Master Equation with time dependent transition rate For a certain physical process, we have a nonconstant transition rate which we will call $\omega(t)$ that is time-dependent. We will suppose that at $t=0$ there were $n$ events that occurred.
The master equation for this process would be:
$$ \dot{P_n} = \omega(t) \left[ P_{n-1}-P_{n} \right] $$
For $n=0$ one would have:
$$ \dot{P_0} = - \omega(t) P_0  $$
since $P_{-1}$ is null, because the probability of going from $0$ events to $-1$ events is zero.
Moreover, for $n=1$ one can write:
$$ \dot{P_1} = \omega(t) \left[ P_0 - P_1 \right]$$
$$ \dot{P_2} = \omega(t) \left[ P_1 - P_2 \right]$$
and so on...
Finally, how can we obtain an expression for $P_n$? And how can I find the transition probability $P_{1|1}(n_0,t_0|n,t)$ for $t>t_0$?
 A: Let us denote
$$D(t)=\omega(t)
\begin{pmatrix}
-1 & 0 & 0 & \cdots & 0 \\
1 & -1 & 0 & \cdots & 0 \\
0 & 1 & -1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots
\end{pmatrix}=\omega(t) W$$
and let $t_0=0$ for convenience of notation. It is not hard to show
\begin{align}
\begin{pmatrix}
P_0(t) \\ P_1(t) \\ P_2(t) \\ \vdots
\end{pmatrix}
& =\bigg(\lim_{h \rightarrow +\infty}{e^{{1 \over h}D(t)}e^{{1 \over h}D(t(1-{1 \over h}))}e^{{1 \over h}D(t(1-{2 \over h}))} \cdots e^{{1 \over h}D({t \over h})}}\bigg)\begin{pmatrix}
P_0(0) \\ P_1(0) \\ P_2(0) \\ \vdots
\end{pmatrix} \\
& = e^{\big(\int_{0}^{t}{\omega(t')dt'}\big)W}\begin{pmatrix}
P_0(0) \\ P_1(0) \\ P_2(0) \\ \vdots
\end{pmatrix}
\end{align}
satisfies the differential equation. The matrix $W$ is non-diagonalizable if you recognize that it is a Joradan matrix. However, it is easy to prove by induction that $(W^k)^{i}_{j}$, which is the $(i,j)$-element of $W^k$, is
\begin{align}
(W^k)^{i}_{j} & = (-1)^{k-i+j}C^{k}_{i-j}, \ & \text{if} \ k \geq i \geq j \\
& = 0, \ & \text{otherwise}
\end{align}
Therefore, letting $\Omega(t)=\int_{0}^{t}{\omega(t')dt'}$, we expand the exponential $e^{\Omega(t)W}$ and have
$$e^{\Omega(t)W}=\begin{pmatrix}
e^{-\Omega(t)} & 0 & 0 & \cdots & 0 \\
\Omega(t)e^{-\Omega(t)} & e^{-\Omega(t)} & 0 & \cdots & 0 \\
{1 \over 2!}\Omega(t)^2e^{-\Omega(t)} & \Omega(t)e^{-\Omega(t)} & e^{-\Omega(t)} & \cdots & 0 \\
{1 \over 3!}\Omega(t)^3e^{-\Omega(t)} & {1 \over 2!}\Omega(t)^2e^{-\Omega(t)} & \Omega(t)e^{-\Omega(t)} & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots
\end{pmatrix}$$
and the $(i,j)$-element of $e^{\Omega(t)W}$ is ${1 \over (i-j)!}\Omega(t)^{i-j}e^{-\Omega(t)}$ if $i \geq j$ and $0$ if $i<j$. This solves the equation. An example of how the distribution behaves can be seen by setting $P_0(0)=1$ and $P_n(0)=0$ for any $n \geq 1$, at time $t$, $P_j(t)$ is Poisson distribution over the variable $j \geq 0$.
