Exit probability and time for asymmetric barriers

Question

A simple symmetric random walk starts at $0$ , stepping $\pm 1$ with probability $\tfrac12$ each. It is absorbed at $-3$ or $+7$ .

Find the probability it exits at $+7$ and the expected number of steps until it stops.

Accepted Answer

Here the distance to the lower barrier is a = 3 and to the upper barrier is b = 7. Exit probability. Since St is a martingale, E[ST] = 0, giving P(ST = +7) = \dfrac{a}{a+b} = \dfrac{3}{10} = 0.3. So the walk exits at +7 only 30\% of the time and hits -3 the other 70\%, it is closer to the lower barrier, so it usually stops there. Expected time. Using Mt = St^2 - t and optional stopping, E[T] = E[ST^2]. With the exit probabilities above, E[T] = 0.3 	imes 7^2 + 0.7 	imes 3^2 = 0.3 	imes 49 + 0.7 	imes 9 = 14.7 + 6.3 = 21. That matches the product form E[T] = ab = 3 	imes 7 = 21 steps. Read…

Exit probability and time for asymmetric barriers

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