Stochastic processes and stopping time Exercises

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1 Stochastic processes and stopping time Exercises Exercise 2.1. Today is Monday and you have one dollar in your piggy bank. Starting tomorrow, every morning until Friday (inclusively), you toss a coin. If the result is tails you withdraw one dollar (if possible) from the piggy bank. If the result is heads you deposit one dollar in it. Model the amount of money in the piggy bank by answering the following questions : a) What is the sample space? b) De ne a stochastic process describing this situation, and explain its signi cation. Do not forget to specify the period of time you are using. c) What -algebra should you use to build your measurable space? d) List the -algebras in the ltration generated by the stochastic process for the following days: Monday, Tuesday, Wednesday and Friday? e) Interpret, as function of the available information, the information structure you built in the previous question for Wednesday. f) What is the distribution of the amount of money in the piggy bank Friday noon? g) Show that the rst time the piggy bank is empty is a stopping time. Exercice 2.2. Let (; F) be a measurable space such that Card () < 1 and equipped with the ltration F = ff t : t 2 f0; 1; :::gg. If the random variables 1 and 2 stopping times with respect to the ltration F, show that 1 _ 2 = max f 1 ; 2 g is also a stopping time. Exercice 2.3 You have 20 dollars and decide to play the following game of chance. You toss a coin four times. On the rst toss, you win 10 dollars if the result is tails and lose 10 dollars if the result is heads. On the subsequent tosses, if the result is identical to that of the previous toss, then you don t win nor lose anything. However, if the result is di erent from the previous toss, you win 10 dollars in the case this result is tails and lose 10 dollars in the case this result is heads. a) De ne the random variables and notation needed to model this situation. b) Give the -algebra re ecting the information available after the second toss. Interpret. c) What is the distribution of the amount of money you own at the end of the game? 1

2 Exercice 2.4 Gambler s ruin. Two gamblers initially own fortunes of r and n r dollars respectively (r and n are positive integers such that r < n). They decide to gamble until one of them is ruined. Say the second player represents the croupier and the rst player always decides of the bets. This rst player wins the amount of money he bets with probability p (0 < p < 1) or loses it with probability q = 1 p; independently from one round to another. The only amounts he can bet are multiples of one dollar and no loans are admitted. Let us study two popular strategies for this game: 1) the bold strategy by which the rst player always bets the minimum between his fortune and the amount required to win the whole game; 2) the shy strategy by which the rst player always bets one dollar. Let X t represent the rst players s fortune after the t-th round when he follows the shy strategy. Let Y t represent the fortune of the rst player after the t-th round when he follows the bold strategy. In this exercise, we assume each round is settled by the toss of a possibly poorly balanced coin. On any given round, the rst player wins if the result of the toss is tails. This happens with probability p. We consider the results of the rst four tosses only. The croupier starts with 4 dollars while the rst player starts with 6 dollars. a) What sample space corresponds to this situation? b) What -algebra should you use to build your measurable space? c) List the -algebras in the ltration generated by the process X = fx t : t 2 f0; 1; 2; 3; 4gg for times 0; 1; 2 et 4. d) List the -algebras in the ltration generated by the process Y = fy t : t 2 f0; 1; 2; 3; 4gg for times 0; 1; 2 et 4. e) Interpret, as function of the available information, the information structure you built in question c) for time t = 2. f) Give the mass functions of X 4 and Y 4. In the next question, we no longer consider only the rst four rounds, but we let the game follow its course until one of the gamblers is ruined. g) Let be a random variable representing the time at which the game stopped, meaning that (!) min ft 2 f0; 1; 2; :::g : X t = 0 or X t = ng : Show that is a stopping time. 2

3 Exercice 2.5. Let 1 and 2 be two (; F; F) stopping times with a sample space containing a nite number of elements, and with F = ff t : t 2 f0; 1; 2; :::gg, a ltration. Is the sum of those two stopping times also a stopping time? If so, show it. If not, give a counterexample. Exercice 2.6. You throw two dices. Let X be the result of the rst dice, and Y the result of the second dice. You receive a payment of amount max (X; Y ) at time = min (X; Y ). Let the process fs t : t 2 f1; 2; :::; 6gg represent the payments made to you throughout time. a) What is the sample space? b) What is the ltration ff t : t 2 f1; 2; :::; 6gg generated by the stochastic process fs t : t 2 f1; 2; :::; 6gg? c) Is a ff t : t 2 f1; 2; :::; 6gg stopping time? Justify your answer. 3

4 Solutions 1 Exercise 2.1 a) What is the sample space? = T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH; HT T T; HT T H; HT HT; HT HH; HHT T; HHT H; HHHT; HHHH b) De ne a stochastic process describing this situation, and explain its signi cation. Do not forget to specify the period of time you are using. Let t = 0 on Monday (today). 8t 2 f0; 1; 2; 3; 4g X t = money in the piggy bank after the transaction on day t happened c) What -algebra should you use to build your measurable space?! X 0 (!) X 1 (!) X 2 (!) X 3 (!) X 4 (!)! 1 = T T T T ! 2 = T T T H ! 3 = T T HT ! 4 = T T HH ! 5 = T HT T ! 6 = T HT H ! 7 = T HHT ! 8 = T HHH ! = HT T T ! 10 = HT T H ! 11 = HT HT ! 12 = HT HH ! 13 = HHT T ! 14 = HHT H ! 15 = HHHT ! 16 = HHHH The -algebra F must ensure that X 0, X 1, X 2, X 3 and X 4 are all random variables. Since all of the 16 trajectories are di erent from one another, F is generated by the partition 4

5 f! 1 ; :::;! 16 g. Hence F = set of all the events in plus?. d) List the -algebras in the ltration generated by the stochastic process for the following days: Monday, Tuesday, Wednesday and Friday? F 0 = f?; g F 1 = f?; A; A c ; g where A = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg F 2 = 8 >< >:?; B 1 ; B 2 ; B 3 ; B 4 ; B 1 [ B 2 ; B 1 [ B 3 ; B 1 [ B 4 ; B 2 [ B 3 ; B 2 [ B 4 ; B 3 [ B 4 ; B 1 [ B 2 [ B 3 ; B 1 [ B 2 [ B 4 ; B 1 [ B 3 [ B 4 ; B 2 [ B 3 [ B 4 ; where B 1 = ft T T T; T T T H; T T HT; T T HHg B 2 = ft HT T; T HT H; T HHT; T HHHg B 3 = fht T T; HT T H; HT HT; HT HHg B 4 = fhht T; HHT H; HHHT; HHHHg >= >; F 4 = F e) Interpret, as function of the available information, the information structure you built in the previous question for Wednesday. F 2 = 8 >< >:?; B 1 ; B 2 ; B 3 ; B 4 ; B 1 [ B 2 ; B 1 [ B 3 ; B 1 [ B 4 ; B 2 [ B 3 ; B 2 [ B 4 ; B 3 [ B 4 ; B 1 [ B 2 [ B 3 ; B 1 [ B 2 [ B 4 ; B 1 [ B 3 [ B 4 ; B 2 [ B 3 [ B 4 ; where B 1 = ft T T T; T T T H; T T HT; T T HHg B 2 = ft HT T; T HT H; T HHT; T HHHg B 3 = fht T T; HT T H; HT HT; HT HHg B 4 = fhht T; HHT H; HHHT; HHHHg >= >; 5

6 The atoms generating F 2 partition according to the results of the rst two coin tosses (those of Tuesday and Wednesday morning), however they cannot distinguish events involving tosses of the last two mornings (those of Thursday and Friday). f) What is the distribution of the amount of money in the piggy bank Friday noon? Say the result is tails with probability p and heads with probability 1 p: Note this means the same coin is used all week long.! P (!) X 4 (!) T T T T p 4 0 T T T H p 3 (1 p) 1 T T HT p 3 (1 p) 0 T T HH p 2 (1 p) 2 2 T HT T p 3 (1 p) 0 T HT H p 2 (1 p) 2 1 T HHT p 2 (1 p) 2 1 T HHH p (1 p) 3 3! P (!) X 4 (!) HT T T p 3 (1 p) 0 HT T H p 2 (1 p) 2 1 HT HT p 2 (1 p) 2 1 HT HH p (1 p) 3 3 HHT T p 2 (1 p) 2 1 HHT H p (1 p) 3 3 HHHT p (1 p) 3 3 HHHH (1 p) 4 5 The density function is : x f X (x) = P f! 2 : X (!) = xg x P! 2 : X (!) = x p = p 4 + 3p 3 (1 p) = 2p 4 + 3p 3 1 p 3 (1 p) + 5p 2 (1 p) 2 = +4p 4 p 3 + 5p 2 2 p 2 (1 p) 2 = p 4 2p 3 + p 2 3 4p (1 p) 3 = 4p 12p p 3 4p (1 p) 4 = 1 4p + 6p 2 4p 3 + p = = =

7 The cumulative distribution function is : x F X (x) = P f! 2 : X (!) xg 0 p 4 + 3p 3 (1 p) = 2p 4 + 3p 3 1 p 4 + 4p 3 (1 p) + 5p 2 (1 p) 2 = 2p 4 6p 3 + 5p 2 2 p 4 + 4p 3 (1 p) + 6p 2 (1 p) 2 = 3p 4 8p 3 + 6p 2 3 p 4 + 4p 3 (1 p) + 6p 2 (1 p) 2 + 4p (1 p) 3 = p 4 + 4p 3 6p 2 + 4p 4 p 4 + 4p 3 6p 2 + 4p 5 p 4 + 4p 3 (1 p) + 6p 2 (1 p) 2 + 4p (1 p) 3 + (1 p) 4 = 1 g) Show that the rst time the piggy bank is empty is a stopping time.! X 0 (!) X 1 (!) X 2 (!) X 3 (!) X 4 (!) (!)! 1 = T T T T ! 2 = T T T H ! 3 = T T HT ! 4 = T T HH ! 5 = T HT T ! 6 = T HT H ! 7 = T HHT ! 8 = T HHH ! = HT T T ! 10 = HT T H ! 11 = HT HT ! 12 = HT HH ! 13 = HHT T ! 14 = HHT H ! 15 = HHHT ! 16 = HHHH

8 f! 2 : (!) = 0g =? 2 F 0 f! 2 : (!) = 1g = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg 2 F 1 f! 2 : (!) = 2g =? 2 F 2 f! 2 : (!) = 3g = fht T T; HT T Hg 2 F 3 f! 2 : (!) = 4g =? 2 F 4 : 2 Exercise 2.2 Since 8k 2 f0; 1; :::g ; f! 2 : 1 (!) _ 2 (!) kg = f! 2 : 1 (!) k et 2 (!) kg = f! 2 : 1 (!) kg \ f! 2 : 2 (!) kg 2 F k ; 2F k 2F k 8

9 then 8t 2 f0; 1; :::g ; f! 2 : 1 (!) _ 2 (!) = tg = f! 2 : t 1 < 1 (!) _ 2 (!) tg = f! 2 : 1 (!) _ 2 (!) tg \ f! 2 : 1 (!) _ 2 (!) > t 1g = f! 2 : 1 (!) _ 2 (!) tg \ f! 2 : 1 (!) _ 2 (!) t 1g 2F t 2F t 1 F t 2 F t : 3 Exercise 2.3 a) De ne the random variables and notation needed to model this situation. The sample space contains 16 elements, all listed in the table below. For i 2 f0; 1; 2; 3; 4g, the random variable X i represents the amount you own after the i-th toss. 2F t 1 F t c! X 0 X 1 X 2 X 3 X 4 tttt ttth ttht tthh thtt thth thht thhh ! X 0 X 1 X 2 X 3 X 4 httt htth htht hthh hhtt hhth hhht hhhh b) Give the -algebra re ecting the information available after the second toss. Interpret. ftttt; ttth; ttht; tthhg ; fthtt; thth; thht; thhhg ; F 2 = : fhttt; htth; htht; hthhg ; fhhtt; hhth; hhht; hhhhg If we observe the stochastic process X until the second toss, we can establish which results were obtained on the rst two tosses, but we are unable to determine the results of the last two tosses.

10 c) What is the distribution of the amount of money you own at the end of the game? Let q be the probability to obtain tails on a given toss. Then P (X 4 = 30) = q 4 + 2q 3 (1 q) + q 2 (1 q) 2 = q 2 P (X 4 = 20) = 2q 3 (1 q) + 4q 2 (1 q) 2 + 2q (1 q) 3 = 2q 2 + 2q = 2q (1 q) 4 Exercise 2.4 P (X 4 = 10) = q 2 (1 q) 2 + 2q (1 q) 3 + (1 q) 4 = 1 2q + q 2 = (1 q) 2 a) What sample space corresponds to this situation? T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH; = HT T T; HT T H; HT HT; HT HH; HHT T; HHT H; HHHT; HHHH In the case the bold strategy is used, one might also describe the sample space by ft; HT T T; HT T H; HT H; HHg since the game can stop only after a few rounds. It s a question of interpretation. We may assume that the coin will be tossed four times, even if ruined has occurred and no bet is on the table. This way we can build the two stochastic processes on the same measurable space. 10

11 b) What -algebra should you use to build your measurable space?! X 0 X 1 X 2 X 3 X 4 Y 0 Y 1 Y 2 Y 3 Y 4! 1 = T T T T ! 2 = T T T H ! 3 = T T HT ! 4 = T T HH ! 5 = T HT T ! 6 = T HT H ! 7 = T HHT ! 8 = T HHH ! = HT T T ! 10 = HT T H ! 11 = HT HT ! 12 = HT HH ! 13 = HHT T ! 14 = HHT H ! 15 = HHHT ! 16 = HHHH The -algebra F must ensure that X 0, X 1, X 2, X 3 and X 4 as well as Y 0, Y 1, Y 2, Y 3 and Y 4 are random variables. Since the 16 trajectories are di erent from one another, F is generated by the partition f! 1 ; :::;! 16 g. Hence Note F contains 2 16 = events. F = set of every event in plus?. c) List the -algebras in the ltration generated by the process X = fx t : t 2 f0; 1; 2; 3; 4gg 11

12 for times 0; 1; 2 and 4. F 0 = f?; g F 1 = f?; A; A c ; g where A = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg F 2 = 8 >< >:?; B 1 ; B 2 ; B 3 ; B 4 ; B 1 [ B 2 ; B 1 [ B 3 ; B 1 [ B 4 ; B 2 [ B 3 ; B 2 [ B 4 ; B 3 [ B 4 ; B 1 [ B 2 [ B 3 ; B 1 [ B 2 [ B 4 ; B 1 [ B 3 [ B 4 ; B 2 [ B 3 [ B 4 ; where B 1 = ft T T T; T T T H; T T HT; T T HHg B 2 = ft HT T; T HT H; T HHT; T HHHg B 3 = fht T T; HT T H; HT HT; HT HHg B 4 = fhht T; HHT H; HHHT; HHHHg >= >; F 4 = F d) List the -algebras in the ltration generated by the process Y = fy t : t 2 f0; 1; 2; 3; 4gg 12

13 for times 0; 1; 2 and 4. G 0 = f?; g G 1 = f?; A; A c ; g where A = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg G 2 = f?; B 1 ; B 3 ; B 4 ; B 1 [ B 3 ; B 1 [ B 4 ; B 3 [ B 4 ; B 1 [ B 3 [ B 4 ; g where B 1 = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg B 3 = fht T T; HT T H; HT HT; HT HHg B 4 = fhht T; HHT H; HHHT; HHHHg G 4 = 8 >< >:?; C 1 ; C 2 ; C 3 ; C 4 ; C 5 ; C 1 [ C 2 ; C 1 [ C 3 ; C 1 [ C 4 ; C 1 [ C 5 ; C 2 [ C 3 ; C 2 [ C 4 ; C 2 [ C 5 ; C 3 [ C 4 ; C 3 [ C 5 ; C 4 [ C 5 ; C 3 [ C 4 [ C 5 ; C 2 [ C 4 [ C 5 ; C 2 [ C 3 [ C 5 ; C 2 [ C 3 [ C 4 ; C 1 [ C 4 [ C 5 ; C 1 [ C 3 [ C 5 ; C 1 [ C 3 [ C 4 ; C 1 [ C 2 [ C 5 ; C 1 [ C 2 [ C 4 ; C 1 [ C 2 [ C 3 ; C 1 [ C 2 [ C 3 [ C 4 ; C 1 [ C 2 [ C 3 [ C 5 ; C 1 [ C 2 [ C 4 [ C 5 ; C 1 [ C 3 [ C 4 [ C 5 ; C 2 [ C 3 [ C 4 [ C 5 ; C 1 = ft T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHHg C 2 = fht T T g C 3 = fht T Hg C 4 = fht HT; HT HHg C 5 = fhht T; HHT H; HHHT; HHHHg >= >; e) Interpret, as function of the available information, the information structure you built in question c) for time t=3. 8?; B 1 ; B 2 ; B 3 ; B 4 ; B 1 [ B 2 ; B 1 [ B 3 ; >< >= B F 2 = 1 [ B 4 ; B 2 [ B 3 ; B 2 [ B 4 ; B 3 [ B 4 ; B 1 [ B 2 [ B 3 ; B 1 [ B 2 [ B 4 ; >: >; B 1 [ B 3 [ B 4 ; B 2 [ B 3 [ B 4 ; where B 1 = ft T T T; T T T H; T T HT; T T HHg B 2 = ft HT T; T HT H; T HHT; T HHHg B 3 = fht T T; HT T H; HT HT; HT HHg B 4 = fhht T; HHT H; HHHT; HHHHg 13

14 The atoms generating F 2 partition according to the results of the rst two coin tosses, however they cannot distinguish events involving the last two tosses. f) Give the density functions of X 4 and Y 4.! X 0 X 1 X 2 X 3 X 4 Y 0 Y 1 Y 2 Y 3 Y 4 P! 1 = T T T T p 4! 2 = T T T H p 3 (1 p)! 3 = T T HT p 3 (1 p)! 4 = T T HH p 2 (1 p) 2! 5 = T HT T p 3 (1 p)! 6 = T HT H p 2 (1 p) 2! 7 = T HHT p 2 (1 p) 2! 8 = T HHH p (1 p) 3! = HT T T p 3 (1 p)! 10 = HT T H p 2 (1 p) 2! 11 = HT HT p 2 (1 p) 2! 12 = HT HH p (1 p) 3! 13 = HHT T p 2 (1 p) 2! 14 = HHT H p (1 p) 3! 15 = HHHT p (1 p) 3! 16 = HHHH (1 p) 4 8 p 2 + p 3 p + 1 if y = 0 >< f Y4 (y) = >: p 2 2p 3 + p 4 = p 2 (1 p) 2 if y = 6 p 4 + p 3 + p if y = 10 0 otherwise >= >; 14

15 8 >< f X4 (y) = >: (1 p) 4 if x = 2 4p (1 p) 3 if x = 4 6p 2 (1 p) 2 if x = 6 4p 3 (1 p) if x = 8 p 4 if x = 10 0 otherwise >= >; g) Let be a random variable representing the time at which the game stopped, meaning that (!) min ft 2 f0; 1; 2; :::g : X t = 0 or X t = ng : Show that is a stopping time. We must show that for every t 2 f0; 1; 2; :::g, the event f! 2 : (!) = tg 2 F t. But, f! 2 : (!) = tg = f! 2 : X 0 (!) ; X 1 (!) ; :::; X t 1 (!) 2 f1; 2; :::; n 1g and X t (!) 2 f0; ngg = t\ 1 k=0 2 F t f! 2 : X k (!) 2 f1; 2; :::; n 1gg 2F k F t 2F t \ f! 2 : X t (!) 2 f0; ngg 2F t A simpler solution would be as follows: is the time of rst contact with the set f0; ng, therefore by a theorem from the course (see end of chapter 2), it is a stopping time. 5 Exercise 2.5 Yes, the sum of those two stopping times is a stopping time, because t 2 f0; 1; 2; :::g. Indeed, let t 2 f0; 1; 2; :::g be arbitrary. Then = f! 2 : = tg 8 t[ >< >: f! 2 : 1 = kg k=0 >= \ f! 2 : 2 = t kg 2 F t : >; 2F k F t 2F t k F t 2F t 15

16 Which is what had to be shown. 16

Conditional Expectation and Independence Exercises

Conditional Expectation and Independence Exercises Exercise 3.1. Show the four following properties : for any random variables and Y and for any real numbers a and b, (E1) E P [a + by ] = ae P [] + be