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I'm looking for a proof of the following statement: Given a sequence of independent random variables $X_n$ satisfying $$ \lim_{n\to \infty} E[X_n] = T, $$ where T is a constant, then
$$ \lim_{n\to \infty} V[X_n] = 0 $$ implies convergence of $X_n$ to $T$ in the mean-square. This statement is supplied without proof or reference in Shreve's Stochastic Calculus book.

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By mean-square, do you mean $E[X_n^2]$? If so, then simply writing out $V[X_n] = E[X_n^2] - E[X_n]^2$ should do the trick, no? – TMM Aug 24 '11 at 14:47
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@t-laarhoven: "Mean-square" convergence means $L^2$ convergence, i.e. we want to show $E[(X_n - T)^2] \to 0$. – Nate Eldredge Aug 24 '11 at 14:59

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I assume $V[X_n]$ is the variance.

Let $\mu_n = E[X_n]$ for convenience, and write $$\begin{align*} E[(X_n - T)^2] &= E[(X_n - \mu_n + \mu_n - T)^2] \\ &= E[(X_n - \mu_n)^2] + (\mu_n - T)^2.\end{align*}$$

(The cross term vanished since $E[X_n - \mu_n]=0$.) Now both terms go to 0 by assumption.

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Hmm, this is very simple. I think my mind was rebelling at the implication that this implies almost-sure convergence for a subsequence of ${X_n}$. The latter fact seems too good to be true... – James Davidoff Aug 24 '11 at 15:16
By the way, this is why $E(X)$ is the value of $x$ which minimizes $E((X-x)^2)$ (and why, as a consequence, the minimal value is the variance of $X$). – Did Aug 28 '11 at 20:45

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