Limiting Distribution of $W_n=\frac{Z_n}{n^2}$ , $Z_n \sim \chi ^2 (n)$ 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.
 A: 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 $X_k\sim N(0,1)$?
A: The other answer here gives a useful hint as to what happens.  I'm going to show you another aspect of the problem.  Using the moment generating functions, it is simple to show that:
$$n W_n = \frac{Z_n}{n} \sim \text{Ga} \bigg( \text{Shape} = \frac{n}{2}, \ \text{Rate} = \frac{n}{2} \bigg).$$
This random variable has mean and variance:
$$\mathbb{E}(n W_n) = 1
\quad \quad \quad \mathbb{V}(n W_n) = \frac{2}{n},$$
and so asymptotically, we have $n W_n \rightarrow 1$ as $n \rightarrow \infty$.  Given that this is true, what do you think happens asymptotically to $W_n$?
A: The limit should be degenerate at 0.
pf: Zn/n2=(Zn/n)(1/n) ;
Zn/n → 1 in probability, 1/n → 0 in probability ;
Thus (Zn/n)(1/n) → 0 in probability and that equals (Zn/n)(1/n) → 0 in distribution such that Zn/n2 is degenerate at 0
