80-646-08StochasticcalculusGeneviŁveGauthier Expectationandconditionalexpectation

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Independence Conditional probability Expectation Moments Conditional expectation

Expectation and conditional expectation

80-646-08 Stochastic calculus

Geneviève Gauthier

HEC Montréal

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Independence Events Random variables Example Conditional probability Expectation Moments Conditional expectation

Independence of two events

De…nition

Let ( _Ω , F ^, P ) be a probability space. Two events A and B are said to be independent if

P ( A \ ^B ) = _P ( A ) _P ( B ) .

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Independence of two random variables

De…nition

The random variables X and Y are both built on the probability space ( Ω , F ^, ^P ) , Card ( Ω ) < ∞ . Let P

^X

= f ^A

¹

^{, ...,} ^A

^m

g ^and P

^Y

= f ^B

¹

^{, ...,} ^B

ⁿ

g be two …nite partitions that respectively generate the sigma-algebras σ ( X ) and σ ( Y ) . The random variables X and Y are said to be independent if

8 ^A 2 P

X

and 8 ^B 2 P

Y

, P ( A \ ^B ) = P ( A ) P ( B ) .

Intuitively, X and Y are independent when having some

information about one of them doesn’t provide us with

any information about the other.

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Independence of two random variables I

Example

Example.

ω X Y P P

1 1 1

¹₆

0 2 1 0

¹₆ ¹₅

3 1 0

¹₆ ¹₅

ω X Y P P

4 0 1

¹₆ ¹₅

5 0 0

¹₆ ¹₅

6 0 0

¹₆ ¹₅

If A = ⁿ 1 , 2 , 3 o

and B = ⁿ 1 , 4 o

then

P

^X

= f ^A, ^A

^c

g ^and P

^Y

= f ^B, ^B

^c

g ^.

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Independence of two random variables II

Example

The random variables X and Y are independent on the probability space ( Ω , F ^, ^P ) since

P

(

A

)

P

(

B

) =

¹ 2

1 3

=

¹

6

=

Pn 1o

=

P

(

A

\

^B

)

, P

(

A^c

)

P

(

B

) =

¹

2 1 3

=

¹

6

=

Pn 4o

=

P

(

A^c

\

^B

)

, P

(

A

)

P

(

B^c

) =

¹

2 2 3

=

¹

3

=

Pn 2, 3o

=

P

(

A

\

^B^c

)

, P

(

A^c

)

P

(

B^c

) =

¹

2 2 3

=

¹

3

=

Pn 5, 6o

=

P

(

A^c

\

^B^c

)

.

Intuitively, the answer to the question a dice is rolled; what

is the probability that the random variable X takes the

value of 1? is the same as the answer to the question a

dice is rolled and the random variable Y takes the value of

1; what is the probability that the random variable X also

takes the value of 1?. The answer is one half.

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Independence of two random variables III

Example

By contrast, the same random variables X and Y are dependent on the probability space ( Ω , F ^, ^P ) since

P

(

A

)

P

(

B

) =

² 5

1 5

=

²

25

6 =

0

=

P n 1o

=

P

(

A

\

^B

)

.

Intuitively, the answer to the question a dice is rolled; what is the probability that the random variable X takes the value of 1? is not the same as the answer to the question a dice is rolled and the random variable Y takes the value of 1; what is the probability that the random variable X also takes the value of 1?. In the …rst case, the answer is

2

5

while in the second case, the answer is 0.

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Independence of two random variables

Remark

Remark. Although random variables may very well exist with

no need at all for a probability space (a measurable space is

su¢ cient), it is necessary to know the probability measure

associated with the measurable space to be able to speak of

independence.

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Independence of random variables I

Example

Example.

ω X Y Z P

ω

1

1 1 0

¹₈

ω

2

1 0 0

¹₈

ω

3

1 0 1

¹₈

ω

4

1 0 1

¹₈

ω X Y Z P

ω

5

0 1 1

¹₈

ω

6

0 0 0

¹₈

ω

7

0 0 0

¹₈

ω

8

0 0 1

¹₈

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Independence of random variables II

Example

Such random variables are pairwise independent since

Pf^X=0gPf^Y=0g = ¹ 2

3 4=³

8=Pfω6,ω₇,ω₈g=P(f^X=0g \ f^Y=0g)^, Pf^X=1gPf^Y=0g = ¹

2 3 4=³

8=Pfω2,ω₃,ω₄g=P(f^X=1g \ f^Y=0g)^, Pf^X=0gPf^Y=1g = ¹

2 1 4=¹

8=Pfω5g=P(f^X=0g \ f^Y=1g)^, PfX=1gPfY=1g = ¹

2 1 4=¹

8=Pfω1g=P(fX=1g \ fY=1g).

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Independence of random variables III

Example

Pf^X=0gPf^Z=0g = ¹ 2

1 2=¹

4=Pfω6,ω7g=P(f^X=0g \ f^Z=0g)^, Pf^X=1gPf^Z=0g = ¹

2 1 2=¹

4=Pfω1,ω2g=P(f^X=1g \ f^Z=0g)^, PfX=0gPfZ=1g = ¹

2 1 2=¹

4=Pfω5,ω8g=P(fX=0g \ fZ=1g), Pf^X=1gPf^Z=1g = ¹

2 1 2=¹

4=_Pfω3,ω4g=_P(f^X=1g \ f^Z=1g)^.

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Independence of random variables IV

Example

Pf^Z=0gPf^Y=0g = ¹ 2

3 4=³

8=Pfω2,ω6,ω7g=P(f^Z=0g \ f^Y=0g)^, PfZ=1gPfY=0g = ¹

2 3 4=³

8=Pfω3,ω4,ω8g=P(fZ=1g \ fY=0g), Pf^Z=0gPf^Y=1g = ¹

2 1 4=¹

8=Pfω1g=P(f^Z=0g \ f^Y=1g), Pf^Z=1gPf^Y=1g = ¹

2 1 4=¹

8=_Pfω5g=_P(f^Z=1g \ f^Y=1g)^.

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Independence of random variables V

Example

But such variables are not mutually independent since P f ^X = 1 g P f ^Y = 1 g P f ^Z = 1 g

= ¹

2 1 4

1 2

= ¹

16

6 = 0

= _P f?g

= P ( f ^X = 1 g \ f ^Y = 1 g \ f ^Z = 1 g ) .

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Independence Conditional probability

Sigma-algebra De…nition Interpretation Example Independence

Expectation Moments Conditional expectation

Conditional probability

De…nition

Let ( _Ω , F ^, P ) be a probability space such that Card ( _Ω ) < _∞ . For all event A 2 F with a positive probability, P ( A ) > 0, the conditional probability given A, denoted P ( j ^A ) , is de…ned as

8 ^B 2 F ^, P ( B j ^A ) = ^P ( B \ ^A )

P ( A ) ^.

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Conditional probability I

Interprétation

The random experiment consists of rolling a dice.

If the dice is well balanced, what is the probability to obtain more than three points?

The correct answer is:

¹₂

.

Now, if after rolling the dice, I inform you that the number

of points obtained is even, what is the probability that the

upper side of the dice shows more than points?

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Conditional probability II

Interprétation

The answer previously given should change in order to take advantage of the information provided. Since, out of three cases where the number of points is even, there are two cases where the number of points is also greater than three, the right answer is

²₃

.

P n

4 , 5 , 6 o n

2 , 4 , 6 o

= ^P

n 4 , 5 , 6 o

\ ⁿ ² ^, ⁴ ^, ⁶ ^o P n

2 , 4 , 6 o

= ^P

n 4 , 6 o P n

2 , 4 , 6 o =

2 6 3 6

= ²

3 .

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Conditional probability III

Interprétation

Why do we require that the probability associated with the event on which we condition be positive?

Mathematically, it’s simply to prevent from making the silly mistake of dividing by zero.

Intuitively, if, after the dice roll, I inform you that the result is negative, the more polite among you will exclaim

”You’re such a joker!” while some other may call me a liar.

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Conditional probability I

Example

Example. Let’s go back to the stochastic process representing the share price of a security, to which we will add a probability measure P on the measurable space ( Ω , F ) . Recall that the sigma-algebra here is F = σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

g ^, f ^ω

⁴

gg ^.

ω X

0

( ω ) X

1

( ω ) X

2

( ω ) X

3

( ω ) _P ( ω )

ω

1

¹₂

1

¹₂

?

ω

2

1

¹₂

1

¹₂

?

ω

3

1 2 1 1

³₈

ω

4

1 2 2 2

²₈

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Conditional probability II

Example

Question. Knowing that the share price is worth one dollar at

time t = 2, do the probabilities associated with the possible

share prices at time t = 3 change?

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Conditional probability III

Example

Answer. Yes. Let

A = f ^ω 2 ^Ω ^: ^X

²

( ω ) = 1 g = f ^ω

¹

^, ^ω

²

^, ^ω

³

g ^.

Since

P

(

A

) =

P

f

^ω¹^,^ω²^,^ω³

g =

_P

f

^ω¹^,^ω²

g +

_P

f

^ω³

g =

³ 8

+

³

8

=

³ 4 alors

P X₃

=

¹

2

jf

^X2

=

1

g

=

^P

( f

^ω1,ω2

g \ f

^ω1,ω2,ω3

g )

P

f

^ω¹^,^ω²^,^ω³

g =

^P

f

^ω1,ω2

g

P

(

A

) =

³

8 4 3

=

¹

2

6 =

³

8

=

P X3

=

¹ 2 ;

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Conditional probability IV

Example

P

( f

^X3

=

1

g jf

^X2

=

1

g )

=

^P

( f

^ω3

g \ f

^ω1,ω2,ω3

g )

P

f

^ω¹^,^ω²^,^ω³

g =

^P

f

^ω3

g

P

(

A

) =

³

8 4 3

=

¹

2

6 =

³

8

=

P

f

^X3

=

1

g

^;

P

( f

^X3

=

2

g jf

^X2

=

1

g )

=

^P

( f

^ω4

g \ f

^ω1,ω2,ω3

g )

P

f

^ω¹^,^ω²^,^ω³

g =

^P

( ? )

P

(

A

) =

0

6 =

²

8

=

P

f

^X3

=

2

g

^.

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Conditional probability V

Example

ω X₀

(

ω

)

X₁

(

ω

)

X₂

(

ω

)

X₃

(

ω

)

P

(

ω

)

ω1 1 ¹₂ 1 ¹₂ ?

ω₂ 1 ¹₂ 1 ¹₂ ?

ω3 1 2 1 1 ³₈

ω4 1 2 2 2 ²₈

We had determined that F = σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

g ^, f ^ω

⁴

gg ^. Then

F

A

= σ ff ^ω

¹

^, ^ω

²

g \ ^A, f ^ω

³

g \ ^A, f ^ω

⁴

g \ ^A g

= σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

g ^, ?g = f ^A, f ^ω

¹

^, ^ω

²

g ^, f ^ω

³

g ^, ?g ^. As a consequence, the measurable space on which

P ( jf ^X

²

= 1 g ) is built, is

( f ^ω

¹

^, ^ω

²

^, ^ω

³

g ^, ^σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

gg ) .

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Conditional probability I

Independence

Let’s go back to the de…nition of independence between two random variables.

Theorem

The random variables X and Y are both built on the probability space ( _Ω , F ^, P ) , Card ( _Ω ) < _∞ . Let P

^X

= f ^A

¹

^{, ...,} ^A

^m

g ^and P

^Y

= f ^B

¹

^{, ...,} ^B

ⁿ

g , be two …nite partitions that respectively generate the sigma-algebras σ ( X ) and σ ( Y ) . If

8 ^A 2 P

^X

^{such that} ^P ( A ) > 0, P ( B j ^A ) = P ( B ) , 8 ^B 2 P

^Y

or again, if

8 ^B 2 P

^Y

^{such that} P ( B ) > 0, P ( A j ^B ) = _P ( A ) , 8 ^A 2 P

^X

^,

then X and Y are independent.

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Conditional probability II

Independence

Proof of the theorem. Let’s assume that

8 ^A 2 P

^X

^{such that} ^P ( A ) > 0, P ( B j ^A ) = P ( B ) , 8 ^B 2 P

^Y

^. We want to show that 8 ^A 2 P

^X

^{such that} P ( A ) > 0 and 8 ^B 2 P

Y

,

P ( B \ ^A ) = _P ( A ) _P ( B ) . But 8 ^A 2 P

^X

^{such that} ^P ( A ) > 0 and 8 ^B 2 P

^Y

^,

P ( B ) = _P ( B j ^A ) = ^P ( B \ ^A )

P ( A ) ) P ( B \ ^A ) = _P ( A ) _P ( B ) et 8 ^A 2 P

^X

^{such that} ^P ( A ) = 0 and 8 ^B 2 P

^Y

^,

0 P ( B \ ^A ) _P ( A ) = 0 ) P ( B \ ^A ) = 0 = _P ( A ) _P ( B ) .

(24)

Conditional probability III

Independence

If we use

8 ^B 2 P

^Y

^{such that} ^P ( B ) > 0, P ( A j ^B ) = P ( A ) , 8 ^A 2 P

^X

^,

as a premise, we can follow a similar reasoning and obtain an

identical result.

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Independence Conditional probability Expectation

De…nition Properties Example Remarks

Moments Conditional expectation

Expectation

De…nition

Let X be a random variable built on the probability space ( Ω , F ^, ^P ) such that Card ( Ω ) < ∞ . The expectation of X , denoted E

^P

[ X ] is

E

^P

[ X ] = ∑

ω2Ω

X ( ω ) P ( ω ) . If the random variable X takes n di¤erent values, say x

1

< ... < x

n

then

E

^P

[ X ] =

∑

n i=1

x

i

P f ^ω 2 Ω : X ( ω ) = x

i

g =

∑

n i=1

x

i

f

_X

( x

i

)

where f

_X

is the probability mass function of X .

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Expectation

Properties

Let X and Y be two random variables built on the probability space ( Ω , F ^, P ) , Card ( Ω ) < _∞ . If a and b represent real numbers, then

E1 E

^P

[ aX + bY ] = aE

^P

[ X ] + bE

^P

[ Y ] ;

E2 If 8 ^ω 2 ^Ω ^, ^X ( ω ) Y ( ω ) 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 ] .

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Expectation I

Example

Example. Let’s take the random variable W again, as well as two probability measures P and Q :

ω W

(

ω

)

P

(

ω

)

Q

(

ω

)

1 5 ¹₆ ₁₂⁴

2 5 ¹₆ ₁₂¹

3 5 ¹₆ ₁₂¹

4 5 ¹₆ ₁₂¹

5 0 ¹₆ ₁₂¹

6 10 ¹₆ ₁₂⁴

x P

f

^W

=

x

g

^Q

f

^W

=

x

g

0 ¹₆ ₁₂¹

5 ⁴₆ ₁₂⁷

10 ¹₆ ₁₂⁴

(28)

Expectation II

Example

E^P

[

W

] = ∑

ω2Ω

W

(

ω

)

P

(

ω

)

=

5 1

6

+

5 1 6

+

5 1

6

+

5 1 6

+

0 1

6

+

10 1 6

=

³⁰

6

=

5 E^P

[

W

] =

∑

n i=1

w_if_W

(

w_i

) =

0 1 6

+

5 4

6

+

10 1 6

=

³⁰

6

=

5 E^Q

[

W

] = ∑

ω2Ω

W

(

ω

)

Q

(

ω

)

=

5 4

12

+

5 1

12

+

5 1

12

+

5 1

12

+

0 1

12

+

10 4 12

=

⁷⁵

12

=

6,25 E^Q

[

W

] =

∑

n i=1

w_if_W

(

w_i

) =

0 1

12

+

5 7

12

+

10 4 12

=

⁷⁵

12

=

6,25

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Expectation

Remarks

The expectation of a random variable is a real number. It is not a random quantity.

From the previous example, we can notice that the

expectation of a random variable depends on the

probability measure that is being used.

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Independence Conditional probability Expectation Moments

De…nition Variance

Properties Covariance Properties Conditional expectation

Moments of a random variable

De…nition

Let X be a random variable built on the probability space ( Ω , F ^, ^P ) such that Card ( Ω ) < ∞ . The kth moment of X , denoted E

^P

X

^k

is the expectation of the variable X

^k

:

E

^P

h X

^k

i

= ∑

ω2Ω

X

^k

( ω ) P ( ω ) =

∑

n i=1

x

_i^k

f

_X

( x

_i

)

where x

1

< ... < x

n

are the values taken by X and f

X

is its

probability mass function.

(31)

Variance I

De…nition

Let X be a random variable built on the probability space ( _Ω , F ^, P ) such that Card ( _Ω ) < _∞ . The variance of X , denoted Var

^P

[ X ] is the expectation of the random variable

X E

^P

[ X ]

²

:

Var

^P

[ X ] = ∑

ω2Ω

X ( ω ) E

^P

[ X ]

²

P ( ω ) (1)

=

∑

n i=1

x

i

E

^P

[ X ]

²

f

_X

( x

i

) (2)

where x

1

< ... < x

n

are the values taken by X and f

_X

is its

probability mass function.

(32)

Variance II

De…nition

Remarks.

The variance is a measure of dispersion of the values x

₁

, ..., x

_n

taken by X around the expectation E

^P

[ X ] . The greater the variance, the more dispersed the values.

Similar to the expectation and the moments, the variance is a real number.

Moreover, whatever the random variable, the variance is never negative.

The standard deviation, commonly used in statistics, is the

square root of the variance.

(33)

Moments of a random variable

Variance properties

Exercise. Show that, if a and b are real numbers, V1 Var

^P

[ X ] 0;

V2 Var

^P

[ X ] = E

^P

X

²

E

^P

[ X ]

²

; V3 8 ^a 2 R , Var

^P

[ aX + b ] = a

²

Var

^P

[ X ] ; V4 If X and Y are independent, then

Var

^P

[ X + Y ] = Var

^P

[ X ] + Var

^P

[ Y ]

(34)

Covariance I

De…nition

Let X and Y be two random variables built on the probability space ( Ω , F ^, ^P ) such that Card ( Ω ) < ∞ . The covariance of X and Y , denoted Cov

^P

[ X , Y ] is the expectation of the random variable X E

^P

[ X ] Y E

^P

[ Y ] :

Cov

^P

[ X , Y ] = ∑

ω2Ω

X ( ω ) E

^P

[ X ] Y ( ω ) E

^P

[ Y ] P ( ω )

(35)

Covariance II

De…nition

Interpretation.

If the covariance is positive, it is because the sum

∑

ω2Ω

X ( ω ) E

^P

[ X ] Y ( ω ) E

^P

[ Y ] P ( ω ) , is dominated by the ω that make the term

X ( ω ) E

^P

[ X ] Y ( ω ) E

^P

[ Y ] positive, which

means that the random variables X and Y tend to be

either greater or smaller than their respective expectations

in the same states of the world ω.

(36)

Covariance III

De…nition

If the covariance is negative, the sum has to be dominated by the ω which make the term

X ( ω ) E

^P

[ X ] Y ( ω ) E

^P

[ Y ] negative, which

means that, when one of the random variables X and Y is

greater than its expectation, the other one tends to be

smaller than its own expectation.

(37)

Covariance

Properties

Exercise. Show that

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 ] ;

C4 Var

^P

[ X + Y ] = Var

^P

[ X ] + Var

^P

[ Y ] + 2Cov

^P

[ X , Y ] .

(38)

Other moments

De…ntion

In addition to the variance, two other centered moments are commonly used in modelling: skewness and kurtosis

coe¢ cients. A description thereof can be found in the appendix

to this document.

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Independence Conditional probability Expectation Moments Conditional expectation De…nition Example Properties

Conditional expectation

De…nition

The random variable X is built on the probability space ( _Ω , F ^, P ) , Card ( _Ω ) < _∞ . Let G F , a sigma-algebra generated by the …nite partition P = f ^A

¹

^{, ...,} ^A

ⁿ

g ^satisfying 8 ⁱ 2 f ^{1, ...,} ⁿ g ^, ^P ( A

_i

) > 0. The conditional expectation of X given G ^{, denoted} ^E

^P

[ X jG ] is

E

^P

[ X jG ] ( ω ) =

∑

n i=1

I

Ai

( ω ) P ( A

i

) ∑

ω 2^Aⁱ

X ( ω ) _P ( ω ) where I

Ai

: Ω ! f ^0, ¹ g is the indicator function

I

Ai

( ω ) = ^{1 si} ^ω 2 ^A

ⁱ

0 si ω 2 / ^A

ⁱ

^.

(40)

Conditional expectation I

Example

Example . Let’s go back to the stochastic process representing the share price of a security, to which we will add a probability measure on the measurable space ( Ω , F ) .

ω X

₀

( ω ) X

₁

( ω ) X

₂

( ω ) X

₃

( ω ) P ( ω ) ω

1

¹₂

1

¹₂ ¹₈

ω

2

1

¹₂

1

¹₂ ²₈

ω

3

1 2 1 1

³₈

ω

4

1 2 2 2

²₈

(41)

Conditional expectation II

Example

We had determined that F

¹

= σ ff ^ω

¹

^, ^ω

²

g ^, f ^ω

³

^, ^ω

⁴

gg ^{. Then}

E^P

[

X₃

jF

1

] (

ω

)

=

∑

2 i=1

IAi

(

ω

)

P

(

Ai

) ∑

ω 2Ai

X3

(

ω

)

P

(

ω

)

=

^I^A¹

(

ω

)

P

(

A1

) ∑

ω2A1

X3

(

ω

)

P

(

ω

) +

^I^A²

(

ω

)

P

(

A2

) ∑

ω2A2

X3

(

ω

)

P

(

ω

)

=

^I^A¹

(

ω

)

3 8

1 2

1 8

+

¹

2 2

8

+

^I^A²

(

ω

)

5 8

1 3 8

+

2 2

8

=

¹

2IA1

(

ω

) +

⁷

5IA2

(

ω

)

.

(42)

Conditional expectation III

Example

Thus,

If ω 2 f ^ω

¹

^, ^ω

²

g ^then E

^P

[ X

₃

jF

¹

] ( ω ) = ¹

2 I

A1

( ω ) + ⁷

5 I

A2

( ω ) = ¹ 2 and if ω 2 f ^ω

³

^, ^ω

⁴

g ^then

E

^P

[ X

₃

jF

¹

] ( ω ) = ¹

2 I

A1

( ω ) + ⁷

5 I

A2

( ω ) = ⁷ 5 . Interpretation. At time t = 1, we will be able to

determine whether the state of the world is an element of

f ^ω

¹

^, ^ω

²

g or an element of f ^ω

³

^, ^ω

⁴

g ^{. If} ^ω 2 f ^ω

¹

^, ^ω

²

g

then the expected value of the random variable X

₃

is

¹₂

.

By contrast, if ω 2 f ^ω

³

^, ^ω

⁴

g , then the expected value of

X

₃

is

⁷₅

.

(43)

Conditional expectation I

Remarks

The conditional expectation is expressed in terms of the conditional probabilities given each of the elements in the partition which generates the sigma-algebra, since

E

^P

[ X jG ] ( ω )

=

∑

n i=1

I

Ai

( ω ) P ( A

_i

) ∑

ω 2^Ai

X ( ω ) P ( ω )

=

∑

n i=1

I

A_i

( ω ) ∑

ω 2^Aⁱ

X ( ω ) ^P ( ω \ ^A

ⁱ

) P ( A

_i

)

=

∑

n i=1

I

Ai

( ω ) ∑

ω 2^Ai

X ( ω ) P ( ω j ^A

ⁱ

)

(44)

Conditional expectation II

Remarks

Important remark. Unlike the expectation, the

conditional expectation is not a real number, but a random

variable. Actually, since it is constant on the atoms which

generate G ^{, it is a} G measurable random variable.

(45)

Conditional expectation I

Properties

Let X and Y be two random variables in the probability space ( Ω , F ^, ^P ) .

The sigma-algebras G ^, G

¹

^and G

²

are respectively

generated by the …nite partitions P = f ^A

¹

^{, ...,} ^A

ⁿ

g ^and

P

¹

= f ^B

¹

^{, ...,} ^B

^m

g ^and P

²

= f ^C

¹

^{, ...,} ^C

ⁿ

g ^.

(46)

Conditional expectation II

Properties

EC1 If X is G measurable, then E

^P

[ X jG ] = X . EC2 If G

¹

G

²

are sigma-algebras, then

E

^P

E

^P

[ X jG

¹

] jG

²

= E

^P

[ X jG

¹

] . EC3 If G

¹

G

²

are sigma-algebras, then

E

^P

E

^P

[ X jG

²

] jG

¹

= E

^P

[ X jG

¹

] . EC4 E

^P

[ X jf? ^, ^Ω g ] = E

^P

[ X ] .

EC5 E

^P

E

^P

[ X jG ] = E

^P

[ X ] .

EC6 If Y is G measurable, then E

^P

[ XY jG ] = Y E

^P

[ X jG ] . EC7 If X and Y are independent, then

E

^P

[ X j ^σ ( Y ) ] = E

^P

[ X ] .

EC8 8 ^a, ^b 2 R , E

^P

[ aX + bY jG ] = aE

^P

[ X jG ] + bE

^P

[ Y jG ] .

(47)

Conditional expectation I

Proof of EC1

To be shown. If X is G measurable, then E

^P

[ X jG ] = X .

Since X is G measurable, then it is constant on the atoms of G . So, let’s set

x

_i

= X ( ω ) 8 ^ω 2 ^A

ⁱ

^, 8 ⁱ 2 f ^{1, ...,} ⁿ g ^.

Let’s consider any ω 2 ^A

ⁱ

^.

(48)

Conditional expectation II

Proof of EC1

E

^P

[ X jG ] ( ω ) =

∑

n j=1

I

A_j

( ω ) P ( A

j

) ∑

ω 2^A^j

X ( ω ) _P ( ω )

= ¹

P ( A

_i

) ∑

ω 2^Aⁱ

X ( ω ) P ( ω )

= ¹

P ( A

i

) ∑

ω 2^Aⁱ

x

i

P ( ω )

= ^x

ⁱ

P ( A

_i

) ∑

ω 2^Aⁱ

P ( ω )

= ^x

ⁱ

P ( A

_i

) ^P ( A

_i

)

= x

_i

= X ( ω ) .

(49)

Conditional expectation III

Proof of EC1

Since ω was arbitrarily chosen, we have that

8 ^ω 2 Ω , E

^P

[ X jG ] ( ω ) = X ( ω ) .

(50)

Conditional expectation

Proof of EC2

To be shown, If G

¹

G

²

F are sigma-algebras,then E

^P

E

^P

[ X jG

¹

] jG

²

= E

^P

[ X jG

¹

] .

Since G

¹

G

²

^{, then any} G

¹

measurable function is also G

²

measurable. Moreover, we know that E

^P

[ X jG

¹

] is G

¹

measurable, therefore E

^P

[ X jG

¹

] is also G

²

measurable.

By using ( EC 1 ) , E

^P

h

E

^P

[ X jG

¹

] jG

²

ⁱ = E

^P

[ X jG

¹

] .

(51)

Conditional expectation I

Proof of EC3

To be shown. If G

¹

G

²

F are sigma-algebras, then E

^P

E

^P

[ X jG

²

] jG

¹

= E

^P

[ X jG

¹

] .

Let B

_k

be a atom of G

¹

^{. Since} ^B

^k

2 G

¹

G

²

^then ^B

^k

2 G

²

and, as a consequence, B

_k

can be represented as the union of some atoms of G

²

, i.e. there exist C

k1

, ..., C

kq

2 P

²

^{such that} B

_k

= ^S

^q_i₌₁

C

_k_i

. Moreover, notice that, E

^P

[ X jG

²

] being G

²

measurable, it is constant on the atoms of G

²

. Let’s set 8 ^k 2 f ^{1, ...,} ⁿ g ^, 8 ^ω 2 ^C

^k

^,

x

_k

E

^P

[ X jG

²

] ( ω )

=

∑

n j=1

I

Cj

( ω ) P ( C

_j

) ∑

ω 2^C^j

X ( ω ) P ( ω )

= ¹

P ( C

_k

) ∑

ω 2^Ck

X ( ω ) P ( ω ) .

(52)

Conditional expectation II

Proof of EC3

Let’s consider any ω 2 ^C

k⁰

B

_k

. E

^P

h

E

^P

[ X jG

²

] jG

¹

ⁱ ( ω )

=

∑

m j=1

I

Bj

( ω ) P ( B

_j

) ∑

ω 2^B^j

E

^P

[ X jG

²

] ( ω ) P ( ω )

= ¹

P ( B

k

) ∑

ω 2^B^k

E

^P

[ X jG

²

] ( ω ) _P ( ω )

= ¹

P ( B

_k

)

∑

q i=1

∑

ω 2^Cki

E

^P

[ X jG

²

] ( ω )

| {z }

x_ki

P ( ω )

= ¹

P ( B

_k

)

∑

q i=1

x

_k_i

∑

ω 2^C_ki

P ( ω )

= ¹

P ( B

_k

)

∑

q i=1

x

_k_i

P ( C

_k_i

)

(53)

Conditional expectation III

Proof of EC3

= ¹

P ( B

k

)

∑

q i=1

0 @ 1

P ( C

k_i

) ∑

ω 2^Cki

X ( ω ) _P ( ω ) 1

A P ( C

_k_i

)

= ¹

P ( B

_k

)

∑

q i=1

∑

ω 2^C_ki

X ( ω ) P ( ω )

= ¹

P ( B

_k

) ∑

ω 2^Bk

X ( ω ) P ( ω )

=

∑

m j=1

I

B_j

( ω ) P ( B

j

) ∑

ω 2^B^j

X ( ω ) _P ( ω )

= E

^P

[ X jG

¹

] ( ω ) .

Thus, 8 ^ω 2 Ω , E

^P

E

^P

[ X jG

²

] jG

¹

( ω ) = E

^P

[ X jG

¹

] ( ω ) .

(54)

Conditional expectation

Proof of EC4

To be shown. E

^P

[ X jf? ^, Ω g ] = E

^P

[ X ] .

Since f? ^, ^Ω g is a sigma-algebra generated by Ω , by applying the de…nition of the conditional expectation of X with respect to such a sigma-algebra, we obtain 8 ^ω 2 Ω ,

80-646-08StochasticcalculusGeneviŁveGauthier Expectationandconditionalexpectation

Expectation and conditional expectation