# How to compute the exit probabilities for a random walk? a question concerning a typo on sinai's paper

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 $$k⁺_{[a,b]}(x) = \frac{1}{\sum_{x= a+1}^b \exp(\log n[w^n((y-1) /\log^2n) - w^n(a /\log^2n)])}$$