Calculate the joint distribution of $X, Y$ and $Z$, following a bivariate normal distribution

Question

Let $X$ and $Y$ follow a bivariate normal distribution with $\mu = (0,0)^T$, $\sigma_x=1$, $\sigma_y=1$ and correlation $\rho =0.5$. Also, suppose that a pair $Z$ and $Y$ follow a bivariate normal distribution with $\mu = (0,0)^T$, $\sigma_z=1$ and $Cov(Y,Z) =0$. Now I would like to calculate the joint distribution of $X, Y$ and $Z$ if $X = \rho Y + \sqrt{1-\rho^2}Z$

My thoughts

The joint probability for $X$ and $Y$ is $$f(x_1,x_2) = \frac{1}{2\pi \sqrt{\sigma_{11}\sigma_{22}(1-\rho_{12}^2})}\exp{\left(-\frac{1}{2(1-\rho_{12}^2)}\left[\left(\frac{x_1-\mu_1}{\sqrt{\sigma_{11}}}\right)^2+\left(\frac{x_2-\mu_2}{\sqrt{\sigma_{22}}}\right)^2-2\rho_{12}\left(\frac{x_1-\mu_1}{\sqrt{\sigma_{11}}}\right)\left(\frac{x_2-\mu_2}{\sqrt{\sigma_{22}}}\right)\right]\right)}$$ But how can I use this to calculate the joint distribution of $X, Y$ and $Z$? That's where I get stuck.

Answer 1

$COV(x,z)=COV(\rho y+\sqrt{1-\rho^2}z,z)=\rho COV(y,z)+\sqrt{1-\rho^2}COV(z,z)=\sqrt{1-\rho^2}$, so the covariance matrix is \begin{align}\Sigma= \begin{bmatrix} 1 & \rho & \sqrt{1-\rho^2}\\ \rho & 1 & 0\\ \sqrt{1-\rho^2} & 0 & 1 \end{bmatrix} \end{align} The joint distribution is $N(\boldsymbol{0},\Sigma)$.