On random walks

4.* Use 3. to show that the average number of visits to a > 0 before returning to the origin is 1 (hint: show that it is closely related to the expectation of some geometric random variable). Solution: let Na be the number of visits to a before returning to 0. Using question 3. one has P(Na = 0) = 1 2 + 1 2 (1− 1/a) (call that probability q). Note that once you’ve made a visit to a (that is given Na ≥ 1), which happens with probability 1− q, the number of returns is a geometric random variable, with probability of success 1− q : the probability of coming back to a before 0 (failure) is 1 2 + 1 2 (1− 1/a) = q. Therefore, E[Na − 1|Na ≥ 1] = 1 1−q .