3
$\begingroup$

What is the derivative of this matrix expression with respect to $\theta_k$ \begin{equation} \begin{aligned} \mathcal{J}(X, \theta) &= {\bf trace}\left( XX^TP(\theta)^{-1} \right) +{\bf trace}\left( (Y-H(\theta)X)(Y-H(\theta)X)^T \Sigma^{-1} \right)\\ & = X^TP(\theta)^{-1}X + (Y-H(\theta)X)^T \Sigma^{-1} (Y-H(\theta)X)^T \end{aligned} \end{equation}

$X$ and $Y$ are vectors

$\theta$ is a vector with entries $\theta_k$

$P(\theta)$ and $H(\theta)$ are matrices constructed using some or all of the entries of $\theta$ and possibly other constants.

The matrix $\Sigma$ is an invertible known constant matrix

All vectors and matrices have compatible dimensions.

I tried to use The Matrix Cookbook to calculate this derivative. Here is my result:

\begin{equation} \begin{aligned} \frac{\partial \mathcal{J}(X,\theta)}{\partial \theta_k } =& - {\bf trace} \left( X X^T P(\theta)^{-1} \frac{\partial P(\theta)}{\partial \theta_k}P(\theta)^{-1} \right) \\ & - 2\; {\bf trace} \left(\frac{\partial H(\theta)}{\partial \theta_k} X Y^T \Sigma_e^{-1}\right)\\ &+ 2\; {\bf trace} \left(\frac{\partial H(\theta)}{\partial \theta_k} \Sigma_e^{-1} H(\theta) X X^T\right) \end{aligned} \end{equation}

Is this result correct? Can you explain if there is a mistake? Also I would like to know if there is a better way to write this derivative.

$\endgroup$
  • $\begingroup$ The last term is wrong. It should be $$2\,{\rm trace}\Bigg(\frac{\partial H}{\partial \theta_k} XX^TH^T\Sigma_e^{-1}\Bigg)$$ You did not define $\Sigma_e^{-1}$ in your question, but it appears to be the symmetric part of $\Sigma^{-1}$ that is $$\Sigma_e^{-1}={\rm sym}(\Sigma^{-1})$$ $\endgroup$ – lynn Nov 18 '15 at 4:06
1
$\begingroup$

Define the vector $z = (Hx-y)$. Then rewrite the function in terms of the Frobenius (:) product and find its differential $$\eqalign{ {\mathcal J} &= \Sigma^{-1}:zz^T + xx^T:P^{-1} \cr\cr d{\mathcal J} &= \Sigma^{-1}:d(zz^T) + xx^T:dP^{-1} \cr &= \Sigma^{-1}:(dz\,z^T+z\,dz^T) - xx^T:P^{-1}\,dP\,P^{-1} \cr &= (\Sigma^{-1}+\Sigma^{-T}):dz\,z^T - P^{-T}xx^TP^{-T}:dP \cr &= (\Sigma^{-1}+\Sigma^{-T})z:dH\,x - P^{-T}xx^TP^{-T}:dP \cr &= (\Sigma^{-1}+\Sigma^{-T})(Hx-y)x^T:dH - P^{-T}xx^TP^{-T}:dP \cr }$$ Now substitute $d_{\theta_k} \rightarrow d$ to obtain the desired derivative $$\eqalign{ \frac{\partial {\mathcal J}}{\partial \theta_k} &= (\Sigma^{-1}+\Sigma^{-T})(Hx-y)x^T:\Big(\frac{\partial H}{\partial \theta_k}\Big) - P^{-T}xx^TP^{-T}:\Big(\frac{\partial P}{\partial \theta_k}\Big) \cr }$$ If you are uncomfortable with the Frobenius product, you can replace it with the equivalent trace expression $$ A:B = {\rm tr}(A^TB)$$

$\endgroup$
0
$\begingroup$

I haven't checked carefully, but your final result "looks" right.

When taking derivatives of matrix, it's always a good idea to put in the indexes and use Einstein convention (i.e. $a_i b_i$ is understood as $\sum_i a_i b_i$).

$X^T P(\theta)^{-1} X$ is the same as $\sum_{ij} X_i P^{-1}_{ij}(\theta) X_j = X_i P^{-1}_{ij}(\theta) X_j$ (Einstein convention in the last step). Thus all we need to do is take the derivative of $P^{-1}(\theta)$. Since $P^{-1}(\theta) P(\theta) = I$ or $P^{-1}_{ij}(\theta) P_{jk}(\theta) = \delta_{ik}$, $\partial_{\theta_k} P^{-1}_{ij}(\theta) P_{jk}(\theta) + P^{-1}_{ij}(\theta) \partial_{\theta_k} P_{jk}(\theta) = 0$, which is equivalent to $\partial_{\theta_k} P^{-1}(\theta) P(\theta) + P^{-1}(\theta) \partial_{\theta_k} P(\theta)= 0$ and thus $\partial_{\theta_k} P^{-1}(\theta) = - P^{-1}(\theta) \partial_{\theta_k} P(\theta) P^{-1}(\theta)$.

The derivative of $H(\theta)$ can be taken straight-forwardly.

By the way, you do not need to "trace" so many times; the expression on the second line does not contain any trace, and it's the most natural form.

$\endgroup$

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.