This is a response to a comment.
The coupon collector's problem is elementary. I don't have a particular scholarly reference in mind, but rather the technique of the proofs. There are a few proofs of the nHn expected time to collect all coupons. One possibility is that you can compute the expected time to collect the kth new coupon,
n/(n−k+1). That uses a lot of symmetry you don't have for a general Markov process. Here, you have transition probabilities and times on (current location, subset visited so far).
Analogous to what I did here, you can use inclusion-exclusion. The expected time to cover everything (with discrete time) is the sum of the probability that you haven't covered everything by time t−1, which you can express as
sumtsumSsubsetV−1|S|+1Prob(XiilttcapS=emptyset)
where V = 1,...,n. You can switch the order of summation to get about 2n analytically solvable problems about avoiding particular subsets.
sumSsubsetV−1|S|+1A(S)
where A(S) = expected time before you first enter S.
The same holds for continuous time.
No comments:
Post a Comment