The multivariate Gaussian PDF of dimension $ N$ with mean $ \mu$ and covariance $ {\bf C}$ is:

$\displaystyle \log p({\bf x}) = -\frac{N}{2} \log (2\pi) - \frac{1}{2}\log \ver...
...$}\right)^\prime \; {\bf C}^{-1} \; \left({\bf x}- \mbox{\boldmath$\mu$}\right)$ (16.3)

For independent zero-mean samples of variance 1, this simplifies to

$\displaystyle \log p({\bf x}) = -\frac{N}{2} \log (2\pi) - \frac{1}{2} \sum_{i=0}^{N-1} \; x_i^2.

Note that if we know the eigendeomposition of $ {\bf C}$, we may write (16.3) in a simpler form. Since $ {\bf C}$ is a symmetric positive definite matrix, the eigenvectors form a complete orthogonal subspace on $ {\cal R}^N$. Let

$\displaystyle {\bf C} = \sum_{i=1}^N \; \lambda_i {\bf u}_i {\bf u}_i^\prime

be the eigendecomposition of $ {\bf C}$. Let the matrix $ {\bf U}$ be formed from the columns $ \{{\bf u}_1 \ldots {\bf u}_N\}$. Let

$\displaystyle \tilde{{\bf x}} = {\bf U}^\prime {\bf x}.$

The covariance matrix of $ \tilde{{\bf x}}$ is clearly the diagonal matrix formed from the eigenvalues $ \{\lambda_1 \ldots \lambda_N\}$. Therefore the above formula simplifies to

$\displaystyle \log p({\bf x}) = -\frac{N}{2} \log (2\pi) - \frac{1}{2} \sum_{i=1}^N
\log \lambda_i - \frac{1}{2} \sum_{i=1}^N \frac{\tilde{x}_i^2}{\lambda_i}.

Baggenstoss 2017-05-19