I have been studying problems from some of my lecture notes and was given this question
A continuous time Markov Chain $X_t$ with state space $\{1,2,3\}$ has infinitesimal generator $$A= \begin{bmatrix}-6 & 2 & 4\\2 &-5 &3 \\2 & 3& -5\end{bmatrix}$$ If the chain starts in state 2, what is expected total time spent in state 2 before the chain first enters state 3
In the solution the answer is given as $$\frac{1}{5}+\frac{1}{5}\frac{2}{15}+\frac{1}{5}(\frac{2}{15})^{2}+.......= \frac{\frac{1}{5}}{1-\frac{2}{15}}$$
I know from my lectures notes that The expected time at state 2 is: $(\frac{1}{5})$
The chain jumps to state 1 with probability $(\frac{2}{5})$ and state 3 with probability $(\frac{3}{5})$
and I know chain jumps from 1 to 2 with probability $(\frac{1}{3})$
so the probability that the chain return to state 2 in two steps, starting at 2 is thus $(\frac{2}{5})(\frac{1}{3})=(\frac{2}{15})$
So after finding these probabilities I can understand the right hand side of the equation given in the solution but im confused about how it becomes $$\frac{\frac{1}{5}}{1-\frac{2}{15}}$$
I have been looking through my notes and haven't been able to find anything. Any help would be great
It is a geometric series. In this particular case you could say:
$$S = \frac{1}{5}+\frac{1}{5}\frac{2}{15}+\frac{1}{5}\left(\frac{2}{15}\right)^{2}+\cdots$$
multiplying by $\frac2{15}$ gives $$\frac2{15}S = \frac{1}{5}\frac{2}{15}+\frac{1}{5}\left(\frac{2}{15}\right)^{2}+\frac{1}{5}\left(\frac{2}{15}\right)^{3}+\cdots$$
and subtracting gives $$\left(1-\frac2{15}\right)S = \frac{1}{5}$$
so $$S = \frac{\frac{1}{5}}{1-\frac2{15}}$$