Hi I came up with a proof and I'm not really sure about it. Can someone help me to check whether is correct or not?


Consider a sequence $(Xn)_{n\geq1}$ of exponentially-distributed random variables with mean $λ>0 \ (i.e.EX_1 =λ)$. Prove or disprove wether

$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg)=1$$


Let us define a sequence of events $(E_n)_{n \geq 1}$ where

$$E_n = \{ \omega \in \Omega: X_n(\omega)< \lambda \log n \}$$

Let us note moreover that $(X_n)_{n \geq 1}$ is a sequence of iid exponentially distributed random variable and hence $(E_n)_{n \geq 1}$ is a sequence of independent events. Let us now compute

$$\forall n\geq 1 \qquad P(E_n)=P(X_n< \lambda \log n)=F_{X_n}( \lambda \log n)=1-e^{-\lambda^2 \log n}$$

Hence we have that

$$\sum_{n=1}^\infty P(E_N)= \infty$$

So by the second Borel-Cantelli lemma we have that

$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)=1$$

We now observe that

$$P \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)=1 \rightarrow P \bigg( \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)^c \bigg)=0$$

Now, since

$$\bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg) \subseteq \bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}<1 \bigg)^c$$

we conclude by monotonicity that

$$P\bigg(\limsup_{n \rightarrow \infty} \frac{X_n}{\lambda \log n}=1 \bigg)=0$$

disproving the initial statement.

  • $\begingroup$ There is probably an issue -- the statement is true. $\endgroup$
    – Clement C.
    Nov 7, 2017 at 0:13
  • $\begingroup$ How can I prove such a result? The question that you are referring to was asked by me previously... But I think that there's an error because we use two contradictory results $\endgroup$ Nov 7, 2017 at 0:26
  • $\begingroup$ See e.g. this solution. $\endgroup$
    – Clement C.
    Nov 7, 2017 at 0:33
  • $\begingroup$ math.stackexchange.com/questions/2491452/… $\endgroup$
    – Math1000
    Nov 7, 2017 at 6:51
  • $\begingroup$ There is a confusion since an exponential variable $X$ with parameter $\lambda$ typically has $E[X]=1/\lambda$ and $P[X>x] = e^{-\lambda x}$, but the problem is defining it as the inverse of that (so the mean is $\lambda$ and $P[X>x] = e^{-x/\lambda}$) and some of your formulas may be inconsistent due to this. $\endgroup$
    – Michael
    Nov 7, 2017 at 7:00

1 Answer 1


There is a confusion between the $\lim \sup$ of real sequences and the set-theoretical $\lim \sup$ of the Borel-Cantelli Lemma.


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