(E4) If X and Y are independent then E P [XY ] = E P [X] E P [Y ] : Exercise 3.2. Show that for any random variable X and any real numbers a and b,

Conditional Expectation and Independence Exercises

Exercise 3.1. Show the four following properties : for any random variables X and Y and for any real numbers a and b,

(E1) E ^P [aX + bY ] = aE ^P [X] + bE ^P [Y ] ;

(E2) If 8 ! 2 , X (!) Y (!) then E ^P [X] E ^P [Y ] ; (E3) In general, E ^P [XY ] 6 = E ^P [X] E ^P [Y ] ;

(E4) If X and Y are independent then E ^P [XY ] = E ^P [X] E ^P [Y ] : Exercise 3.2. Show that for any random variable X and any real numbers a and b,

(V 1) Var ^P [X] 0;

(V 2) Var ^P [X] = E ^P [X ² ] E ^P [X] ² ; (V 3) 8 a 2 R ; Var ^P [aX + b] = a ² Var ^P [X] :

(V 4) Var ^P [X + Y ] = Var ^P [X] + Var ^P [Y ] + 2Cov ^P [X; Y ]

Exercise 3.3. Show that for any random variables X and Y and for any real numbers a and b,

(C1) Cov ^P [X; Y ] = E ^P [XY ] E ^P [X] E ^P [Y ] ; (C2) If X and Y are independent then Cov ^P [X; Y ] = 0;

(C3) 8 a; b 2 R ; Cov ^P [aX ₁ + bX ₂ ; Y ] = aCov ^P [X ₁ ; Y ] + bCov ^P [X ₂ ; Y ] :

Exercise 3.4. Let us revisit a problem treated in chapter 2. 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. What is the conditional expectation of the amount of money in the bank Friday noon, given the information by Wednesday noon?

Exercise 3.5. Recall exercise 2.4 on the gambler’s ruin. Calculate E [X ₄ j F ⁰ ] ; E [ X ₄ j F ² ] ; E [ Y ₄ j G ² ] and E [Y ₂ j G ⁴ ] for an arbitrary p and for the speci…c case p = ¹ ₂ .

Exercise 3.6. Justify all your answers.

Let the stochastic processes X = f X _t : t 2 f 0; 1; 2; 3; 4 gg represent the price of a share throughout time. We consider a time interval of 6 months.

X ₀ (!) X ₁ (!) X ₂ (!) X ₃ (!) X ₄ (!) P (!)

! ₁ 11 13 15 16 16 0; 06

! ₂ 11 13 15 16 14 0; 065

! ₃ 11 13 15 15 16 0; 06

! ₄ 11 13 15 15 14 0; 065

! ₅ 11 13 15 14 16 0; 06

! ₆ 11 13 15 14 14 0; 065

! ₇ 11 13 12 13 14 0; 0625

! ₈ 11 13 12 13 18 0; 08

! ₉ 11 13 12 13 12 0; 04

! ₁₀ 11 13 12 13 11 0; 0675

! ₁₁ 11 11 12 13 14 0; 0625

! ₁₂ 11 11 12 13 12 0; 0625

! ₁₃ 11 11 12 12 12 0; 0625

! ₁₄ 11 11 11 14 20 0; 0625

! ₁₅ 11 11 11 14 12 0; 0625

! ₁₆ 11 11 11 12 12 0; 0625

a) What …ltration is generated by the process X ? b) Give the conditional distribution of X ₄ , knowing X ₂ . c) Calculate the conditional expectation of X ₄ , knowing X ₂ . d) Give the conditional distribution of X ₄ , knowing F ² . e) Calculate the conditional expectation of X ₄ , knowing F ²

Problem 3.7. Let us remain in the context of problem 3.6. Two investors, call them A

at the end of the next year (t = 4). If an investor chooses to sell his share at time t = 2, he will then place the money in a bank account yielding 15% in interest per year (the interest is compounded annually at times t = 2 and t = 4). If instead he chooses not to sell the share at time t = 2, he will have to sell it at time t = 4.

Those two investors will be away for the next …ve years because they are doing a (possibly metaphorical, this is up to you) trip to Venus. Hence they both appoint prosecutors (talented and trustworthy) to manage the operations in their place. But they’re out of luck: both prosecutors are unfamiliar with money issues (we said talented ?), therefore the two investors must give precise instructions today to their respective prosecutors on what to do in the next year, depending on how the share will have evolved during that period. It goes without saying that both investors aim at maximizing their pro…t. The prosecutor of investor A will check the price of the stock at time t = 1, but the prosecutor of investor B will not (he might not be the most suited candidate for this job, don’t you think?). Said otherwise, observation times for prosecutor A are t = 1 and t = 2, while for prosecutor B the only observation time is t = 2:

a) What instructions does investor A give to his prosecutor ? b) What instructions does investor B give to his prosecutor ?

To facilitate comparisons, use the holding period yield (HPY). The HPY of a share on a given time interval (say we buy the share at the beginning of period t ₁ and sell it at the end of period t ₂ ) is

R _t

_;t

= X t

X _t

:

Problem 3.8. This problem uses notations de…ned in problems 3.6 and 3.7.

a) Let A and B represent the random times at which the shares are sold, respectively by prosecutors A and B. Are A and B stopping times ?

b) Say we replace the probability measure P by Q where Q (!) = ₁₆ ¹ for every ! 2 . Are A and B stopping times in this case ?

Problem 3.9. Let X ₁ ; X ₂ ; X ₃ be independent random variables such that P (X _i = 1) = P (X _i = 1) = ¹ ₂ : Let us de…ne A ₁ = f X ₂ = X ₃ g ; A ₂ = f X ₁ = X ₃ g and A ₃ = f X ₁ = X ₂ g . Show that the events A 1 ; A 2 and A 3 are pairwise independent, but not mutually independent.

Problem 3.10. Let X and Y be two independent random variables, both of standard normal distribution. Let " be a random variable independent of Y and such that: P f " = 1 g = P f " = 1 g = ¹ ₂ :

a) Show that Z = "Y is also normal but that Y + Z is not.

b) Show that Y and Z are not independent, although Cov(Z; Y ) = 0.

Problem 3.11. In the case X and Y are independent and identically distributed random

Solutions

1 Exercise 3.1

Proof of (E 1) E [aX + bY ] = aE [X] + bE [Y ].

Since

X

y 2S

f _X;Y (x; y) = X

y 2S

P [X = x and Y = y]

= P

"

[

y 2S

f X = x and Y = y g

#

because those events are disjoint

= P

"

[

y 2S

ff X = x g \ f Y = y gg

#

= P

"

f X = x g \

( [

y 2S

f Y = y g )#

= P [ f X = x g \ ] = P [ f X = x g ] = f _X (x) (1) hence

E [aX + bY ] = X

x 2S

X

y 2S

(ax + by) f _X;Y (x; y)

= X

x 2S

X

y 2S

axf X;Y (x; y) + X

y 2S

X

x 2S

byf X;Y (x; y)

= a X

x 2S

x X

y 2S

f _X;Y (x; y) + b X

y 2S

y X

x 2S

f _X;Y (x; y )

= a X

x 2S

xf _X (x) + b X

y 2S

yf _Y (y)

= aE [X] + bE [Y ] .

To begin with, note that if W is a non negative random variable then E [W ] 0. Indeed, E [W ] = X

x 2S

|{z} x

0

f _W (x)

| {z }

0

0:

Let W = Y X. Since X Y , then W 0. Hence E [W ] 0:

But using (E1), we have that

E [W ] = E [Y X] = E [Y ] E [X] : Therefore,

0 E [Y ] E [X] , E [X] E [Y ] :

2 Exercise 3.3

Proof of (C1) Cov [X; Y ] = E [XY ] E [X] E [Y ].

Cov [X; Y ]

= X

x 2S

X

y 2S

(x E [X]) (y E [Y ]) f _X;Y (x; y)

= X

x 2S

X

y 2S

(xy xE [Y ] yE [X] + E [X] E [Y ]) f _X;Y (x; y)

= X

x 2S

X

y 2S

xy f _X;Y (x; y ) E [Y ] X

x 2S

x X

y 2S

f _X;Y (x; y) E [X] X

y 2S

y X

x 2S

f _X;Y (x; y) + E [X] E [Y ] X

x 2S

X

y 2S

f _X;Y (x; y)

= X

x 2S

X

y 2S

xy f _X;Y (x; y ) E [Y ] X

x 2S

x f _X (x) E [X] X

y 2S

yf _Y (y) + E [X] E [Y ] X

x 2S

f _X (x)

| {z }

=1 by de…nition of a

density function

= E [XY ] E [Y ] E [X] E [X] E [Y ] + E [X] E [Y ]

= E [XY ] E [X] E [Y ] .

Proof of ^(C2) If X and Y are independent then Cov [X; Y ] = 0.

Since independence of X and Y implies E [XY ] = E [X] E [Y ], to obtain the result we only have to invoke (C1) :

Cov [X; Y ] = E [XY ] E [X] E [Y ]

= E [X] E [Y ] E [X] E [Y ]

= 0.

Proof of (C3) Cov [a 1 X 1 + a 2 X 2 ; Y ] = a 1 Cov [X 1 ; Y ] + a 2 Cov [X 2 ; Y ].

By (C1), we have that

Cov [a ₁ X ₁ + a ₂ X ₂ ; Y ]

= E [(a ₁ X ₁ + a ₂ X ₂ ) Y ] E [a ₁ X ₁ + a ₂ X ₂ ] E [Y ]

= E [a ₁ X ₁ Y + a ₂ X ₂ Y ] (a ₁ E [X ₁ ] + a ₂ E [X ₂ ]) E [Y ] by (E1)

= a ₁ E [X ₁ Y ] + a ₂ E [X ₂ Y ] a ₁ E [X ₁ ] E [Y ] + a ₂ E [X ₂ ] E [Y ] by (E1)

= a ₁ (E [X ₁ Y ] E [X ₁ ] E [Y ]) + a ₂ (E [X ₂ Y ] E [X ₂ ] E [Y ])

= a ₁ Cov [X ₁ ; Y ] + a ₂ Cov [X ₂ ; Y ]

Proof of (C4) V ar [X + Y ] = V ar [X] + V ar [Y ] + 2Cov [X; Y ].

V ar [X + Y ]

= E (X + Y ) ² (E [X + Y ]) ² by (V 2)

= E X ² + 2XY + Y ² (E [X] + E [Y ]) ² by (E1)

= E X ² + 2E [XY ] + E Y ² (E [X]) ² 2E [X] E [Y ] (E [Y ]) ²

= E X ² (E [X]) ² + E Y ² (E [Y ]) ² + 2 (E [XY ] E [X] E [Y ])

= V ar [X] + V ar [Y ] + Cov [X; Y ] using (V 2) and (C1).

3 Exercise 3.4

The blocks of F ² are

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 : Since

! 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

then

P (B _i )

B ₁ p ⁴ + 2p ³ (1 p) + p ² (1 p) ² = p ²

B ₂ p ³ (1 p) + 2p ² (1 p) ² + p (1 p) ³ = p p ² = p (1 p) B ₃ p ³ (1 p) + 2p ² (1 p) ² + p (1 p) ³ = p p ² = p (1 p) B ₄ p ² (1 p) ² + 2p (1 p) ³ + (1 p) ⁴ = 1 2p + p ² = (1 p) ² Recall that if ! 2 B _i , then

E ^P [X ₄ jF ² ] (!) = 1 P (B _i )

X

! 2 B

X (! ) P (! )

Therefore,

If ! 2 B ₁ then E ^P [X ₄ jF ² ] (!)

= 1

P (B ₁ ) X

! 2 B

X (! ) P (! )

= 1

p ² 0p ⁴ + 1p ³ (1 p) + 0p ³ (1 p) + 2p ² (1 p) ²

= (1 p) (2 p) If ! 2 B ₂ then E ^P [X ₄ jF ² ] (!)

= 1

P (B ₂ ) X

! 2 B

X (! ) P (! )

= 1

p (1 p) 0p ³ (1 p) + 1p ² (1 p) ² + 1p ² (1 p) ² + 3p (1 p) ³

= (3 p) (1 p) If ! 2 B ₃ then E ^P [X ₄ jF ² ] (!)

= 1

P (B ₃ ) X

! 2 B

X (! ) P (! )

= 1

p (1 p) 0p ³ (1 p) + 1p ² (1 p) ² + 1p ² (1 p) ² + 3p (1 p) ³

= (3 p) (1 p) If ! 2 B ₄ then E ^P [X ₄ jF ² ] (!)

= 1

P (B ₄ ) X

! 2 B

X (! ) P (! )

= 1

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

= 5 4p

In the speci…c case of p = ¹ ₂ ,

If ! 2 B ₁ then E ^P [X ₄ jF ² ] (!) = 3 4 If ! 2 B ₂ then E ^P [X ₄ jF ² ] (!) = 5 4 If ! 2 B 3 then E ^P [X 4 jF ² ] (!) = 5 4 If ! 2 B ₄ then E ^P [X ₄ jF ² ] (!) = 3

4 Exercise 3.5

E [ X ₄ j F ⁰ ] = 2 (1 p) ⁴ + 4 4p (1 p) ³

+6 6p ² (1 p) ² + 8 4p ³ (1 p) + 10 p ⁴

= 2 + 8p

If p = 1

2 , then E [X ₄ j F ⁰ ] = 6:

If ! 2 B ₁ = f T T T T; T T T H; T T HT; T T HH g ; then

E [ X ₄ j F ² ] (!) = 10 p ⁴ + 8 p ³ (1 p) + 8 p ³ (1 p) + 6 p ² (1 p) ² p ⁴ + p ³ (1 p) + p ³ (1 p) + p ² (1 p) ²

= 4p + 6

If ! 2 B ₂ = f T HT T; T HT H; T HHT; T HHH g ; then

E [ X ₄ j F ² ] (!) = 8 p ³ (1 p) + 6 p ² (1 p) ² + 6 p ² (1 p) ² + 4 p (1 p) ³ p ³ (1 p) + p ² (1 p) ² + p ² (1 p) ² + p (1 p) ³

= 4p + 4

If ! 2 B ₃ = f HT T T; HT T H; HT HT; HT HH g ; then

E [ X ₄ j F ² ] (!) = 8 p ³ (1 p) + 6 p ² (1 p) ² + 6 p ² (1 p) ² + 4 p (1 p) ³ p ³ (1 p) + p ² (1 p) ² + p ² (1 p) ² + p (1 p) ³

= 4p + 4

If ! 2 B 4 = f HHT T; HHT H; HHHT; HHHH g ; then

E [ X 4 j F ² ] (!) = 6 p ² (1 p) ² + 4 p (1 p) ³ + 4 p (1 p) ³ + 2 (1 p) ⁴ p ² (1 p) ² + p (1 p) ³ + p (1 p) ³ + (1 p) ⁴

= 4p + 2

If p = 1 2 , then E [X ₄ j F ² ] (!) =

8 >

> <

> >

:

8 if ! 2 B ₁ = f T T T T; T T T H; T T HT; T T HH g

6 if ! 2 B ₂ = f T HT T; T HT H; T HHT; T HHH g

6 if ! 2 B 3 = f HT T T; HT T H; HT HT; HT HH g

4 if ! 2 B ₄ = f HHT T; HHT H; HHHT; HHHH g

If ! 2 B ₁ = T T T T; T T T H; T T HT; T T HH;

T HT T; T HT H; T HHT; T HHH ; then E [ Y ₄ j G ² ] (!) = 10

If ! 2 B ₃ = f HT T T; HT T H; HT HT; HT HH g ; then

E [ Y ₄ j G ² ] (!) = 10 p ³ (1 p) + 6 p ² (1 p) ² + 0 p ² (1 p) ² + 0 p (1 p) ³ p ³ (1 p) + p ² (1 p) ² + p ² (1 p) ² + p (1 p) ³

= 2p (2p + 3)

If ! 2 B ₄ = f HHT T; HHT H; HHHT; HHHH g ; then E [ Y 4 j G ² ] (!) = 0

If p = 1 2 , then E [ Y ₄ j G ² ] (!) =

8 >

> <

> >

:

10 if ! 2 B ₁ = T T T T; T T T H; T T HT; T T HH;

T HT T; T HT H; T HHT; T HHH 4 if ! 2 B ₃ = f HT T T; HT T H; HT HT; HT HH g 0 if ! 2 B ₄ = f HHT T; HHT H; HHHT; HHHH g If ! 2 C ₁ = T T T T; T T T H; T T HT; T T HH;

T HT T; T HT H; T HHT; T HHH ; then E [ Y ₂ j G ⁴ ] (!) = 10

If ! 2 C ₂ = f HT T T g ; then E [ Y ₂ j G ⁴ ] (!) = 4

If ! 2 C ₃ = f HT T H g ; then E [ Y ₂ j G ⁴ ] (!) = 4

If ! 2 C ₄ = f HT HT; HT HH g ; then E [ Y ₂ j G ⁴ ] (!) = 4

If ! 2 C ₅ = f HHT T; HHT H; HHHT; HHHH g ; then

E [ Y ₂ j G ⁴ ] (!) = 0

5 Exercise 3.6

a) What …ltration is generated by the process X ? F ⁰ = f? ; g

F ¹ = ff ! ₁ ; :::; ! ₁₀ g ; f ! ₁₁ ; :::; ! ₁₆ gg

F ² = ff ! ₁ ; :::; ! ₆ g ; f ! ₇ ; :::; ! ₁₀ g ; f ! ₁₁ ; :::; ! ₁₃ g ; f ! ₁₄ ; ! ₁₅ ; ! ₁₆ gg

F ³ = f ! 1 ; ! 2 g ; f ! 3 ; ! 4 g ; f ! 5 ; ! 6 g ; f ! 7 ; :::; ! 10 g ; f ! ₁₁ ; ! ₁₂ g ; f ! ₁₃ g ; f ! ₁₄ ; ! ₁₅ g f ! ₁₆ g

F ⁴ = f ! ₁ g ; f ! ₂ g ; f ! ₃ g ; f ! ₄ g ; f ! ₅ g ; f ! ₆ g ; f ! ₇ g ; f ! ₈ g f ! ₉ g ; f ! ₁₀ g ; f ! ₁₁ g ; f ! ₁₂ g ; f ! ₁₃ g ; f ! ₁₄ g ; f ! ₁₅ g ; f ! ₁₆ g

b) Give the conditional distribution of X ₄ , knowing X ₂ .

P [X ₄ = x j X ₂ = 15 ] = 8 >

<

> :

3 0:06

3 0:06+3 0:065 = 0:48 if x = 16

3 0:065

3 0:06+3 0:065 = 0:52 if x = 14

0 otherwise

P [X ₄ = x j X ₂ = 12 ] = 8 >

> >

> >

> <

> >

> >

> >

:

0:08

4 0:0625+0:08+0:04+0:0675 = 0:18286 if x = 18

2 0:0625

4 0:0625+0:08+0:04+0:0675 = 0:28571 if x = 14

0:04+2 0:0625

4 0:0625+0:08+0:04+0:0675 = 0:37714 if x = 12

0:0675

4 0:0625+0:08+0:04+0:0675 = 0; 15429 if x = 11

0 otherwise

P [X ₄ = x j X ₂ = 11 ] = 8 >

<

> :

0;0625

3 0:0625 = ¹ ₃ if x = 20

2 0;0625

3 0:0625 = ² ₃ if x = 12

0 otherwise

c) Calculate the conditional expectation of X 4 , knowing X 2 . If ! 2 f ! ₁ ; :::; ! ₆ g

E ^P [X ₄ j (X ₂ ) ] (!) = 16 3 0:06

3 0:06 + 3 0:065 +14 3 0:065 3 0:06 + 3 0:065

= 14:96

If ! 2 f ! ₇ ; :::; ! ₁₃ g

E ^P [X ₄ j (X ₂ ) ] (!) = 18 0:08

4 0:0625 + 0:08 + 0:04 + 0:0675

+14 2 0:0625

4 0:0625 + 0:08 + 0:04 + 0:0675 +12 0:04 + 2 0:0625

4 0:0625 + 0:08 + 0:04 + 0:0675

+11 0:0675

4 0:0625 + 0:08 + 0:04 + 0:0675

= 13:514

If ! 2 f ! ₁₄ ; ! ₁₅ ; ! ₁₆ g E ^P [X ₄ j (X ₂ ) ] (!) = 20 1

3 +12 2 3

= 14:667

d) Give the conditional distribution of X 4 , knowing F ² .

P [X ₄ = x jf ! ₁ ; :::; ! ₆ g ] = 8 >

<

> :

3 0:06

3 0:06+3 0:065 = 0:48 if x = 16

3 0:065

3 0:06+3 0:065 = 0:52 if x = 14

0 otherwise

P [X ₄ = x jf ! ₇ ; :::; ! ₁₀ g ] = 8 >

> >

> >

> <

> >

> >

> >

:

0:08

0:0625+0:08+0:04+0:0675 = 0:32 if x = 18

0:0625

0:0625+0:08+0:04+0:0675 = 0:25 if x = 14

0:04

0:0625+0:08+0:04+0:0675 = 0:16 if x = 12

0:0675

0:0625+0:08+0:04+0:0675 = 0:27 if x = 11

0 otherwise

P [X ₄ = x jf ! ₁₁ ; :::; ! ₁₃ g ] = 8 >

<

> :

0:0625

3 0:0625 = ¹ ₃ if x = 14

2 0:0625

3 0:0625 = ² ₃ if x = 12

0 otherwise

P [X ₄ = x jf ! ₁₄ ; ! ₁₅ ; ! ₁₆ g ] = 8 >

<

> :

0;0625

3 0:0625 = ¹ ₃ if x = 20

2 0;0625

3 0:0625 = ² ₃ if x = 12

0 otherwise

e) Calculate the conditional expectation of X ₄ , knowing F ² .

if ! 2 f ! ₁ ; :::; ! ₆ g ; E ^P [X ₄ jF ² ] (!) = 16 3 0:06

3 0:06 + 3 0:065 +14 3 0:065 3 0:06 + 3 0:065

= 14:96

if ! 2 f ! ₇ ; :::; ! ₁₀ g ;

E ^P [X ₄ jF ² ] (!) = 18 0:08

0:0625 + 0:08 + 0:04 + 0:0675 + 14 0:0625

0:0625 + 0:08 + 0:04 + 0:0675

+12 0:04

0:0625 + 0:08 + 0:04 + 0:0675 + 11 0:0675

0:0625 + 0:08 + 0:04 + 0:0675

= 14:15

if ! 2 f ! ₁₁ ; ! ₁₂ ; ! ₁₃ g ; E ^P [X ₄ jF ² ] (!) = 14 1

3 + 12 2 3

= 12:667

if ! 2 f ! ₁₄ ; ! ₁₅ ; ! ₁₆ g ;

1 2

6 Problem 3.7

The conditional expectations of the yields are :

E ^P [R _2;4 jf ! ₁ ; :::; ! ₆ g ] = 14:96

15 = :997 33 E ^P [R _2;4 jf ! ₇ ; :::; ! ₁₀ g ] = 14:15

12 = 1:1792 E ^P [R _2;4 jf ! ₁₁ ; ! ₁₂ ; ! ₁₃ g ] = 12:67

12 = 1:0558 E ^P [R _2;4 jf ! ₁₄ ; ! ₁₅ ; ! ₁₆ g ] = 14:667

11 = 1:3334 E ^P [R _2;4 jf ! ₇ ; :::; ! ₁₃ g ] = 13:514

12 = 1:1262 Possible values for the random variable R _2;4 are :

16

15 = 1:0667 and 14

15 = 0:93333 18

12 = 1:5 and 14

12 = 1:1667 and 12

12 = 1 and 11

12 = 0:91667 20

11 = 1:8182 and 12

11 = 1:0909:

Recall that the bank account has a yield of 1.15 for times t = 2 to t = 4.

P [R _2;4 < 1:15 jf ! ₁ ; :::; ! ₆ g ] = P [X ₄ = 16 or X ₄ = 14 jf ! ₁ ; :::; ! ₆ g ]

= 1

P [R 2;4 < 1:15 jf ! 7 ; :::; ! 10 g ] = P [X 4 = 12 or X 4 = 11 jf ! 7 ; :::; ! 10 g ]

= 0:43

P [R _2;4 < 1:15 jf ! ₁₁ ; ! ₁₂ ; ! ₁₃ g ] = P [X ₄ = 12 jf ! ₁₁ ; ! ₁₂ ; ! ₁₃ g ]

= 0:66667

P [R _2;4 < 1:15 jf ! ₁₄ ; ! ₁₅ ; ! ₁₆ g ] = P [X ₄ = 12 jf ! ₁₄ ; ! ₁₅ ; ! ₁₆ g ]

= 0:66667

P [R _2;4 < 1:15 jf ! ₇ ; :::; ! ₁₃ g ] = P [X ₄ = 12 or X ₄ = 11 jf ! ₇ ; :::; ! ₁₃ g ]

= 0:37714 + 0:15429 = 0:53143

a) What instructions does investor A give to his prosecutor ? Justify your answer.

If we observe the process from time t = 0, at time t = 2 we will be able to determine which of the following four events happened : f ! ₁ ; :::; ! ₆ g ; f ! ₇ ; :::; ! ₁₀ g ; f ! ₁₁ ; :::; ! ₁₃ g ; f ! ₁₄ ; ! ₁₅ ; ! ₁₆ g .

If event f ! ₁ ; :::; ! ₆ g happened (the price of the share followed trajectories 11, 13, 15), then the probability to obtain a holding period yield (for the period from t = 2 to t = 4) superior to that of the bank account is zero. Hence in that case we would sell the share and invest the 15 dollars in the bank account. Then, the value of our portfolio at time t = 4 would be 15 1:15 = 17:25:

If event f ! ₇ ; :::; ! ₁₀ g happened (the price of the share followed trajectories 11, 13, 12), then the probability to obtain a HPY (for the period from t = 2 to t = 4) inferior to that of the bank account is around 43%. However, the expected HPY is 1:18 which is superior to that of the bank account. Hence we would keep the share.

If event f ! ₁₁ ; :::; ! ₁₃ g happened (the price of the share followed trajectories 11, 12, 12),

then the probability to obtain a HPY (for the period from t = 2 to t = 4) inferior to that of

the bank account is around 67%. In addition, the expected HPY is 1:0558 which is inferior

to the 1.15 of the bank account. Hence we would sell the share and invest the 12 dollars in

the bank account. Then, the value of our portfolio at time t = 4 would be 12 1:15 = 13:8.

of the bank account is around 67%. However, the expected HPY is 1:3334 which is superior to the 1.15 of the bank account. This is because when the HPY exceeds the yield of the bank account, it does it by a huge margin (1:82 versus 1.15). People with a taste for risky investment would choose to keep the share, especially because even when the HPY is inferior to the yield of the bank account, it is not drastically inferior (1.09 versus 1.15).

Trajectory Decision at time t = 2

Value of the portfolio at time t = 4 if the share is sold

Value of the portfolio at time t = 4 if the share is kept 11; 13; 15 sell the share 17:25

17:25

16 with prob. 18%

14 with prob. 19:5%

11; 13; 12 keep the share

13:8 13:8 13:8 13:8

18 with prob. 8%

14 with prob. 6:25%

12 with prob. 4%

11 with prob. 6:75%

11; 11; 12 sell the share 13:8 13:8

14 with prob. 6:25%

12 with prob. 12:5%

11; 11; 11 keep the share 12:65 12:65

20 with prob. 6:25%

12 with prob. 12:5%

b) What instructions does investor B give to his prosecutor ? Justify your answer.

If we observe the process only at time t = 2, we will be able to determine which of the following three events happened : f ! ₁ ; :::; ! ₆ g ; f ! ₇ ; :::; ! ₁₃ g ; f ! ₁₄ ; ! ₁₅ ; ! ₁₆ g .

If event f ! ₁ ; :::; ! ₆ g happened (the price of the share at time t = 2 is 15), then the probability to obtain a HPY (for the period from t = 2 to t = 4) superior to that of the bank account is zero. Hence in that case we would sell the share and invest the 15 dollars in the bank account. Then, the value of our portfolio at time t = 4 would be 15 1:15 = 17:25:

If event f ! ₇ ; :::; ! ₁₃ g happened (the price of the share at time t = 2 is 12), then the probability to obtain a HPY (for the period from t = 2 to t = 4) inferior to that of the bank account is around 53% (0:37714 + 0:15429). In addition, the expected HPY is 1:1262, which is lower then the yield of the bank account. Hence in that case we would sell the share and invest the 12 dollars in the bank account. Then, the value of our portfolio at time t = 4 would be 12 1:15 = 13:8:

If event f ! ₁₄ ; ! ₁₅ ; ! ₁₆ g happened (the price of the share at time t = 2 is 11), then the

probability to obtain a HPY (for the period from t = 2 to t = 4) inferior to that of the bank

account is around 67%. However, the expected HPY is 1:3334 which is superior to the 1.15

choose to keep the share, especially because even when the HPY is inferior to the yield of the bank account, it is not drastically inferior (1.09 versus 1.15).

Value of X ₂

Decision at time t = 2

Value of the portfolio at time t = 4 if the share is sold

Value of the portfolio at time t = 4 if the share is kept

15 sell the share 17:25

17:25

16 with prob. 18%

14 with prob. 19:5%

12 sell the share

13:8 13:8 13:8 13:8

18 with prob. 8%

14 with prob. 12:5%

12 with prob. 16:5%

11 with prob. 6:75%

11 keep the share 12:65

12:65

20 with prob. 6:25%

12 with prob. 12:5%

7 Problem 3.8

a) Let A and B represent the random times at which the shares are sold, respectively by prosecutors A and B. Are A and B stopping times ? Justify your answer.

A is a stopping time because

f ! 2 : _A (!) = 0 g = ? 2 F ⁰ f ! 2 : _A (!) = 1 g = ? 2 F ¹

f ! 2 : _A (!) = 2 g = f ! ₁ ; :::; ! ₆ g [ f ! ₁₁ ; :::; ! ₁₃ g 2 F ² f ! 2 : _A (!) = 3 g = ? 2 F ³

f ! 2 : _A (!) = 4 g = f ! ₇ ; :::; ! ₁₀ g [ f ! ₁₄ ; :::; ! ₁₅ g 2 F ⁴

B is a stopping time because

f ! 2 : _B (!) = 0 g = ? 2 F ⁰ f ! 2 : _B (!) = 1 g = ? 2 F ¹

f ! 2 : _B (!) = 2 g = f ! ₁ ; :::; ! ₁₃ g 2 F ² f ! 2 : B (!) = 3 g = ? 2 F ³

f ! 2 : _B (!) = 4 g = f ! ₁₄ ; :::; ! ₁₅ g 2 F ⁴

b) Say we replace the probability measure P by Q where Q (!) = ₁₆ ¹ for every ! 2 . Are A and B stopping times in this case ? Justify your answer.

Yes since being (or not being) a stopping time does not depend in any way on the

probability measure used on the …ltrated measurable space.