TITLE: In what sense is the Bayesian posterior mean a “convex combination”?
QUESTION [9 upvotes]: I asked this on math.stackexchange with no response, I'm hoping someone here might have something.
Suppose I want to estimate $x \in \mathbb{R}^n$ from two signals with zero mean, normally distributed noise: $y_1 = x + \epsilon_1$ and $y_2 = x + \epsilon_2$, where $\epsilon_1 \sim N(0, \Sigma_1)$, $\epsilon_2 \sim N(0, \Sigma_2)$. Using Bayesian estimation, the posterior mean is $\hat{x} =  (\Sigma_1^{-1} + \Sigma_2^{-1})^{-1} (\Sigma_1^{-1} y_1 + \Sigma_2^{-1} y_2) $, which is a "convex combination" of the data points $y_1$ and $y_2$. 
In the univariate case, the posterior mean is a convex combination of the data points in the usual sense. In the multivariate case, the "weights" are matrices that add up to $I$. What can we say about the posterior mean (e.g. is the set of possible $\hat{x}$ convex and in what sense, what are the extreme points, how does it vary with the parameters, etc)? A search for "matrix convex combination" gives this result: http://www.math.uni-sb.de/ag/wittstock/OperatorSpace/node73.html which seems to be talking about something different.

REPLY [3 votes]: Answering my own question: the posterior mean will lie on the locus of tangency between the ellipsoids centered at $y_1, y_2$ and with shape matrices $\Sigma_1^{-1}, \Sigma_2^{-1}$ respectively.  See Chamberlain & Leamer (1976), "Matrix Weighted Averages and Posterior
Bounds," Journal of the Royal Statistical Society. Series B (Methodological),
38(1), 73-84 for more.