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




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.