The drunkard who wants to climb one step uphill

Question

A walker stands at position 0 on a number line. Each second he steps +1 (uphill) with probability p = 0.6 and -1 (downhill) with probability q = 0.4, independently. Because he drifts uphill, he will eventually stand at position +1. On average, how many seconds pass before the first time he reaches +1?

Accepted Answer

Let T be the expected number of steps to raise the position by exactly one net unit. The key trick is that climbing one level is self-similar: whatever it takes to go up one level, it takes the same in distribution to go up any single level, and the times for separate levels are independent. Condition on the first step: - With probability p, he goes straight to +1 in one step. - With probability q, he first drops to -1. Now he must climb two levels to reach +1: from -1 up to 0, then 0 up to +1. Each single climb costs T on average, so this branch costs 1 + 2T. Therefore T = p(1) + q(1 + 2T) =…

The drunkard who wants to climb one step uphill

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