2
$\begingroup$

A random vector $X$ is defined to be multivariate normal, if any linear combination of its components is a 1-dimensional normal distributed.

I wonder when partitioning $X$ into $(X_1, X_2)$, if $X_1$ conditional on $X_2$ equal any value is multivariate normal and $X_2$ is multivariate normal, is $X$ multivariate normal?

Thanks!

1 Answers 1

1

No, not in general. Consider the example wehere $X_1$ and $X_2$ are scalars, $X_1|X_2$ is normal with zero means and variance $\exp(X_2)$.

However, if you add the additional condition that $X_1|X_2$ is multivariate normal with the mean a linear transformation of $X_2$ and a constant (i.e. non depending on $X_2$) covariance matrix, then the joint of $X_1$ and $X_2$ is multivariate normal. To prove that, you need to complete the squares of two quadratic forms in the exponential and use the properties of the determinant of block matrices. (The proof is straightforward but tedious. Maybe there is a simpler proof!)

  • 0
    Use $E[\exp(i u^T X)]=E[\exp(i u_1^T X_1+ i u_2^T X_2)]$. Under the integral sign, write the joint density $f(x_1,x_2)=f(x_1|x_2)f(x_2)$. Integrate first over $X_1|X_2$ by applying the formula for the characteristic function of that multivariate normal, etc...2012-11-21