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

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.

• Work out the covariance matrix of the triple $(X,Y,Z)$. They are jointly Gaussian, with mean 0, so you should be able to read the answer off. Note the distribution is supported on a 2 dimensional subspace of $\mathbb R^3$, so there will not be a density function. Sep 30, 2018 at 18:57

$$COV(x,z)=COV(\rho y+\sqrt{1-\rho^2}z,z)=\rho COV(y,z)+\sqrt{1-\rho^2}COV(z,z)=\sqrt{1-\rho^2}$$, 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 $$N(\boldsymbol{0},\Sigma)$$.