= 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 $\widehat P_{ij}=q_{ij}/q_i$ for $j\ne i$, with $\widehat P_{ii}=0$. The holding time in state $i$ is exponential with mean $1/q_i$.
If $\pi Q=0$, set $C=\sum_i\pi_iq_i$ and $\widehat\pi_i=\pi_iq_i/C$. Then
$$
(\widehat\pi\widehat P)_j=\frac1C\sum_{i\ne j}\pi_iq_{ij}=\frac{\pi_jq_j}{C}=\widehat\pi_j.
$$
Conversely, $\widehat\pi\widehat P=\widehat\pi$ implies that $\pi_i=c\widehat\pi_i/q_i$, normalized to sum one, satisfies $\pi Q=0$. Therefore
$$
\boxed{\widehat\pi_i=\frac{\pi_iq_i}{\sum_j\pi_jq_j},\qquad \pi_i=\frac{\widehat\pi_i/q_i}{\sum_j\widehat\pi_j/q_j}.}
$$
Jump 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 $q_i=0$.
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