Limiting Distribution of the given function

Question

Can someone please help me in finding the limiting distribution of $$\frac{n(X_1X_2 + X_3X_4+\cdots+X_{2n-1}X_{2n})^2}{(X_1^2 + X_2^2+\cdots+X_{2n}^2)^2}$$ where $X_i$ are iid standard normal $\forall i\ge1$. I guess it has to be done using delta method, which I did considering $(X_1,X_3,\ldots,X_{2n-1})$ as $(Z_1,Z_2,\ldots,Z_n)$ and $(X_2,X_4,\ldots,X_{2n})$ as $(Y_1,Y_2,\ldots,Y_n)$ and tried applying delta method

$$\left(\frac{1}{n}\sum_{i=1}^nY_iZ_i,\frac{1}{n}\sum_{i=1}^nY_i^2,\frac{1}{n}\sum_{i=1}^nZ_i^2\right)$$

But in doing so I am missing out on that $n$ in the numerator, as the function $g4 should not depend on $n$. The calculations will be long, so can someone please help me just with what I should take the function as and on what should I apply it? I am sure, I will be able to do the calculations after that.

Answer 1

Forget the delta method, the ratio you consider is $U_n^2/V_n^2$ where $$ U_n=\frac1{\sqrt{n}}\sum\limits_{k=1}^nY_k,\qquad Y_k=X_{2k-1}X_{2k},\qquad V_n=\frac1n\sum\limits_{k=1}^nX_k^2. $$ Furthermore:

• The sequence $(Y_k)$ is i.i.d. and centered hence, by the CLT, $U_n$ converges in distribution to a centered normal distribution with variance $\sigma^2$ the variance of each $Y_k$, thus $\sigma^2=E[X_1^2]^2$.
• The sequence $(X_k^2)$ is i.i.d. hence by the LLN, $V_n\to E[X_1^2]$ almost surely.

Finally, $U_n^2/V_n^2$ converges in distribution to the square of a standard normal random variable.