Math 180B Homework 4 Solutions

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1 Math 80B Homework 4 Solutions Note: We will make repeated use of the following result. Lemma. Let (X n ) be a time-homogeneous Markov chain with countable state space S, let A S, and let T = inf { n 0 : X n A } be the first time that the Markov chain reaches a state in A. Then for each i S \ A, we have E[T X 0 = i] = + S\A P (X = X 0 = i)e[t X 0 = ]. Proof. Suppose i S \ A and S. By the Markov property and time homogeneity, we have E[T X 0 = i, X = ] = P (T > n X 0 = i, X = ) = P (X 0 / A,..., X n / A X 0 = i, X = ) = + P (X / A,..., X n / A X 0 = i, X = ) n= = + P (X / A,..., X n / A X = ) n= = + P (X 0 / A,..., X n / A X 0 = ) n= = + P (T n X = ) n= = + E[T X 0 = ]. Also, note that if A, then E[T X 0 = ] = 0. Therefore, by first-step analysis, if i / A, then E[T X 0 = i] = np (T = n X 0 = i)

2 Math 80B Homework 4 Solutions = n P (T = n, X = X 0 = i) S = np (T = n X 0 = i, X = )P (X = X 0 = i) S = S P (X = X 0 = i)e[t X 0 = i, X = ] = P (X = X 0 = i) ( + E[T X 0 = ] ) S = S P (X = X 0 = i) + S P (X = X 0 = i)e[t X 0 = ] = + S\A P (X = X 0 = i)e[t X 0 = ]. Problem (Pinsky & Karlin, Exercise 3.4.). Find the mean time to reach state 3 starting from state 0 for the Markov chain whose transition probability matrix is P = Solution. Let T be the time it takes for the Markov chain to reach state 3. That is, if (X n ) is the Markov chain in question, then T = inf { n 0 : X n = 3 }. We need to compute E[T X 0 = 0]. Let g i = E[T X 0 = i] for i {0,,, 3}. By Lemma, we have the following linear system of equations: g 0 = + 0.4g g + 0.g, g = + 0.7g + 0.g, g = + 0.9g. Solving this system, we get E[T X 0 = 0] = g 0 = 0.

3 Math 80B Homework 4 Solutions Problem (Pinsky & Karlin, Exercise 3.4.4). A coin is tossed repeatedly until two successive heads appear. Find the mean number of tosses required. Hint: Let X n be the cumulative number of successive heads. The state space is 0,,, and the transition probability matrix is P = Determine the mean time to reach state starting from state 0 by invoking a first step analysis. Solution. Let T = inf { n 0 : X n = }. We wish to compute E[T X 0 = 0]. Let g i = E[T X 0 = i] for i {0,, }. By Lemma, we get the following linear system of equations: Therefore, E[T X 0 = 0] = g 0 = 6. g 0 = + g 0 + g, g = + g 0. Problem 3 (Pinsky & Karlin, Problem 3.4.). Which will take fewer flips, on average: successively flipping a quarter until the pattern HHT appears, i.e., until you observe two successive heads followed by a tails; or successively flipping a quarter until the pattern HT H appears? Can you explain why these are different? Solution. Let (X n ) be the Markov chain whose states are the possible patterns of three consecutive coin tosses. Then (X n ) has the following 3

4 Math 80B Homework 4 Solutions transition probability matrix (blank entires represent 0): P = T T T T T H T HT T HH HT T HT H HHT HHH T T T / / T T H / / T HT / / T HH / / HT T / / HT H / / HHT / / HHH / / Define T HHT = inf { n 0 : X n = HHT }, T HT H = inf { n 0 : X n = HT H }. We wish to compare E[T HHT ] and E[T HT H ]. Let g i = E[T HHT X 0 = i] for each i. Then by Lemma, we have the following linear system of equations: g T T T = + g T T T + g T T H, g HT T = + g T T T + g T T H, g T T H = + g T HT + g T HH, g HT H = + g T HT + g T HH, g T HT = + g HT T + g HT H, g T HH = + g HHH, g HHH = + g HHH. Solving this system yields g HHH = g T HH =, g HT H = g T T H = 6, g T T T = g HT T = g T HT = 8, and hence E[T HHT ] = P (X 0 = )g = 8( ) = 5. Next, let h i = E[T HT H X 0 = i] for each i. Then by Lemma, we have the following linear system of equations: h T T T = + h T T T + h T T H, h HT T = + h T T T + h T T H, 4

5 Math 80B Homework 4 Solutions h T T H = + h T HT + h T HH, h T HT = + h HT T, h HHT = + h HT T, h T HH = + h HHT + h HHH, h HHH = + h HHT + h HHH. Solving this system yields h T HT = h HHT = 6, h T T H = h T HH = h HHH = 8, h HT T = h T T T = 0, and hence E[T HT H ] = P (X 0 = )h = 8( ) = 7. Thus, on average, it will take fewer flips to observe HHT than HT H. Problem 4 (Pinsky & Karlin, Problem 3.4.). A zero-seeking device operates as follows: If it is in state m at time n, then at time n +, its position is uniformly distributed over the states 0,,..., m. Find the expected time until the device first hits zero starting from state m. Note: This is a highly simplified model for an algorithm that seeks a maximum over a finite set of points. Solution. This device can be modeled by a Markov chain (X n ) taking values in the nonnegative integers with transition probabilities given by Let, if m > 0 and 0 < m, m P (X n+ = X n = m) =, if m = = 0, 0, otherwise. T = inf { n 0 : X n = 0 }. We wish to compute E[T X 0 = m] for all m 0. If m = 0, then clearly E[T X 0 = m] = 0. Suppose m > 0. By Lemma, we have E[T X 0 = m] = + P (X = X 0 = m)e[t X 0 = ] = 5

6 Math 80B Homework 4 Solutions = + m m = E[T X 0 = ] We will prove by strong induction that E[T X 0 = m] = m k=, and the base k case m = is clear. Now suppose m > and that E[T X 0 = ] = k= k when < m. Then E[T X 0 = m] = + m = + m = + m m = m = k= m k= E[T X 0 = ] k = + m m k k = + m k= m k= m =k k k m m = m k= k. Problem 5 (Pinsky & Karlin, Problem 3.4.7). Let X n be a Markov chain with transition probabilities P i,. We are given a discount factor β with 0 < β < and a cost function c(i), and we wish to determine the total expected discounted cost starting from state i, defined by [ ] h i = E β n c(x n ) X 0 = i. Using a first step analysis show that h i satisfies the system of linear equations h i = c(i) + β P i, h. Solution. Assuming c(i) 0 for all i (so that we may interchange summation and expectation), we have [ ] h i = E β n c(x n ) X 0 = i = β n E[c(X n ) X 0 = i] = E[c(X 0 ) X 0 = i] + β β n E[c(X n ) X 0 = i] n= 6

7 Math 80B Homework 4 Solutions = c(i) + β β n E[c(X n+ ) X 0 = i] = c(i) + βe [ ] β n c(x n+ ) X 0 = i. Thus, it suffices to show that [ ] E β n c(x n+ ) X 0 = i = P i, h. (5.) To see this, we use time-homogeneity and the Markov property: P (X = X 0 = i)e[c(x n ) X 0 = ] = P (X = X 0 = i) c(k)p (X n = k X 0 = ) k =,k =,k =,k = k c(k)p (X = X 0 = i)p (X n+ = k X = ) c(k)p (X = X 0 = i)p (X n+ = k X 0 = i, X = ) c(k)p (X n+ = k, X = X 0 = i) c(k)p (X n+ = k X 0 = i) = E[c(X n+ ) X 0 = i], and so P i, h = [ ] P (X = X 0 = i)e β n c(x n ) X 0 = = β n P (X = X 0 = i)e[c(x n ) X 0 = ] = β n E[c(X n+ ) X 0 = i] [ ] = E β n c(x n+ ) X 0 = i. Therefore, (5.) holds. 7

8 Math 80B Homework 4 Solutions Problem 6 (Pinsky & Karlin, Problem 3.4.8). An urn contains five red and three green balls. The balls are chosen at random, one by one, from the urn. If a red ball is chosen, it is removed. Any green ball that is chosen is returned to the urn. The selection process continues until all of the red balls have been removed from the urn. What is the mean duration of the game? Solution. Let X 0 be the initial number of red balls in the urn, and for n, let X n denote the number of red balls in the urn after the nth draw. Then (X n ) is a Markov chain with transition matrix P = /4 3/ /5 3/ /6 3/ /7 3/ /8 3/8. Let T = inf { n 0 : X n = 0 } be the time until all red balls are removed from the urn and the game ends. We wish to compute E[T X 0 = 5]. Let g i = E[T X 0 = i]. By Lemma, we have the following linear system of equations: g = g, g = + 5 g g, g 3 = g g 3, g 4 = g g 4, g 5 = g g 5. Solving this system, we get E[T X 0 = 5] = g 5 = 77/0. 8

9 Math 80B Homework 4 Solutions Problem 7 (Pinsky & Karlin, Problem 3.4.5). A simplified model for the spread of a rumor goes this way: There are N = 5 people in a group of friends, of which some have heard the rumor and the others have not. During any single period of time, two people are selected at random from the group and assumed to interact. The selection is such that an encounter between any pair of friends is ust as likely as between any other pair. If one of these persons has heard the rumor and the other has not, then with probability α = 0. the rumor is transmitted. Let X n denote the number of friends who have heard the rumor at the end of the nth period. Assuming that the process begins at time 0 with a single person knowing the rumor, what is the mean time that it takes for everyone to hear it? Solution. The stochastic process (X n ) is a Markov chain with state space {,, 3, 4, 5} and transition probabilities given by P (X n+ = 5 X n = 5) =, and if i {,, 3, 4}, then and P (X n+ = i + X n = i) = α P (X n+ = i X n = i) = All other transition probabilities are zero. Let ( )( ) i 5 i ( ) 5 = T = inf { n 0 : X n = 5 } i(5 i) 00. i(5 i) 00, be the time until all five people have heard the rumor. We wish to compute E[T X 0 = ]. Let g i = E[T X 0 = i] for i =,, 3, 4, 5. By Lemma, we have the following linear system of equations: ( g = + 4 ) g + 4 ( g, g = + 3 ) 00 ( g 3 = + 3 ) g ( g 4, g 4 = Solving this system, we get E[T X 0 = ] = g = 50/3. g g 3, ) g 4. 9

10 Ex The transition probability matrix of the Markov chain (of states 0,,, 3) is , which is of the form of a success runs Markov chain with all probabilities p i = q i =. Ex The state space of the Markov chain is the nonnegative integers. The transition probabilities are as follows: α = 0 = k α = 0, k = ( α)β > 0, k = P(X n+ = k X n = ) =. αβ + ( α)( β) = k > 0 α( β) > 0, k = + 0 otherwise So corresponding to (3.38), r 0 = α, p 0 = α, q i = ( α)β, r i = αβ + ( α)( β), p i = α( β) for i. Pr 3.5. We are taking M = 3 in equation (3.35) in section 3.5., with P(ξ = i) = a i for i = 0,,, 3 and a 0 + a + a + a 3 =. All the a i = 0 with i 4. Putting the corresponding a 3, a 4,... into equation (3.35), we have µ = /a 3. Since in this case, T = min{n : X n 3} = min{n : X n = 3}, µ = ν 0, the mean time until absorption starting from 0. Pr We let x i, i =,,... to be i.i.d. discrete random variable have uniform distribution on the integers,,..., 6. X 0 = 0 and X n+ = X n + ξ n+ is the Markov chain we are interested in. Set T = min{n 0 : X n > 0}. We wil solve u i = P(X T = 3 X 0 = i) for 0 i 0. Clearly, u i = 6 P(X T = 3 X i + = i + k)p(ξ i+ = k) = k= 6 k= u i+k 6.

11 So we have u 0 = 6 u 9 = 6 u u 8 = 6 u u u 7 = 6 u u u u 6 = 6 u u u u 0 u 5 = 6 u u u u u 0 u 4 = 6 u u u u u u 0 u 3 = 6 u u u u u u 9 u = 6 u u u u u u 8 u = 6 u + 6 u u u u u 7 u 0 = 6 u + 6 u + 6 u u u u 6. Solving the system of linear equations yields u 0 = Pr (a) X n is clearly nonnegative. We observe that for k, E[X n+ X n = k] = (k ) + (k + ) = k and Thus E[X n+ X n ] = X n and since E[X n+ X n = 0] = 0. E[X n+ ] = E[X n+ X n = k]p(x n = k) = kp(x n = k) = E[X n ], k=0 k= we have E[ X n ] < for all n. (b) Using the maximal inequality, if X 0 < N, and if X 0 N. P(max n X n N) E[X 0] N P(max n X n N) =

Expected Value 7/7/2006

Expected Value 7/7/2006 Definition Let X be a numerically-valued discrete random variable with sample space Ω and distribution function m(x). The expected value E(X) is defined by E(X) = x Ω x m(x), provided