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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 $\alpha$?

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    Do I have to set $\$b$et$a$=(E(X_n)-\$a$lph$a$)/\sqrt{\text{V$a$r}(X_n)}$?2012-05-29

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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.

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    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.2012-05-29