# Distribution of right jumps conditional of hitting time for a random walk with possibility of inaction.

Suppose we have a random walk that moves in discrete time. It starts at zero and in each period it jumps one unit to the right with probability $$\alpha$$, it jumps to the left one unit with probability $$\beta$$, and it does not move with probability $$1-\alpha-\beta$$. Obviously $$\alpha,\beta>0$$ and $$\alpha+\beta<1$$.

Conditional on the stopping time at which the random walk hits 1 being $$k$$ periods, what is the distribution of the number of right jumps?

• – Mike Earnest Mar 7 at 20:47

I'll give you a hint how I would approach the problem to help you get started. Consider the case it takes 1 right jump to get to x=1. There is only one way that could have happened - it took k-1 periods of standing still each period it had a probability of $$1-\alpha-\beta$$ of doing so. The final probability of such an event is $$\alpha\times(1-\alpha-\beta)^{k-1}$$. Next consider the case it took 2 right jumps to get to x=1. There are now many ways this could have happened, but it boils down to it standing still for k-3 steps and going left 1 step. Write the probability for this happening. Just keep going until you get to the case where it took $$(k-1)/2$$ left, and $$(k+1)/2$$ steps to the right.