0
$\begingroup$

I have a matrix $\Sigma$ with element $(i,j)$

$$\Sigma_{i,j}= \exp(-h_{i,j}\rho).$$

The matrix is positive definite and symmetric (it is a covariance matrix). Now I need to evaluate $$\frac{\partial \log(\det(\Sigma))}{\partial \rho} \text{ and } \frac{\partial \Sigma^{-1}}{\partial \rho}.$$

Someone can help me?

$\endgroup$
  • 1
    $\begingroup$ orion.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf $\endgroup$ – Piyush Grover Apr 10 '13 at 21:33
  • $\begingroup$ I have this pdf, but i don't understand how use it to solve my problem. $\endgroup$ – niandra Apr 10 '13 at 23:00
1
$\begingroup$

[modified according to the clarification given in comments].

In general, for an invertible square matrix $\Sigma=\Sigma(\rho)$, differentiably depending on the real variable $\rho$, we have: $(\Sigma^{-1})'=-\Sigma^{-1} \Sigma' \Sigma^{-1}$, and $\big(\det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')\det(\Sigma)$, so that $\big(\log \det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')$.

Here, $\Sigma'$ is the matrix with entries $-h_{ij}e^{-\rho h_{ij}}$, the Hadamard product of the matrices $-(h_{ij})_{ij}$ and $\Sigma$; and since both $\Sigma^{-1}$ and $\Sigma'$ are symmetric, $\operatorname{tr} (\Sigma^{-1} \Sigma')$ is just their Frobenius scalar product. I do not see other simplifications, unfortunately.

$\endgroup$
  • $\begingroup$ No! it'isn't.It is for work. I have some problem with your answer. Why $\Sigma^{-1}=exp(\rho H)$. For what i know, this is true for a diagonal matrix, but this is not the case. P.S. $H$ is positive definite, symmetric and its diagonal has all value equals to 1 $\endgroup$ – niandra Apr 10 '13 at 21:58
  • $\begingroup$ Think on the power series of exp(x) and exp(-x), and why exp(x)exp(-x)=1. $\endgroup$ – O.R. Apr 10 '13 at 22:21
  • $\begingroup$ Sorry, maybe i didn't understand. Let me use an example. If H is a 2×2 matrix with element (i don't know how to write a matrix so i write its elements) (0, 1 ,1, 0) (before i made a mistake, the diagonal elements are 0 and not 1). If $\rho=2$, $\Sigma$ is (1, 0.1353353, 0.1353353 ,1 ). Using the formula $\Sigma^{-1}=exp(\rho H)$ I obtain (1, 7.389056 ,7.389056 ,1), while using a software to obtain the inversion, i have (1.0186574, -0.1378603, -0.1378603 ,1.0186574). Where is the problem? $\endgroup$ – niandra Apr 10 '13 at 22:59
  • $\begingroup$ In the original question, $\Sigma\neq \exp(\rho H)$. Here, each entry is exponentiated, while the matrix exponential is defined by inserting a matrix into the power series expansion of the exponential. The problem as stated, Niandra, is a bit harder. I don't have an answer, and doubt there is a clean one. Is it possible you meant/need the matrix exponential instead? $\endgroup$ – D. Kelleher Apr 11 '13 at 4:11
  • $\begingroup$ No, what I wrote is what I have. I know i can use, for a generic matrix the following: $\frac{\partial log(det(\Sigma))}{\partial \rho}= TR(\Sigma^{-1}\frac{\partial \Sigma}{\partial \rho})$ and $\frac{\partial \Sigma^{-1}}{\partial \rho}=-\Sigma^{-1}\frac{\partial \Sigma}{\partial \rho}\Sigma^{-1}$. I' not sure if these can be applied to covariance matrix or if with symmetric matrix there is another formulation $\endgroup$ – niandra Apr 11 '13 at 10:45

Your Answer

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

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