Let $Y_1,\ldots,Y_n \overset{\text{iid}}{\sim} N(0,\sigma^2)$, where $\sigma > 0$, and let $Z_1,\ldots,Z_n \overset{\text{iid}}{\sim} N(0,1)$. Define
$$U_n = \frac{\sum_{i=1}^{n} (Y_i^2 – \sigma Z_i^2)}{\sum_{i=1}^{n} Z_i^2}. $$
What is the limiting distribution of $\sqrt{n} \,U_n$ as $n \rightarrow \infty$?
My attempt:
I started by writing $U_n$ as
\begin{align*}
\frac{\frac{1}{n} \sum_{i=1}^{n} (Y_i^2 – \sigma Z_i^2)}{ \frac{1}{n}\sum_{i=1}^{n} Z_i^2},
\end{align*}
and then attempted to treat the numerator and denominator individally. First computing some relevant information: From the fact that the mean and variance of a chi-squared random variable with $p$ degrees of freedom are $p$ and $2p$, respectively, we have $E(Z_i^2) = 1$ and $\text{Var}(Z_i^2) = 2$. We also have $E(Y_i^2) = V(Y_i) + (E(Y_i))^2 = \sigma^2 + 0^2 = \sigma^2$. Noting that $\frac{1}{\sigma}Y_i \sim N(0,1)$, we have $(\frac{1}{\sigma} Y_i)^2 = \frac{1}{\sigma^2} Y_i^2 \sim \chi^2_1$. Hence, $\text{Var}(\frac{1}{\sigma^2} Y_i^2) = \frac{1}{\sigma^4} \text{Var}(Y_i^2) = 2$, and so $\text{Var}(Y_i^2) = 2\sigma^4 < \infty$. Therefore, by the Weak Law of Large Numbers,
$$\frac{1}{n} \sum_{i=1}^{n} Y_i^2 \overset{p}{\longrightarrow} \sigma^2 \text{ as } n \to \infty $$
(where $\overset{p}{\longrightarrow}$ denotes convergence in probability). Then by Slutsky's Theorem, if $\sqrt{n}\left(\frac{1}{n} \sum_{i=1}^{n} (Y_i^2 – \sigma Z_i^2)\right)$ converges to $W$ in distribution, then $U_n$ converges to $\sigma^2 W$ in distribution. I'm currently stuck on how to find the limiting distribution of the numerator. My idea is to apply the Central Limit Theorem somehow, maybe to the random variables $X_i := Y_i^2 – \sigma Z_i^2$. I've computed that $E(X_i) = \sigma^2 – \sigma$ and $\text{Var}(X_i) = 2\sigma^4 + 2\sigma^2$. Then by the CLT,
$$ \frac{\frac{1}{n} \sum_{i=1}^{n} (Y_i^2 – \sigma Z_i^2) – (\sigma^2 – \sigma)}{\sqrt{\frac{2\sigma^4 + 2 \sigma^2}{n}} } \longrightarrow N(0,1) \text{ in distribution}.$$
But I'm not sure where to go from here…Any suggestions for how to progress?
Best Answer
I suspect the numerator should be $(\sigma^2Z_i^2-Y_i^2)$, otherwise the numerator will have non-zero mean and thus escapes to infinity.
By SLLN we have $$\frac{\sum Z_i^2}{n}\to 1,a.s.$$
and by CLT we have $$\sum(\sigma^2Z_i^2-Y_i^2)/\sqrt{n}\Rightarrow N(0,4\sigma^4)$$
since $(\sigma^2Z_i^2-Y_i^2)$ has mean zero and variance $4\sigma^4$.
Thus by Slutsky $$\sqrt{n}U_n=\frac{\sum(Z_i^2-\sigma Y_i^2)/\sqrt{n}}{\sum Z_i^2/n}\Rightarrow N(0,4\sigma^4).$$