# Hitting time for a random walk where step probabilities linearly depend on the distance to an absorbing wall

Consider the instance where I have a distinct random walk on the integers over an interval $[0, M]$, where I start at some placement $0 < k < M$, and also both endpoints (i.e. $0$ and also $M$) are totally soaking up. The caution, to this or else straightforward random walk trouble, is that the probability of taking an action relies on the range in between the pedestrian and also the soaking up borders. Allowing $x_i$ stand for the placement of the pedestrian, we have $P[+1] = \frac{x_i}{M}$, and also $P[-1] = 1 - P[+1]$.

To put it simply, the probability of the pedestrian taking a $+1$ action amounts to the probability of tasting from the integers over the interval $[1, M]$, with consistent probability, and also seleting an integer $j$ such that $j \leq x_i$.

Can we mention the probability that the pedestrian gets to the soaking up target $M$?

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2022-06-07 14:37:42
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The common means: write $u(x)$ for the probability to get to $M$ prior to $0$ for the random walk beginning with $x$, and also note that $(u(x))_{0\leqslant x\leqslant M}$ is the one-of-a-kind remedy of the straight system $u(0)=0$, $u(M)=1$, and also, for every single $1\leqslant x\leqslant M-1$, $$Mu(x)=xu(x+1)+(M-x)u(x-1).$$ Hence, for every single $0\leqslant x\leqslant M$, $$u(x)=\frac1{2^{M-1}}\sum_{k=0}^{x-1}{M-1\choose k}.$$