Radon-Nikodym theorem suggests, in simplified terms, that if we have two measures $$\mu,\lambda$$ of the same space with $$\mu$$ absolutely continuous with respect to $$\lambda$$ and a function $$\phi$$ is $$\mu$$-integrable, then $$\int_A \phi d\mu = \int_A \phi \frac{d\mu}{d\lambda} d\lambda$$ which the term $$d\mu/d\lambda$$ is called the Radon-Nikodym derivative.

The Radon-Nikodym derivative is commonly used as a tool for change-of-measure in probability. Girsanov theorem tells you how to find one in case of Brownian motion. However, we have something closely related in probability distribution called the variable transformation formula: If we have a function $$Y=\phi(X)$$ and $$X$$ is a random variable with distribution function $$F_X(x)$$, then

$f_Y(y)=f_X(\phi^{-1}(y))\left\lvert\frac{1}{\phi'(\phi^{-1}(y))}\right\rvert$

or

$f_X(x)=f_Y(\phi(x))\lvert\phi'(x)\rvert.$

This can be derived from using the CDF of $$Y$$, assuming $$\phi(x)$$ is an increasing function:

\begin{align} F_Y(y) &= \Pr[Y<y] = \Pr[\phi(X)<y] \\ &= \Pr[X<\phi^{-1}(y)] \\ &= F_X(\phi^{-1}(y)) \\ f_Y(y) = \frac{d}{dy}F_Y(y) &= \frac{d}{dy}F_X(\phi^{-1}(y)) \\ &= f_X(\phi^{-1}(y))\frac{d}{dy}\phi^{-1}(y) \\ &= f_X(\phi^{-1}(y))\frac{1}{\phi'(g^{-1}(y))}. \\ \end{align}

which the above applied the inverse function theorem

$\frac{d}{dy}h^{-1}(y) = \frac{1}{\frac{d}{dx}h(x)} = \frac{1}{h'(h^{-1}(y))}.$

In case of $$\phi(x)$$ a descreasing function, we have $$\Pr[\phi(X)<y]=\Pr[X>\phi^{-1}(y)]=1-F_X(\phi^{-1}(y))$$ but the derivative $$\phi'(x)$$ is negative. Therefore the absolute value in the formula above.

The variable transformation formula is useful to derive new distribution function. For example if $$X\sim N(\mu, \sigma^2)$$, we have $$f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2}),$$ and with $$Y=\phi(X)=e^X$$, we get the distribution for log-normal as

\begin{align} f_Y(y) &= f_X(x)\frac{1}{\phi'(x)} \\ &= \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2}) \frac{1}{e^x} \\ &= \frac{1}{y\sqrt{2\pi\sigma^2}}\exp(-\frac{(\ln y-\mu)^2}{2\sigma^2}). \end{align}

Radon-Nikodym derivative, however, is useful in reverse. Assume that we have two random variables, both in the domain of $$[0,1]$$, and with the distribution functions

\begin{align} F_X(x) &= x\mathbb{I}_{0\le x\le 1} \\ F_Y(y) &= \sqrt{y}\mathbb{I}_{0\le y\le 1} \end{align}

then we can find the function $$Y=\phi(X)$$ according to the Radon-Nikodym theorem and the variable transformation formula:

\begin{align} \frac{dF_Y}{dF_X} &= \left\lvert\frac{1}{\phi'(x)}\right\rvert \\ \frac{1/(2\sqrt{y})}{1} &= \frac{1}{\lvert y'\rvert} \\ y' &= 2\sqrt y \\ y^{-1/2} dy &= 2dx \end{align}

and the solution to the above separable differential equation is $$y=(x+c)^2$$. Plug in the boundary condition of $$X,Y\in[0,1]$$, we get $$c=0,-1$$. Hence $$Y=X^2$$ or $$Y=(X-1)^2$$ will both have the same distribution function as above.