# Limsup of independent exponential random variables

In the lecture, we are given an example on independent and identically distributed random variables, but I am not quite sure what is the idea of this question. The exercise is as following:

Let (Xn) be a sequence of independent and identically distributed random variables defined on a probability space (Ω,F,P). Assume that X1 is exponentially distributed with parameter λ>0. The distribution of X1 is absolutely continuous with density function $f(x)=λe^{-λx}1_{(0,∞)}(x)$, x∈R.

Show that $P(limsup_{(n→∞)}X_n/logn=1/λ)=1$.

I know that for the exponential distribution, the mean E[x]=1/λ, it says X1 is exponentially distributed, is that mean all of the Xn are also exponentially distributed? $limsup_{(n→∞)}(X_n)$ means that Xn happens for infinitely often times as n close to infinity. so $limsup_{(n→∞)}(X_n)=⋂_{n∈N}⋃_{m>n}X_n$, and also as $n→∞$, $log(n)→∞$ . I am not sure about what this question is asking about? Is it something with 0/1 law? like if an event A happens infinitely times, then P(A)∈{0,1}

Can any one please help me with this question? details are preferred. Thanks!

• please use Latex – Alex May 4 '14 at 6:49
• I am not sure how to use Latex in here, sorry – xz330 May 4 '14 at 7:00
• @xz330 I have submitted a revision to your question that fixes the latex formatting. Please check that I haven't altered the meaning of your question. – senshin May 4 '14 at 7:49
• @senshin thanks! any idea about this question? – xz330 May 4 '14 at 7:54
• Changed the title to something more accurate. Do you understand why it fits better the content of the question? – Did May 8 '14 at 9:09

## 1 Answer

Show that $P(lim(n→∞)supXn/logn=1/λ)=1$.

(...) $lim(n→∞)(supXn)$ means that Xn happens for infinitely often times as n close to infinity. so $lim(n→∞)(supXn)=⋂_n⋃_mXn$, where m>n and n∈N.

This is confusing limsup of events and limsup of random variables. Here, each $X_n$ is a random variable hence $$\limsup\limits_{n\to\infty}X_n/\log n$$ is the random variable $Y$ such that, for every $\omega$ in $\Omega$, $$Y(\omega)=\limsup\limits_{n\to\infty}X_n(\omega)/\log n.$$ And the task is to prove that the event $[Y=1/\lambda]$ has probability $1$.

Note that the question most probably does not reproduce faithfully the text of the exercise asked. For example, $$\text{lim(n→∞)supXn/logn}$$ should read $$\limsup\limits_{n\to\infty}X_n/\log n,$$ where $\limsup\limits_{n\to\infty}$ acts as a single operation, or, equivalently, $$\limsup\limits_{n\to\infty}\frac{X_n}{\log n}.$$

• oh right, I got what it means now, but how do you prove that $P[Y=1/λ]=1$? – xz330 May 4 '14 at 7:58
• Yeah, I have changed all the equations to the proper formatting now. – xz330 May 4 '14 at 8:00
• That is a different question. – Did May 4 '14 at 8:01
• isn't it what the question is asking about? – xz330 May 4 '14 at 8:01
• No it is not. And if you want an answer to that other question, you should add your thoughts on it. – Did May 4 '14 at 8:02