Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\{X_n\}$ is a sequence of non-negative random variables on $(\Omega,\mathcal{F},\mathbf{P})$, such that $\mathbf{E}X_n\to\infty$ as $n\to\infty$ and $\text{Var} X_n=c$ for all $n$. How can I use Chebyshev's inequality to prove that $\mathbf{P}(X_n>\alpha)\to 1$ as $n\to\infty$ for all $\alpha$?

share|cite|improve this question
Compare the events $[X_n\lt\alpha]$ and $[|X_n-E(X_n)|\gt\beta\sqrt{\text{Var}(X_n)}]$. – Did May 29 '12 at 7:32
Do I have to set $\beta=(E(X_n)-\alpha)/\sqrt{\text{Var}(X_n)}$? – bob May 29 '12 at 8:24
up vote 2 down vote accepted

Let $\alpha\in\Bbb R$ fixed. We can write $$\mathbf P(X_n\leq \alpha)=\mathbf P(X_n-EX_n\leq \alpha-EX_n).$$ Since $\lim_{n\to +\infty}-EX_n=-\infty$, for $n$ large enough we have that $\alpha-EX_n<0$ hence $$\mathbf P(X_n\leq \alpha)=\mathbf P((X_n-EX_n)^2\geq (\alpha-EX_n)^2)$$ and by Chebyshev's inequality $$\mathbf P(X_n\leq \alpha)\leq \frac c{(\alpha-EX_n)^2}$$ and we are done.

share|cite|improve this answer
To make it clear that in the step where Xn-EXn is squared the inequality is reversed because both sides of the inequality are negative. – Michael Chernick May 29 '12 at 12:09

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.