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Say I have multivariate normal $N(\mu, \Sigma)$ density. I want to get the second (partial) derivative w.r.t. $\mu$. Not sure how to take derivative of a matrix.

Wiki says take the derivative element by element inside the matrix.

I am working with Laplace approximation $$\log{P}_{N}(\theta)=\log {P}_{N}-\frac{1}{2}{(\theta-\hat{\theta})}^{T}{\Sigma}^{-1}(\theta-\hat{\theta}) \>.$$
The mode is $\hat\theta=\mu$.

I was given $${\Sigma}^{-1}=-\frac{{{\partial }^{2}}}{\partial {{\theta }^{2}}}\log p(\hat{\theta }|y),$$ how did this come about?

What I have done:
$$\log P(\theta|y)=-\frac{k}{2}\log 2\pi-\frac{1}{2} \left| \Sigma \right|-\frac{1}{2}{(\theta-\hat \theta)}^{T}{\Sigma}^{-1}(\theta-\hat\theta)$$

So, I take derivative w.r.t to $\theta$, first off, there is a transpose, secondly, it is a matrix. So, I am stuck.

Note: If my professor comes across this, I am referring to the lecture.

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part of your problem may be that your expression for the log-likelihood has an error - you have $|\Sigma|$ where you should have $\log(|\Sigma|)$. Also, by any chance did you mean ${\Sigma}^{-1}=-\frac{{{\partial }^{2}}}{\partial {{\theta }^{2}}}\log p(\theta|y)$? – Macro May 1 '12 at 12:19
Yes, you are right, sorry. Why is there negative sign in front of the partial derivative? – user1061210 May 1 '12 at 14:38
I was just clarifying about the negative sign because, the negative second derivative is the observed fisher information, which is usually of interest. Also, by my own calculation, I'm finding that $\frac{{{\partial }^{2}}}{\partial {{\theta }^{2}}}\log p(\theta|y) = -\Sigma^{-1}$ – Macro May 1 '12 at 14:48
So, what is the general procedure for discrete/continuous function? Take log, write in Taylor expansion form, differentiate twice w.r.t. $\theta$. Fisher info is not generally true most other densities, right? – user1061210 May 1 '12 at 15:19
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@user As I pointed out, the second derivative of the logarithm must have non-positive eigenvalues. Yes, there are links between variances and negative second partial derivatives, as the theory of maximum likelihood estimation, Fisher information, etc., reveals--Macro has referred to that earlier in these comments. – whuber♦ May 1 '12 at 19:19
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1 Answer

up vote 7 down vote accepted

In chapter 2 of the Matrix Cookbook there is a nice review of matrix calculus stuff that gives a lot of useful identities that help with problems one would encounter doing probability and statistics, including rules to help differentiate the multivariate Gaussian likelihood.

If you have a random vector ${\boldsymbol y}$ that is multivariate normal with mean vector ${\boldsymbol \mu}$ and covariance matrix ${\boldsymbol \Sigma}$, then use equation (70) in the matrix cookbook to find that the gradient of the log likelihood ${\bf L}$ with respect to ${\boldsymbol \mu}$ is

$$\begin{align} \frac{ \partial {\bf L} }{ \partial {\boldsymbol \mu}} &= -\frac{1}{2} \left( \frac{\partial \left( {\boldsymbol y} - {\boldsymbol \mu} \right)' {\boldsymbol \Sigma}^{-1} \left( {\boldsymbol y} - {\boldsymbol \mu}\right) }{\partial {\boldsymbol \mu}} \right) \nonumber \\ &= -\frac{1}{2} \left( -2 {\boldsymbol \Sigma}^{-1} \left( {\boldsymbol y} - {\boldsymbol \mu}\right) \right) \nonumber \\ &= {\boldsymbol \Sigma}^{-1} \left( {\boldsymbol y} - {\boldsymbol \mu} \right) \end{align}$$

I'll leave it to you to differentiate this again and find the answer to be $-{\boldsymbol \Sigma}^{-1}$.

As "extra credit", use equations (51) and (55) to find that the gradient with respect to ${\boldsymbol \Sigma}$ is

$$ \begin{align} \frac{ \partial {\bf L} }{ \partial {\boldsymbol \Sigma}} &= -\frac{1}{2} \left( \frac{ \partial \log(|{\boldsymbol \Sigma}|)}{\partial{\boldsymbol \Sigma}} + \frac{\partial \left( {\boldsymbol y} - {\boldsymbol \mu}\right)' {\boldsymbol \Sigma}^{-1} \left( {\boldsymbol y}- {\boldsymbol \mu}\right) }{\partial {\boldsymbol \Sigma}} \right)\\ &= -\frac{1}{2} \left( {\boldsymbol \Sigma}^{-1} - {\boldsymbol \Sigma}^{-1} \left( {\boldsymbol y} - {\boldsymbol \mu} \right) \left( {\boldsymbol y} - {\boldsymbol \mu} \right)' {\boldsymbol \Sigma}^{-1} \right) \end{align} $$

I've left out a lot of the steps, but I made this derivation using only the identities found in the matrix cookbook, so I'll leave it to you to fill in the gaps.

I've used these score equations for maximum likelihood estimation, so I know they are correct :)

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Great reference - was going to recommend it myself. Not a good pedagogical reference for someone who doesn't know matrix algebra though. The real challenge comes from actually working out $\Sigma$. A real pain. – probabilityislogic May 1 '12 at 11:11
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Another good source on matrix calculus is Magnus & Neudecker, amazon.com/… – StasK May 1 '12 at 15:32
Thanks Marco, very helpful. I accepted it as the answer! – user1061210 May 2 '12 at 15:57

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