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?

  • 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

[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.

  • $\begingroup$ 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

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