my problem is that I obviously want to consider the complementary event, but that will give me the statement that $$P ( \liminf \lbrace X_n/\ln(n) \leq 1+ \epsilon \rbrace ) =1 $$ Ultimately, following the same procedure as above, I would also show that $$ \liminf X_n/ \ln(n) \leq 1 \text{ almost surely} $$ But I have doubts about this being true. This would give me statement that $$ \lim X_n/ \ln(n) =1 \text{ almost surely} $$ of course I could also then say that $\lim, \limsup, \liminf$ are all equal in this case, but I don't think that's what I am supposed to show.

Question 2: What am I supposed to do in order to show that $\limsup X_n/ \ln(n) \leq 1$ ? (Partially answered by myself, see update)

Update: Given that $$ P(\limsup \lbrace X_n/\ln(n) > 1+ \epsilon \rbrace )=0, \ \forall \epsilon \in (0,1) $$ We can set that $$A_k= \limsup \lbrace X_n/ \ln(n) > 1+ 1/k \rbrace $$ Which is a nullset. Then we get by the subaddivitiy of $P$ that $$P(\cup A_k) \leq \sum P(A_k)=0 \implies \cup A_k = \limsup \lbrace X_n/\ln(n) > 1 \rbrace \text{ is nullset} $$

So I believe what is left for me to understand to complete and fully grasp this exercise is:

Question 3: Why do we have $$ \lbrace \limsup X_n/ \ln (n) >1 \rbrace \subset \bigcup_k \limsup \lbrace X_n/ \ln(n) > 1 + 1/k \rbrace \\ \bigcap_k \ \limsup \lbrace X_n/\ln(n) \geq 1 - 1/k \rbrace \subset \lbrace \limsup X_n/ \ln(n) \geq 1 \rbrace $$

Question 3 is more general than 1 and 2 and would also answer them. I think this statement might be general. So if I am lucky I might find it in a measure theory/probability theory book.

  • You haven't quite understood Borel-Cantelli, and the difference helps with your question 1. The $\lim \sup$ is as $n \to \infty$, so you don't need to define $A_k$ at all; it's already part of your previous equation. "Almost surely" is the same as "with probability 1", so you're done at the $P(\lim_{n \to \infty} \sup \dots)$ line. This might help you with part 2 as well. – jbowman Apr 27 '16 at 17:51
  • Thanks for your comment @jbowman. I do know that by definition $$P ( \lbrace \omega \in \Omega: : \lim X_n(\omega) = X(\omega)\rbrace ) = 1 \iff X_n \to X \text{ almost surely} $$ However I just have shown that $$ P( \limsup \lbrace X_n/\ln(n)-1 \geq -\epsilon \rbrace ) = 1 $$ I think some extra steps are required (for a beginner like me) to go from the above to $$ \limsup X_n/\ln(n) = 1 \text{ almost surely}$$ or do I understand you wrong? Also what "difference" do you mean? – Spaced Apr 27 '16 at 18:08
  • +1 for showing admirable effort! However, this is a very, very, long question. Would you be able to break it up into several questions? Each one would possibly require decent post. – user75138 Apr 28 '16 at 4:12
  • Thanks a lot @Bey, I do absolutely agree with you! I got lost in equations and during the process my post became way too lengthy. I will break this question apart and most likely split it into two separate questions. – Spaced Apr 28 '16 at 14:51

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