# Convergence of iid exponentially distributed random var

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?

Statement:

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

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

Proof:

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.

• There is probably an issue -- the statement is true. – Clement C. Nov 7 '17 at 0:13
• 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 – Random-newbie Nov 7 '17 at 0:26
• See e.g. this solution. – Clement C. Nov 7 '17 at 0:33
• math.stackexchange.com/questions/2491452/… – Math1000 Nov 7 '17 at 6:51
• 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. – Michael Nov 7 '17 at 7:00

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