On Sinai's paper "The Limiting Behavior of a One-Dimensional Random Walk in a Random Medium" (1982) one considers a random walk on $Z^1$ 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 paper one reads
$\ \ $ Let $[a,b]$ be a segment with end-points belonging to $Z^1$. We shall denote by $k^+_{[a,b]}(x)$ the $P$-probability of paths starting at $x$ which hit $b$ before $a$; $k^-_{[a,b]}(x)=1-k^+_{[a,b]}(x)$ is the $P$-probability of paths leaving $x$ and hitting $a$ before $b$. Then $$k^+_{[a,b]}(x)=p(x)k^+_{[a,b]}(x+1)+q(x)k^+_{[a,b]}(x-1), \\ k^+_{[a,b]}(a)=0,\qquad k^+_{[a,b]}(b)=1$$
$$w^{(n)}(t)=\left\{\begin{array}{lr}\dfrac1{\log n}\sum\limits_{0\leqq x\leqq k}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2n}, & k=1,2,\ldots,\\ \dfrac1{\log n}\sum\limits_{k\leqq x\leqq 0}\log\dfrac{q(x)}{p(x)}\text{ for }t=\dfrac k{\log^2 n}, & \qquad k=-1,-2,\ldots.\end{array}\right.$$
$\ \ $ Lemma 1. The following equations hold: $$k^+_{[a,b]}(x)=\left(\sum\limits_{y=a+1}^x\exp\{\log n[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right) \qquad\qquad\\ \qquad\qquad\cdot\left(\sum_{y=a+1}^b\exp\{\log n[w^{(n)}(y\log^{-2}n)-w^{(n)}(a\log^{-2}n)]\}\right)^{-1},$$
Let's prove Lemma 1:
denote starting from the one step equation $$ k⁺_{[a,b]}(x) = p(x)k⁺_{[a,b]}(x + 1) + q(x) k⁺_{[a,b]}(x-1) $$
define $\Delta(x) = k⁺_{[a,b]}(x) - k⁺_{[a,b]}(x-1) $ for $x = a+1, \ldots, b$ rewriting the one step equation, we obtain $$q(x)\Delta(x) = p(x)\Delta(x+1) $$
Which means that $\Delta(x+1) = \frac{q(x)}{p(x)}\Delta(x)$
Now, summing the increments $\Delta(x)$ and using the fact that $k⁺_{[a,b]}(a) = 0$ and $k⁺_{[a,b]}(b) = 1$ $$\sum_{x= a+1}^b \Delta(x) = k⁺_{[a,b]}(b) - k⁺_{[a,b]}(a) = 1$$
Since $\Delta(x) = \Pi_{l = a+1}^{x-1} \frac{q(l)}{p(l)}\Delta (a+1)$ this gives $$1 = \sum_{x= a+1}^b\Delta(x) = \Delta(a+1) (1 + \sum_{x= a+2}^b \Pi_{l = a+1}^{x-1} \frac{q(l)}{p(l)}) $$ which implies that $\Delta(a+1) = 1 + \sum_{x= a+2}^b \Pi_{l = a+1}^{x-1} \frac{q(l)}{p(l)}$
Since $ k⁺_{[a,b]}(x) = \sum_{y = a+1}^x \Delta(y)$ we obtain that
$$ k⁺_{[a,b]}(x) = \frac{1 + \sum_{y = a+2}^x \Pi_{l = a+1}^{y-1} \frac{q(l)}{p(l)}}{1+\sum_{x= a+2}^b \Pi_{l = a+1}^{x-1} \frac{q(l)}{p(l)}} $$
If we use the notation $w^{n}(t)$ as above, we see that $$\exp(\log n[w^n(y /\log^2n) - w^n(a /\log^2n)]) = \Pi_{l =a+1}^y \frac{q(l)}{p(l)} $$ so the formula becomes $$ k⁺_{[a,b]}(x) = \frac{ \sum_{y = a+1}^x \exp(\log n[w^n((y-1) /\log^2n) - w^n(a /\log^2n)])}{\sum_{x= a+1}^b \exp(\log n[w^n((y-1) /\log^2n) - w^n(a /\log^2n)])} $$
Is there a typo in the in formula in the paper?
This seems to have been carried further, see
the expression for $k^+_{[a,b]}(x)$ for $x = a+1$ should be $$ k⁺_{[a,b]}(x) = \frac{1}{\sum_{x= a+1}^b \exp(\log n[w^n((y-1) /\log^2n) - w^n(a /\log^2n)])} $$
