0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – kimchi lover Sep 30 '18 at 18:57
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.