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I'm reading through some notes on Probability, and the statement is made that:

If random variables $X_1, \ldots, X_n$ converge to $X$ in mean square, then they also converge in probability.

Can someone please explain why this is the case? Regards.

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1 Answer 1

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Fix $\delta>0$. Then $$\delta^2 P(|X_n-X|\geq \delta)=\delta^2 P(|X_n-X|^2\geq \delta^2)\leq \int_{\Omega}|X_n-X|^2dP,$$ so $P(|X_n-X|\geq \delta)\leq \frac 1{\delta^2}\int_{\Omega}|X_n-X|^2dP$ and we can conclude since the las integral converges to $0$.

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Many thanks. What rule are you invoking to say that $\delta^2 P(|X_n-X|^2\geq \delta^2)\leq \int_{\Omega}|X_n-X|^2dP$? –  Mathmo Feb 9 '12 at 19:18
    
You integrate over the set $\{|X_n-X|\geq \delta\}$ the constant $\delta^2$. On this set it's smaller than $|X_n-X|^2$, and the integral over this set is small than the integral on $\Omega$. –  Davide Giraudo Feb 9 '12 at 19:21
    
Excellent, thanks for that. –  Mathmo Feb 9 '12 at 19:23

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