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I am trying to convince myself that if $X_i$ are iid random variables with $\mathbb{E}(X_i)=\mu$ and $Var(X_i)=\sigma^2<\infty$ then $\frac{1}{n}\sum_{i=1}^n X_i^2$ converges in probability to $\mathbb{E}(X_i^2)$. I was wondering if we need an assumption on finiteness of fourth moments of the RVs for the proof, if there is a proof that doesnt use finite fourth moments or if finite fourth moments hold for the given problem without assuming them separately. Any help would be highly appreciated.

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This is just the law of large numbers applied on the iid sequence $(Y_n)_{n\geq 1}=(X_n^2)_{n\geq 1}$, so you do not need any additional assumptions. – Stefan Hansen Jan 7 '13 at 13:51
In response to the other question: no, finite fourth moments won't exist in general, see for example a t distribution with 3 degrees of freedom. – guy Jan 7 '13 at 14:24
up vote 1 down vote accepted

Following up on @StefanHansen comment, you need a law of large numbers for i.i.d. variables that only assumes the first moment exists (or in your case $E[X_i^2]<\infty$, because you are applying it to $X_i^2$ as @SetefanHansen noted).

Such a theorem with just the two assumptions of finite mean and i.i.d. structure could be found with proof in Billingsley's "Probability and Measure" (theorem 22.1 p.282 of the thrid edition), or Rosenthal's "A First look at rigorous probability theory" (theorem 5.4.4 p.62 of the second edition).

The two theorems prove almost sure convergence. You could deduce from them convergence in probability.

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Thanks for the references. In my notes I had a version of WLLN that shows convergence in means square (and hence convergence in P). However the proof used the existence of second moments and this led me to think that applying the same WLLN to a setting with $Y_i=X_i^2$ would create problems as we dont know the existence of $Var(X_i^2)$. – mathemagician Jan 7 '13 at 14:40
@mathemagician In most introductory courses, the assumptions of existence of higher moments when proving laws of large numbers are there just to make the proofs simpler (this is pedagogic at the beginning). These are not the weakest possible conditions under which LLNs hold as you could see from the two theorems in the references. – Learner Jan 7 '13 at 14:49

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