= Stationary distribution of a regenerative countdown chain
{title2=$\pi_j=pq^{j-1}/(1+p),\quad j\geq1$}
For the <geometric-jump countdown Markov chain>, regeneration at zero gives
$$
\pi_0=\frac p{1+p},\qquad \pi_j=\frac p{1+p}q^{j-1}\quad(j\geq1).
$$
A cycle visits positive state $j$ precisely when its jump reaches at least $j$, with probability $q^{j-1}$. Dividing this expected occupation by the mean cycle length proves the formula. Mean recurrence times are $1/\pi_j$, and the limiting probability away from zero is $1/(1+p)$.
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