Sum of random variables that are normally distributed and correlated

Question

Suppose that $X$, $Y$ and $W$ are jointly normal random variables, such that $X\sim N(\mu_x,\sigma_x^2)$, $Y\sim N(\mu_y,\sigma_y^2)$ and $W\sim N(\mu_w,\sigma_w^2)$ where $\sigma_{x,y}\neq 0$ and $\sigma_{x,w}= 0 = \sigma_{w,y}$. Furhtermore, suppose that $S=X+u$, where $X$, $u$ are independently disributed and $u\sim N(0,\sigma_u^2)$. If $(X,Y,W,S)$ are jontly normally distributed, then how can we show that the random variable $Z=X+WY$ conditional on $F=\{S,W\}$, namely $Z|F$ is also a Gaussian normal?

Answer 1

$W$ is independent of $X$ and $Y$ and $S$, so if you are conditioning on it, you can just treat it as a constant throughout the problem. For visual convenience I will denote it as a lower case $w$.

$(Z,S) = (X + wY, X+u)$ is jointly normal since any linear combination $aZ+bS$ is a linear combination of independent normals $X,Y,u$. (Definition.)

Then it is a well-known fact that the conditional distribution $Z \mid S$ is normal, which you can find on Wikipedia and elsewhere online. One proof approach involves explicitly writing out the density $f_{Z \mid S}(z) = \frac{f_{Z,S}(z,s)}{f_S(s)}$.

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Response to comments:

If $Z=X+wY$ then $Z \sim N(\mu_z, \sigma_z^2)$ where $\mu_z = \mu_x + w \mu_y$ and $\sigma_z^2 = \sigma_x^2 + w^2 \sigma_y^2 + 2 w \sigma_{x,y})$. Meanwhile $S = X+u \sim N(\mu_s, \sigma^2_s)$ where $\mu_s = \mu_x$ and $\sigma_s^2 = \sigma_x^2+ \sigma_u^2$. Finally, $\sigma_{z,s} := \text{Cov}(Z, S) = \text{Cov}(X+wY, X+u) = \sigma_x^2 + w \sigma_{x,y}$.

$(Z,S)$ is bivariate normal with the above parameters. At this point you can ignore $X,Y,u$.

Let $V$ be a standard normal random variable independent of all the other random variables. Then $Z$ has the same distribution as $$\mu_z + \rho \sigma_z \frac{S-\mu_s}{\sigma_s} + \sigma_z \sqrt{1-\rho^2} \; V$$ where $\rho = \frac{\sigma_{z,s}}{\sigma_z \sigma_s}$ is the correlation between $Z$ and $S$.

\begin{align} E[e^{tZ} \mid S=s] &= E\left[\exp\left(t \left(\mu_z + \frac{\rho \sigma_z}{\sigma_s} (S-\mu_s) + \sigma_z \sqrt{1-\rho^2} \; V\right)\right) \;\Big|\; S=s\right] \\ &= e^{t(\mu_z + \frac{\rho \sigma_z}{\sigma_s} (s-\mu_s))} E[e^{t\sigma_z\sqrt{1-\rho^2} V}] \\ &= e^{t(\mu_z + \frac{\rho \sigma_z}{\sigma_s} (s-\mu_s))} e^{t^2 \sigma_z^2 (1-\rho^2)/2} \end{align} which is the MGF of a normal distribution with mean $(\mu_z + \frac{\rho \sigma_z}{\sigma_s} (s-\mu_s))$ and variance $\sigma_z^2 (1-\rho^2)$.