Show that $Y\sim N(a+A\mu,AVA^T)$

Question

Consider $Z=(Z_1,\ldots,Z_n)^T\sim N(\mu,V)$. Show:

If $a\in\mathbb{R}^m$ and $A$ is a $(m\times n)$-matrix with $\text{rang}(A)=m$ then $$ Y=a+AZ\sim N(a+A\mu,AVA^T). $$

My questions is, what do I have to show for this?

I think I have to show that $Y$ has the density function $$ f(Y)=(2\pi)^{-m/2}\text{det}(AVA^T)^{-1/2}\exp\left\{-\frac{1}{2}(Y-(a+A\mu))^T(AVA^T)^{-1}(Y-(a+A\mu)\right\}. $$

But how can I show that? Don't know how to start.

Answer 1

I wonder whether by "rang" you mean "rank"? If the rank of $A$ is the number of its columns and $m\le n$, then $AVA^T$ will be nonsingular if $V$ is nonsigular, but if $m>n$ that won't work, and if $V$ is singular then so is $AVA^T$. So either $V$ or $AVA^T$ may be singular, and then there is no density function. Your statement about the density is correct when the matrices involved are nonsingular. But the result you're trying to prove is true even when the variances are singular.

$$ \operatorname{var}(a+AZ) = \mathbb E((a+AZ-(a+A\mu))(a+AZ-(a+a\mu))^T) $$ $$ = \mathbb E(A(Z-\mu)(A(Z-\mu))^T) = A\mathbb E((Z-\mu)(Z-\mu)^T) A^T = A(\operatorname{var}(Z))A^T $$ $$ = AVA^T. $$ And you can do a similar thing with the mean.

That $Z$ is multivariate normal can be defined by saying that for every column vector $c\in \mathbb R^n$ we have $c^T Z$ is univariate normal. So is it true that for every column vector $d\in\mathbb R^m$, the vector $d^T(a+AZ)$ is univariate normal? $$ d^T(a+AZ) = d^T a + (d^T A)Z. $$ Letting $b=d^T A\in\mathbb R^n$, we see that $b^T Z$ is univariate normal since $Z$ is multivariate normal. And $d^T a$ is a constant, i.e. non-random, scalar. So constant plus univariate normal r.v. is a univariate normal r.v.

Answer 2

For any invertible function $g$ and continuous random variable $Z$ with density $f$, $Y=g(Z)$ has density function

$$h(y) = \frac{ f(g^{-1}(y)) }{ \bigl|\det\bigl(g'(g^{-1}(y))\bigr)\bigr| }.$$

This is a general property. Then just use $g(z)=a+Az$. If you want $g$ to be a function of lower dimension then just integrate out the unwanted dimensions.

Answer 3

There are two results here:

a) linear combinations of multivariate normals are themselves multivariate normal (can be shown using moment generating functions).

b) For the mean 0 random variables, $Cov(Y) = E(Y Y^\top)$, therefore you can show that $Cov(AZ) = A Cov(Z) A^\top$

The mean $\mu$ can be subtracted beforehand, so mean 0 requirement is not that restrictive.

I'm not sure that it's easy to produce the density formula for $f(y)$ directly by integration.