What is the joint density of 2 random variables that Linear combinations of the same random variables? Suppose we have random variable $W$ and $M$ that are independent standard normal random variables. If we were to define $X$ and $Y$ as:
$X=aW +bM$ and $Y=cW+bM$
How do we find the joint density of $X$ and $Y$? ie $f_{X,Y}$.
I found the pdf of $X$ and the pdf of $Y$ (linear combination of Normals are normal with new mean and variance) but im not sure where to go from there. I believe you can't just multiply $f_U$ and $f_V$ together because they share $W$ and $M$ which make them dependent. However, when you find $f_X$ and $f_Y$ those $M$ and $W$ terms just disappear leaving you with only $a$'s, $b$'s etc.
 A: Linear combinations of jointly Gaussian random variables are also jointly Gaussian.
Independent Gaussians are jointly Gaussian, so $(X,Y)$ follow a joint Gaussian distribution. This is specified by the $E[X],E[Y], \sigma_X^2, \sigma_Y^2, \sigma_{XY}$. 
$E[X] = E[aW+bM] = a E[W] + b E[M] =0$ and similarly $E[Y]=0$.
$\sigma_X^2 = var(aW+bM) = a^2 var(W) + b^2 var(M) = a^2+b^2$ and similarly $\sigma_Y^2 = c^2+b^2$.
$\sigma_{XY} = E[XY] - E[X]E[Y] = E[XY] - 0 = E[(aW+bM)(cW+bM)] = E[acW^2+b^2M + (ab+bc) MW]= ac+b^2 + (ab+bc)E[MW]=ac+b^2$ since $E[MW]=E[M]E[W]=0$.
Thus, $(X,Y)$ follows a normal distribution with mean zero (vector) and covariance matrix $\begin{bmatrix} \sigma_X^2 & \sigma_{XY} \\ \sigma_{XY} & \sigma_Y^2 \end{bmatrix}$. 
A: Summarizing @Batman succinctly, let $A=\begin{pmatrix}a&b\\c&b\end{pmatrix}$. From $\begin{pmatrix}W\\M \end{pmatrix} \sim N(0, I)$, we have 
\begin{align*}
\begin{pmatrix}X\\Y \end{pmatrix} &= A\begin{pmatrix}W\\M \end{pmatrix}\\
&\sim N(A0, AIA^T)\\
&= N\left(0, \begin{pmatrix}a^2+b^2 & ac+b^2 \\ ac+b^2 & b^2+c^2\end{pmatrix}\right)
\end{align*}
Note there is implicit assumption that $A$ is full rank, i.e. $a\neq c$. To prove that independent linear combination of normal is normal, you could use the moment generating function. Linear combination of normal distribution
