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I'm having some trouble with an exercise. Suppose $(X_n)_n$ is a sequence of independent r.v with $\mathbb{P}(X_n > x) = e^{-x}$ for positive x and one otherwise. The exercise first asks for what values of $\alpha>0$ is it almost surely the case the $X_n > \alpha \log \ n$ i.o. which I have managed.

Then it asks if $L= \limsup_n \frac{X_n}{\log \ n}$ show that $\mathbb{P}(L=1)=1$. So far I can show $\mathbb{P}(L \geq 1)=1$ and now want to show $\mathbb{P}(L > 1)=0$. But this part is giving me some problems.

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It is natural that you have some trouble: what you want to prove is false (with probability $1$, the r.v. $X_1$ is larger than $\ln (1) = 0$). If you put $L := \limsup_{n \to + \infty} X_n / \ln (n)$, on the other hand... – D. Thomine Feb 9 '12 at 11:44
And now, the hint : try to prove that for all $n \geq 1$, one has $\mathbb{P} (L \geq 1 + 1/n) = 0$ (with my definition of $L$). – D. Thomine Feb 9 '12 at 11:47
in the definition of the distribution, didn't you mean and one otherwise? – Ilya Feb 9 '12 at 12:31
Sorry about that. It was meant to be limsup and one as you both rightly said. – user17904 Feb 9 '12 at 13:26
What does "i.o." stand for? – joriki Feb 9 '12 at 15:26

I think it is time to give an answer. Warning: I give a full answer, so you might want to spot reading halfway if you just want a good hint.

We can write:

$$\{L > 1\} = \bigcup_{x > 1} \{L > x\},$$

but since we have an uncountable number of event on the right, we won't be able to conclude that the left side has measure $0$ even if each of the $\{L > x\}$ has measure $0$. So we use the usual trick (discretization):

$$\{L > 1\} = \bigcup_{k \in \mathbb{N}^*} \{L > 1+1/k\},$$

and all we have to show is that, for all $k \in \mathbb{N}^*$, we have $\mathbb{P} (L > 1+1/k) = 0$. Let us choose any positive $k$. Then:

$$\{L > 1+1/k\} = \{ \exists \varepsilon > 0: X_n \geq (1+1/k+\varepsilon) \ln (n) \ \text{i.o.} \} \subset \{ X_n \geq (1+1/k) \ln (n) \ \text{i.o.} \}.$$

If you have the good answer for the first question, then you know that $\mathbb{P} (X_n \geq (1+1/k) \ln (n) \ \text{i.o.}) = 0$ because $1+1/k>1$. Hence, $\mathbb{P} (L > 1+1/k) = 0$, and this holds for all $k \geq 1$. By the argument I gave in the beginning, $\mathbb{P} (L > 1) = 0$.


Just a quick remark: I did not do exactly what I said in my comment up there. Indeed, if we choose to use the events $\{L \geq 1+1/k\}$, we don't benefit from the inclusion:

$$\{L \geq 1+1/k\} \subset \{ X_n \geq (1+1/k) \ln (n) \ \text{i.o.} \},$$

which is false. Instead, we must use something like:

$$\{L \geq 1+1/k\} \subset \{ X_n \geq (1+1/(2k) \ln (n) \ \text{i.o.} \},$$

which is messy. So, for the sake of elegance, the strict inequality in $\{L > 1+1/k\}$ is quite useful.

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