1. Settings
Two measurable spaces:
\[(X,\mathcal{A},\mu)\quad\text{and}\quad (Y,\mathcal{B},\mu_T)\]
Consider a transform \(T\) such that:
\[T:X\rightarrow Y\]
where T is a measurable map. i.e., \(T^{-1}(B)\in \mathcal{A}\quad \forall B\in\mathcal{B}\).
\(T\) induces a pushforward measure \(\mu_T\):
\[\mu_T(B)=\mu(T^{-1}(B))\]
2. Change-of-variables
Let \(\nu\) and \(\mu_T\) are measures defined on \(Y\) and \(\nu\) is dominated by \(\mu_T\) (\(\nu\ll\mu_T\)). Then there exists a Radon-Nikodym derivative \(\rho\) such that:
\[\forall B \in \mathcal{B},\quad \nu(B)=\int_B \rho(y)d\mu_T(y) \]
where \(\rho:Y\rightarrow [0,\infty]\). (I missed this point before: The Radon-Nikodym assumes the measures are over the same measurable space.)
Let \(f\) be a nonnegative measurable function on \(Y\). Then,
\[\int_Y f(y)d\mu_T(y)=\int_Xf(T(x))d\mu(x)\]
\[\begin{aligned} \int_Y f(y)d\nu(y)&=\int_Y f(y)\frac{d\nu}{d\mu_T}d\mu_T(y)\\&=\int_Yf(y)\rho(y)d\mu_T(y)\\&=\int_X f(T(x))\rho(T(x))d\mu(x) \end{aligned}\]
3. Examples
- Random variable \(Z: \Omega \rightarrow \mathbb{R}\)
- A random variable defines a pushforward measure \(\mu_Z\) over the real number line. Assume that the pushforward measure \(\mu_Z\) is dominated by Lebesgue measure\(\lambda\) then \(f=\frac{d\mu_Z}{d\lambda}\). The Radon-Nikodym theorem is applied between the measures \(\mu_T, \lambda\) over the same measurable space \((\mathbb{R}, \mathcal{R})\)
- Inverse transform sampling
- Normalizing flows
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