We have the pdf of matrix normal distribution for the random matrix $X$ (https://en.wikipedia.org/wiki/Matrix_normal_distribution): enter image description here

However here in my case, $X$ is of a parameter, say $\theta$. So my question is how to get the derivative of the log of this pdf with respect to $\theta$?. i.e.,

What's the derivative with respect to $\theta$ for this:




They are all in matrix multiplication form so I am confused. The result is used in an EM algorithm.

Thanks in advance!!

  • $\begingroup$ How can the result not contain $\partial_{\theta}X(\theta)$ ? $\endgroup$ – Thomas Mar 29 at 8:16
  • $\begingroup$ To me the result is the second, but with an overall derivative over $\theta$, which can be also brought inside the trace if desired. To simplify further I guess that one should specify the dependence on $\theta$ $\endgroup$ – Thomas Mar 29 at 8:22
  • $\begingroup$ I wrote an answer showing a type of calculations that may help for this last question. Have a look to see if it is what you need... $\endgroup$ – Thomas Mar 29 at 14:50

So using the convention that repeated indexes are summed we have:


and (see comments) the task is to find:

$\frac{\partial f(Z)}{\partial Z_{x,y}}$

Using linearity of the derivative:

$\frac{\partial f(Z)}{\partial Z_{x,y}}=V^{-1}_{a,b}\frac{\partial Z_{c,b}}{\partial Z_{x,y}}U^{-1}_{c,d}Z_{d,a}+V^{-1}_{a,b}Z^+_{b,c}U^{-1}_{c,d}\frac{\partial Z_{d,a}}{\partial Z_{x,y}}$

But now:

$\frac{\partial Z_{c,b}}{\partial Z_{x,y}}=\delta_{c,x}\delta_{b,y}$

and similar, where $\delta$ is the kroenecker symbol. Therefore:

$\frac{\partial f(Z)}{\partial Z_{x,y}}=V^{-1}_{a,y}U^{-1}_{x,d}Z_{d,a}+V^{-1}_{y,b}Z^+_{b,c}U^{-1}_{c,x}$

and in matrix form this can be rewritten:

$\frac{\partial f(Z)}{\partial Z}=U^{-1}ZV^{-1}+[V^{-1}Z^+U^{-1}]^+$


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