Geometric return count before absorption ($K\sim \mathrm{Geom}({\mathbb{P}}_i(\text{absorption before return to }i))$)

Let $i$ be a state outside an absorbing set $A$ of a Markov chain. Write $T_A$ for its first hitting time of $A$ and $T_i^{+}=\inf \{n\ge 1:X_n=i\}$. Assume each excursion ends: ${\mathbb{P}}_i(T_A\wedge T_i^{+}<\infty )=1$, and put $p={\mathbb{P}}_i(T_A<T_i^{+})>0$. Then the number $K$ of visits to $i$ before absorption, including the initial visit, has the positive-support geometric distributionAfter a return, the Strong Markov property restarts the same experiment. Thus $\mathbb{P}(K\ge k)=(1-p{)}^{k-1}$ and $\mathbb{E}K=1/p$.

The excursion-ending assumption matters: on an infinite state space, an excursion may escape forever without either absorption or return. In that case the geometric parameter for the number of visits is ${\mathbb{P}}_i(T_i^{+}=\infty )$, which need not equal the absorption-before-return probability.

For a continuous-time Markov chain, use the embedded jump chain and assume independent holding times at $i$ with exponential distribution of rate ${\lambda }_i>0$, independent of the jump destinations. The total occupation time before absorption is a geometric sum of exponential variables, with exponential distribution of rate $p{\lambda }_i$ and expected value $1/(p{\lambda }_i)$.