 Solution

The jump chain records the successive states visited at actual jumps of the continuous-time Markov chain. Put $q_i=-q_{ii}$. For an irreducible chain with at least two states, all $q_i>0$, and its transition probabilities are ${\hat{P}}_{ij}=q_{ij}/q_i$ for $j\ne i$, with ${\hat{P}}_{ii}=0$. The holding time in state $i$ is exponential with mean $1/q_i$.

If $\pi Q=0$, set $C={\sum }_i{\pi }_i{q}_i$ and ${\hat{\pi }}_i={\pi }_i{q}_i/C$. ThenConversely, $\hat{\pi }\hat{P}=\hat{\pi }$ implies that ${\pi }_i=c{\hat{\pi }}_i/q_i$, normalized to sum one, satisfies $\pi Q=0$. ThereforeJump observations overrepresent states with short holding times. A one-state chain has no actual jumps; its trivial equilibrium can be handled separately, without using this division by $q_i=0$.