0

For a multivariate normal vector $\mathbf{x} = (\mathbf{x}_1, \mathbf{x}_2, \mathbf{x}_3, \mathbf{x}_4)$ with mean 0 and variance $\mathbf{\Sigma} = \begin{bmatrix} \mathbf{\Sigma}_{11} & \mathbf{\Sigma}_{12} & \mathbf{\Sigma}_{13} & \mathbf{\Sigma}_{14} \\ \mathbf{\Sigma}_{21} & \mathbf{\Sigma}_{22} & \mathbf{\Sigma}_{23} & \mathbf{\Sigma}_{24} \\ \mathbf{\Sigma}_{31} & \mathbf{\Sigma}_{32} & \mathbf{\Sigma}_{33} & \mathbf{\Sigma}_{34} \\ \mathbf{\Sigma}_{41} & \mathbf{\Sigma}_{42} & \mathbf{\Sigma}_{43} & \mathbf{\Sigma}_{44} \end{bmatrix}$, is it true that

  1. $(\mathbf{x}_1, \mathbf{x}_3)$ is normally distributed with mean 0 and variance $\begin{bmatrix} \mathbf{\Sigma}_{11} & \mathbf{\Sigma}_{13} \\ \mathbf{\Sigma}_{31} & \mathbf{\Sigma}_{33} \end{bmatrix}$?
  2. $\mathbf{x}_3 | \mathbf{x}_1$ is normally distributed with mean 0 and variance $\mathbf{\Sigma}_{33} - \mathbf{\Sigma}_{31} \mathbf{\Sigma}_{11}^{-1} \mathbf{\Sigma}_{13}$?

I believe (1) is true because of the accepted answer at Deriving the conditional distributions of a multivariate normal distribution. I think (2) follows from (1), if (1) is true. However, I wonder why (1) is true - I guess (1) could be verified mathematically by integrating out $\mathbf{x}_2, \mathbf{x}_4$?

Yang
  • 113
  • 2
  • 1
    The computational part of $(1)$ is trivial: it is obtained by ignoring the other two components $x_2$ and $x_4.$ In short, you just write down what you know about the variable $(x_1,x_3).$ This would work regardless of the 4-variate distribution. The only content of $(1)$ is to assert that $(x_1,x_3)$ has a binormal distribution. One demonstration exhibits $(x_1,x_3)$ as a linear transformation of $(x_1,x_2,x_3,x_4),$ because linear transformations of Normal variables have Normal distributions. – whuber Dec 01 '21 at 17:48
  • @whuber Thank you so much! The idea of using linear transformation (projection) for (1) is very smart. – Yang Dec 01 '21 at 17:52

0 Answers0