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For a discrete-time Markov chain and a set $A$ of absorbing states, its absorption time is the hitting time $T_A=\inf \{m\ge 0:X_m\in A\}$. It is zero when the initial state already lies in $A$, and infinite on a sample path that never enters $A$. Finiteness almost surely need not imply a finite expected value. For a finite-state model, a uniform positive probability of absorption within a fixed number of steps gives a geometric series bound on the survival probabilities and proves that the mean is finite. The mean absorption time of a lazy symmetric random walk is an explicit example.

At an absorbing state, the remaining absorption time is zero, so $e_0=e_n=0$. Multiplying the linear recurrence relation by three givesThe second finite difference of $k^2$ is two. A particular solution is therefore $-3k^2/2$, while every solution of the corresponding homogeneous linear recurrence relation has the form $A+Bk$. The boundary conditions give $A=0$ and $B=3n/2$, henceThis is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.

The absorbing states give $e_0=e_n=0$. Multiplying the difference equation in (i) by $3$ givesThe second finite difference of $k^2$ is $2$, so a particular solution is $-3k^2/2$; the homogeneous solutions are $A+Bk$. Imposing the two boundary conditions gives $A=0,B=3n/2$. ThereforeUniqueness follows because the difference of any two solutions has zero second finite difference, hence is affine function, and vanishes at both endpoints. This is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.