Jacobian

Obtaining the pdf of a transformed variable (using a one-to-one transformation) is simple using the Jacobian (Jacobian of inverse) $$\begin{array}{rl}Y& =g(X)\\ X& =g^{-1}(Y)\\ f_Y(y)& =f_X(g^{-1}(y))|\frac{dx}{dy}|\end{array}$$

The modulus ensures that the probability density is positive when the transformation is either increasing or decreasing.

For the two variable case, let $\mathbf{X}=(X_1,X_2)$ denote the random vector. Suppose we have a one-to-one transform (and its inverse) defined as follows $$\begin{array}{rl}\mathbf{Y}& =T(\mathbf{X})\\ \mathbf{Y}& =(Y_1,Y_2)\\ Y_1& =u_1(X_1,X_2)\\ Y_2& =u_2(X_1,X_2)\\ T^{-1}& =(w_1(Y_1,Y_2),w_2(Y_1,Y_2))\\ X_1& =w_1(Y_1,Y_2\\ X_2& =w_2(Y_1,Y_2\end{array}$$

Then, the pdf is given by the expression $$\begin{array}{rl}{f}_Y(y)& =f_X(w_1(y_1,y_2),w_2(y_1,y_2))|\mathbf{J}|\\ \mathbf{J}& =\left|\begin{array}{cc}\frac{\partial x_1}{y_1}& \frac{\partial x_1}{y_2}\\ \frac{\partial x_2}{y_1}& \frac{\partial x_2}{y_2}\end{array}\right|\end{array}$$

If $A$ denotes the support of $\mathbf{X}$, then $B=T(A)$ will denote the support of $\mathbf{Y}$. Everywhere else, the value of pdf for $Y$ will be zero. Refer to the exercises section for a demonstration of this method.

This method is powerful and applies to change of variables when doing integrals as well. $$\begin{array}{r}\int \int f_X(x_1,x_2)d{x}_{1}d{x}_2=\int \int f_Y(w_1(y_1,y_2),w_2(y_1,y_2))|J|d{y}_{1}d{y}_2\end{array}$$ where $J$ is the same as defined earlier.

The two variable expression extends to multiple variables as well. Suppose the transformations are given by $$\begin{array}{rl}{Y}_i& =u_i(X_1,\dots X_n)\mathrm{\forall }i=1,2,\dots ,n\\ X_i& =w_i(Y_1,\dots Y_n)\mathrm{\forall }i=1,2,\dots ,n\\ ⟹f_{\mathbf{Y}}(\mathbf{y})& =f_{\mathbf{X}}(w_1(\mathbf{y}),\dots ,w_n(\mathbf{y}))\left|\mathbf{J}\right|\\ \text{with}\mathbf{J}& =\left|\begin{array}{ccc}\frac{\partial x_1}{\partial y_1}& \cdots & \frac{\partial x_1}{\partial y_n}\\ ⋮& \ddots & ⋮\\ \frac{\partial x_n}{\partial y_1}& \cdots & \frac{\partial x_n}{\partial y_n}\end{array}\right|\end{array}$$

However, we may not be always have a one-to-one transformation. The simplest case is $Y=X^2$. In such scenarios, we will partition the space so that each has a one-to-one transform.

For instance, in the above example the partitions will be $-\mathrm{\infty }<x<0$ and $0\le x<\mathrm{\infty }$. Then, we individually calculate the pdf for $Y$ in all the partitions. If the partitions have an overlap (from the view of support of $y$), we will sum up the corresponding pmf to get the final pmf.

Continuing with the same example, both the partitions map to $0<\le y<\mathrm{\infty }$. Hence, after finding the pmf in each partition, we can simply sum it up (due to symmetry of the transform). Refer to the exercises for an illustration of this principle.