conditional expectation under change of measure

conditional expectation under change of measure

Let

ℙ

be a given

probability measure

on some

σ

-

algebra

ℱ

.
Suppose a new probability measure

ℚ

is defined
by

d

⁢

ℚ

=

Z

⁢

d

⁢

ℙ

, using some

ℱ

-measurable

random variable

Z

as the

Radon-Nikodym derivative

.
(Necessarily we must have

Z

≥

0

almost surely, and

𝔼

⁢

Z

=

1

.)

We denote with

𝔼

the

expectation

with respect to the

measure

ℙ

,
and with

𝔼

ℚ

the expectation with respect to the measure

ℚ

.

Theorem 1

.

If

Q

is restricted to a sub-

σ

-algebra

G

⊆

F

,
then the

restriction

has the

conditional expectation

E

[

Z

∣

G

]

as its Radon-Nikodym derivative:

d

Q

∣

G

=

E

[

Z

∣

G

]

d

P

∣

G

.

In other words,

d

⁢

ℚ

∣

𝒢

d

⁢

ℙ

∣

𝒢

=

(

d

⁢

ℚ

d

⁢

ℙ

)

∣

𝒢

.

Proof.

It is required to prove that, for all

B

∈

𝒢

,

ℚ

(

B

)

=

𝔼

[

𝔼

[

Z

∣

𝒢

]

1

B

]

.

But this follows at once from the law of iterated conditional expectations:

𝔼

[

𝔼

[

Z

∣

𝒢

]

1

B

]

=

𝔼

[

𝔼

[

Z

1

B

∣

𝒢

]

]

=

𝔼

[

Z

1

B

]

=

ℚ

(

B

)

.

∎

Theorem 2

.

Let

G

⊆

F

be any sub-

σ

-algebra.
For any

F

-measurable random variable

X

,

𝔼

[

Z

∣

𝒢

]

𝔼

ℚ

[

X

∣

𝒢

]

=

𝔼

[

Z

X

∣

𝒢

]

.

That is,

(

d

⁢

ℚ

d

⁢

ℙ

)

∣

𝒢

𝔼

ℚ

[

X

∣

𝒢

]

=

𝔼

[

d

⁢

ℚ

d

⁢

ℙ

X

∣

𝒢

]

.

Proof.

Let

Y

=

𝔼

[

Z

∣

𝒢

]

, and

B

∈

𝒢

. We find:

𝔼

ℚ

[

1

B

𝔼

[

Z

X

∣

𝒢

]

]

=

𝔼

[

Y

1

B

𝔼

[

Z

X

∣

𝒢

]

]

(since

d

⁢

ℚ

∣

𝒢

=

Y

⁢

d

⁢

ℙ

∣

𝒢

)

=

𝔼

[

𝔼

[

Y

1

B

Z

X

∣

𝒢

]

]

=

𝔼

⁢

[

Y

⁢

1

B

⁢

Z

⁢

X

]

=

𝔼

ℚ

⁢

[

Y

⁢

1

B

⁢

X

]

(since

d

⁢

ℚ

=

Z

⁢

d

⁢

ℙ

)

=

𝔼

ℚ

[

1

B

𝔼

ℚ

[

Y

X

∣

𝒢

]

]

.

Since

B

∈

𝒢

is arbitrary, we can equate the

𝒢

-measurable integrands:

𝔼

[

Z

X

∣

𝒢

]

=

𝔼

ℚ

[

Y

X

∣

𝒢

]

=

Y

𝔼

ℚ

[

X

∣

𝒢

]

.

∎

Observe that if

d

⁢

ℚ

/

d

⁢

ℙ

>

0

almost surely,
then

𝔼

ℚ

[

X

∣

𝒢

]

=

𝔼

[

d

⁢

ℚ

d

⁢

ℙ

X

∣

𝒢

]

/

(

d

⁢

ℚ

d

⁢

ℙ

)

∣

𝒢

.

Theorem 3

.

If

X

t

is a

martingale

with respect to

Q

and some

filtration

{

F

t

}

,
then

X

t

⁢

Z

t

is a martingale with respect to

P

and

{

F

t

}

,
where

Z

t

=

E

[

Z

∣

F

t

]

.

Proof.

First observe that

X

t

⁢

Z

t

is indeed

ℱ

t

-measurable.
Then, we can apply

Theorem

2,
with

X

in the statement of that theorem replaced by

X

t

,

Z

replaced by

Z

t

,

ℱ

replaced by

ℱ

t

, and

𝒢

replaced by

ℱ

s

(

s

≤

t

),
to obtain:

𝔼

[

X

t

Z

t

∣

ℱ

s

]

=

Z

s

𝔼

ℚ

[

X

t

∣

ℱ

s

]

=

Z

s

X

s

,

thus proving that

X

t

⁢

Z

t

is a martingale under

ℙ

and

{

ℱ

t

}

.
∎

Sometimes the random variables

Z

t

in Theorem 3
are written as

(

d

⁢

ℚ

d

⁢

ℙ

)

t

.
(This is a Radon-Nikodym derivative

process

;
note that

Z

t

defined as

Z

t

=

𝔼

[

Z

∣

ℱ

t

]

is always a martingale
under

ℙ

and

{

ℱ

t

}

.)

Under the

hypothesis

Z

t

>

0

,
there is an alternate restatement of Theorem 3
that may be more easily remembered:

Theorem 4

.

Let

Z

t

=

(

d

⁢

Q

/

d

⁢

P

)

t

>

0

almost surely.
Then

X

t

is a martingale with respect to

P

,
if and only if

X

t

/

Z

t

is a martingale with respect to

Q

.

Title

conditional expectation under change of measure

Canonical name

ConditionalExpectationUnderChangeOfMeasure

Date of creation

2013-03-22 16:54:21

Last modified on

2013-03-22 16:54:21

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

9

Author

stevecheng (10074)

Entry type

Derivation

Classification

msc 60A10

Classification

msc 60-00

Related topic

Martingale

Related topic

ConditionalExpectation

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