The Radon-Nikodym Theorem
Let (ω, F , P) be a probability space. Suppose a random variable X ≥ 0 a.s. has the property E{X| = 1. Then if we define a set function Q on F by $$ Q\left( \wedge \right) = E\left\{ {1_ \wedge X} \right\} $$ (28.1) then it is easy to see that Q defines a
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Let (Ω, F, P ) be a probability space. Suppose a random variable X ≥ 0 a.s. has the property E{X} = 1. Then if we define a set function Q on F by Q(∧) = E{1∧ X}
(28.1)
then it is easy to see that Q defines a new probability (see Exercise 9.5). Indeed Q(Ω) = E{1Ω X} = E{X} = 1 and if A1 , A2 , A3 , . . . are disjoint in F then ∞ 6 Ai = E{1∪∞ Q A X} i =1 i i=1
=E
∞
1Ai X
i=1
= =
∞ i=1 ∞
E {1Ai X} Q(Ai )
i=1
and we have countable additivity. The interchange of the expectation and the summation is justified by the Monotone Convergence Theorem (Theorem 9.1(d)). Let us consider two properties enjoyed by Q: (i) If P (∧) = 0 then Q(∧) = 0. This is true since Q(∧) = E{1∧ X}, and then 1∧ is a.s. 0, and hence 1∧ X = 0 a.s. (ii) For every ε > 0 there exists δ > 0 such that if ∧ ∈ F and P (∧) < δ, then Q(∧) < ε. Indeed property (ii) follows from Property (i) in general. We state it formally. Theorem 28.1. Let P, Q be two probabilities such that P (∧) = 0 implies Q(∧) = 0 for all ∧ ∈ F. Then for each ε > 0 there exists δ > 0 such that if ∧ ∈ F and P (∧) < δ, then Q(∧) < ε. J. Jacod et al., Probability Essentials © Springer-Verlag Berlin Heidelberg 2004
244
28 The Radon-Nikodym Theorem
Proof. Suppose the result were not true. Then there would be a sequence ∧n ∈ F with P (∧n ) < 21n (for example) and Q(∧n ) ≥ ε, all n, for some ε > 0. Set ∧ = lim supn→∞ ∧n . By Borel-Cantelli Lemma (Theorem 10.5) we have P (∧) = 0. Fatou’s lemma has a symmetric version for limsups, which we established in passing during the proof of Theorem 9.1(f); this gives Q(∧) ≥ lim sup Q(∧n ) ≥ ε, n→∞
and we obtain a contradiction.
It is worth noting that conditions (i) and (ii) are actually equivalent. Indeed we showed (i) implies (ii) in Theorem 28.1; that (ii) implies (i) is simple: suppose we have (ii) and P (∧) = 0. Then for any ε > 0, P (∧) < δ and so P (∧) < ε. Since ε was arbitrary we must have Q(∧) = 0. Definition 28.1. Let P, Q be two finite measures. We say Q is absolutely continuous with respect to P if whenever P (∧) = 0 for ∧ ∈ F, then Q(∧) = 0. We denote this Q P . Examples: We have seen that for any r.v. X ≥ 0 with E{X} = 1, we have Q(∧) = E{1∧ X} gives a probability measure with Q P . A naturally occurring example is Q(∧) = P (∧ | A), where P (A) > 0. It is trivial to check that P (∧) = 0 implies Q(∧) = 0. Note that this example 1 is also of the form Q(∧) = E{1∧ X}, where X = P (A) 1A . The Radon–Nikodym theorem characterizes all absolutely continuous probabilities. Indeed we see that if Q P , then Q must be of the form (28.1). Thus our original class of examples is all that there is. We first state a simplified version of the theorem, for separable σ-fields. Our proof follows that of P. A. Meyer [15]. Definition 28.2. A sub σ-algebra G of F is separable if G = σ(A1 , . . ., An , . . .), with Ai ∈ F, all i. That is, G is generated by a countable sequence of events. Theorem 28.2 (Radon-Nikodym). Let (Ω, F, P ) be a probability space with a separable σ-algebra F. If Q is a finite measure on F and if P (∧) = 0 implie
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