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

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

$X_i$, and $Y_j$ are dependent for $i =j$.

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. –  Did Jan 27 '11 at 21:33
    
@Didier Piau: Isn't this just $\text{Cov}(X_i, Y_i) \neq 0$? –  PEV Jan 28 '11 at 6:26
    
@PEV As you see from mpiktas' answer, the limit in distribution depends on the covariance matrix mpiktas called $\Sigma$, as was to be expected (and on the unwritten hypothesis that everything is square integrable). –  Did Jan 31 '11 at 7:23

1 Answer 1

up vote 3 down vote accepted

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\sim N(0,4(EX_1,EY_1)\Sigma(EX_1,EY_1)') \end{align}

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 21:25

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