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Let $x$ be a multivariate normal distributed variable with mean $\mu$ and covariance $\Sigma$, and let $y = Ax+b$.

I have to show that $y$ is distributed as $\mathcal{N}(A\mu+b, A\Sigma A^T)$, therefore $p(y=z) = p(Ax+b=z) = p(x=A^{-1}z-A^{-1}b)$.

After some transormations i end up with $p(y) = \frac{1}{(2\pi)^{D/2}} \frac{1}{|\Sigma|^{1/2}} e^{-\frac{1}{2}(x-(A\mu+b))^T(A\Sigma A^T)^{-1}(x-(A\mu+b))}$.

The exponent seems to be correct, but in the second factor it should be $\frac{1}{|A\Sigma A^T|^{1/2}}$ or am i wrong?

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    you are right that the new covariance matrix should be $A \Sigma A^{T}$ (in fact it is straight forward to derive what the mean and covariance of the linear transformation would be, it is showing that the new transformation is still Gaussian that is more involved), it would be hard to know what went wrong with your transformations without seeing them. One possible route of solving this problem is to consider linear transformations of the characteristic function. – 2017-01-28

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