Markov Chains: Finding the Embedded DTMC (transition probability matrix) $P(t)$ from generator matrix $Q$ where the sample space $S=(0,1,2)$
$Q=\begin{pmatrix}
-4 & 2 & 2 \\
3 & -5 & 2 \\
0 & 3 & -3 \\
\end{pmatrix}$
I found the forward Kolmogorov equations to be:
$$p'_{00}(t)=-4p_{00}(t)+3p_{01}(t)$$
$$p'_{01}(t)=2p_{10}(t)-5p_{11}(t)+3p_{12}(t)$$
$$p'_{02}(t)=2p_{20}(t)+2p_{21}(t)-3p_{21}(t)$$
Also I solved for the stationary probability to get:
$$(\pi_0 , \pi_1 , \pi_2 )=(9/35, 12/35 , 14/35)$$
I think I'm supposed to find $e^{Qt}$ in which case I'll need to diagonalise a $3\times 3$ which I'm not sure I can do or somehow integrate the equations.
Best Answer
The transition probabilities $p_{i,j}$ are given by
\begin{equation} p_{i,j} = \frac{q_{i,j}}{-q_{i,i}}, \quad i \neq j \end{equation}
and $0$ otherwise, with $q_{i,j}$ the entry in the $i$-th row and $j$-th column of the transition rate matrix $Q$ (numbering of rows and columns starts at $0$ in this case). See the $s_{i,j}$ at Wikipedia. The interpretation of $p_{i,j}$ is the probability that the process jumps to state $j$ given that the process is currently in state $i$ and a jump occurs.
Entirely unrelated, but the Kolmogorov forward equations that you have derived are incorrect. Define $p_{i,j}(t) := \mathbb{P}(X(t) = j \mid X(0) = i)$ and let $S$ be the state space of the continuous time Markov chain. Then the Kolmogorov forward equations are
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}t} p_{i,j}(t) = \sum_{k \in S} p_{i,k}(t) q_{k,j}. \end{equation}
In your case the Kolmogorov forward equations for $i = 0$ are
\begin{align} \frac{\mathrm{d}}{\mathrm{d}t} p_{0,0}(t) &= -4 p_{0,0}(t) + 3 p_{0,1}(t), \\ \frac{\mathrm{d}}{\mathrm{d}t} p_{0,1}(t) &= 2 p_{0,0}(t) - 5 p_{0,1}(t) + 3 p_{0,2}(t), \\ \frac{\mathrm{d}}{\mathrm{d}t} p_{0,2}(t) &= 2 p_{0,0}(t) + 2 p_{0,1}(t) - 3 p_{0,2}(t). \end{align}