Consider the random walk on Z with jump probabilities from n...

Consider the random walk on Z with jump probabilities from n given by Pn,n+1 = p > 1/2 and Pn,n-1 = 1 - p. Calculate the probability that, starting at 0, the walk returns to 0 at some later time. One possible method: Let q be the probability that if we are at n we ever hit n - 1. Argue why this does not depend on n. Condition on the first step from n and find an equation satisfied by q. Use the value of q to solve the original problem. Another method: Calculate the expected number of returns and use its relation to the return probability.

Transcript

00:01 So in this question we're told that we have a random walk on the integers, with the probability of going from n to n plus 1 is p, which is greater than a half, and the probability of going from n to n to n minus 1 is 1.

00:20 So now what we're going to do is we're going to calculate the probability.

00:23 So if we start at 0, what's the probability of returning to 0 at some other time? well, what we're going to do is we're going to use the first method that is suggested.

00:40 So let's have q be the probability of hitting n minus 1, hit n minus 1, given that we are at n.

00:53 So we hit n minus 1 at some future time.

01:02 So let's work out what this is.

01:05 So q is, so if we go to the left with probability 1 minus p, then we hit n minus 1 for certain.

01:17 But if we go to the right with probability p, then we will come back to point n with probability q, because q is the probability that we go to point n minus 1 given we start at n.

01:35 So if we're at point n plus 1, so at n plus 1 now, then q takes us back to n, and then q again takes us back to n minus 1.

01:48 So that means that q is equal to 1 minus p plus p q squared.

02:01 So that is going to be the equation satisfied by q.

02:11 So then let's use this q.

02:13 So let's call this equation star.

02:17 Now we can use q to work out the probability of returning to zero.

02:21 So if we start at zero, then what we have is that if we go, right to start with, so the probability of returning to zero, is the probability that we go right.

02:48 So the probability of going right is p, and then we're at step one, and the probability we ever return to zero is then q.

02:59 But if we go left with probability 1 minus p, we're now at step minus 1, and then the probability that we ever return to 0 will be q, but we have a probability, to send p to 1 minus p in the definition of q because sending p to 1 minus p has the effect of flipping left and right so and so basically if we can solve for q of p so star gives us q of p then this is going to be q of 1 minus p.

03:45 So this is seen as a function, not a product...

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