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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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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
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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.

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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. –  Michael Chernick May 29 '12 at 12:09
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