the bridge frog and groom frog perform Markov chains simultaneously on the state space { 1,2}. Call the bride's chain Xo, X1, . .. and the groom's chain Yo, Y1, .... At the start of the ceremony (time 0), the bride starts in state 1 and the groom starts in state 2; that is, Xo = 1 and Yo = 2. The X and Y chains are performed independently with transition matrices Px = .3 .7 .2 .8 .2 .8 and Py = .9 . 1 , respectively. The ceremony ends when the bridge and groom meet, visiting the same state at the same time. That is, the duration of the ceremony is the random variable T = inf{t : X =Y,}, where the "inf" of a nonempty set of numbers is the minimum of the numbers ("inf" is for "infimum"). Find the expected value of T. [ Here are some hints: this is an elaboration of the idea used in class to find the expected value of the geometric distribution, using conditional expectation. Define E12(. . . ) = E(. . . [ Xo =1, Yo = 2) and E21 (. . . ) = E(. . . |Xo = 2, Yo = 1). We want to find E12(7). We will do this by finding two linear equations satisfied by E12(T) and E21 (T) and solving those simultaneously to find both E12(7) and E21 (7). To derive an equation for E127, think of conditioning on the first hop of the bride and the first hop of the groom, that is, we condition on the random vector (X1, Y1 ). This vector has 4 possible values. For example, given (X1, Y1 ) = (1, 1), of course then we know T = 1, that is, E12(7 | X1 = 1, Y1 = 1) = 1. As another example, given (X1, Y1 ) = (2, 1), you can argue (some persuasive words are enough) that E12(T | X1 = 2, Y1 = 1) = 1 + E21(7). So using the law of total expectation you can write an equation for E12(7), having 4 terms involving E12(7) and E21 (7). And by the same process you can write an eqution for E21 (7). Then you can solve those two equations simultaneously. ]I