I have a question about a continuous-time Markov process on the discrete space.
I am given the generator and asked for find the expected time the Markov process needs to get back to state 3, given that it started in state 3.
Is there a particular method to do this? I was thinking I would find the stationary distribution and then solve for the mean return time, but I am not sure if this is correct.
Thank you in advance
Yes, if you find a the strictly positive left eigenvector $\pi$ for $G$ (normalized so that it sums to $1$) corresponding to the eigenvalue $0$ then $\pi$ will be the unique stationary distribution for the chain, and it is well known that if $T_i$ is the return time to $i$ then $$ {\bf E}[T_i] = \frac{1}{\pi(i)}. $$ If you are curious about why this is true, it is a consequence of the fact (easily checked) that $$ x_i := {\bf E}_0\left[ \sum_{n \geq 1} I_{\{ X_n = i \}} I_{\{n \leq T_0 \}}\right] \in (0,\infty) $$ is an invariant measure for the kernel (this applies to the discrete setting of course, but you can show it holds in the continuous setting with some work). It gives the expected number of visits to $i$ before returning to $0$. Since $X_n=0$ if and only if $n=T_0$, $x_0=1$. Also $$ \sum_{i \in S} x_i = {\bf E}_0\left[\sum_{n \geq 1} \sum_{i \in S}I_{\{ X_n = i \}} I_{\{n \leq T_0 \}} \right] = {\bf E}_0\left[\sum_{n \geq 1} I_{\{n \leq T_0 \}} \right] = {\bf E}_0[T_0] $$ (here $S$ is the state space). Now, since both $\pi$ and $x$ are invariant measures, and there is a unique invariant probability measure, it must be that $$ \pi(i) = \frac{x_i}{\sum_{j \in S} x_j}; $$ in particular $$ \pi(0) = \frac{x_0}{{\bf E}_0[T_0]} = \frac{1}{{\bf E}_0[T_0]}. $$ Since $0$ was not special (we could have defined $x$ using any other point) it follows that $\pi(i) = 1/{\bf E}_i[T_i]$ for all $i \in S$. Therefore $$ {\bf E}_i[T_i] = \frac{1}{\pi(i)}. $$