Markov Chains. Chapter Stochastic Processes

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1 Chapter 4 Markov Chains 4 Stochastic Processes Often, we need to estimate probabilities using a collection of random variables For example, an actuary may be interested in estimating the probability that he is able to buy a house in The Hamptons before his company bankrupt To study this problem, the actuary needs to estimate the evolution of reserves of his company over a period of time For example, consider the following insurance model U t is the surplus at time t P t is the earned premium up to time t S t is the paid losses until time t The basic equation is U t = U 0 + P t S t We may be interested in the probability of ruin, ie the probability that at some moment U(t) < 0 The continuous time infinite horizon survival probability is given by Pr{U t 0 for all t 0 U 0 = u} The continuous time finite horizon survival probability is given by Pr{U t 0 for all 0 t τ U 0 = u}, where τ > 0 Probabilities like the one before are called ruin probabilities We also are interested in the distribution of the time of the ruin Another model is the one used for day traders Suppose that a day trader decides to make a series of investment Wisely, he wants to cut his losses So, he decides to sell when either the price of the stock is up 50% or down 0% Let P (k) be the price of the stock, k days after begin bought For each nonnegative integer k, P (k) is a rv To study the considered problem, we need to consider the collection of rv s {P (k) : k {0,,, } We may be interested in the probability that the stock goes up 50% before going down 0% We may be interested in the amount of time it takes the stock to go either 50% up or 0% down To estimate these probabilities, we need to consider probabilities which depend on the collection of rv s {P (k) : k {0,,, } Recall that ω is the sample space, consisting by the outcomes of a possible experiment A probability P is defined in the (subsets) events of Ω A rv X is a function from Ω into R 45

2 46 CHAPTER 4 MARKOV CHAINS Definition 4 A stochastic process is a collection of rv s {X(t) : t T } defined in the same probability space T is a parameter set Usually, T is called the time set If T is discrete, {X(t) : t T } is called a discrete-time process Usually, T = {0,,, } If T is an interval, {X(t) : t T } is called a continuous-time process Usually, T = [0, ) Let R T be the collection of functions from T into R A stochastic process {X(t) : t T } defines a function from Ω into R T as follows ω Ω X( )(ω) R T In some sense, we have that a rv associates a number to each outcome and a stochastic processes associates a function to each outcome A stochastic process is a way to assign probabilities to the set of functions R T To find the probability of ruin, we need to study stochastic processes A stochastic process can be used to model: (a) the sequence of daily prices of a stock; (b) the surplus of an insurance company; (c) the price of a stock over time; (d) the evolution of investments; (e) the sequence of scores in a football game; (f) the sequence of failure times of a machine; (g) the sequence of hourly traffic loads at a node of a communication network; (h) the sequence of radar measurements of the position of an airplane 4 Random Walk Let {ɛ j } j= be a sequence of iidrv s with P(ɛ j = ) = p and P(ɛ j = ) = p, where 0 < p < Let X 0 = 0 and let X n = n j= ɛ j for n The stochastic process {X n : n 0} is called a random walk Imagine a drunkard coming back home at night We assume that the drunkard goes in a straight line, but, he does not know which direction to take After giving one step in one direction, the drunkard ponders which way to take home and decides randomly which direction to take The drunkard s path is a random walk Let s j = ɛ j+ Then, P(s j = ) = p and P(s j = 0) = p = q S n = n j= s j has a binomial distribution with parameters n and p We have that for each m n, P (S n = k) = ( n k) p k ( p) k, for 0 k n, E[S n ] = np, Var(S n ) = np( p), S m and S n S m are independent Cov(S m, S n S m ) = 0, Cov(S m, S n ) = Cov(S m, S m ) = Var(S m ) = mp( p), S n S m has the distribution of S n m Since S n = Xn+n and X n = S n n, for each m n, P (X n = j) = P (S n = n+j ) = ( ) n n+j n+j p ( p) n j, for n j n, with n+j integer E[X n ] = E[S n n] = E[S n ] n = n(p ), Var(X n ) = Var(S n n) = n4p( p), X m and X n X m are independent Cov(X m, X n X m ) = 0, Cov(X m, X n ) = Cov(X m, X m ) = Var(X m ) = m4p( p), X n X m has the distribution of X n m

3 4 RANDOM WALK 47 In general, a random walk can start anywhere, ie we will assume that X 0 = i and X n = i + n j= ɛ j Example 4 Suppose that {X n : n } is a random walk with X 0 = 0 and probability p of a step to the right, find: (i) P{X 4 = } (ii) P{X 3 =, X 6 =, } (iii) P{X 5 =, X 0 = 4, X 6 = } Solution: (i) (ii) P{X 4 = } = P{Binom(4, p) = (4 + ( ))/} = ( 4 ) p q 3 = 4p q 3 P{X 3 =, X 6 =, } = P{X 3 =, X 6 X 3 = 3, } = P{X 3 = }P{X 6 X 3 = 3} = ( ) 3 p q ( 3 3) p 3 q 0 = 3p 4 q (iii) P{X 5 =, X 0 = 4, X 6 = } = P{X 5 =, X 0 X 5 = 3, X 0 = 4, X 6 X 0 = } = P{X 5 = }P{X 0 X 5 = 3}P{X 6 X 0 = } = ( ) 5 3 p q 4( ) 5 4 p 4 q ( ) 6 p q 4 = 750p 7 q 9 Example 4 Suppose that {X n : n } is a random walk with X 0 = 0 and probability p of a step to the right Show that: (i) If m n, then Cov(X m, X n ) = m4p( p) (ii) If m n, then Var(X n X m ) = (n m)4p( p) (iii) Var(aX n + bx m ) = (a + b) n4p( p) + b (m n)4p( p) Solution: (i) Since X m and X n X m are independent rv s, Cov(X m, X n ) = Cov(X m, X m + X n X m ) = Cov(X m, X m ) + Cov(X m, X n X m ) = Var(X m ) = m4p( p) (ii) Since X n X m has the distribution of X n m, Var(X n X m ) = Var(X n m ) = (n m)4p( p) (iii) Since X m and X n are independent rv s, Var(aX m + bx n ) = Var(aX m + b(x m + X n X m )) = Var((a + b)x m + b(x n X m )) = Var((a + b)x m ) + Var(b(X n X m ) = (a + b) m4p( p) + b (n m)4p( p) Example 43 Suppose that {X n : n } is a random walk with X 0 = 0 and probability p of a step to the right, find: (i) Var( 3 + X 4 ) (ii) Var( + 3X X 5 ) (iii) Var(3X 4 X 5 + 4X 0 )

4 48 CHAPTER 4 MARKOV CHAINS Solution: (i) (ii) (iii) Var( 3 + X 4 ) = () Var(X 4 ) = () (4)p( p) = 6p( p) Var( + 3X X 5 ) = Var(3X X 5 ) = Var((3 )X (X 5 X )) = Var(X (X 5 X )) = Var(X ) + ( ) Var(X 5 X ) = ()(4)p( p) + ( ) (3)(4)p( p) = 56p( p) Var(3X 4 X 5 + 4X 0 ) = Var((3 + 4)X 4 + ( + 4)(X 5 X 4 ) + 4(X 0 X 5 )) = Var(5X 4 + (X 5 X 4 ) + 4(X 0 X 5 )) = Var(5X 4 ) + Var((X 5 X 4 ) + Var(4(X 0 X 5 )) = (5) Var(X 4 ) + () Var(X 5 X 4 ) + (4) Var(X 0 X 5 ) = (5) (4)(4)p( p) + () ()(4)p( p) + (4) (5)(4)p( p) = 736p( p) Exercise 4 Let {S n : n 0} be a random walk with S 0 = 0 and P [S n+ = i+ S n = i] = p and P [S n+ = i S n = i] = p Find: (i) P [S 0 = 4] (ii) P [S 4 =, S 0 = ] (iii) P [S 3 =, S 9 = 3, S 4 = 6] (iv) Var(S 7 ) (v) Var(5 S 3 S 0 + 3S 0 ) (vi) Cov( S 3 + 5S 0, 5 S 4 + S 7 ) Given k > 0, let T k be the first time that S n = k, ie T k = inf{n 0 : S n = k} Then, T k has negative binomial distribution, ( ) n P (T k = n) = p k ( p) n k, n k k We have that E[T k ] = k p and Var(T k) = kq p Exercise 4 The probability that a given driver stops to pick up a hitchhiker is p = 0004; different drivers, of course, make their decisions to stop or not independently of each other Given that our hitchhiker counted 30 cars passing her without stopping, what is the probability that she will be picked up by the 37-th car or before? Solution: You need to find the probability that any of the seven cars stops This is the complementary that none of the cars stops So, it is ( 0004) 7 = Gambler s ruin problem Imagine a game played by two players Player A starts with $k and player B starts with $(N k) They play successive games until one of them ruins In every game, they bet $, the probability that A wins is p and the probability that A losses

5 4 RANDOM WALK 49 is p Assume that the outcomes in different games are independent Let X n be player A s money after n games After, one of the player ruins, no more wagers are done If X n = N, then X m = N, for each m n If X n = 0, then X m = 0, for each m n For k N, P (X n+ = k + X n = k) = p, P (X n+ = k X n = k) = p Let P k be the probability that player A wins (player B gets broke) We will prove that ( p) q k if p P k = ( p) q N k if p = N P k is the probability that a random walk with X 0 = k, reaches N before reaching 0 P k is the probability that a random walk goes up N k before it goes down k If N, ( ) k q if p > p P k = 0 if p Now, P k is the probability that a random walk with X 0 = k, where k > 0, never reaches 0 Exercise 43 Two gamblers, A and B make a series of $ wagers where B has 55 chance of winning and A has a 45 chance of winning on each wager What is the probability that B wins $0 before A wins $5? Solution: Here, p = 055, k = 5, N k = 0 So, the probability that B wins $0 before A wins $5 is ( ) k q p ( ) = ( ) N ( ) q p Exercise 44 Suppose that on each play of a game a gambler either wins $ with probability p or losses $ with probability p, where 0 < p < The gamble continuous betting until she or he is either winning n or losing m What is the probability that the gambler quits a winner assuming that she/he start with $i? Solution: We have to find the probability that a random walk goes up n before it goes down m So, N k = n and k = m, and the probability is ( q p) m ( q p) m+n, if p m m+n, if p = Problem 4 (# 7, November 00) For an allosaur with 0,000 calories stored at the start of a day: (i) The allosaur uses calories uniformly at a rate of 5,000 per day If his stored calories reach 0, he dies (ii) Each day, the allosaur eats scientist (0,000 calories) with probability 045 and no

6 50 CHAPTER 4 MARKOV CHAINS scientist with probability 055 (iii) The allosaur eats only scientists (iv) The allosaur can store calories without limit until needed Calculate the probability that the allosaur ever has 5,000 or more calories stored (A) 054 (B) 057 (C) 060 (D) 063 (E) 066 Solution: This is the gamblers ruin problem with p = 045, q = 055 and X 0 = We need to find P(X k is up before going down ) = ( ( ) = ) 43 Markov Chains Definition 43 A Markov chain {X n : n = 0,,, } is a stochastic process with countably many states such that for each states i 0, i,, i n, j, P(X n+ = j X 0 = i 0, X = i,, X n = i n, X n = i n ) = P(X n+ = j X n = i n ) In words, for a Markov chain the conditional distribution of any future state X n+ given the past states X 0, X,, X n and the present state X n is independent of the past values and depends only on the present state Having observed the process until time n, the distribution of the process from time n + on depends only on the value of the process at time n Let E the set consisting the possible states of the Markov chain Usually, E = {0,,, } or E = {,,, m} We will assume that E = {0,,, } We define P ij = P (X n+ = j X n = i n ) Note that P (X n+ = j X n = i n ) is independent of n P ij is called the one-step transition probability from state i into state j We define P (k) ij = P (X n+k = j X n = i n ) P (k) ij is called the k-step transition probability from state i into state j We denote P = (P ij ) i,j to the matrix consisting by the one-step transition probabilities, ie P 00 P 0 P 0 P P = 0 P P P 0 P P We denote P (k) = (P (k) ij ) i,j to the matrix consisting by the k-step transition probabilities We have that P () = P Every row of the matrix P (k) is a conditional probability mass function So, the sum of the elements in each row of the matrix P (k) adds, ie for each i, j=0 P (k) ij =

7 43 MARKOV CHAINS 5 Kolmogorov-Chapman equations Let α i = P(X 0 = i), then Proof P(X 0 = i 0, X = i, X n = i n, X n = i n ) = α i0 P i0 i P i i P in i n = k=0 P ikp kj = P () ij P (n) ij P (n+m) ij k=0 P (n ) ik P kj k=0 P (n) ik P (m) kj = P (n+m) = P n P (m) P (n) = P n P(X n = j) = i=0 α ip (n) ij P(X 0 = i 0, X = i,, X n = i n, X n = i n ) = P(X 0 = i 0 )P (X = i X 0 = i 0 )P (X = i X 0 = i 0, X = i ) P (X n = i n X 0 = i 0, X = i, X n = i n ) = P(X 0 = i 0 )P (X = i X 0 = i 0 )P (X = i X = i )P (X n = i n X n = i n ) = α(i 0 )P i0,i P i,i P in,i n For 0 < n < m, P(X 0 = i, X n = j, X m = k) = P(X 0 = i)p (X n = j X 0 = i)p (X m = k X 0 = i, X n = k) = P(X 0 = i)p (X n = j X 0 = i)p (X m = k X n = k) = α(i)p (n) i,j P (m n) j,k Finally, P (n+m) ij = P(X n+m = j X 0 = i) = P(X 0=i,X n+m =j) = α(i)p (n) i,k P (m) k,j k=0 α(i) = k=0 P (n) ik P (m) kj P(X 0 =i) = P(X 0 =i,x n=k,x n+m =j) k=0 P(X 0 =i) P(X n = j) = i=0 P(X 0 = i)p(x n = j X 0 = i) = i=0 α ip (n) ij QED Exercise 45 Let E = {,, 3}, transition matrix /3 /3 0 P = / 0 /, /4 /4 / and α = (/, /3, /6) Find: (i) P, (ii) P 3, (iii) P (X = ), (iv) P (X 0 =, X 3 = 3), (v) P (X =, X = 3, X 3 = X 0 = ), (vi) P (X = 3 X = 3), (vii) P (X = X 5 = 3, X 0 = ), (viii) P (X 3 = 3, X 5 = X 0 = ), (ix) P (X 3 = 3 X 0 = )

8 5 CHAPTER 4 MARKOV CHAINS Solution: (i) /3 /3 0 /3 / P = / 0 / / 0 / = /4 /4 / /4 /4 / (ii) = P 3 /3 / = / 0 / /4 /4 / (iii) αp = (/, /3, /6) = (0375, 035, 035) So, P (X = ) = 035 (iv) P (X 0 =, X 3 = 3) = P (X 0 = )P (X 3 = 3 X 0 = ) = P (X 0 = )P (),3 = (05)(033333) = (v) P (X =, X = 3, X 3 = X 0 = ) = P (X 0 = )P, P,3 P 3, = (/)(/3)(/)(/4) = /4 (vi) (vii) P (X = 3 X = 3) = P 3,3 = / P (X = X 5 = 3, X 0 = ) = P (X = X 0 = ) = P (), = (viii) P (X 3 = 3, X 5 = X 0 = ) = P (X 3 = 3 X 0 = )P (X 5 = X 3 = 3) == P (),3 P () 3, = (03333)(03333) = 0 (ix) P (X 3 = 3 X 0 = ) = P (),3 = Exercise 46 Let P be a three state Markov Chain with transition matrix P =

9 43 MARKOV CHAINS 53 Suppose that chain starts at time 0 in state (a) Find the probability that at time 3 the chain is in state (b) Find the probability that the first time the chain is in state is time 3 (c) Find the probability that the during times,,3 the chain is ever in state Exercise 47 Suppose that 3 white balls and 5 black balls are distributed in two urns in such a way that each contains 4 balls We say that the system is in state i is the first urn contains i white balls, i = 0,,, 3 At each stage ball is drawn from each urn at random add the ball drawn from the first urn is placed in the second, and conversely the ball of the second urn is placed in the first urn Let X n denote the state of the system after the n th stage Prove {X n } n= is a Markov chain Find the matrix of transition probabilities Exercise 48 Let E = {, }, transition matrix ( ) p p P = p p Prove by induction that: P n = ( ) + (p )n (p )n (p )n + (p )n Problem 4 (# 38, May, 000) For Shoestring Swim Club, with three possible financial states at the end of each year: (i) State 0 means cash of 500 If in state 0, aggregate member charges for the next year are set equal to operating expenses (ii) State means cash of 500 If in state, aggregate member charges for the next year are set equal to operating expenses plus 000, hoping to return the club to state 0 (iii) State means cash less than 0 If in state, the club is bankrupt and remains in state (iv) The club is subject to four risks each year These risks are independent Each of the four risks occurs at most once per year, but may recur in a subsequent year (v) Three of the four risks each have a cost of 000 and a probability of occurrence 05 per year (vi) The fourth risk has a cost of 000 and a probability of occurrence 00 per year (vii) Aggregate member charges are received at the beginning of the year (viii) i = 0 Calculate the probability that the club is in state at the end of three years, given that it is in state 0 at time 0 (A) 04 (B) 07 (C) 030 (D) 037 (E) 056 Solution: We may joint states 0 and in a unique state Then, P = P (no big risk and at most one small risk) = (09)(75) 3 + (09)(3)(5)(75) =

10 54 CHAPTER 4 MARKOV CHAINS So, P (3) = P 3 = (07594) 3 = P (3) = P 3 = 056 Problem 43 (#6, Sample Test) You are given: The Driveco Insurance company classifies all of its auto customers into two classes: preferred with annual expected losses of 400 and standard with annual expected losses of 900 There will be no change in the expected losses for either class over the next three years The one year transition matrix between driver classes is given by: with states E = {preferred, standard} i = 005 Losses are paid at the end of each year There are no expenses ( ) T = All drivers insured with Driveco at the start of the period will remain insured for three years Calculate the 3-year term insurance single benefit premium for a standard driver Solution: Let T be the transition matrix Then: ( ) T = Hence, the 3-year insurance single benefit premium for a standard driver is P(X = X 0 = )(900)(05) + P(X = X 0 = )(400)(05) + P(X = X 0 = )(900)(05) + P(X = X 0 = )(400)(05) + P(X 3 = X 0 = )(900)(05) 3 + P(X 3 = X 0 = )(400)(05) 3 = ()(900)(05) + (0)(400)(05) + (085)(900)(05) + (05)(400)(05) + (085)(900)(05) 3 + (0875)(400)(05) 3 = Problem 44 (#, May 00) The Simple Insurance Company starts at time t = 0 with a surplus of S = 3 At the beginning of every year, it collects a premium of P = Every year, it pays a random claim amount: Claim Amount Probability of Claim Amount

11 43 MARKOV CHAINS 55 Claim amounts are mutually independent If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3 If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business Simple has no administrative expenses and its interest income is 0 Calculate the expected dividend at the end of the third year (A) 05 (B) 0350 (C) 044 (D) 0458 (E) 0550 Solution: The Markov chain has states E = {0,,, 3} and transition matrix is P = We have that α = (0, 0, 0, ), αp = (0, 00, 0, 090) and αp = (00, 04, 005, 085) The expected dividend at the end of the third year is (005)()(05) + (085)()(05) + (085)()(05) = Problem 45 (# 3, November 00) The Simple Insurance Company starts at time 0 with a surplus of 3 At the beginning of every year, it collects a premium of Every year, it pays a random claim amount as shown: Claim Amount Probability of Claim Amount If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3 If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business Simple has no administration expenses and interest is equal to 0 Calculate the probability that Simple will be in business at the end of three years (A) 000 (B) 049 (C) 059 (D) 090 (E) 00 Solution: Let states 0,,, 3 correspond to surplus of 0,,, 3 The probability that Simple will be in business at the end of three years is P [X 3 0 X 0 = 3] The one year transition matrix is P = We have that α = (0, 0, 0, ), αp = (0, 0, 0, 08), αp = (004, 04, 005, 067) and αp 3 = (0098, 030, 0080, 059) So, P [X 3 0 X 0 = 3] = 0098 = 090 Problem 46 (#, November 00) An insurance company is established on January (i) The initial surplus is

12 56 CHAPTER 4 MARKOV CHAINS (ii) At the 5th of every month a premium of is collected (iii) At the middle of every month the company pays a random claim amount with distribution as follows x p(x) (iv) The company goes out of business if its surplus is 0 or less at any time (v) i = 0 Calculate the largest number m such that the probability that the company will be out of business at the end of m complete months is less than 5% (A) (B) (C) 3 (D) 4 (E) 5 Solution: Let states 0,,, correspond to surplus of 0,, The one year transition matrix is 0 0 P = We have that α = (0, 0, ), αp = (00, 009, 090), αp = (008, 06, 08) and αp 3 = (0053, 087, 079) So, the answer is Problem 47 (# 9, November 00) A machine is in one of four states (F, G, H, I) and migrates annually among them according to a Markov process with transition matrix: F G H I F G H I At time 0, the machine is in State F A salvage company will pay 500 at the end of 3 years if the machine is in State F Assuming ν = 090, calculate the actuarial present value at time 0 of this payment (A) 50 (B) 55 (C) 60 (D) 65 (E) 70 Solution: We need to find P (3), We have that α = (, 0, 0, 0) and P = So, αp = (0, 08, 0, 0), αp = (044, 06, 04, 0), and αp 3 = (0469, 035, 008, 0) Hence, P (3), = 0468 The actuarial present value at time 0 of this payment is (0468)(500)(09) 3 = 7058

13 44 CLASSIFICATION OF STATES 57 Random Walk: A random walk Markov chain with states E = {0, ±, ±, } and one-step transition probabilities, P i,i+ = p and P i,i = p Absorbing Random Walk with one barrier: The absorbing random walk with one barrier is the Markov chain with states E = {0,,, } and one-step transition probabilities, P i,i+ = p and P i,i = p for i, and P 0,0 = Absorbing Random Walk with two barriers: The absorbing random walk with two barriers is the Markov chain with states E = {0,,,, N} and one-step transition probabilities, P i,i+ = p and P i,i = p for i N, and P 0,0 = P N,N = Reflecting Random Walk with one barrier: The reflecting Random Walk with one barrier is the Markov chain with states E = {0,,, } and one-step transition probabilities, P i,i+ = p and P i,i = p for i, and P 0, = 44 Classification of States 44 Accessibility, communication State j is said to be accessible form state i if P (n) ij > 0, for some n 0 This is equivalent to P((X n ) n 0 ever enters j X 0 = i) > 0 If this happens, we say that i j We say that i j if i j and j i Since P (0) ii =, i i We can put the information about accessibility in a graph We put an arrow from i to j if P ij > 0 Usually, we do not write the arrows from an element to itself Exercise 49 Let E = {,, 3} and let 0 P = 0 0 Represent using graphs the one-step accessibility of states Exercise 40 Let E = {,, 3} and let the transition matrix be /3 /3 0 P = / 0 / /4 / / Represent using graphs the one-step accessibility of states The relation i j is an equivalence relation, ie it satisfies (i) For each i, i i (ii) For each i, j, i j if and only if j i (iii) For each i, j, k, if i j and j k, then i k 3 So, we may decompose E into disjoint equivalent classes A class consists of elements which communicate between themselves and no element from outside A Markov chain is said to be irreducible if there is only one class, that is, if all state communicate with each other 3 3

14 58 CHAPTER 4 MARKOV CHAINS Exercise 4 Prove that if the number of states in a Markov chain is M, where M is a positive integer, and if state j can be reached from state i, where i j, then it can be reached in M steps or less Exercise 4 Consider the Markov chain with states E = {,, 3, 4, 5} and transition matrix / / 0 / 0 / 0 P = /4 /4 /4 /4 / / Represent using graphs the one-step accessibility of states Identity the communicating classes Exercise 43 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3, 4, 5, 6, 7, 8} Represent using graphs the one-step accessibility of states Find the communicating classes 44 Recurrent and transient states Let f i denote the probability that, starting in state i, the process will ever reenters state i, ie f i = P(X n = i, for some n X 0 = i) State i is said to be recurrent if f i = State i is said to be transient if f i < A state i is absorbing if P ii = An absorbing state is recurrent If state i is recurrent, If state i is transient, P(X n = i, infinitely often n X 0 = i) = P((X n ) n is never i X 0 = i) = ( f i ) P((X n ) n enters i exactly k times n X 0 = i) = fi k ( f i ), k 0 P((X n ) n 0 enters i exactly often k times X 0 = i) = f k i ( f i ), k 0 Let N i be the number of times, which the Markov chain stays at state i, ie N i = n=0 I(X n = i) Then,

15 44 CLASSIFICATION OF STATES 59 (i) If state i is recurrent, (ii) If state i is transient, We have that P(N i = X 0 = i) =, E[N i X 0 = i] = P(N i = k X 0 = i) = f k i ( f i ), k 0, E[N i X 0 = i] = f i E[N i X 0 = i] = E[ n=0 I(X n = i) X 0 = i)] = n=0 P (n) ii Theorem 4 State i is recurrent, if n=0 P (n) ii = State i is transient, if n=0 P (n) ii < If state i is transient, the averate number of times spent at i is f i = n=0 P (n) ii Corollary 4 If state i is recurrent and states i and j communicate, then state j is recurrent Proof If states i and j communicate, then there are integers k and m such that P (k) ij P (m) ji > 0 For each integer n, P (m+n+k) jj P (m) ji P (n) ii P (k) ji > 0 and So, n=0 P (m+n+k) jj n=0 P (m) ji P (n) ii P (k) ji = QED All states in the same communicating class have the same type, either all are recurrent or all are transient Theorem 4 A finite state Markov chain has at least one recurrent class Theorem 43 A communicating class C is closed if it cannot leave C, ie for each i C, each j C c and each n 0, P (n) ij = 0 Theorem 44 A recurrent class is closed We may have elements i of a transient class such that i j, for some j C We may have elements i of a recurrent class such that j i, for some j C Theorem 45 A closed finite class is recurrent So, if E has finite many states, a class C is recurrent if and only if it is closed

16 60 CHAPTER 4 MARKOV CHAINS Exercise 44 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {0,,, 3, 4} Find the communicating classes Classify the states as transient or recurrent Exercise 45 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3} Find the communicating classes Classify the states as transient or recurrent Exercise 46 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3, 4, 5} Find the communicating classes Classify the states as transient or recurrent Exercise 47 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3, 4, 5, 6, 7} Find the communicating classes Classify the states as transient or recurrent Exercise 48 Let {X n } be a Markov chain with one step transition probability matrix P =

17 44 CLASSIFICATION OF STATES 6 and states E = {,, 3, 4, 5} Find the communicating classes Classify the states as transient or recurrent Exercise 49 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3, 4, 5, 6, 7, 8} Find the communicating classes Classify the states as transient or recurrent Exercise 40 Let {X n } be a Markov chain with one step transition probability matrix P = and states E = {,, 3, 4, 5, 6, 7} Find the communicating classes Classify the states as transient or recurrent Problem 48 (#, May 00) The Simple Insurance Company starts at time t = 0 with a surplus of S = 3 At the beginning of every year, it collects a premium of P = Every year, it pays a random claim amount: Claim Amount Probability of Claim Amount Claim amounts are mutually independent If, at the end of the year, Simple s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3 If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business Simple has no administrative expenses and its interest income is 0 Determine the probability that Simple will ultimately go out of business (A) 000 (B) 00 (C) 044 (D) 056 (E) 00

18 6 CHAPTER 4 MARKOV CHAINS Solution: The Markov chain has states E = {0,,, 3} and transition matrix is P = State 0 is recurrent So, the probability that Simple will ultimately go out of business is Random walk: The random walk is the Markov chain with state space {0, ±, ±, } and transition probabilities P i,i+ = p and P i,i = p The Markov chain is irreducible X By the law of the large numbers, given X 0 = i, with probability one, lim n n n = p So, if p > /, with probability one, lim X n = If p < /, with probability one, lim X n = So, if p /, the Markov chain is transient We have that P (n) 00 = ( ) n n p n ( p) n By the Stirling formula, So, lim n lim n n! n n+ e n π P00 n (4p( p)) n πn Hence, for p = /, lim n P00 n πn and n=0 P n 00 = The random walk is recurrent for p = / 443 Periodicity State i has period d if P (n) ii = 0 whenever n is not divisible by d and d s the largest integer with this property Starting in i, the Markov chain can only return at i at the times d, d, 3, d and d is the largest integer satisfying this property A state with period is said to be aperidodic Theorem 46 If i has period d and i j, then j has period d Example 44 Consider the Markov chain with state space E = {,, 3} and transition matrix P = Find the communicating classes Classify the communicating classes as transient or recurrent Find the period d of each communicating class

19 45 LIMITING PROBABILITIES 63 Example 45 Consider a random walk on E = {0,,, 3, 4, 5, 6} with transition matrix P = Find the communicating classes Classify the communicating classes as transient or recurrent Find the period d of each communicating class Example 46 Consider a random walk on E = {0,,, 3, 4, 5, 6} with transition matrix P = Find the communicating classes Classify the communicating classes as transient or recurrent Find the period d of each communicating class Example 47 Consider the Markov chain with state space E = {,, 3, 4} and transition matrix P = Find the communicating classes Classify the communicating classes as transient or recurrent Find the period d of each communicating class 45 Limiting Probabilities A recurrent state i is said to be positive recurrent is starting at i, the expected time until the process returns to state i is finite A recurrent state i is said to be null recurrent is starting at i, the expected time until the process returns to state i is infinite A positive recurrent aperiodic state is called ergodic Theorem 47 In a finite state Markov chain all recurrent states are positive recurrent Theorem 48 For an irreducible ergodic Markov chain, lim n P (n) ij exists and is independent of i Furthermore, letting π j = lim n P (n) ij, j 0, then π j is the unique nonnegative solution of π j = i=0 π ip ij, j 0 = i=0 π i

20 64 CHAPTER 4 MARKOV CHAINS The equation above can be written as π 0 π π = ( π 0 π π P 00 P 0 P 0 ) P 0 P P P 0 P P π i is called the long run proportion of time spent at state i Let a j (N) = N n= I(X n = j) be the amount of time the Markov chain spends in state j during the periods,, N Then, with probability one, lim n N N I(X n = j) = π j n= Let T i = inf{x n : n, X n = i} Given X 0 = i, T i is the time to return to state i We have that f i = P(T i < X 0 = i) A state is recurrent if f i = A state is positive recurrent if E[T i X 0 = i] < A state is null recurrent if E[T i X 0 = i] = We have that π i = E[T i X 0 =i] π i is the average time to return to state i Definition 45 A distribution (α 0, α, α, ) of initial probabilities is called stationary distribution If given that X 0 has this distribution, then P (X N = j) = α j, for each j 0 (α 0, α, α, ) has a stationary distribution if P 00 P 0 P 0 P (α 0, α, α, ) = (α 0, α, α, ) 0 P P P 0 P P The long run probabilities (π 0, π, π, ) determines a stationary distribution Theorem 49 Let {X n : n 0} be an irreducible Markov chain with long run probabilities (π 0, π, π, ), and let r be a bounded function on the state space Then, with proability one, lim N N N r(x n ) = n= r(j)π j j=0 Exercise 4 Consider the Markov chain with state space E = {,, 3, 4, 5, 6} and transition matrix P =

21 45 LIMITING PROBABILITIES 65 Compute the matrix F Compute lim n P n Compute where r = (,, 3, 4, 5, 6) lim n n + n E[r(X m ) X 0 = i] m= Exercise 4 Let E = {, } and let ( α P = β ) α β Show that if 0 < α, β <, the Markov chain is irreducible and ergodic Find the limiting probabilities Exercise 43 Find the following limit ( lim n a a b b ) n, where 0 < a, b < Exercise 44 Consider the Markov chain with state space E = {,, 3} and let P = Show that the Markov chain is irreducible and aperiodic Find the limiting probabilities Exercise 45 A particle moves on a circle through points which have been marked 0,,,3, and 4 (in a clockwise order) At the each step it has probability p of moving to the right (clockwise) and p to the left (counter wise) Let X n denote its location on the circle after the n th step The process {X n : n 0} is a Markov chain (a) Find the transition probability matrix (b) Calculate the limiting probabilities Exercise 46 Automobile buyers of brands A, B and C, stay with or change brands according to the following matrix: P = After several years, what share of the market does brand C have? Exercise 47 Harry, the semipro Our hero, Happy Harry, used to play semipro basketball where he was a defensive specialist His scoring productivity per game fluctuated between three state states: (scored 0 or points), (scored between and 5 points), 3 (scored more than 5 points) Inevitably, if Harry scored a lot of points in one game, his jealous teammates

22 66 CHAPTER 4 MARKOV CHAINS refused to pass him the ball in the next game, so his productivity in the next game was nil The team statistician, Mrs Doc, upon observing the transition between states concluded these transitions could be modeled by a Markov chain with transition matrix P = What is the long run proportion of games that our hero had high scoring games? Check that the Markov chain is irreducible and ergodic Find the long run proportion vector (π, π, π 3 ) Exercise 48 Harry, the semipro Our hero, Happy Harry, used to play semipro basketball where he was a defensive specialist His scoring productivity per game fluctuated between three state states: (scored 0 or ponts), (scored between and 5 points), 3 (scored more than 5 points) Invevitaby, if Harry scored a lot of points in one game, his jealous teammates refused to pass him the ball in the next game, so his productivity in the next game was nill The team statistician, Mrs Doc, upon observing the transition between states concluded these transitions could be modelled by a Markov chain with transition matrix P = What is the long run proportion of games that our hero had high scoring games? The salary structure in the the semipro leagues includes incentives for scoring Harry was paid $40/game for a high scoring performance, $30/game when he scored between and 5 points and only $0/game when he scored one or less What was the long run earning rate of our hero? Exercise 49 Suppose that whether or not it rains tomorrow depends on past weather conditions only through the last two days Specifically, suppose that if it has rained yesterday and today then it will rain tomorrow with probability 8; if it rained yesterday but not today then it will rain tomorrow with probability 3; if it rained today but not yesterday then it will rain tomorrow with probability 4; and if it has not rained whether yesterday or today then it will rain tomorrow with probability What proportion of days does it rain? Exercise 430 Harry, the semipro Our hero, Happy Harry, used to play semipro basketball where he was a defensive specialist His scoring productivity per game fluctuated between three state states: (scored 0 or ponts), (scored between and 5 points), 3 (scored more than 5 points) Invevitaby, if Harry scored a lot of points in one game, his jealous teammates refused to pass him the ball in the next game, so his productivity in the next game was nill The team statistician, Mrs Doc, upon observing the transition between states concluded these transitions could be modelled by a Markov chain with transition matrix P =

23 45 LIMITING PROBABILITIES 67 What is the long run proportion of games that our hero had high scoring games? The salary structure in the the semipro leagues includes incentives for scoring Harry was paid $40/game for a high scoring performance, $30/game when he scored between and 5 points and only $0/game when he scored one or less What was the long run earning rate of our hero? Exercise 43 At the start of a business day an individual worker s computer is one of the three states: () operating, () down for the day, (3) down for the second consecutive day Changes in status take place at the end of a day If the computer is operating at the beginning of a day there is a 5 % chance it will be down at the start of the next day If the computer is down for the first day at the start of day, there is a 90 % chance that it sill be operable for the start of the next day If the computer is down for two consecutive days, it is automatically replaced by a new machine for the following day Calculate the limiting probability that a worker will have an operable computer at the start of a day Assume the state transitions satisfy the Markov chain assumptions Problem 49 (# 33, May, 000) In the state of Elbonia all adults are drivers It is illegal to drive drunk If you are caught, your driver s license is suspended for the following year Driver s licenses are suspended only for drunk driving If you are caught driving with a suspended license, your license is revoked and you are imprisoned for one year Licenses are reinstated upon release from prison Every year, 5% of adults with an active license have their license suspended for drunk driving Every year, 40% of drivers with suspended licenses are caught driving Assume that all changes in driving status take place on January, all drivers act independently, and the adult population does not change Calculate the limiting probability of an Elbonian adult having a suspended license (A) 009 (B) 000 (C) 008 (D) 0036 (E) 0047 Solution: We have a Markov chain with states E = {,, 3}, where = {ok license}, = {suspended license}, and 3 = {prison for one year} The transition matrix is To find the limiting probabilities π, π, π 3, we solve π = 095π + 06π + π 3 π = 005π = π + π + π 3 We get π = 00 07, π = 5 07, π 3 = 07 So, the answer is π = 5 07 = Problem 40 (# 40, May, 000) Rain is modeled as a Markov process with two states: (i) If it rains today, the probability that it rains tomorrow is 050 (ii) If it does not rain today, the probability that it rains tomorrow is 030 Calculate the limiting probability that it rains on two consecutive days (A) 0 (B) 04 (C) 06 (D) 09 (E) 0

24 68 CHAPTER 4 MARKOV CHAINS Solution: We have a Markov chain with states E = {, }, where = {rain} and = {not rain} The transition matrix is ( ) To find the limiting probabilities π, π, we solve We get π = 3, π 8 = 5 So, the answer is 8 π = 05π + 03π = π + π lim P [X n =, X n+ = ] = lim P [X n = ]P = 3 n n 8 = 3 6 = 0875 Problem 4 (# 34, November 000)The Town Council purchases weather insurance for their annual July 4 picnic (i) For each of the next three years, the insurance pays 000 on July 4 if it rains on that day (ii) Weather is modeled by a Markov chain, with two states, as follows: The probability that it rains on a day is 050 if it rained on the prior day The probability that it rains on a day is 00 if it did not rain on the prior day (iii) i = 00 Calculate the single benefit premium for this insurance purchased one year before the first scheduled picnic (A) 340 (B) 40 (C) 540 (D) 70 (E) 760 Solution: We have a Markov chain with states E = {, }, where = {rain} and = {not rain} The transition matrix is ( ) To find the limiting probabilities π, π, we solve π = 05π + 0π = π + π We get π =, π 7 = 5 So, the benefit premium is 7 ( 7 (000)() + 7 (000)() + 7 (000)() 3 = () 3 7 (000)() 3 ) = 7053 Problem 4 (# 7, May 00) A coach can give two types of training, light or heavy, to his sports team before a game If the team wins the prior game, the next training is equally

25 45 LIMITING PROBABILITIES 69 likely to be light or heavy But, if the team loses the prior game, the next training is always heavy The probability that the team will win the game is 04 after light training and 08 after heavy training Calculate the long run proportion of time that the coach will give heavy training to the team (A) 06 (B) 064 (C) 067 (D) 070 (E) 073 Solution: We have the following options Option Probability light win light (04)(5) = 0 light win heavy (04)(5) = 0 light lose heavy (06)() = 06 heavy win light (08)(5) = 04 heavy win heavy (08)(5) = 04 heavy lose heavy (0)(5) = 0 We have a Markov chain with states E = {, }, where = {light} and = {heavy} The transition matrix is ( 0 ) To find the limiting probabilities π, π, we solve We get π = 3, π = 3 π = 0π + 04π π + π = Problem 43 (#, November 00) A taxi driver provides service in city R and city S only If the taxi driver is in city R, the probability that he has to drive passengers to city S is 08 If he is in city S, the probability that he has to drive passengers to city R is 03 The expected profit for each trip is as follows: a trip within city R: 00 a trip within city S: 0 a trip between city R and city S: 00 Calculate the long-run expected profit per trip for this taxi driver (A) 44 (B) 54 (C) 58 (D) 70 (E) 80 Solution: We have a Markov chain with states E = {R, S, }, where R means the taxi is in city R and S means the taxi is in city S The transition matrix is ( ) To find the limiting probabilities π, π, we solve π + π = π = 0π + 03π π = 08π + 07π

26 70 CHAPTER 4 MARKOV CHAINS We get π = 3, π = 8 The long expected profit is 3 (0)() + 3 (08)() + 8 (03)() + 8 (07)() = 54 Problem 44 (# 9, November 00) Homerecker Insurance Company classifies its insureds based on each insured s credit rating, as one of Preferred, Standard or Poor Individual transition between classes is modeled as a discrete Markov process with a transition matrix as follows: Preferred Standard Poor Preferred Standard Poor Calculate the percentage of insureds in the Preferred class in the long run (A) 33% (B) 50% (C) 69% (D) 75% (E) 9 Solution: We have a Markov chain with states E = {Preferred, Standard, Poor}, The transition matrix is To find the limiting probabilities π, π, π 3, we solve We get π + π + π 3 = π = 9π π 0 π = π 5 + 4π π 4 3 π = 75 08, π = 5 55, π 3 = 8 08 The percentage of insureds in the Preferred class in the long run is π = 75 = 6944% 08 Example 48 Let X be an irreducible aperiodic Markov chain with m < states, and suppose its transition matrix P is doubly Markov Show that then π(i) =, i E is the m limiting distribution 46 Hitting Probabilities Given a set A E, let h i = P((X n ) n 0 is ever in A X 0 = i) h i is the probability that the process hits A given that it starts at i Theorem 40 (h 0, h, ) is the minimal nonnegative solution to the system of linear equations, h i =, for i A, h i = j=0 P ijh j, for i A

27 46 HITTING PROBABILITIES 7 (Minimality means that if (x 0, x, ) is another solution with x i 0, for each i, then x i h i for each i) Given a set A E, let H = inf{n 0 : X n A} H is the hitting time, ie it is the time the Markov chain hits A We have that h i = P(H < X 0 = i) Let k i = E[H X 0 = i] = k i is the mean hitting time np(h = n X 0 = i) + P(H = X 0 = i) n= Theorem 4 (k 0, k, ) is the minimal nonnegative solution to the system of linear equations, k i = 0, for i A, k i = + j A P ijk j, for i A We may also consider hitting times, requiring that a hits only occurs for times bigger than Given a set A E, let H = inf{n : X n A} Let k i = E[ H X 0 = i] = We can find k i, using the formula np( H = n X 0 = i) + P( H = X 0 = i) n= k i = j A P i,j + j A P i,j ( + k j ), where k i = E[H X 0 = i] and H = inf{n 0 : X n A} If E is an irreducible ergodic Markov chain and A{i}, then, k i = π i Theorem 4 (k 0, k, ) is the minimal nonnegative solution to the system of linear equations, k i = 0, for i A, k i = + j A P ijk j, for i A Exercise 43 For the random walk, ie the Markov chain with states E = {0, ±, ±,, } the transition probabilities are P i,i+ = p, and P i,i = p The Markov chain is irreducible If p, the chain is transient If p =, the chain is null recurrent Using the previous theorem, it is possible to prove that the mean hitting time to {0} is Note that there is no solution to k 0 = 0, k i = + j 0 P ijk j, for i 0

28 7 CHAPTER 4 MARKOV CHAINS Exercise 433 For the random walk with a reflecting barrier, ie the Markov chain with states E = {0,,, } and transition probabilities P i,i+ = p, and P i,i = p for i, and P 0 = The Markov chain is irreducible For A = {0}, ( ) i q if p > p h i = if p, if i > 0 If follows from this that for a regular Markov chain, { ( p) if p > f(0, 0) = P((X n ) n ever transits to 0 X 0 = 0) = p if p Exercise 434 A mouse is in a maze, form a room, the mouse can go any available room which is either up, or down or to the right or to the left at random There is a cat in room 4 If the mouse gets to either room 4, stays there forever Find the average life of the mouse assuming that it starts at room i 3 4 CAT Solution: The states are E = {,, 3, 4} The transition matrix is P = For A = {}, we need to find k i We setup the equations The solutions are k = k 3 = 3 and k = 4 k 4 = 0 k = + (05)k + (05)k 3 k = + (05)k k 3 = + (05)k Exercise 435 A mouse is in a 3 maze, form a room, the mouse can go any available room which is either up, or down or to the right or to the left at random Room 6 is full of cheese and there is a cat in room 3 If the mouse gets to either room 3 or 6, stays there forever Find the probability that the mouse end up in room 6 assuming that it starts at room i 3 CAT CHEESE

29 46 HITTING PROBABILITIES 73 Solution: The states are E = {,, 3, 4, 5, 6} The transition matrix is /3 0 /3 0 /3 0 P = For A = {}, we need to find h i We setup the equations h 6 =, h 3 = 0 h = h + h 4 h = 3 h + 3 h 5 h 4 = h + h 5 h 5 = 3 h h + 3 The solutions are h = 5, h = 4, h 4 = 6, h 5 = 7 Exercise 436 A mouse is in a 3 3 maze, form a room, the mouse can go any available room which is either up, or down or to the right or to the left at random Room 9 is full of cheese and there is a cat in room 3 If the mouse gets to either room 3 or 9, stays there forever Find the probability that the mouse end up in room 3 assuming that it starts at room i Hint: By symmetry h 4 = h 5 = h 6 = 3 CAT CHEESE Exercise 437 A student goes to an Atlantic City casino during a spring break to play a roulette table The student begins with $50 and wants to simply win $50 (or go broke) and go home The student bets on red each play In this case, the probability of winning her bet (and her doubling her ante) at each play is 8/38 To achieve her goal of reaching a total of $00 of wealth quickly, the student will bet her entire fortune until she gets $00 (a) What is the probability the student will go home a winner? (b) How many plays will be needed, on average, to reach a closure at the roulette table? Solution: E = {0, 50, 00, 00} and /9 0 9/9 0 P = 0 0/9 0 9/ (i) With A = {00}, we need to find h 50 We have that h 0 = 0 and h 00 =, h 50 = (9/9)h 00, h 00 = h 00

30 74 CHAPTER 4 MARKOV CHAINS So, h 50 = (9/9) = 04 (i) With A = {0, 00}, we need to find k 50, the equations to solve are k 0 = 0 and k 00 = 0, So, k 50 = (8/9) = k 50 = + (9/9)k 00, k 00 = Exercise 438 In a certain state, a voter is allowed to change his or her party affiliation (for primary elections) only by abstaining from the next primary election Experience shows that, in the next primary, a former Democrat will abstain /3 of the time, whereas a Republican will abstain /5 of the time A voter who abstains in a primary election is equally likely to vote for either party in the next election (regardless of their past party affiliation) (i) Model this process as a Markov chain by writing and labeling the transition probability matrix (ii) Find the percentage of voters who vote Democrat in the primary elections What percentage vote Republican? Abstain? For the remaining questions, suppose Eileen Tudor-Wright is a Republican (iii) What is the probability that, three elections from now, Eileen votes Republican? (iv) How many primaries, on average, does it take before Eileen will again vote Republican? Until she abstains? Solution: (i) We have a Markov chain with state space E = {,, 3}, where,,3 correspond to Democrat, Republican and Independent, respectively The transition matrix is /3 0 /3 P = 0 4/5 /5 / / 0 (ii) We find the long run probabilities, solving π = 3 π + π 3 π = 4 5 π + π 3 π 3 = 3 π + 5 π = π + π + π 3 We get π = 3, π 0 =, π 3 = 0 (iii) We need to find P (3), We find (0,, 0)P 3 We have that /3 0 /3 (0,, 0)P = (0,, 0) 0 4/5 /5 = (0, 4, ) 5 5 / / 0 /3 0 /3 (0,, 0)P = (0, 4, )P = (0, 4, ) /5 /5 = (, 37 0 / / 0 /3 0 /3 (0,, 0)P 3 = (, 37, 4 )P = (, 37, 4 ) /5 /5 = ( / / 0, 4 ) 50 5, 84, 68 )

31 46 HITTING PROBABILITIES 75 The probability that, three elections from now, Eileen votes Republican is 84 5 (iv) We take A = {}, we need to find k and k 3 First, we find k, k, k 3 We have that k = 0, k = + 3 k + 3 k 3, k 3 = + k, We get that k = 8 and k 3 = 5 Hence, the average time until she votes Republican again is k = (4/5)() + (/5)( + 5) = The average time until she votes abstain is k 3 = 5 Using Theorem 40, it is possible to prove the following theorem: Theorem 43 (Gambler s ruin probability) Consider the Markov chain with states E = {0,,,, N} and one-step transition probabilities, P i,i+ = p and P i,i = p for i N, and P 0,0 = P N,N = Let h i be the probability that the Markov chain hits 0 Then, ( p) q i if p h i = ( p) q N i if p = N Theorem 44 (Gambler s ruin probability against an adversary with infinite amount of money) Consider the Markov chain with states E = {0,,, } and one-step transition probabilities, P i,i+ = p and P i,i = p for i, and P 0,0 = Let h i be the probability that the Markov chain hits N Then, ( ) k q if p > p P k = 0 if p Exercise 439 A gambler has $ and he needs $0 He play successive plays until he is broke or he gets $0 In every play, he stakes $ The probability that the wins is and the probability that the losses is Find the probability that the gambler gets $0 Find the average amount of time that the game last Solution: The states are E = {0,,, 3, 4, 5, 6, 7, 8, 9, 0} For A = {0, 0}, we need to find k i We setup the equations k 0 = k 0 = 0 k 9 = k 0 + k 8 k 8 = k 9 + k 7 k 7 = k 8 + k 6 k 6 = k 7 + k 5 k 5 = k 6 + k 4 k 4 = k 5 + k 3 k 3 = k 4 + k k = k 3 + k k = k + k 0

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