Limiting distribution of $W_n = Z_n/n^2$ where $Z_n\sim \chi^{2}(n).$

Question

The goal is to find the limiting distribution of $W_n = Z_n/n^2$ where $Z_n\sim \chi^{2}(n).$

For this, we consider the characteristic function of $W_n$ which is, $$C_{W_n}(t) = E[e^{itW_n}]=E[e^{i\cdot \frac{t}{n^2}\cdot Z_n}] = C_{Z_n}\left(\frac{t}{n^2}\right) = \left(1-2i\cdot \frac{t^2}{n}\right)^{-n/2}\to 1.$$

And so $W_n$ converges to a degenerate distribution whose characteristic function is $1.$ Is this reasoning correct?

Answer 1

Yes, that's one possibility. Alternatively you can use the strong law of large numbers to show that $n^{-2} Z_n \to 0$ in distribution. Let $X_j$, $j \geq 1$, be independent standard Gaussian random variables, then $Y_n := X_1^2+\ldots+X_n^2$ satisfies $Y_n \sim \chi^2(n)$. By the strong law of large numbers,

$$\frac{1}{n} Y_n = \frac{1}{n} \sum_{j=1}^n X_j^2 \xrightarrow[]{n \to \infty} \mathbb{E}(X_1^2)=1,$$

and so

$$\frac{1}{n^2} Y_n \xrightarrow[]{n \to \infty} 0 \quad \text{a.s.}$$

In particular, $n^{-2} Y_n \to 0$ in distribution. Now if $Z_n \sim \chi^2(n)$, then $Y_n =Z_n$ in distribution, and hence $n^{-2} Z_n \to 0$ in distribution.