On Sinai's paper "The Limiting Behavior of a One-Dimensional Random Walk in a Random Medium" (1982) one considers a random walk on Z1 that moves from x to x+1 with probability p(x) and moves from x to x−1 with probability q(x)=1−p(x). In the pg 21 one reads (I couldn't upload image, maybe a bug, I will check that)
I did the computations and found a different result.
To compute $h_{[a,b]}(x) = g(x)$ we first relate to the one step equation: $$g(x) = p(x) g(x+1) + q(x)g(x-1) + 1 $$ which then, writting $g(x) - g(x-1)= \Delta(x)$ becomes $$\Delta(x+1) = \frac{q(x)}{p(x)}\Delta(x) -\frac{1}{p(x)} $$ This relation can be iterated and yields $$\Delta(x) = \lambda(x) \Delta(1) -\mu(x) $$ where $$\lambda(x) = \frac{q(x-1)}{p(x-1)}\ldots \frac{q(a+1)}{p(a+1)}$$ and $$\mu(x) = \frac{1}{p(x-1)} + \frac{1}{p(x-2)}\frac{q(x-1)}{p(x-1)} +\ldots + \frac{1}{p(a+1)}\frac{q(a+2)}{p(a+2)}\ldots \frac{q(x-1)}{p(x-1)}\\ =\sum_{a+1\leq y \leq x-1} \frac{1}{p(y)}\Pi_{z = y+1}^{x-1} \frac{q(z)}{p(z)}$$
By summing all the differences $$\sum_{x = a}^{b}\Delta(x) = 0$$ $$ \sum_{x = a+1}^{b}\lambda(x) \Delta(1) -\mu(x) =0$$ $$ \sum_{x = a+1}^{b}\lambda(x) \Delta(1) = \sum_{x = a+1}^b\mu(x) $$ Thus $$ \Delta(1) =\frac{\sum_{x = a+1}^b\mu(x)}{\sum_{x = a+1}^{b}\lambda(x)} $$ Then
$$g(x)=\sum_{i = a+1}^x\Delta(i) = \bigg(\sum_{i = a+1}^x \lambda(i)\bigg) \frac{\sum_{x = a+1}^b\mu(x)}{\sum_{x = a+1}^{b}\lambda(x)} - \sum_{i = a+1}^x\mu(i) $$
which becomes $$h_{[a,b]}(x) = \sum_{y = a+1}^x \bigg(\Pi_{z = a+1}^{y-1} \frac{q(z)}{p(z)}\bigg)\\ \cdot \bigg(\sum_{a+1\leq y_1<y_2\leq b} \frac{1}{p(y_1)}\Pi_{z = y_1 +1}^{y_2-1} \frac{q(z)}{p(z)}\bigg)\\ \cdot \bigg(\sum_{a+1\leq y \leq b} \Pi_{z = a+1}^{y-1} \frac{q(z)}{p(z)}\bigg)^{-1}\\ -\sum_{a+1\leq y_1\leq y_2 < x}\frac{1}{p(y_1)} \Pi_{z = y_1 + 1}^{y_2 -1}\frac{q(z)}{p(z)} $$
since $$\sum_{i = a+1}^x \sum_{a+1\leq y \leq i-1} \frac{1}{p(y)}\Pi_{z = y}^{i-1} \frac{q(z)}{p(z)} = \sum_{a+1\leq y \leq i <x} \frac{1}{p(y)}\Pi_{z = y}^{i-1} \frac{q(z)}{p(z)}.$$
So instead of $\sum_{a+1\leq y_1 < y_2 < x}$ should we have $\sum_{a+1\leq y_1 \leq y_2 < x}$?
Also, instead of $$ \bigg(\sum_{a+2\leq y \leq b} \Pi_{z = a+1}^{y-1} \frac{q(z)}{p(z)}\bigg)^{-1}$$ I found $$ \bigg(\sum_{a+1\leq y \leq b} \Pi_{z = a+1}^{y-1} \frac{q(z)}{p(z)}\bigg)^{-1}.$$