I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variabl

Erin Lozano

Erin Lozano

Answered question

2022-06-20

I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let ( Ω , F ) be a measurable space, and P and Q be two probability measures on this space such that Q P . Radon-Nikodym theorem tells us that there exists a F -measurable function f : Ω [ 0 , ) denoted by d Q d P , such that
Q ( E ) = E f ( ω ) d P ( ω )
for all E F .
If somehow we know the expression for f ( ω ) already, and X : Ω R d is a random vector, is there a way to derive the Radon-Nikodym derivative d Q X d P X for the induced distribution measures Q X and P X on ( R d , B ( R d ) )? My rough guess was d Q X d P X ( X ( ω ) ) = f ( ω ) but am not sure how to prove/disprove it.

Answer & Explanation

Lisbonaid

Lisbonaid

Beginner2022-06-21Added 22 answers

Your guess that
d Q X d P X X = f
holds is correct but to define properly that Radon-Nikodym derivative we need one more restriction on X, namely, hat X is invertible.
By definition:
(1) Q X ( B ) = Q ( X B ) = Q ( X 1 ( B ) ) = X 1 ( B ) f ( ω ) d P ( ω ) .
When X is invertible then R = f X 1 is a map from R d to [ 0 , ) .
This R is the Radon-Nikodym derivative you are looking for
(2) d Q X d P X = R = f X 1 .
To see this observe first that for every B B ( R d ),
(3) P X ( B ) = R d 1 B d P X = Ω 1 X 1 ( B ) d P = Ω ( 1 B X ) d P .
The function R is the supremum of simple functions of the form
i = 1 n R i 1 C i , R i R , C i B ( R d ) .
By monotone convergence, and using (3),
B R d P X = sup n i = 1 n R i B 1 C i d P X = sup n i = 1 n R i Ω ( 1 B C i X ) d P = sup n i = 1 n R i X 1 ( B ) ( 1 C i X ) d P
Note however that
sup n i = 1 n R i ( 1 C i X ) = R X = f .
Therefore, using (1)
B R d P X = X 1 ( B ) f d P = Q X ( B ) .
This shows (2).

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