Report

ST3236: Stochastic Process Tutorial 4 TA: Mar Choong Hock Email: [email protected] Exercises: 5 Question 1 An urn initially contains a single red and single green ball. A ball is drawn at random, removed and replaced by a ball of the opposite color and this procedure repeats so that there are always two balls in the urn. Let Xn be the number of red balls in the urn after n draws, with X0 = 1. Specify the transitions probabilities for MC {X}. Question 1 Case (Xn=0): Both balls are green. One ball will certainly be replaced with red after a ball is drawn. P(Xn+1 = 1 | Xn = 0) = 1, P(Xn+1 = 0 | Xn = 0) = 0, P(Xn+1 = 2 | Xn = 0) = 0 Case (Xn=2): Both balls are red. One ball will certainly be replaced with green after a ball is drawn. P(Xn+1 = 1 | Xn = 2) = 1, P(Xn+1 = 0 | Xn = 2) = 0, P(Xn+1 = 2 | Xn = 2) = 0 Question 1 Case (Xn = 1): One ball is red. Outcome dependent on the colour of the ball drawn. Note: P(Green is drawn|Xn=1) = P (Red is drawn|Xn=1) = 0.5 P(Xn+1 = 0 | Xn = 1) = P(Red is drawn|Xn = 1) = 0.5 P(Xn+1 = 2 | Xn = 1) = P(Green is drawn|Xn = 1) = 0.5 P(Xn+1 = 1 | Xn = 1) = 0 Question 1 The transition matrix is given as follows: Question 2 Find the mean time to reach state 3 starting from state 0 for the MC whose transition probability matrix is Question 2 Let T = min{n : Xn = 3} and vi = E(T | X0 = i). The mean time to reach state 3 starting from state 0 is v0. We apply first step analysis. Recursive Formula for average time step required: vo 1 p0i vi i 0 i 1 step 3 vi Question 2 Therefore, v0 = 1 + 0.4v0 + 0.3v1 + 0.2v2 + 0.1v3 v1 = 1 + 0v0 + 0.7v1 + 0.2v2 + 0.1v3 v2 = 1 + 0v0 + 0v1 + 0.9v2 + 0.1v3 v3 = 0 Solving the equations, we have v0 = 10. Question 3 Consider the MC with transition probability matrix (a) Starting in state 1, determine the probability that the MC ends in state 0 (b) Determine the mean time to absorption. Question 3a Let T = min{n : Xn = 0 and Xn = 2}, ui = P(XT = 0|X0 = i) and vi = E(T|X0 = i). u0 = 1 u1 = 0.1u0 + 0.6u1 + 0.3u2 u2 = 0 we have u1 = 0.25. Question 3b Let T = min{n : Xn = 0 and Xn = 2}, ui = P(XT = 0|X0 = i) and vi = E(T|X0 = i). v0 = 0 v1 = 1 + 0.1v0 + 0.6v1 + 0.3v2 v2 = 0 we have v1 = 2.5. Question 4 Consider the MC with transition probability matrix (a) Starting in state 1, determine the probability that the MC ends in state 0 (b) Determine the mean time to absorption. Question 4a Let T = min{n : Xn = 0 and Xn = 2}, ui = P(XT = 0|X0 = i) and vi = E(T|X0 = i). u0 = 1 u1 = 0.1u0 + 0.6u1 + 0.1u2 + 0.2u3 u2 = 0.2u0 + 0.3u1 + 0.4u2 + 0.1u3 u3 = 0 we have, u1 = 0.3810, u2 = 0.5238 Starting in state 1, the probability that the MC ends in state 0 is u1 = 0.3810 Question 4b Let T = min {n : Xn = 0 and Xn = 2}, ui = P(XT = 0|X0 = i) and vi = E(T|X0 = i). v0 = 0 v1 = 1 + 0.1v0 + 0.6v1 + 0.1v2 + 0.2v3 v2 = 1 + 0.2v0 + 0.3v1 + 0.4v2 + 0.1v3 v3 = 0 we have v1 = v2 = 3.33. Question 5 A coin is tossed repeatedly until two successive heads appear. Find the mean number of tosses required. [Hint: Let Xn be the cumulative number of successive heads. Then the state space is 0,1,2 before stop] Question 5 – Method 1 Let Yn be the outcome {H, T} of each toss and (Yn-1, Yn) denotes the sample point for the sucessive tosses. There are 4 possible sample points. Denote : 0 1 Zn 2 3 T , T T , H H ,T H , H Denote: States = {(Yn-1, Yn)} Where Yn = H if head at nth toss and T if otherwise. 0 (T,T) P(Yn=H)=0.5 1 P(Yn=H)=0.5 3 (T,H) (H,H) P(Yn=T)=0.5 P(Yn=H)=0.5 P(Yn=T)=0.5 P(Yn=T)=0.5 2 (H,T) 1 Question 5 – Method 1 The transition probability matrix is 0 0.5 0.5 0 0 0 0.5 0.5 P 0 0.5 0.5 0 0 0 0 1 Question 5 – Method 1 Let vi denote the mean time to each state 3 starting from state i. By the first step analysis, we have the following equations: v0 = 1 + 0.5v0 + 0.5v1 v1 = 1 + 0.5v2 + 0.5v3 v2 = 1 + 0.5v0 + 0.5v1 v3 = 0 Therefore, v0 = 6 (v1 = 4, v2 = 6) Question 5 – Method 2 Let Xn be the cumulative number of successive heads. The 3-state state space now is {0,1,2}. Example: n Yn 0 T 1 H 2 T 3 H 4 H Xn 0 1 0 1 2 Question 5 – Method 2 Case (Xn = 0) :Previous two tosses are tails Or a head followed by tail. Note: P(Head) = P(Tail) = 0.5 P(Xn+1 = 1 | Xn = 0) = P(Head) = 0.5, P(Xn+1 = 0 | Xn = 0) = P(Tail) = 0.5, P(Xn+1 = 2 | Xn = 0) = 0 Case (Xn = 1): A tail is followed by a head P(Xn+1 = 2 | Xn = 1) = P(Head) = 0.5 P(Xn+1 = 0 | Xn = 1) = P(Tail) = 0.5 P(Xn+1 = 1 | Xn = 1) = 0 Question 5 – Method 2 Case (Xn = 2): Previous two tosses are heads. We make this state absorbing. P(Xn+1 = 1 | Xn = 2) = 0 P(Xn+1 = 2 | Xn = 2) = 1, P(Xn+1 = 0 | Xn = 2) = 0 Question 5 – Method 2 The transition probability matrix is Question 5 – Method 2 Let vi denote the mean time to each state 2 starting from state i. By the first step analysis, we have the following equations: v0 = 1 + 0.5v0 + 0.5v1 v1 = 1 + 0.5v0 + 0.5v2 v2 = 0 Thus, we have, v0 = 6, v1 = 4