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1.6 Change of Measure. 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
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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 then it will division by zero. We could rewrite this equation as • But the equation tells us nothing about the relationship among and Z. Because , the value of Z(ω) • However, we should do this set-by-set, rather than ω-by-ω
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 defineThen 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.
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
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 , thenwhich is • When Z>0 almost surely, is defined and we may replace X inby to obtain
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.
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 impliesOn the other hand, suppose satisfies . Then almost surely under , soEquation (1.6.5) implies P(B)=EIB=0. This shows that P and agree which sets have probability zero.
Example 1.6.4 (1) • Let Ω=[0,1], P is the uniform (i.e., Lebesgue) measure, andUse the fact that dP(ω)=dω, thenB[0,1] is a σ-algebra generated by the close intervals.Since [a,b] [0,1] impliesThis is with Z(ω)=2ω
Example 1.6.4 (2) • Note that Z(ω)=2ω is strictly positive almost surely (P{0}=0), andAccording to Theorem 1.6.1, for every nonnegative random variable X(ω), we have the equationThis suggests the notation
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 asA 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:
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
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 SolFind Z>0, EZ=1, Let , and Z>0 is obvious because Z is defined as an exponential. • And EZ=1 follows from the integration
Example 1.6.6 (2) • Where we have made the change of dummy variable y=x+θ in the last step. • But , being the integral of the standard normal density, is equal to one. 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。
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