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My try ended in an awkward result. I thought it best to use the moment generating function (MGF) technique. We can derive the MGF of $W_n$ as follows:

$$ E \left[ e^{tZ /n^2} \right]= \left(1-\frac{2t}{n^2} \right)^{-n/2}$$ from the chi-squared MGF. But the problem is that the limit of that as $n \to \infty$ leaves $1$ and I am left puzzled whether I did everything right. Have I missed something? Thanks.

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Your calculation is correct. You simply need to interpret it. Which distribution has an MGF identically equal to 1?

Alternatively, your problem can be approached without using MGFs. Recall that $\chi^2(n)$ has the distribution of a sum of $n$ squares of $N(0,1)$ random variables. What can you say about the limiting distribution of $$\frac{1}{n}\sum_{k=1}^n X_k^2,$$ if $X_k\sim N(0,1)$?

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  • $\begingroup$ Why are you dividing by $n$ in the above? The sum of iid chi-squared RVs is chi-squared with degrees of freedom equal to the respective sum. It could be a degenerate distribution at 0. $\endgroup$ – JohnK Dec 18 '13 at 19:31
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    $\begingroup$ @JohnK You should think about the question phils is asking you to consider. There's a good reason to think about what a $\chi^2_n/n$ is in the limit; if you can see what it is, the answer to the original question becomes obvious, and the meaning of the MGF result you got (if it wasn't already clear) is readily seen. $\endgroup$ – Glen_b Dec 18 '13 at 19:46
  • $\begingroup$ @Glen_b It is $n^2$ .I think it's a degenerate distribution at 0, like I said above. I know that the mgf of a degenerate distribution at some $k$ is $e^{kt}$. Now put $k=0$ and it is equal to 1. $\endgroup$ – JohnK Dec 18 '13 at 19:49
  • $\begingroup$ Yes it's correct, but I can't tell whether you think it because you already have the MGF or whether you actually thought about what the limiting distribution of a chi-square on its degrees of freedom is (which as soon as I read your question was how I knew what it would be without doing any calculation). $\endgroup$ – Glen_b Dec 18 '13 at 19:54
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    $\begingroup$ @JohnK Your answer is correct. The idea behind my suggestion is that the law of large numbers tells us that if we divide by just $n$, we obtain in the limit the expectation of $X_1^2$, which is 1. Thus dividing by anything that grows faster, such as $n^2$, we obtain $0$ almost surely, hence the limit is degenerate at $0$. $\endgroup$ – phils Dec 18 '13 at 19:55

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