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.



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 $A$.

  • $\begingroup$ 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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