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.

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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.

Lisbonaid · Accepted Answer

Your guess that            d                        Q                X                    d                        P                X              ∘  X  =  fholds 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).

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

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