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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?


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up vote 1 down vote accepted

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!)

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simpler proof... Use characteristic functions. – Did Nov 21 '12 at 6:22
- did is right ! – Learner Nov 21 '12 at 6:27
Thanks! How is it done by characteristic functions? – Tim Nov 21 '12 at 6:30
I think the steps are as follows: 1) Compute the characteristic function of $X_1,X_2$. 2) Apply Fubini by first integrating over $X_1|X_2$ and then $X_2$. This will result in exponential of a quadratic form plus a linear form multiplied by the imaginary number (both in the argument of the characteristic function). Look at for how to manipulate the characteristic function of a multivariate normal. – Learner Nov 21 '12 at 6:49
You still have to complete squares in the exponential function but no need to handle determinants. – Learner Nov 21 '12 at 6:56

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