Consider a two-state Markov process in continuous time, with states labelled $A$ and $B$. The transition rates for going from state $A$ to $B$, and state $B$ to $A$ are $\alpha$ and $\beta$ respectively. We are interested in the quantity $p_t(A,x,t_A|A)$, which denotes the probability that, at time $t$, the Markov process is in state $A$, has made a total of $x$ jumps, and has spent time equal to $t_A$ in state $A$, given that it started from state $A$.

I was thinking of two approaches (and both are likely to be equivalent). But I seem to be running into hurdles in both of them, and would be grateful for some help.

My first approach was to write a set of coupled recurrence relations in the following way: \begin{align} p_t(A,x,t_A|A) &= \int_0^t e^{-\alpha t'}\cdot \alpha \cdot p_{t-t'}(A,x-1,t_A-t'|B)\cdot dt'\\ p_t(A,x,t_A|B) &= \int_0^t e^{-\beta t'}\cdot \beta \cdot p_{t-t'}(A,x-1,t_A|A)\cdot dt' \end{align} The basic idea behind the above equations is that the first jump takes place at time $t'$, and thus, the rest of the $x-1$ jumps have to be made in the remaining $t-t'$ amount of time. Similarly, depending on what the initial state was, the time needed to be spent in state $A$ in the remaining $t-t'$ time is stated in the argument (e.g. if the first $t'$ amount of time was spent in state $A$, then in the remaining amount of time, only $t_A-t'$ amount of time needs to be spent there). There is a minor issue here that the above equation does not necessarily restrict the possibility of $t'>t_A$, which is unphysical. But that can be taken care of by changing the upper limit of the integral in the first equation to $t_A$.

Naively, one would expect that taking a Laplace transform of the two equations with respect to $t$ should simplify matters a lot. For the second equation, it indeed does, as the equation is a simple convolution in $t$. However, the first is not **(please correct me if I am wrong)** and thus, naive Laplace transform with respect to $t$ might not be helpful. Is there any other approach which makes the problem easier?

My second approach was to derive the differential equation satisfied by the quantity $p_t(A,x,t_A|A)$, by working in discrete time-steps of $\Delta t$, and finally taking the limit $\Delta t \to 0$:

\begin{align} p_{t+\Delta t}(A,x,t_A|A) = p_t(A,x,t_A-\Delta t|A)\cdot (1-\alpha \Delta t) + p_t(B,x-1,t_A-\Delta t|A)\cdot \beta \Delta t \end{align} which upon rearranging gives us \begin{align} \frac{p_{t+\Delta t}(A,x,t_A|A) - p_{t}(A,x,t_A- \Delta t|A)}{\Delta t} = -\alpha\cdot p_t(A,x,t_A-\Delta t|A) + \beta \cdot p_t(B,x-1,t_A - \Delta t|A). \end{align} In the limit of $\Delta t \to 0$, we get: $$\frac{\partial{p_{t}(A,x,t_A|A)}}{\partial{t}}+\frac{\partial{p_{t}(A,x,t_A|A)}}{\partial{t_A}} = -\alpha\cdot p_t(A,x,t_A|A) + \beta \cdot p_t(B,x-1,t_A|A). $$

Similarly, one could write equations for $p_t(B,x,t_A|A)$. Here (again naively), it seems like one could take a Laplace transform with respect to $t$ and $t_A$, and define a generating function for $x$ to simplify the problem. However, one thing that scares me is that $t_A$ cannot take any value. It must always take values less than $t$. Thus, independently taking Laplace transforms with respect to $t$ and $t_A$ seems shady.

Is there a way to overcome one (or both) of the hurdles?