2
$\begingroup$

I was wondering if anyone would be able to help me determine whether the following quantity is log concave or not with respect to $\alpha$?

$$\left[\det(\textbf Y^\top \textbf P \textbf G \textbf P^\top \textbf Y + \textbf Q)\right]^{-\frac{q}{2}}$$

In the above, $\textbf Y \in \mathbb{R}^{n \times q}$, $\textbf P \in \mathbb{R}^{n \times n}$ is orthogonal, $\textbf G \in \mathbb{R}^{n \times n}$ is diagonal and $\textbf Q \in \mathbb{R}^{q \times q}$ is positive definite. Furthermore, the diagonal elements of $\textbf G$ are: $$(\textbf G)_{ii} = \frac{g}{g \lambda_{i}^{\alpha} + \lambda_{i}^{2}} \forall i \in [1,n]$$ Where $g > 0$, $\alpha \in [0,1]$ and the $\{\lambda_i\}_{i=1}^{n}$ are in the reals.

Thanks in advance!

$\endgroup$
  • $\begingroup$ Are you sure that $\lambda_i\in\mathbb{R}$? It seems to me that for fraction $\alpha$ you need $\lambda_i$ to be nonnegative; and because of the form of $G_{ii}$, $\lambda_i\neq 0$. $\endgroup$ – Michael Grant Jan 25 '15 at 14:12
  • $\begingroup$ They are strictly positive and real, yes $\endgroup$ – user202654 Jan 25 '15 at 15:15
  • $\begingroup$ To be perfectly honest I seriously doubt that it is either log-concave or log-convex. Is it even the case when $n=q=1$, $Y=P=Q=1$? $\endgroup$ – Michael Grant Jan 26 '15 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.