# Radon-Nikodym Derivative and Push Forward of Measures

I am wondering how the Radon-Nikodym derivative is affected by push-forwarding with a random variable. Formally, let $$(\Omega, \mathcal{F})$$ be a measurable space, and $$\mathbb{P}$$ and $$\mathbb{Q}$$ be two probability measures on this space such that $$\mathbb{Q} \ll \mathbb{P}$$. Radon-Nikodym theorem tells us that there exists a $$\mathcal{F}$$-measurable function $$f: \Omega \rightarrow [0, \infty)$$ denoted by $$\frac{d\mathbb{Q}}{d\mathbb{P}}$$, such that

$$\mathbb{Q}(E) = \int_E f(\omega) d\mathbb{P}(\omega)$$

for all $$E \in \mathcal{F}$$.

If somehow we know the expression for $$f(\omega)$$ already, and $$X: \Omega \rightarrow \mathbb{R}^d$$ is a random vector, is there a way to derive the Radon-Nikodym derivative $$\frac{d\mathbb{Q}^{X}}{d\mathbb{P}^X}$$for the induced distribution measures $$\mathbb{Q}^X$$ and $$\mathbb{P}^X$$ on $$(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d))$$? My rough guess was $$\frac{d\mathbb{Q}^{X}}{d\mathbb{P}^X}(X(\omega)) = f(\omega)$$ but am not sure how to prove/disprove it.

• I don't think there is a general formula for what you are asking for especially if $\Omega$ is just a set. Just apply the standard change of variables for the push forward and that should be the best you can get. Commented Mar 26, 2022 at 4:29

Your guess that $$\frac{d\mathbb Q^X}{d\mathbb P^X}\circ X=f$$ holds is correct but to define properly that Radon-Nikodym derivative we need one more restriction on $$X$$, namely, that $$X$$ is invertible.
By definition: \begin{align}\tag{1} {\mathbb Q}^X(B)&=\mathbb Q(X\in B)=\mathbb Q(X^{-1}(B))=\int_{X^{-1}(B)}f(\omega)\,d\mathbb P(\omega)\,. \end{align} When $$X$$ is invertible then $$R=f\circ X^{-1}$$ is a map from $$\mathbb R^d$$ to $$[0,\infty)\,.$$
This $$R$$ is the Radon-Nikodym derivative you are looking for $$\tag{2}\boxed{ \quad\frac{d\mathbb Q^X}{d\mathbb P^X}=R=f\circ X^{-1}\,.\quad}$$ To see this observe first that for every $$B\in{\cal B}(\mathbb R^d)$$, $$\tag{3} \mathbb P^X(B)=\int_{\mathbb R^d} 1_B\,d\mathbb P^X=\int_\Omega 1_{X^{-1}(B)}\,d\mathbb P=\int_\Omega (1_B\circ X)\,d\mathbb P\,.$$ The function $$R$$ is the supremum of simple functions of the form $$\sum_{i=1}^nR_i1_{C_i}\,,\quad\quad R_i\in\mathbb R\,,\quad C_i\in{\cal B}(\mathbb R^d)\,.$$ By monotone convergence, and using (3), \begin{align} &\int_B R\,d\mathbb P^X=\sup_n\sum_{i=1}^nR_i\int_B1_{C_i}\,d\mathbb P^X=\sup_n\sum_{i=1}^nR_i\int_\Omega (1_{B\cap C_i}\circ X)\,d\mathbb P\\ &=\sup_n\sum_{i=1}^nR_i\int_{X^{-1}(B)} (1_{C_i}\circ X)\,d\mathbb P \end{align} Note however that $$\sup_n\sum_{i=1}^nR_i(1_{C_i}\circ X)=R\circ X=f.$$ Therefore, using (1) $$\int_{B} R\,d\mathbb P^X=\int_{X^{-1}(B)} f\,d\mathbb P=\mathbb Q^X(B)\,.$$ This shows (2).
• Thank you for the answer! I've been wondering whether using simple functions is needed: provided that there exists a measurable left inverse $a$ of $X$, the change of variables theorem should give: \begin{align*} \int_B f\circ a \, \mathrm{d}X_\sharp P &= \int_{X^{-1}(B)} f\circ (a\circ X) \, \mathrm{d}P \\ &= \int_{X^{-1}(B)} f\, \mathrm{d}P \\ &= \int_{X^{-1}(B)} \mathrm{d}Q \\ &= Q(X^{-1}(B)) = X_\sharp Q(B), \end{align*} so that $f\circ a = \mathrm{d}X_\sharp Q / \mathrm{d}X_\sharp P$ from the Radon–Nikodym theorem. Please, let me know if I'm wrong! Commented May 12 at 22:52
• @PawełCzyż looks right. Just one thing: I know that in one dimension $X$ can have no inverse but a left inverse. I have never thought about the higher dimensional case. Commented May 13 at 4:34
• @guest1 Yes. It is a fairly strong assumption. It assumes that $X$ is a bijection between its domain $\Omega$ and its range (here $\mathbb R^n$). A typical probability space $\Omega$ is far larger than $\mathbb R^n\,.$ To handle a space more general than $\mathbb R^n$ you can use Pawel Czyz' remark. Commented Jun 4 at 10:27
• @guest1 . Pawel's proof in his first comment should go through for any codomain of $X$ that carries a $\sigma$-algebra. Please think of it and do not throw around terms that you learned in some measure theory class. That $X$ needs to be a bijection won't get less restrictive if we change its domain and codomain. Commented Jun 4 at 11:35
• BTW: exercise for you: the question can be solved purely with the measure theoretic theorem that $$\tag1 \int f'dT(\mu)=\int f'\circ T\,d\mu$$ where $T(\mu)$ is the push-forward (image measure) of $\mu$ on $(\Omega,\mathscr A)$ by $T:(\Omega,\mathscr A)\to (\Omega',\mathscr A')$ and $f'\ge 0$ on $\Omega'\,,$ H. Bauer, Measure and Integration Theory Thm. 19.1. plus the assumption that $X$ ($T$ in Bauer's notation) is a bijection. Commented Jun 4 at 11:38