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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.