A.2 Absolutely continuous probability measures

Suppose that P and Q are two probability measures on a measurable space (Ω,F).

Definition A.10. Q is said to be absolutely continuous with respect to P on the σ-algebra F, and we write Q P, if for all A ∈ F,

If both Q P and P Q hold, we will say that Q and P are equivalent, and we will write Q ≈ P.

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The following characterization of absolute continuity is known as the Radon–Nikodym theorem:

Theorem A.11 (Radon–Nikodym). Q is absolutely continuous with respect to P on F if and only if there exists an F-measurable function φ ≥ 0 such that

for all F-measurable functions F ≥ 0.

Proof. See, e.g., §17 of [21].

The function φ is called the density or Radon–Nikodym derivative of Q with respect to P, and we will write

Clearly, the Radon–Nikodym derivative is uniquely determined through (A.8) up to a P-nullset and it satisfies

Remark A.12. If Q P, then

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Corollary A.13. If Q P on F, then

In this case, the density of P with respect to Q is given by

Proof. Suppose that Q P, let φ := dQ/dP, and take an F-measurable function F ≥ 0. By Remark A.12, we have Q[ φ = 0 ] = 0. Therefore,

The rightmost integral equals ∫ F dP for all F if and only if P[ φ = 0 ] = 0. This proves the result.

The preceding result admits the following extension.

Exercise A.2.1. Let P, Q, and R be three probability measures on (Ω,F) such that R Q and Q P. Show that R P and that the corresponding Radon-Nikodym derivative can be obtained in the following way:

Note that we can obtain Corollary A.13 as a special case by choosing R := P.

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Remark A.14. Let us stress that absolute continuity depends on the underlying σ-field F. For example, let P be the Lebesgue measure on Ω := [0, 1). Then every probability measure Q is absolutely continuous with respect to P on a σ-algebra F0 which is generated by finitely many intervals [ai−1, ai) with 0 = a0 < a1 < ·· · < an = 1. However, if we take for F the Borel σ-algebra on Ω, then for instance a Dirac point mass Q = δx is clearly not absolutely continuous with respect to P on F.

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While the preceding example shows that, in general, absolute continuity is not preserved under an enlargement of the underlying σ-algebra, the next proposition states that it is safe to take smaller σ-algebras. This proposition involves the notion of a conditional expectation

E[ F | F0 ]

of an F-measurable function F ≥ 0 with respect to a probability measure P and a σ-algebra F0 ⊂ F. Recall that E[ F | F0 ] may be defined as the P-a.s. unique F0-measurable random variable F 0 such that

see, e.g., §15 of [20]. Note also our convention of writing

Clearly, we can replace in (A.9) the class of all indicator functions of sets in F0 by the class of all bounded F0-measurable functions or by the class of all nonnegative F0-measurable functions.

Proposition A.15. Suppose that Q and P are two probability measures on the measurable space (Ω,F) and that Q P on F with density φ. If F0 is a σ-algebra contained in F, then Q P on F0, and the corresponding density is given by

Proof. Q P on F0 follows immediately from the definition of absolute continuity. Since φ is the density on F ⊇ F0, it follows for A ∈ F0 that

Therefore the F0-measurable random variable E[ φ | F0 ] must coincide with the density on F0.

Now we prove a formula for computing a conditional expectation EQ[ F | F0 ] under a measure Q in terms of conditional expectations with respect to another measure P with Q P.

Proposition A.16. Suppose that Q P on F with density φ, and that F0 ⊆ F is another σ-algebra. Then, for any F-measurable F ≥ 0,

Proof. We must show that

for all F0-measurable G0 ≥ 0, where φ0 := E[φ | F0] is the density of Q with respect to P on F0 by Proposition A.15. Since φ0 > 0 Q-a.s. by Remark A.12, we may assume without loss of generality that G0 = 0 on {φ0 = 0}. Then, on the one hand,

On the other hand,

Comparing (A.11) with (A.12) yields (A.10).

If neither Q P nor P Q holds, one can use the following Lebesgue decomposition of P with respect to Q.

Theorem A.17. For any two probability measures Q and P on (Ω,F), there exists a set N ∈ F with Q[ N ] = 0 and an F-measurable function φ ≥ 0 such that

One writes

Proof. Let Then both Q and P are absolutely continuous with respect to R with respective densities dQ/dR and dP/dR. Let

Then Q[ N ] = 0. We define

Then, for A ∈ F,

where we have used the fact that Q[ N ] = 0 in the last step.