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I'm trying to work out an expression for a correlation of the weighted sums of two r.v.'s with a third r.v. To be precise, I have a trivariate normal distribution:

$$\{X,Y,Z\}\approx \mathcal{N}(\vec{\mu},\vec{\Sigma}); \; \vec{\mu} = (\mu_x,\mu_y, \mu_z), \; \;\vec{\Sigma} = \left( \begin{array}{ccc} \sigma_x^2 & \rho_{x,y}\sigma_x\sigma_y & \rho_{x,z}\sigma_x\sigma_y \\ \rho_{x,y}\sigma_x\sigma_y & \sigma_y^2 & \rho_{y,z}\sigma_y\sigma_z \\ \rho_{x,z}\sigma_x\sigma_z & \rho_{y,z}\sigma_y\sigma_z & \sigma_y^2 \end{array} \right)$$

Now, I define the auxiliary variable $S = wX+(1-w)Y$. Clearly, $S$ is normal with $\mu_S=w\mu_x+(1-w)\mu_y$ and $\sigma_S^2=w^2\sigma_x^2+(1-w)^2\sigma_y^2+2w(1-w)\rho_{x,y}\sigma_x\sigma_y$. Now, it seems like $S$ and $Z$ should themselves have a bivariate normal distribution with: $$\vec{\mu}=(\mu_S,\mu_Z),\; \vec{\Sigma}=\left( \begin{array}{cc} \sigma_S^2 & \rho_{S,z}\sigma_S\sigma_z \\ \rho_{S,z}\sigma_S\sigma_z & \sigma_z^2 \\ \end{array} \right) $$

for some correlation $\rho_{S,z}$. However, I can't find out what on earth $\rho_{S,z}$ would be as a function of the other terms in the first correlation matrix, or if, in fact, they are even bivariate normal at all. Any help or comments are greatly appreciated.

Thanks!

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I'm going to follow standard practice and take $\mathcal N(\mu,\Sigma)$ to mean $\mu$ is a column vector and so is the random vector that has that distribution.

Now $\mu$ is a $3\times1$ vector and $\Sigma$ is $3\times3$.

Say we have a $2\times3$ matrix $A$, so that $A\mu$ is $2\times1$.

And $\begin{bmatrix} X \\ Y \\ Z \end{bmatrix}\sim\mathcal N(\mu,\Sigma)$.

Then $$ AX \sim \mathcal N\left(A\mu, A\Sigma A^T\right). $$ That's what you need here. You multiply $\Sigma$ on the left by $A$ and on the right by the transpose of $A$. That generalizes the rule that with scalar-valued random variables, you multiply the variance by the square of the scalar factor involved. If $A$ is $k\times n$ and $\Sigma$ is $n\times n$ then $A\Sigma A^T$ is $k\times k$.

You have \begin{align} \begin{bmatrix} S \\ Z \end{bmatrix} = \begin{bmatrix} w, & 1-w, & 0 \\ 0, & 0, & 1 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix}. \end{align} There's your $2\times3$ matrix $A$.

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  • $\begingroup$ Yes, I meant that $\vec{\mu}$ is a column vector. I tried out your solution, and it works perfectly. Thank you. $\endgroup$ – logosintegralis Aug 31 '13 at 21:07

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