Transcript 1.6 Change of Measure

1.6 Change of Measure
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Introduction
• We used a positive random variable Z to change probability measures on a
space Ω.
•
is risk-neutral probability measure
• P(ω) is the actual probability measure
• When Ω is uncountably infinite and
division by zero. We could rewrite this equation as
then it will
• But the equation tells us nothing about the relationship among
Because
, the value of Z(ω)
and Z.
• However, we should do this set-by-set, rather than ω-by-ω
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Theorem 1.6.1
• Let (Ω, F, P) be a probability space and let Z be an almost surely
nonnegative random variable with EZ=1. For A F define
Then is a probability measure. Furthermore, if X is a nonnegative
random variable, then
If Z is almost surely strictly positive, we also have
for every nonnegative random variable Y.
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Concept of Theorem 1.6.1
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Proof of Theorem 1.6.1 (1)
• According to Definition 1.1.2, to check that is a probability measure, we
must verify that
and that is countably additive. We have by
assumption
• For countable additivity, let A1, A2,… be a sequence of disjoint sets in F,
and define
,
. Because
and
, we may use the Monotone Convergence Theorem,
Theorem 1.4.5, to write
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Proof of Theorem 1.6.1 (2)
• But
, and so
• Now support X is a nonnegative random variable. If X is an indicator
function X=IA , then
which is
• When Z>0 almost surely, is defined and we may replace X in
by
to obtain
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Definition 1.6.3
• (測度等價)
• Let Ω be a nonempty set and F a σ-algebra of subsets of Ω. Two probability
measures P and on (Ω, F) are said to be equivalent if they agree which
sets in F have probability zero.
•
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Definition 1.6.3 - Description
• Under the assumptions of Theorem 1.6.1, including the assumption that
Z>0 almost surely, P and are equivalent. Support
is given and
P(A)=0. Then the random variable IAZ is P almost surely zero, which implies
On the other hand, suppose
almost surely under , so
satisfies
. Then
Equation
(1.6.5) implies P(B)=EIB=0. This shows that P and
agree which sets have probability zero.
•
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Example 1.6.4 (1)
• Let Ω=[0,1], P is the uniform (i.e., Lebesgue) measure, and
Use the fact that dP(ω)=dω, then
B[0,1] is a σ-algebra generated by the close intervals.
Since [a,b] [0,1] implies
This is
with Z(ω)=2ω
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Example 1.6.4 (2)
• Note that Z(ω)=2ω is strictly positive almost surely (P{0}=0), and
According to Theorem 1.6.1, for every nonnegative random variable X(ω),
we have the equation
This suggests the notation
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Example 1.6.4 - Description
• In general, when P, , and Z are related as in Theorem 1.6.1, we may
rewrite the two equations
and
as
A good way to remember these equations is to formally write
Equation
is a special case of this notation
that captures the idea behind the nonsensical equation
that Z
is somehow a “ratio of probabilities.”
• In Example 1.6.4, Z(ω) is in fact a ration of densities:
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Definition 1.6.5
• (Radon-Nikodým derivative)
• Let (Ω, F, P) be a probability space, let be another probability measure
on (Ω, F) that is equivalent to P, and let Z be an almost surely positive
random variable that relates P and via
(1.6.3). Then Z is
called the Radon-Nikodým derivative of with respect to P, and we write
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Example 1.6.6 (1)
• (Change of measure for a normal random variable)
• X~N(0,1) with respect to P, Y=X+θ~N(θ,1) with respect to P.
Find such that Y~N(0,1) with respect to
Sol
Find Z>0, EZ=1,
Let
, and Z>0 is obvious because Z is defined as an exponential.
• And EZ=1 follows from the integration
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Example 1.6.6 (2)
• Where we have made the change of dummy variable y=x+θ in the last
step.
• But
density, is equal to one.
, being the integral of the standard normal
where y=x+θ.
• It shows that Y is a standard normal random variable under the
probability .
• 當機率分配函數定義下來之後，一切特性都定義下來了。
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Theorem 1.6.7
• (Radon-Nikodým)
• Let P and be equivalent probability measures defined on (Ω, F). Then
there exists an almost surely positive random variable Z such that EZ=1
and
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