For the random walk, \(\pi_{k, m}\) has been computed in (2.3). Find \(3_{R . m}\)
and subsequently the conditional mean first-passage time*)
$$
\tau_{R, m}=\frac{1}{6}(R-m)(R+m-2 L) \quad(L \leqslant m \leqslant R)
$$
The equations we have found for \(\pi_{R, m}, \tau_{R, m}, 3_{R, m}\) can be
solved explicitly. First take equation (2.2) for \(\pi_{R, m}\). To solve it set
\(\pi_{R_{t m+1}}-\pi_{R, m}=\Delta_{m}\) so that the equation reduces to a one-
step recursion:
$$
g_{m} \Delta_{m}=r_{m} \Delta_{m-1} \quad(L+1 \leqslant m \leqslant R-1)
$$
This can be solved to give
$$
\Delta_{m}=\frac{r_{m}}{g_{m}} \frac{r_{m-1}}{g_{m-1}} \frac{r_{m-2}}{g_{m-2}}
\cdots \frac{r_{L+1}}{g_{L+1}} \Delta_{L^{-}}
$$
Note that \(\Delta_{L}=\pi_{R, L+1}\), but it is still undetermined.
Subsequently
$$
\pi_{k, m}=\sum_{n=1}^{m-1} \Delta_{g}=\pi_{R, L+1}+\sum_{u=L+1}^{m-1}
\frac{r_{u} r_{u-1} r_{\mu-2} \cdots r_{L+1}}{g_{\mu} g_{\mu-1} g_{p-2} \cdots
g_{L+1}} \pi_{R, L+1}
$$
The condition that this must be compatible with \(\pi_{\mathbb{q},
\boldsymbol{R}}=1\) determines \(\pi_{R, L+1}\). Inserting the result one finally
gets
$$
\pi_{R, m}=\frac{1+\sum_{\mu=L+1}^{m-1} \frac{r_{p} r_{\mu-1} \cdots
r_{L+1}}{g_{\mu} g_{\mu-1} \cdots g_{L+1}}}{1+\sum_{\mu-L+1}^{n-1} \frac{r_{u}
r_{j-1} \cdots r_{L+1}}{g_{\mu} g_{\mu-1} \cdots g_{L+1}}}
$$