6
$\begingroup$

I develop open-source statistical software (http://openmx.psyc.virginia.edu/), but matrix calculus is not my strong point. I need the 1st and 2nd derivatives of the log multivariate normal density. I was happy to find the 1st derivatives here on CrossValidated,

How to take derivative of multivariate normal density?

However, the 2nd derivatives are left as an exercise to the reader. I am sure the 2nd derivatives have been independently derived many times. However, I cannot find them exhibited anywhere. Is there a calculus expert out there who can detail the 2nd derivatives?

Many thanks.

$\endgroup$
3
  • 1
    $\begingroup$ What parameters do you want to calculate the 2nd derivative with respect to? $\endgroup$ Commented Feb 4, 2014 at 21:58
  • $\begingroup$ The mean vector and covariance matrix. $\endgroup$ Commented Feb 5, 2014 at 0:19
  • 1
    $\begingroup$ A way to derive the formulas is by Matrix Differential Calculus. If you look at Abadir, Magnus, "Matrix Algebra", Cambridge, pag. 388 you will see the expression of the 2nd order differential for the multivariate normal. If you set ${\rm d}\,\mu=0$ and ${\rm d}\,\Omega=0$ and use the identification formula at the beginning of the chapter you should be able to find expressions for the Hessians $H\, l(\mu)$ and $H\, l(\Omega)$. $\endgroup$ Commented Feb 5, 2014 at 1:29

1 Answer 1

1
$\begingroup$

Alright, the negative information matrix for $L(\mu,K)$ is

$$\frac{\partial^2 L}{\partial\mu\,\partial\mu'} = -\left(\frac{1}{N}K\right)^{-1}$$

$$\frac{\partial^2 L}{\partial K\partial \mu} = 0$$

$\frac{\partial^2 L}{\partial K\, \partial K'}$, in a more general formulation, is given at http://en.wikipedia.org/wiki/Fisher_information#Multivariate_normal_distribution

Thanks to Mike Hunter for finding it.

$\endgroup$
1
  • $\begingroup$ Derivatives with respect to matrices are poorly defined mathematically. See the wiki entry on matrix calculus for cleaner definitions. $\endgroup$
    – StasK
    Commented Apr 22, 2015 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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