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 $cov(W_1,W_2)=0$

  • $\begingroup$ I have added some discussion of what I think may be an answer. Could you confirm whether it is correct? $\endgroup$ – TEX 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 $\endgroup$ – TEX Sep 16 '17 at 17:44

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.