How to find the rate of the Poisson process asociated with a continious Markov chain

Question

Suppose we are given the generator matrix Q of a continious, homogeneous Markov chain X how do we find the Poisson process $N_t$, such that $Y_{N_t}=X_t$, where $Y$ denotes the jump chain. I know how to find the jump chain asociated with $X$ however, how to find the Poisson process, I do not. If possible, could you explain how to?

Answer 1

The transition function P(t) can be computed from the generator Q by solving a system of Kolmogorov forward or backward differential equations. The Kolmogorov backward equation $P'(t) = QP(t)$ is a matrix equation, which bears a striking resemblance to the nonmatrix differential equation $p′(t) = qp(t)$, where $p$ is a differentiable function and $q$ is a constant. If $p(0) = 1$, the latter has the unique solution: $$p(t) =e^{tQ} \quad \text{for} \quad t\geq0$$ If you are not familiar with solutions to matrix differential equations it might be tempting to try to solve the backward equation by analogy, and write $$P(t) =e^{tQ} \quad (1)$$ since $P(0) = I.$

Given now that that you know matrix $Q$ and do the calculation in (1) you will obtain the transition matrix $P$.

\begin{align} P_{ij}(t) &= \sum_{n=j-i}^\infty \frac{t^n}{n!}Q_{ij}^{(n)}\\ &=\sum_{n=j-i}^\infty \frac{t^n}{n!}\binom n{j-i} \lambda^n (-1)^{n-(j-i)}\\ &=\sum_{n=j-i}^\infty \frac{t^{j-i}}{(j-i)!}\lambda^{j-i}\frac{t^{n-(j-i)}}{(n-(j-i))!}(-\lambda)^{n-(j-i)}\\ &= \frac{(\lambda t)^{j-i}}{(j-i)!}\sum_{n=j-i}^\infty \frac{(-\lambda t)^{n-(j-i)}}{(n-(j-i))!}\\ &=\frac{(\lambda t)^{j-i}}{(j-i)!} \sum_{k=0}^\infty \frac{(-\lambda t)^k}{k!}\\ &=\frac{(\lambda t)^{j-i}}{(j-i)!} e^{-\lambda t}. \end{align} As expected, the probability of transitioning from state $i$ to state $j$ (with $j\geqslant i$) in $(0,t]$ is the probability of there being $j-i$ arrivals in $(0,t]$.