# Limiting distribution of a squared sum of random variables

Let $S_n = \frac{1}{n}\sum_{i=1}^n X_i$, and $T_n = \frac{1}{n}\sum_{j=1}^nY_i$, where

The $X_i$ are iid, the $Y_i$ are iid (with a different law)

$X_i$, and $Y_i$ are dependent

For $i\neq j$, $X_i$ and $Y_j$ are independent.

Is there a central limit type result for $S_n^2 - T_n^2$?

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Whether $E(X_1)=\pm E(Y_1)$ or not would yield quite different results. And a hypothesis more precise than "$X_i$ and $Y_i$ are dependent" might be needed. (Duplicate from math.stackexchange... You might want to choose one of the two sites and stick to it.) –  Did Jan 27 '11 at 21:38
@Ian Langmore, I've answered the question in math.SE (math.stackexchange.com/questions/19274/…) –  mpiktas Jan 28 '11 at 12:53
hm, what is the usual practice for questions duplicated in two SE sites? Should I post the answer in both sites, and then gather the feedback from two different communities? –  mpiktas Jan 28 '11 at 13:01
@mpiktas Just flag it for mod attention (which is what I've done) –  chl Jan 28 '11 at 13:17
Sorry for the duplicate question...I'm claiming newbie over-enthusiasm...In any case, if people agree that the question is better on stats.SE, I happy with it being closed on math.SE. Not sure if I can do this though...I could vote to delete it. –  Ian Langmore Jan 29 '11 at 21:24

If $X_i$ and $Y_j$ are dependent for $i=j$, but independent for $i\neq j$ we have an iid sample from bivariate distribution $Z_i=(X_i,Y_i)$. Then central limit theorem gives us

\begin{align} \sqrt{n}\left(\frac{1}{n}\sum_{i=1}^nZ_i -EZ_1\right)\xrightarrow{D}N(0,\Sigma) \end{align} with $\Sigma=cov(Z_1)$ and $\xrightarrow{D}$ indicating convergence in distribution. Note that

$$(S_n,T_n)=\frac{1}{n}\sum_{i=1}^nZ_i.$$

Now we can use delta method, which states that if $r_n(U_n-\theta)\xrightarrow{D}U$ for numbers $r_n\to\infty$, then $r_n(\phi(U_n)-\phi(\theta))\xrightarrow{D}\phi'_\theta(U)$. This statement can be found here. Delta method is also described here.

In our case now we have $r_n=\sqrt{n}$, $\phi(x,y)=x^2-y^2$, $U_n=(S_n,T_n)$ and $\theta=(EX_1,EY_1)$. We have

$$\phi_{\theta}'=(2EX_1,-2EY_1)$$

and

$$U=(U_1,U_2)\sim N(0,\Sigma)$$

Finally we get

\begin{align} \sqrt{n}\left(S_n^2-T_n^2-(EX_1)^2+(EY_1)^2\right)\xrightarrow{D} 2U_1EX_1-2U_2EY_1:=V \end{align}

Since we know that $(U_1,U_2)\sim N(0,\Sigma)$ we get that

$$V\sim N(0,4(EX_1,EY_1)\Sigma(EX_1,EY_1)').$$

Note that I basically redid the example after the theorem in the link.

The final answer then depends on $\Sigma=cov((X_1,Y_1))$, but this should be known to original poster.

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Thanks for the response! –  Ian Langmore Jan 29 '11 at 21:16
@Ian Langmore, you are welcome, I had some fun in answering it. If this answer was useful, you might want to accept it, this is the way this site works. To cite the faq you can accept the answer by clicking on the check box outline to the left of the answer. –  mpiktas Jan 30 '11 at 7:53