TITLE: Convergence in mean square from expected value/variance
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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.