The goal is to find the distribution of the transformed random variable or joint distribution
of the random vector.
Theorem 8.1.1 (Change of variables in one dimension). Let X be a continuous r.v. with PDF \(f_X\), and let Y = g(X), where g is differentiable and strictly increasing (or strictly decreasing). Then the PDF of Y is given by: \[ f_Y(y) = f_X(x)\left| \dfrac{dx}{dy} \right|\\ \] proof: \[ \begin{align} F_Y(y) &= P(Y ≤ y)\\ &= P(g(X) ≤ y) \\ &= P(X ≤ g^{-1}(y)) \\ &= F_X(g^{-1}(y)) \\ &= F_X(x) && (y = g(x)) \\ \end{align} \]
So we can have the formula: \(f_Y(y) = \dfrac{dF_Y(y)}{dy} = \dfrac{dF_X(x)}{dy} = \dfrac{dF_X(x)}{dx}\dfrac{dx}{dy} = f_X(x)\left| \dfrac{dx}{dy} \right|\)
Theorem 8.1.5 (Change of variables). Let X = (X1, . . . ,Xn) be a continuous random vector with joint PDF X, and let Y = g(X) where g is an invertible function from \(R^n\) to \(R^n\). (The domain of g does not actually have to be all of \(R^n,\) but it does need to be large enough to contain the support of X, else g(X) could be undefined!) \[ \begin{align} f_Y(y) = f_X(x)\left|\dfrac{\partial x}{\partial y}\right| \end{align} \] The \(\dfrac{\partial x}{\partial y}\) is the Jacobian matrix.
The Box-Muller example: Let U ~ Unif(0, \(2\pi\)), and let T ~ Exp(1) be independent of U. Define X = \(\sqrt{2T}cosU\) and Y = \(\sqrt{2T}sinU\). Find the joint PDF of (X, Y ). Are they independent? What are their marginal distributions? \[ \begin{align} f_{X,Y}(x, y) &= f_{T,U}(t,u) \left| \begin{array}{cc} \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y} \\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \end{array} \right|\\ &= 1 \cdot e^{-t} \frac{1}{2\pi} \cdot 1 \\ &= \frac{e^{\frac{x^2 + y^2}{-2}}}{2\pi} \\ &= \frac{1}{\sqrt{2\pi}}e^{\frac{x^2}{-2}} \cdot \frac{1}{\sqrt{2\pi}}e^{\frac{y^2}{-2}} \end{align} \]