b If the flea is at vertex i which is the case π i fraction of the time then it

B if the flea is at vertex i which is the case π i

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(b) If the flea is at vertex i (which is the case π i fraction of the time), then it makes a counterclockwise move followed by 5 consecutive clockwise moves with probability q i p i - 1 p i p i +1 p i +2 p i +3 . (Here, arithmetic in the subscripts is done (sort of) modulo 3: i.e., subscript 0 means 3, 4 means 1, 5 means 2, 6 means 3.) Thus, by conditioning on the vertex at which the flea is in the long run, we get that the proportion of time there is a counterclockwise move followed by 5 consecutive clockwise moves is 3 X i =1 π i q i p i - 1 p i p i +1 p i +2 p i +3 = π 1 q 1 p 3 p 1 p 2 p 3 p 1 + π 2 q 2 p 1 p 2 p 3 p 1 p 2 + π 3 q 3 p 2 p 3 p 1 p 2 p 3 . Problem 4. (a) Let X n = i mean that there are exactly i umbrellas stored at his present position, just before he is about to depart on his journey to the other position. The state space is S = { 0 , 1 , 2 , . . . , r } . The single-step transition matrix has entries: P 0 ,r = 1 , P i,r - i = 1 - p, P i,r - i +1 = p, for i = 1 , . . . , r. Specifically, for r = 4, we get P = 0 0 0 0 1 0 0 0 q p 0 0 q p 0 0 q p 0 0 q p 0 0 0 . (b) In order to verify the given stationary probabilities, we just have to put them into the equations π = πP, i π i = 1 and verify they work. The equations π = πP, i π i = 1 amount to: π 0 = r π j = r - j + r - j +1 π r = π 0 + 1 1 = π 0 + π 1 + · · · + π r . It is easy to verify that these are satisfied by π 0 = q/ ( r + q ) , π j = 1 / ( r + q ) , for j = 1 , 2 , . . . , r . (c) Ross gets wet whenever he is in state 0 (no umbrella at his current location) and it rains. The long-run fraction of time he is in state 0 is π 0 = q/ ( r + q ), given that he is in state 0, the probability he gets wet (i.e., that it rains) is p. Thus, the probability he gets wet on any one day far in the future (in steady state) is
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