# Joint trivariate normal density

I now introduce the problem:

Let assume $\mathbf{z} = (z_1, z_2,z_3)$ be a trivariate normal variable. I want to find the covariance matrix of $\mathbf{z}$. I now that the density of $(z_1, z_2)$ is a bivariate normal with mean vector $(\mu_1, \mu_2)$ and covariance matrix $$\left( \begin{array}{cc} \sigma_1^2 & \sigma_1\sigma_2 \rho \\ \sigma_1\sigma_2 \rho & \sigma_2^2 \\ \end{array} \right)$$ and i now that $z_3 | z_1, z_2 = \beta_0+\beta_1z_1+\beta_2 z_2+\epsilon$ where $\epsilon \sim N(0, \sigma_3^2)$ and then the distribution of $z_3 | z_1, z_2$ is normal with mean $\beta_0+\beta_1z_1+\beta_2 z_2$ and variance $\sigma_3^2$.

I want to find the distribution of $z_1, z_2, z_3$. I think that is a trivariate normal with mean $\mu_1, \mu_2, \beta_0$ and covariance matrix $$\left( \begin{array}{ccc} \sigma_1^2 & \sigma_1\sigma_2 \rho& \beta_1\sigma_1^2+\beta_2 \sigma_1\sigma_2 \rho \\ \sigma_1\sigma_2 \rho & \sigma_2^2 & \beta_1\sigma_1\sigma_2 \rho +\beta_2\sigma_2^2 \\ \beta_1\sigma_1^2+\beta_2 \sigma_1\sigma_2 \rho & \beta_1\sigma_1\sigma_2 \rho +\beta_2\sigma_2^2 & \beta_1^2\sigma_1^2+\beta_2^2\sigma_2^2+\sigma_3^2 \end{array} \right)$$

I think that this is right but with $\sigma_1^2=0.4$, $\sigma_2^2=1$, $\sigma_3^2=0.4$, $\rho=-0.1$, $\beta_1=-2$ and $\beta_2 = 3$ the matrix is not positive definite definite, i.e if i try to simulate a trivariate normal variable in R i get the following message:

Error in chol.default(V):
the leading minor of order 3 is not positive definite

Where is the error?

The Law of Total Covariance applied to $z_3$ asserts

$$\text{Var}(z_3) = \mathbb{E}(\text{Var}(z_3\ |\ (z_1,z_2)) + \text{Var}\left(\mathbb{E}(z_3\ |\ (z_1, z_2))\right)$$

whence, because $\mathbb{E}(\varepsilon)=0$ and $\text{Var}(\varepsilon)=\sigma_3^2$,

\eqalign{ \text{Var}(z_3) &= \mathbb{E}(\sigma_3^2) + \text{Var}(\mathbb{E}(\beta_0+\beta_1z_1+\beta_2z_2+\varepsilon\ |\ (z_1,z_2))) \\ &= \sigma_3^2 + \text{Var}(\beta_0 + \beta_1z_1 + \beta_2z_2) \\ &= \sigma_3^2 + \beta_1^2\sigma_1^2 + \beta_2^2\sigma_2^2 + 2\beta_1\beta_2\sigma_1\sigma_2\rho. }

That is what belongs in the lower right entry of the covariance matrix.

One can also use symmetry and linearity of the covariance operator, i.e. \eqalign{\text{cov}(x+y,z)&=\text{cov}(x,z)+\text{cov}(y,z), \hspace{0.5em}\text{cov}(\alpha x,y) &= \alpha \,\text{cov}(x,y), \hspace{0.5em} \text{cov}(x,y) = \text{cov}(y,x)} The missing element $\text{var}(z_3)$ can be found by \eqalign{\text{var}(z_3)=\text{cov}(z_3,z_3)&=\text{cov}(\beta_0+\beta_1 z_1 + \beta_2 \,z_2 + \varepsilon,\beta_0+\beta_1 z_1 + \beta_2 \,z_2 + \varepsilon)\\ &=\beta_1^2 \text{cov}(z_1,z_1) + \beta_2^2 \text{cov}(z_2,z_2) + 2\beta_1 \beta_2 \text{cov}(z_1,z_2) + \text{cov}(\varepsilon,\varepsilon)\\ &= \beta_1^2 \sigma_1^2 + \beta_2^2\sigma_2^2 + 2\beta_1\beta_2\sigma_1\sigma_2\rho + \sigma_3^2} where the covariances with $\beta_0$ and $\beta_1$ as arguments were dropped, because the covariance where one argument is constant yield zero. Furthermore $\text{cov}(\varepsilon,z_1)=0$ and $\text{cov}(\varepsilon,z_2)=0$

• +1 I like this solution because it is straightforward and elementary. The somewhat cumbersome "Cov" notation (instead of "Var", which you introduce earlier) actually makes it look more complicated than it really is.
– whuber
Mar 4, 2014 at 17:04