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 $(X_n)$ be a sequence of independent and identically distributed random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Assume that $X_1$ is exponentially distributed with parameter $\lambda > 0$. The distribution of $X_1$ is absolutely continuous with density function $f(x)=\lambda e^{-\lambda x}1_{(0,\infty)}(x)$, $x\in\mathbb{R}$.

Show that $\mathbb{P}(\lim\sup_{n\to \infty}X_n/\log(n)=1/\lambda)=1$.

I know that for the exponential distribution, the mean $\mathbb{E}[x]=1/\lambda$, it says $X_1$ is exponentially distributed, is that mean all of the $X_n$ are also exponentially distributed? $\lim\sup_{n\to\infty}(X_n)$ means that $X_n$ happens for infinitely often times as $n$ close to infinity. so $\lim\sup_{n\to\infty}(X_n)=\bigcap_{n∈N}\bigcup_{m>n}X_n$, and also as $n\to\infty$, $\log(n)\to\infty$ . 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 $\mathbb{P}(A)\in\{0,1\}$

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

  • $\begingroup$ please use Latex $\endgroup$
    – Alex
    May 4, 2014 at 6:49
  • $\begingroup$ I am not sure how to use Latex in here, sorry $\endgroup$
    – xz330
    May 4, 2014 at 7:00
  • $\begingroup$ @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. $\endgroup$
    – senshin
    May 4, 2014 at 7:49
  • $\begingroup$ @senshin thanks! any idea about this question? $\endgroup$
    – xz330
    May 4, 2014 at 7:54
  • $\begingroup$ Changed the title to something more accurate. Do you understand why it fits better the content of the question? $\endgroup$
    – Did
    May 8, 2014 at 9:09

1 Answer 1


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, $$ \limsup\limits_{n\to\infty}\frac{X_n}{\log n}. $$

  • $\begingroup$ oh right, I got what it means now, but how do you prove that $P[Y=1/λ]=1$? $\endgroup$
    – xz330
    May 4, 2014 at 7:58
  • $\begingroup$ Yeah, I have changed all the equations to the proper formatting now. $\endgroup$
    – xz330
    May 4, 2014 at 8:00
  • $\begingroup$ That is a different question. $\endgroup$
    – Did
    May 4, 2014 at 8:01
  • $\begingroup$ isn't it what the question is asking about? $\endgroup$
    – xz330
    May 4, 2014 at 8:01
  • 1
    $\begingroup$ "If you want an answer to that other question, you should add your thoughts on it" (bis). $\endgroup$
    – Did
    May 8, 2014 at 9:07

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