= Solution
At a state with positive departure rate $\lambda_i=R-r_i$, the next-jump probabilities are
$$
\boxed{\widehat p_{ij}=\frac{r_j}{R-r_i}\ (j\ne i),\qquad\widehat p_{ii}=0.}
$$
If at least two rates are positive, every state has positive departure rate and the normalizing constant $Z=\sum_i r_i(R-r_i)=R^2-\sum_i r_i^2$ is positive. The jump-chain equilibrium law is
$$
\boxed{\widehat\pi_i=\frac{r_i(R-r_i)}{Z}.}
$$
Indeed $\widehat\pi_i\widehat p_{ij}=r_ir_j/Z$ is symmetric, proving both stationarity and reversibility. Jump observations weight the continuous-time equilibrium by the departure rate, so $\widehat\pi$ is generally different from $\pi$.
If exactly one rate, $r_k$, is positive, state $k$ is absorbing and the denominator $R-r_k$ vanishes. There is no next jump from that state: the genuine embedded jump sequence terminates there. Under the usual extended-chain convention set $\widehat p_{kk}=1$; every other state jumps to $k$ with <probability> one. Its equilibrium law is $\delta_k$ and it is trivially reversible there. The normalized $r_i(R-r_i)$ formula cannot be used in that absorbing case.
Back to article page
