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.
Best Answer
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 $X_k\sim N(0,1)$?