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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?

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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.

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