Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X_1,X_2,\dotsc$ be defined jointly. I'm not entirely sure what this means, but I think it means they're all defined on the same space. Let $E[X_i]=0, E[X_i^2]=1 \;\forall\; i$. Show $P(X_n\geq n \text{ infinitely often}=0$. The condition that an event occurs infinitely often is equivalent to saying the $\lim\sup (X_n\geq n)$ (at least I'm pretty sure that's the correct way to write it). The $\lim\sup$ is defined as $$ \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (X_k \geq n) $$ Now since the mean for each $X_i$ is $0$ intuitively it makes sense that the probability would go to $0$. My concern though is that I'm not using the other hypothesis and I can't really see how it fits in. Thanks!

share|improve this question

1 Answer 1

Hint: According to the first Borel-Cantelli lemma, the limsup of the events has probability zero as soon as the series $(*)$ $\sum\limits_n\mathrm P(X_n\geqslant n)$ converges. Hence if one shows $(*)$ converges, the proof is over.

How to show that $(*)$ converges? Luckily, one is given only one hypothesis on $X_n$, hence one knows that one must use it somehow. Since the hypothesis is that $\mathrm E(X_n)=0$ and $\mathrm E(X_n^2)=1$ for every $n$, the problem is to bound $\mathrm P(X\geqslant n)$ for any random variable $X$ such that $\mathrm E(X)=0$ and $\mathrm E(X^2)=1$. Any idea?

One might begin with the obvious inclusion $[X\geqslant n]\subseteq[|X-\mathrm E(X)|\geqslant n]$ and try to use one of the not-so-many inequalities one knows which allow to bound $\mathrm P(|X-\mathrm E(X)|\geqslant n)$...

share|improve this answer
This gives $Var[X]=1$ but I'm not quite sure how to apply that. It's incredibly late here and I know there's a mistake in my logic, but this is what I'm thinking: $E[X]=P(X_n=1)=\lim_{n\to\infty} \frac{1}{n}=0$ $\Rightarrow P(X_n=k)=0 \;\forall\; k$. Then $P(X\geq n)=P(X_n=n)+P(X_n=n+1)+\cdots = 0$ so the sum is $<\infty$. –  bret Oct 12 '11 at 8:41
@bret, right, let us continue this once you will have had some sleep. –  Did Oct 12 '11 at 10:13
@Bret: Consider how $\int_{|x|>\alpha}\alpha^2\;\mathrm{d}\mu(x)$ compares to $\int_{|x|>\alpha}x^2\;\mathrm{d}\mu(x)$. –  robjohn Oct 12 '11 at 14:50
@Didier: So does my idea for showing the sum is $<\infty$ works? And I'm not entirely sure how to apply your suggestion rob; we haven't used any integrals yet so I'm not sure it's the right approach. –  bret Oct 12 '11 at 16:04
@bret, sorry but I see no upper bound of $\mathrm P(X\geqslant n)$ in what you wrote... Question: what probabilistic inequalities do you know? –  Did Oct 12 '11 at 20:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.