Geometric-jump countdown Markov chain ($p_{i,i-1}=1,p_{0j}=p{q}^{j-1}$)

On the nonnegative integers, every positive state decreases by one, while zero jumps to $j\ge 1$ with probability $p{q}^{j-1}$, $p=1-q\in (0,1)$. Its return time from zero is one plus this geometric jump, so it has mean $(1+p)/p$. Return lengths two and three occur, establishing aperiodicity; deterministic downward motion and positive jumps establish irreducibility.

For the geometric-jump countdown Markov chain, regeneration at zero givesA 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 $1/(1+p)$.