# Convergence in mean squared?

I'm working the question below: $X_1, X_2, \dots$ are uncorrelated random variables with $E(X_i) = \mu_i$ and $\operatorname{var}(X_i)/i \rightarrow 0$ as $i \rightarrow \infty$. Now, let $S_n = X_1 + X_2 +\dots + X_n$ and $\nu_n = E(S_n)/n$.

The goal is to show that as $n \rightarrow \infty$ then $S_n/n - \nu_n\rightarrow 0$ in mean squared and in probability.

Could you please explain what do we mean by convergence in mean squared? Also can you guide me on solving this question?

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Convergence in mean squared of a sequence $\{V_n\}$ to $V$ means that $E[|V_n-V|^2]\to 0$.

• Show that mean squared convergence implies convergence in probability.
• Hence we just have to show mean squared convergence. To see that, show that $$E\left[\left(\frac{S_n}n-\frac{ES_n}n\right)^2\right]=\frac 1{n^2}\sum_{j=1}^n\operatorname{var}(X_j).$$
• Conclude by a "Cesaro-like" argument.
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I appreciate it @Davide Giraudo. –  Eli Nov 9 '12 at 16:54
Thanks. If you have some problems, don't hesitate. –  Davide Giraudo Nov 9 '12 at 16:55