a) What is the sample space ?

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 f 0; 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 ?

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 = f X _t : t 2 f 0; 1; 2; 3; 4 gg for times 0; 1; 2 et 4.

d) List the -algebras in the …ltration generated by the process Y = f Y _t : t 2 f 0; 1; 2; 3; 4 gg 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 f t 2 f 0; 1; 2; ::: g : X _t = 0 or X _t = n g :

Show that is a stopping time.

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 f 0; 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 f S t : t 2 f 1; 2; :::; 6 gg represent the payments made to you throughout time.

a) What is the sample space ?

b) What is the …ltration fF ^t : t 2 f 1; 2; :::; 6 gg generated by the stochastic process f S t : t 2 f 1; 2; :::; 6 gg ?

c) Is a fF ^t : t 2 f 1; 2; :::; 6 gg stopping time ? Justify your answer.

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). 8 t 2 f 0; 1; 2; 3; 4 g

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 1 0 0 0 0

! ₂ = T T T H 1 0 0 0 1

! ₃ = T T HT 1 0 0 1 0

! ₄ = T T HH 1 0 0 1 2

! ₅ = T HT T 1 0 1 0 0

! ₆ = T HT H 1 0 1 0 1

! ₇ = T HHT 1 0 1 2 1

! ₈ = T HHH 1 0 1 2 3

! ₉ = HT T T 1 2 1 0 0

! ₁₀ = HT T H 1 2 1 0 1

! ₁₁ = HT HT 1 2 1 2 1

! ₁₂ = HT HH 1 2 1 2 3

! ₁₃ = HHT T 1 2 3 2 1

! ₁₄ = HHT H 1 2 3 2 3

! ₁₅ = HHHT 1 2 3 4 3

! ₁₆ = HHHH 1 2 3 4 5

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

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 = f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g

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 ;

9 >

> =

> >

; where B ₁ = f T T T T; T T T H; T T HT; T T HH g

B ₂ = f T HT T; T HT H; T HHT; T HHH g B ₃ = f HT T T; HT T H; HT HT; HT HH g B ₄ = f HHT T; HHT H; HHHT; HHHH g 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 ;

9 >

> =

> >

; where B ₁ = f T T T T; T T T H; T T HT; T T HH g

B ₂ = f T HT T; T HT H; T HHT; T HHH g

B ₃ = f HT T T; HT T H; HT HT; HT HH g

B ₄ = f HHT T; HHT H; HHHT; HHHH g

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 (!) = x g 0 p ⁴ + 3p ³ (1 p) = 2p ⁴ + 3p ³

1 p ³ (1 p) + 5p ² (1 p) ² = +4p ⁴ 9p ³ + 5p ² 2 p ² (1 p) ² = p ⁴ 2p ³ + p ²

3 4p (1 p) ³ = 4p 12p ² + 12p ³ 4p ⁴

4 0

5 (1 p) ⁴ = 1 4p + 6p ² 4p ³ + p ⁴

x P ! 2 : X (!) = x p = ¹ ₂ 0 ₁₆ ⁴ = ¹ ₄

1 ₁₆ ⁶ = ³ ₈

2 ₁₆ ¹

3 ₁₆ ⁴ = ¹ ₄

4 0

5 ₁₆ ¹

The cumulative distribution function is :

x F _X (x) = P f ! 2 : X (!) x g 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 1 0 0 0 0 1

! ₂ = T T T H 1 0 0 0 1 1

! 3 = T T HT 1 0 0 1 0 1

! ₄ = T T HH 1 0 0 1 2 1

! ₅ = T HT T 1 0 1 0 0 1

! 6 = T HT H 1 0 1 0 1 1

! ₇ = T HHT 1 0 1 2 1 1

! ₈ = T HHH 1 0 1 2 3 1

! 9 = HT T T 1 2 1 0 0 3

! ₁₀ = HT T H 1 2 1 0 1 3

! ₁₁ = HT HT 1 2 1 2 1 1

! ₁₂ = HT HH 1 2 1 2 3 1

! ₁₃ = HHT T 1 2 3 2 1 1

! ₁₄ = HHT H 1 2 3 2 3 1

! ₁₅ = HHHT 1 2 3 4 3 1

! ₁₆ = HHHH 1 2 3 4 5 1

f ! 2 : (!) = 0 g

= ? 2 F ⁰

f ! 2 : (!) = 1 g

= f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g 2 F ¹ f ! 2 : (!) = 2 g

= ? 2 F ²

f ! 2 : (!) = 3 g

= f HT T T; HT T H g 2 F ³ f ! 2 : (!) = 4 g

= ? 2 F ⁴ :

2 Exercise 2.2

Since 8 k 2 f 0; 1; ::: g ;

f ! 2 : ₁ (!) _ ² (!) k g

= f ! 2 : ₁ (!) k et 2 (!) k g

= f | ! 2 : {z ₁ (!) k g }

2F

\ f | ! 2 : {z ₂ (!) k g }

2F

2 F ^k ;

then 8 t 2 f 0; 1; ::: g ;

f ! 2 : ₁ (!) _ ² (!) = t g

= f ! 2 : t 1 < ₁ (!) _ ² (!) t g

= f ! 2 : ₁ (!) _ ² (!) t g \ f ! 2 : ₁ (!) _ ² (!) > t 1 g

= f | ! 2 : ₁ (!) {z _ ² (!) t g }

2F

\ f | ! 2 : ₁ (!) {z _ ² (!) t 1 g }

2F

F

c

| {z }

2F

F

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 f 0; 1; 2; 3; 4 g , the random variable X _i represents the amount you own after the i-th toss.

! X ₀ X ₁ X ₂ X ₃ X ₄ tttt 20 30 30 30 30 ttth 20 30 30 30 20 ttht 20 30 30 20 30 tthh 20 30 30 20 20 thtt 20 30 20 30 30 thth 20 30 20 30 20 thht 20 30 20 20 30 thhh 20 30 20 20 20

! X ₀ X ₁ X ₂ X ₃ X ₄ httt 20 10 20 20 20 htth 20 10 20 20 10 htht 20 10 20 10 20 hthh 20 10 20 10 10 hhtt 20 10 10 20 20 hhth 20 10 10 20 10 hhht 20 10 10 10 20 hhhh 20 10 10 10 10

b) Give the -algebra re‡ecting the information available after the second toss. Interpret.

F ² = f tttt; ttth; ttht; tthh g ; f thtt; thth; thht; thhh g ; f httt; htth; htht; hthh g ; f hhtt; hhth; hhht; hhhh g :

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.

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)

P (X ₄ = 10) = q ² (1 q) ² + 2q (1 q) ³ + (1 q) ⁴

= 1 2q + q ² = (1 q) ²

4 Exercise 2.4

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 f T; HT T T; HT T H; HT H; HH g

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.

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 6 7 8 9 10 6 10 10 10 10

! ₂ = T T T H 6 7 8 9 8 6 10 10 10 10

! ₃ = T T HT 6 7 8 7 8 6 10 10 10 10

! ₄ = T T HH 6 7 8 7 6 6 10 10 10 10

! ₅ = T HT T 6 7 6 7 8 6 10 10 10 10

! ₆ = T HT H 6 7 6 7 6 6 10 10 10 10

! ₇ = T HHT 6 7 6 5 6 6 10 10 10 10

! ₈ = T HHH 6 7 6 5 4 6 10 10 10 10

! ₉ = HT T T 6 5 6 7 8 6 2 4 8 10

! ₁₀ = HT T H 6 5 6 7 6 6 2 4 8 6

! ₁₁ = HT HT 6 5 6 5 6 6 2 4 0 0

! ₁₂ = HT HH 6 5 6 5 4 6 2 4 0 0

! ₁₃ = HHT T 6 5 4 5 6 6 2 0 0 0

! ₁₄ = HHT H 6 5 4 5 4 6 2 0 0 0

! ₁₅ = HHHT 6 5 4 3 4 6 2 0 0 0

! ₁₆ = HHHH 6 5 4 3 2 6 2 0 0 0

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

F = set of every event in plus ? . Note F contains 2 ¹⁶ = 65 536 events.

c) List the -algebras in the …ltration generated by the process X = f X _t : t 2 f 0; 1; 2; 3; 4 gg

for times 0; 1; 2 and 4 . F ⁰ = f? ; g F ¹ = f? ; A; A ^c ; g

where A = f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g

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 ;

9 >

> =

> >

; where B ₁ = f T T T T; T T T H; T T HT; T T HH g

B ₂ = f T HT T; T HT H; T HHT; T HHH g B ₃ = f HT T T; HT T H; HT HT; HT HH g B ₄ = f HHT T; HHT H; HHHT; HHHH g F ⁴ = F

d) List the -algebras in the …ltration generated by the process Y = f Y _t : t 2 f 0; 1; 2; 3; 4 gg

for times 0; 1; 2 and 4 . G ⁰ = f? ; g G ¹ = f? ; A; A ^c ; g

where A = f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g 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 = f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g B ₃ = f HT T T; HT T H; HT HT; HT HH g

B ₄ = f HHT T; HHT H; HHHT; HHHH g

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 ;

9 >

> >

> >

> >

> >

> =

> >

> >

> >

> >

> >

; C ₁ = f T T T T; T T T H; T T HT; T T HH; T HT T; T HT H; T HHT; T HHH g C 2 = f HT T T g

C ₃ = f HT T H g

C ₄ = f HT HT; HT HH g

C ₅ = f HHT T; HHT H; HHHT; HHHH g

e) Interpret, as function of the available information, the information structure you built in question c) for time t=3.

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 ;

9 >

> =

> >

; where B ₁ = f T T T T; T T T H; T T HT; T T HH g

B ₂ = f T HT T; T HT H; T HHT; T HHH g

B ₃ = f HT T T; HT T H; HT HT; HT HH g

B ₄ = f HHT T; HHT H; HHHT; HHHH g

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 6 7 8 9 10 6 10 10 10 10 p ⁴

! ₂ = T T T H 6 7 8 9 8 6 10 10 10 10 p ³ (1 p)

! ₃ = T T HT 6 7 8 7 8 6 10 10 10 10 p ³ (1 p)

! ₄ = T T HH 6 7 8 7 6 6 10 10 10 10 p ² (1 p) ²

! 5 = T HT T 6 7 6 7 8 6 10 10 10 10 p ³ (1 p)

! ₆ = T HT H 6 7 6 7 6 6 10 10 10 10 p ² (1 p) ²

! ₇ = T HHT 6 7 6 5 6 6 10 10 10 10 p ² (1 p) ²

! ₈ = T HHH 6 7 6 5 4 6 10 10 10 10 p (1 p) ³

! ₉ = HT T T 6 5 6 7 8 6 2 4 8 10 p ³ (1 p)

! ₁₀ = HT T H 6 5 6 7 6 6 2 4 8 6 p ² (1 p) ²

! ₁₁ = HT HT 6 5 6 5 6 6 2 4 0 0 p ² (1 p) ²

! ₁₂ = HT HH 6 5 6 5 4 6 2 4 0 0 p (1 p) ³

! ₁₃ = HHT T 6 5 4 5 6 6 2 0 0 0 p ² (1 p) ²

! ₁₄ = HHT H 6 5 4 5 4 6 2 0 0 0 p (1 p) ³

! 15 = HHHT 6 5 4 3 4 6 2 0 0 0 p (1 p) ³

! ₁₆ = HHHH 6 5 4 3 2 6 2 0 0 0 (1 p) ⁴

f _Y

(y) = 8 >

> >

> >

> >

> <

> >

> >

> >

> >

:

p ² + p ³ p + 1 if y = 0 p ² 2p ³ + p ⁴ = p ² (1 p) ² if y = 6 p ⁴ + p ³ + p if y = 10

0 otherwise

9 >

> >

> >

> >

> =

> >

> >

> >

> >

;

f _X

(y) = 8 >

> >

> >

> >

> >

> >

> >

> >

> <

> >

> >

> >

> >

> >

> >

> >

> >

:

(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

9 >

> >

> >

> >

> >

> >

> >

> >

> =

> >

> >

> >

> >

> >

> >

> >

> >

;

g) Let be a random variable representing the time at which the game stopped, meaning that

(!) min f t 2 f 0; 1; 2; ::: g : X _t = 0 or X _t = n g : Show that is a stopping time.

We must show that for every t 2 f 0; 1; 2; ::: g , the event f ! 2 : (!) = t g 2 F ^t . But, f ! 2 : (!) = t g

= f ! 2 : X ₀ (!) ; X ₁ (!) ; :::; X _{t 1} (!) 2 f 1; 2; :::; n 1 g and X _t (!) 2 f 0; n gg

=

t \ 1 k=0

f ! 2 : X _k (!) 2 f 1; 2; :::; n 1 gg

| {z }

2F

F

| {z }

2F

\ f | ! 2 : X _t {z (!) 2 f 0; n gg }

2F

2 F ^t

A simpler solution would be as follows: is the time of …rst contact with the set f 0; n g , 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 f 0; 1; 2; ::: g . Indeed, let t 2 f 0; 1; 2; ::: g be arbitrary. Then

f ! 2 : ₁ + ₂ = t g

= [ t k=0

8 >

<

> : f | ! 2 {z : ₁ = k g }

2F

F

\ f | ! 2 : {z ₂ = t k g }

2F

F

9 >

=

> ;

| {z }

2 F ^t :

Which is what had to be shown.