Solving for a stationary point of a multivariate Gaussian parameter (CCA)

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I am reading a paper, probabilistic CCA, and am stuck on a particular derivation. Given the following multivariate Gaussian:

$$ x \sim N(u, W W^{\top} + P) $$

where $x \in \mathbb{R}^p$. Let $S = W W^{\top} + P$ where $S \in \mathbb{R}^{p \times p}$, $W \in \mathbb{R}^{p \times n}$ and $P \in \mathbb{R}^{p \times p}$.

Given the negative log-likelihood, $L$:

$$ L = \frac{1}{2} \sum_{i=1}^{n}(x_i - u)^{\top} S^{-1} (x_i - u) + \frac{n}{2} \ln |S| + \text{const} $$

Taking the derivative, I have a stationary point $\frac{dL}{dW} = 0$, and I would like to solve for $W$ (Equation 3 in the paper):

$$ (S^{-1} - S^{-1} Z Z^{\top} S^{-1}) W = 0 $$

where $Z = (x - u)$. How can I solve for $W$? I can reduce it to this form:

$$ \begin{align} (S^{-1} - S^{-1} Z Z^{\top} S^{-1}) W &= 0 \\ S^{-1} W - S^{-1} Z Z^{\top} S^{-1} W &= 0 \\ S^{-1} W &= S^{-1} Z Z^{\top} S^{-1} W \\ W &= Z Z^{\top} S^{-1} W \end{align} $$

This looks like an eigenvector problem, $A v = \lambda v$, but $S^{-1}$ contains $W$:

$$ W = Z Z^{\top} (WW^{\top} + P) W $$

I'm not sure how to simplify this further.