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Is the average of i.i.d. random variables with zero mean and finite variance convergent to 0 in probability or $L^2$? How do I show this? I am just beginning to learn convergence of random variables, so an example would be helpful.

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Law of large numbers. –  Marek Nov 11 '10 at 10:52

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Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with ${\rm E}(X_i)=0$ and ${\rm Var }(X_i) = \sigma^2$. Then, by the strong law of large numbers, the average $\bar X_n = \frac{{\sum\nolimits_{i = 1}^n {X_i } }}{n}$ converges to $0$ almost surely, hence, in particular, in probability. However, this can also be shown directly as follows. By definition of converges in probability, we need to show that, for any $\varepsilon > 0$, ${\rm P}(|\bar X_n - 0| \geq \varepsilon) \to 0$ as $n \to \infty$. We have ${\rm E}(\bar X_n) = 0$ and ${\rm Var}(\bar X_n) = \sigma^2/n$. Hence, by Chebyshev's inequality, $$ {\rm P}(|\bar X_n - 0| \ge \varepsilon ) \le \frac{{{\rm Var}(\bar X_n )}}{{\varepsilon ^2 }} \to 0. $$ The convergence in probability is thus proved. As for the convergence to $0$ in $L^2$, this follows straight from the definition. Indeed, $$ {\rm E}(|\bar X_n - 0|^2 ) = {\rm Var}(\bar X_n ) \to 0. $$

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An assumption of finite variance ${\rm Var }(X_i) = \sigma^2 < \infty$ is not necessary for the convergence almost surely or in probability of $\bar X_n$ to $\mu$. –  Shai Covo Dec 8 '10 at 4:00

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