# How to find Kolmogorov Forward Equations, given generator matrix Q?

I am having difficulty in forming Kolmogorov Forward Equations. I understand how the KFE is derived and that $$\frac {d}{ds} p_{ij} (s) = \sum_{k \neq j} p_{ik} (s) \lambda_{k} r_{kj} - p_{ij} (s)\lambda_j$$ is the formula to be used [i.e., $P'(s) = P(s)Q$] but I don't understand how to form the KFEs for particular scenarios in the first place using these formulae. For example, could you please take me through the process involved in forming the KFE for a continuous time Markov chain with generator matrix as follows (I believe having the generator matrix, Q is sufficient to finding the KFE): $$Q= \begin{pmatrix} -0.1 & 0.1 & 0 \\ 0 & -0.08 & 0.08 \\ 0 & 0.2 & -0.2 \\ \end{pmatrix}$$

and state space, $S = \{0,1,2\}$ Once I know how to form the KFE, I believe solving it wouldn't be especially difficult.

Thank you !

## 1 Answer

Firstly, just a notation issue. Say $q_{ij}$ is the general element in matrix $Q$, and that, $\lambda_j$ is the rate of transition out of state $j$, as you seem to have. Then the KFEs can be written as:

\begin{align} p_{ij}^{'}(s) &= \sum_{k\neq j} q_{kj}p_{ik}(s) - \lambda_j p_{ij}(s). \\ \end{align}

In the example, we have:

\begin{align} \lambda_0 &= 0.1 \\ \lambda_1 &= 0.08 \\ \lambda_2 &= 0.2. \\ \end{align}

The specific KFEs for the first row ($i=0$) are:

\begin{align} p_{00}^{'}(s) &= q_{10}p_{01}(s) + q_{20}p_{02}(s) - \lambda_0 p_{00}(s) \\ &= -0.1 p_{00}(s) \\ & \\ p_{01}^{'}(s) &= q_{01}p_{00}(s) + q_{21}p_{02}(s) - \lambda_1 p_{01}(s) \\ &= 0.1 p_{00}(s) + 0.2 p_{02}(s) - 0.08 p_{01}(s) \\ & \\ p_{02}^{'}(s) &= q_{02}p_{00}(s) + q_{12}p_{01}(s) - \lambda_2 p_{02}(s) \\ &= 0.08 p_{01}(s) - 0.2 p_{02}(s). \end{align}

The first equation is easily solved. The other two in turn can be solved using:

$$p_{00}(s) + p_{01}(s) + p_{02}(s) = 1\quad\forall\; s\geq 0.$$

The KFEs for remaining rows (for $i=1,2$) are done similarly.