# stochastic control / geometric mean

Consider the following problem:

Given $\Omega$ and $U$ two symmetric definite positive matrices, choose a matrix $K$ to minimize the expectation $x' \Omega x + x'K'UKx$ when $x$ follows the invariant distribution of the Ornstein-Uhlenbeck process:

$dx_t = -K x_t dt + \Sigma dW_t$.

The intuition of the problem is straightforward: there is a penalty on $x$ for the metric associated to $\Omega$ and a penalty on $Kx$ for the metric associated to $U$.

The solution of this problem is $K^* = U^{-1}(U \sharp \Omega )$ with $U \sharp \Omega$ the geometric mean of $U$ and $\Omega$.

I can get this result by a convoluted way: expressing the variance of the invariant distribution of $x$ as a function of $K$ and $\Sigma$ (it gives the Algebraic Riccati equation as a constraint) and then solving the optimization problem under this constraint.

Given that the solution involves the mean of my two matrices $U$ and $\Omega$ for the Riemannian metric on positive definite matrices, I am looking for more elegant proofs of this result: something along the lines of "the total cost is the sum of two costs which correspond to some distances in a well defined space and by minimizing the sum of these distances, we end up with a geometric mean..."

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Perhaps you meant to write the SDE as $dx_{t} = -Kx_{t} dt + \Sigma dW_{t}$ – Abhishek Halder Jan 25 at 17:27
@AbhishekHalder yes thank you, edit made. – Bernard Jan 26 at 18:27
Did you figure out an answer to your question? I would also be interested to learn about the "convoluted" derivation that you mentioned, so if possible, please feel free to point me if there is a document I can take a look at. – Abhishek Halder Jan 26 at 19:40