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For a nearest-neighbour random walk on $\{0,\dots ,N\}$ with right-step Probability $p$, left-step Probability $q=1-p$, and Absorption time $\tau$, first-step conditioning gives $u_k=s(p{u}_{k+1}+q{u}_{k-1})$, $u_0=1$, $u_N=0$. For $0<s<1$, the two positive distinct characteristic roots are $r_{\pm }=(1\pm \sqrt{1-4pq{s}^2})/(2ps)$. Solving the boundary equations gives $u_k=(r_{+}^N{r}_{-}^k-r_{-}^N{r}_{+}^k)/(r_{+}^N-r_{-}^N)$. The maximum principle $\Vert u-v{\Vert }_{\infty }\le s\Vert u-v{\Vert }_{\infty }$ proves uniqueness for equal boundary data. This is the joint generating function of the Absorption time and the lower-end absorption event.

For the chain with $0$ absorbing, equal-probability nearest-neighbour moves between $1,2,3$, and a probability-$1/2$ self-loop at $3$, starting from $1$ the expected Absorption time is $6$, the Probability of visiting $3$ before $0$ is $1/3$, and the expected number of visits to $3$ before absorption is $2$.

Consider a random walk on $\{0,1,\dots ,n\}$ with absorbing states $0,n$. At each interior state it stays in place with probability $q$, and moves one position in either direction with probability $(1-q)/2$, where $0\le q<1$. Its mean absorption time from $k$ isProof. A sequence of at most $n$ consecutive leftward moves ensures absorption, with probability at least $((1-q)/2{)}^n>0$. Dividing time into blocks of length $n$ gives a geometric series bound on the survival probabilities, so the expected value is finite. Let $e_k={\mathbb{E}}_k[T]$. The law of total expectation and first-step analysis giveThus $e_{k+1}-2e_k+e_{k-1}=-2/(1-q)$. A particular solution of this linear recurrence relation is $-k^2/(1-q)$; its homogeneous solutions are $A+Bk$. The two boundary conditions give $A=0$, $B=n/(1-q)$, proving both the formula and its uniqueness. Holding slows the usual symmetric random walk by the factor $1/(1-q)$.

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 $1/3$.

First the expected value of the absorption time is finite. From any interior state, consecutive leftward moves lead to absorption within at most $n$ steps, with probability at least $p=3^{-n}$. By independence of the subsequent steps, the probability of surviving $jn$ steps is at most $(1-p{)}^j$. Summing this geometric series bounds the expected value of the absorption time by $n/p$.

Now apply first-step analysis. Starting from $k$ in the interior, one step is always used. Conditional on that step, the remaining expected values are $e_{k+1}$, $e_k$, and $e_{k-1}$, each with probability $1/3$. The law of total expectation therefore givesThe term $1$ remains even when the particle stays at $k$, because a step has elapsed. Rearranging gives

Let $\tau$ be the absorption time. Its expectation is finite: from every interior state, a string of at most $n$ successive left moves reaches $0$, and the probability of such a string is at least $p=3^{-n}$. Hence the Markov property gives ${\mathbb{P}}_k(\tau >jn)\le (1-p{)}^j$, a geometric tail bound, and ${\mathbb{E}}_k[\tau ]<\infty$.

For $1\le k\le n-1$, one step is always spent before absorption or the remaining walk. Apply first-step analysis and the law of total expectation to the three possible first moves. The conditional expectations of the remaining absorption time are $e_{k+1},e_k,e_{k-1}$, respectively; independence of future moves gives the same law as a fresh random walk from the new position. Thuswhich rearranges toThe term $1$ also appears when the walk stays at $k$, because staying still consumes a step.