Change of Variables
Last modified — 13 Nov 2026
By the end of this lecture, students are anticipated to be able to:
Use the Distribution method to transform discrete or continuous CDFs
Use the Jacobian method to transform continuous PDFs
Sometimes we want to perform transformations on a random variable. This is when we can turn to change of variables.
There are two ways to do this: the distribution method and the Jacobian method. We will present both.
The distribution method just uses the definition:
\[\mathbb{P}(Y \in A) = \mathbb{P}(g(X) \in A) = \mathbb{P}(\{x : g(x) \in A\}).\]
If we can characterize these sets, we can find the distribution of \(Y\). This method always works, and is easy for discrete random variables.
Let \(X\) be a discrete random variable with PMF \(p_X(x)\). Let \(Y = g(X)\) for some function \(g : {\mathbb{R}}\to {\mathbb{R}}\). Then the PMF of \(Y\) is given by \[p_Y(y) = \sum_{x : g(x) = y} p_X(x) = \sum_{x \in g^{-1}(y)} p_X(x).\]
Let’s make this method a little more algorithmic.
If you want to find the distribution of \(Y = g(X)\), then you can do the following: \[\mathbb{P}(Y \in A) = \mathbb{P}(g(X) \in A) = \mathbb{P}(\{x : g(x) \in A\}) = \cdots\]
Let’s start with a straightforward example:
Find the PMF of \(Y\).
Let \(X \sim {\mathrm{Exp}}(\lambda)\), find the distribution of \(Y = 3X\).
If \(X \sim {\mathrm{Exp}}(\lambda)\), what is the CDF of \(Y = X + 5\)?
If \(X \sim {\mathrm{Unif}}(0,1)\), what is the CDF of \(Y = X + 3\)?
The following useful results follow from the previous examples
If \(X\) be a random variables with CDFs \(F_{X}\) Then the the following hold:
For continuous random variables, we can often use the Jacobian method in addition to the distribution method.
Let \(X\) be an (absolutely) continuous random variable, with density function \(f_X\). Let \(Y = h(X)\), where \(h : {\mathbb{R}}\to {\mathbb{R}}\) is a function that is differentiable and monotonic.
Then \(Y\) is also absolutely continuous, and its density function \(f_Y\) is given by \[f_Y(y) = f_X(h^{-1}(y)) \left| \frac{\mathsf{d}}{\mathsf{d}y} (h^{-1}(y)) \right|,\] where \(h^{-1}(y)\) is the unique number \(x\) such that \(h(x) = y\).
Find the PDF of \(Y\).
Let \(X \sim {\mathrm{Unif}}\left( -1,1\right)\). Use the distribution method to find the PDF of \(Z = X^2\).
Let \(X \sim {\mathrm{Gam}}(\alpha, \lambda)\). Use the Jacobian method to find the PDF of \(Y = 1/X\).
Recall the Gamma PDF: \[f_X(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\lambda x} I_{(0, \infty)}(x).\]
Stat 302 - Winter 2025/26