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If $X_i$, $i=1,\ldots,n$ are iid standard normal distributed, what is the limiting distribution of $S_n=\sum X^4 / (\sum X^2)^2$?

After finding the moments and since $Cov(X^4, X^2)=0$, I have the bivariate normal

$\sqrt{n} (\frac{1}{n}\sum X^4-\mu_1)\rightarrow N(0,\sigma_1^2)$ $\sqrt{n} (\frac{1}{n}\sum X^2 - \mu_2)\rightarrow N(0,\sigma_2^2)$ with $Cov(X^2,Cov^4)=0$ (not sure how to type latex matrices)

and using multivariate delta method $h(x,y)=\frac{x}{y^2}$,

I get

$\sqrt{n}(nS_n-\frac{\mu_1}{\mu_2^2})\rightarrow N(0,\nabla h^T\sigma \nabla h)$.

My question is - is this the right form for finding the limiting distribution of $S_n$? I thought the form should be $\sqrt{n}(S_n-\mu)\rightarrow N(0,\sigma^2)$, but here I have an 'extra' $n$ as the coefficient of $S_n$).

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    $\begingroup$ Of course one should consider $nS_n$ and not $S_n$ since $S_n\to0$. And you might want to explain more precisely the result you reach through the delta method, I believe one gets $\sqrt{n}(nS_n-3)\to N(0,24)$ (or something similar). $\endgroup$
    – Did
    Jun 21, 2014 at 21:17
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    $\begingroup$ I find that $Cov(X^4,X^2)=12.$ $\endgroup$
    – Mr.Spot
    Jun 21, 2014 at 22:35

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I agree with the result of @Did. The "extra n" appears in the right place by carefully following the delta method. From the given info, $\sqrt n[B_n-\beta ]\to N(0,\Sigma)$ where $B_n= \begin{bmatrix} (1/n)\sum_1^n{X_i^4}\\ (1/n)\sum_1^n{X_i^2} \\ \end{bmatrix}$, $\beta=(3,1)^T$ and $ \Sigma=\begin{bmatrix} 96 & 12 \\ 12 & 2 \\ \end{bmatrix} $.

Then with $h(x,y)=\frac x {y^2}$, we obtain $\sqrt n[h(B_n)-h(\beta) ]\to N(0,\nabla h(\beta)^T\Sigma\nabla h(\beta))$ $$\sqrt n \left [\frac{(1/n)\sum_1^nX_i^4} {[(1/n)\sum_1^n{X_i^2}]^2}-3\right]\to N(0,24)$$The extra $n$ comes from the square in the denominator.

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