Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@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
up vote 6 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.