I am preparing for an MLC certification, and have gone far afield into my own little world.
We are introduced to Kolmogorov's forward equation, which essentially has the form
$$P'=PM.$$
In my case, $P$ is a matrix of transition probabilities, and $M$ is a matrix of transition intensities. For one application, we have two lives and two states for each, and mnemonically, the states can be associated as binary, 0=00, both alive, 1=01, wife dead, 2=10, husband dead, and 3=11, both dead. In most of the relevant material, we assume the distributions are independent. At this point, I wondered if I could simplify the matrix somehow, and I thought it looked a lot like a tensor product, and it seems to work. So I write
$$P=P_x\otimes P_y.$$
In other words, it seems okay to write
$$ P_x\otimes P_y = \begin{pmatrix} p_x & q_x \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} p_y & q_y \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} p_xp_y & p_xq_y & q_xp_y & q_xq_y \\ 0 & p_x & 0 & q_x \\ 0&0&p_y&q_y\\ 0&0&0&1\\ \end{pmatrix} $$
But then I try to do the same for the mortality matrix, I write
$$ M_m\otimes M_f = \begin{pmatrix} -\mu_m & \mu_m \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} -\mu_f & \mu_f \\ 0 & 0 \end{pmatrix} \overset?\ne \begin{pmatrix} -\mu_f - \mu_m- \mu_{mf} & \mu_f & \mu_m & \mu_{mf}\\ 0 & -\mu_m & 0 & \mu_m\\ 0 & 0 & -\mu_f & \mu_f\\ 0 & 0 & 0 & 0 \end{pmatrix} $$
Is there a rule I don't know about? Something to do with derivatives?
EDIT: I did notice a trick that seems to work, If I replace the diagonal with 1s, then I get a better match, that is
$$ M_m\otimes M_f \overset?\ne\begin{pmatrix} 1 & \mu_m \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & \mu_f \\ 0 & 1 \end{pmatrix} \overset?= \begin{pmatrix} 1 & \mu_f & \mu_m & \mu_{mf}\\ 0 & 1 & 0 & \mu_m\\ 0 & 0 & 1 & \mu_f\\ 0 & 0 & 0 & 1 \end{pmatrix} $$
Why does this work out the way it does?
2nd EDIT:
If I write $M$ as a sum of a nilpotent matrix and a diagonal one, ie $M=N+D$, we note it is easy to recover $D$, so we discard it.
And above, we note we have the form
$$\big((I+N_x)\otimes(I+N_y)\big)-I=N.$$
So we can easily construct $M$ from $M_x$ and $M_y$. But I've no idea why this works.
Transition matrices for paired, independent continuous time Markov chains are, as you expect given by the Kronecker product $e^{tM_m}\otimes e^{t M_f}$. Differentiating and setting $t=0$ we see that the paired process has generator $M_m\oplus M_f:=(M_m\otimes I)+(I\otimes M_f)$, the Kronecker sum. That is, $$e^{t(M_m\oplus M_f)}=e^{tM_m}\otimes e^{t M_f}\neq e^{t(M_m\otimes M_f)}.$$
For the matrices you give, the Kronecker sum works out to $$\pmatrix{-\mu_m-\mu_f&\mu_f&\mu_m&0\cr0&-\mu_m&0&\mu_m\cr0&0&-\mu_f&\mu_f\cr0&0&0&0}.$$