Let $X_n = \frac{1}{n} \sum_{i=1}^{n} Y_i^2 $ where $Y_i$ are standard normals. How do I show that $X_n$ converges to 1 in probability?
Does the CLT not imply that $X_n$ converges to a normal(0,2) in distribution? This has been a point of confusion.
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... $Y_i$ are independent standard normals (add independent). – Jimmy R. Mar 07 '14 at 19:51
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The central limit theorem is offtopic to describe the convergence in probability of $X_n$. Note instead that $X_n$ is the mean of the i.i.d. random variables $Y_k^2$ hence, by the law of large numbers, $X_n$ converges to $E(Y_1^2)=1$ (in probability, using the weak law of large numbers, and in fact almost surely, using the strong law of large numbers).
The CLT is a second order result since, refining the law of large numbers, it describes the asymptotics of $X_n$ minus its limit. Here, $Z_n=\sqrt{n}(X_n-1)$ converges in distribution to a centered normal random variable with variance $\mathrm{var}(Y_1^2)=2$.
Three modes of convergence were mentioned in this answer: in probability, almost surely, in distribution. It is highly recommended to master their respective definitions and the implications between them.
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