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Consider two random vectors $X\equiv(X_1, X_2),Y\equiv(Y_1, Y_2)$ distributed as below

1) $X\sim N(\overbrace{\begin{pmatrix} \mu_{x,1}\\ \mu_{x,2}\\ \end{pmatrix}}^{\mu_x}, \overbrace{\begin{pmatrix} v_{x,1} & 0\\ 0 & v_{x,2} \end{pmatrix}}^{\Sigma_x})$

2) $Y\sim N(\begin{pmatrix} \mu_{y,1}\\ \mu_{y,2}\\ \end{pmatrix}, \begin{pmatrix} v_{y,1} & 0\\ 0 & v_{y,2} \end{pmatrix})$

Consider now the random vector $W\equiv(W_1, W_2)$ whose probability distribution is obtained by mixing $X,Y$ with equal weights $1/2$, i.e. $$ f_W=\frac{1}{2}f_X+ \frac{1}{2}f_Y $$ where $f$ denotes the pdf.

(a) Could you help me to show that $W_1$ could be correlated with $W_2$ despite $X_1\perp X_2$, and $Y_1\perp Y_2$?

(b) Also, how does the correlation between $W_1, W_2$ depend on $\mu_x, \Sigma_x, \mu_y, \Sigma_y$? Is there any result saying that the higher $|\mu_{x,1}-\mu_{y,1}|$ or $|\mu_{x,2}-\mu_{y,2}|$, the higher the correlation between $W_1, W_2$?


My thoughts so far $$ cov(W_1,W_2)\equiv E(W_1 W_2)-E(W_1)E(W_2)=\frac{1}{4}E(X_1X_2)+\frac{1}{4}E(X_1Y_2)+\frac{1}{4}E(Y_1 X_2)+\frac{1}{4}E(Y_1 Y_2)- [\frac{1}{2}E(X_1)+\frac{1}{2}E(Y_1)][\frac{1}{2}E(X_2)+\frac{1}{2}E(Y_2)]=\frac{1}{4}cov(X_1, Y_2)+\frac{1}{4}cov(Y_1, X_2) $$

Is this correct? If yes, it does not seem to depend on $|\mu_{x,1}-\mu_{y,1}|$ or $|\mu_{x,2}-\mu_{y,2}|$.

Moreover, if $X\perp Y$ then $cov(W_1,W_2)=0$

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  • $\begingroup$ I have added some discussion of what I think may be an answer. Could you confirm whether it is correct? $\endgroup$ – STF Sep 13 '17 at 17:04
  • $\begingroup$ For clarity it might be great if you would spell out the relation between $W_1$, $W_2$ and $X$ and $Y$ in detail. $\endgroup$ – g g Sep 16 '17 at 17:31
  • $\begingroup$ I don't specify the relation between $X,Y$. I have explained better how $W$ comes out of $X,Y$. Thank you $\endgroup$ – STF Sep 16 '17 at 17:44
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Let $\mu_{W,i}=\frac12 (\mu_{X,i}+\mu_{Y,i})$. Notice that $$E[(X_1-\mu_{W,1})(X_2-\mu_{W,2})]=(\mu_{X,1}-\mu_{W,1})(mu_{X,2}-\mu_{W,2})=\frac14 (mu_{X,1}-\mu_{Y,1})(mu_{X,2}-\mu_{Y,2})$$

Then, letting $Z$ be the rv that indicates if $W$ comes from $X$ or $Y$: $$\begin{align} Cov(W_1,W_2) &=E[(W_1- \mu_{W,1})(W_2- \mu_{W,2})]\\ &=E[E[(W_1- \mu_{W,1})(W_2- \mu_{W,2})]\mid Z]\\ &=\frac12 E[(X_1- \mu_{W,1})(X_2- \mu_{W,2})]+\frac12 E[(Y_1- \mu_{W,1})(Y_2- \mu_{W,2})]\\ &=\frac14 (\mu_{X,1}-\mu_{Y,1})(\mu_{X,2}-\mu_{Y,2})=\frac14 \Delta_1 \Delta_2 \end{align}$$

where $\Delta_i = mu_{X,i}-\mu_{Y,i}$.

Also, $$ \begin{align} Var(W_1)&=\frac12 E[ (X_1 - \mu_{W,1})^2] + \frac12 E[ (Y_1 - \mu_{W,1})^2]\\ &=\frac12 \left( Var(X_1) + (\mu_{W,1}-\mu_{X_1})^2 + Var(Y_1) + (\mu_{W,1}-\mu_{Y_1})^2 \right)\\ &=\frac12 \left(v_{X_1} +v_{Y_1} + \frac12 (mu_{X,1}-\mu_{Y,1})^2 \right) \end{align} $$

$$ \rho=\frac{\Delta_1 \Delta_2}{2\sqrt{(v_{X_1} +v_{Y_1}+\Delta_1^2/2)(v_{X_2} +v_{Y_2}+\Delta_2^2/2) }}$$

This result should answer both points.

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