Coordinate Transformation

Jacobian Matrix

If $\mathbf{f}:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is a function that maps from n-dimensional Euclidean space to m-dimensional Euclidean space. The Jacobian matrix of $\mathbf{f}$ is defined to be a $m\times n$ matrix, denoted by $\mathbf{J}$, with $\mathbf{J}_{ij} = \frac{\partial f_i}{\partial x_j}$, or

$$\mathbf{J^f_x} = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n}\\
\vdots &\ddots &\vdots\\
\frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n} \end{bmatrix} $$

where $\mathbf{f} = \mathbf{J_x^f}\cdot \mathbf{x}$. Similarly, $\mathbf{x} = \mathbf{J_f^x}\times \mathbf{f}$ and $\mathbf{J_f^x} = \mathbf{J_x^f}^{-1}$

Coordination Transformation

Given a function $f_X(\mathbf{x})$ and two vectors withe the same dimension $\mathbf{x}$ and $\mathbf{z}$, how to find $f_Z(\mathbf{z})$?

one-to-one map

Find Jacobian matrix $J^z_x$ or $J^x_z$, so that $z = J^z_x \cdot x$ and $x = J_z^x \cdot z$

$$f_Z(\mathbf{z}) = \left|\mathbf{J^x_z}\right| \cdot f_X(\mathbf{\mathbf{J_z^x}\cdot\mathbf{x}}) = f_X(\mathbf{\mathbf{J_z^x}\cdot\mathbf{x}})/|\mathbf{J^z_x}| $$

many-to-one map

$$\begin{aligned}f_Z(\mathbf{z}) &= \sum_{x: g(x)=z} \left|\mathbf{J^x_z}\right| \cdot f_X(\mathbf{\mathbf{J_z^x}\cdot\mathbf{x}}) \\
&= \sum_{x: g(x)=z} f_X(\mathbf{\mathbf{J_z^x}\cdot\mathbf{x}})/|\mathbf{J^z_x}| \end{aligned}$$

example

Question

Let X, Y be continuous random variables with the following joint probability distribution function (p.d.f.)
$$f_{X,Y} = 2e^{-(x+y)},\; 0<x<y$$
Define $z=x+y$, $w=\frac{y}{x}$, determine the joint p.d.f. of z, w.

Answer

$$ x=\frac{z}{w+1}, \; y= \frac{zw}{w+1}\\$$

  • To find the joint p.d.f.

$$\begin{aligned}
J_{xy}^{zw} &= \begin{bmatrix} 1 & 1\ -\frac{y}{x^2} & \frac{1}{x}\end{bmatrix} \\
|J_{xy}^{zw}| &= \frac{1}{x}+\frac{y}{x^2} = \frac{(w+1)^2}{z}\\
f_{Z,W} &= 2e^{-z} \times |J^{xy}_{zw}| = \frac{2ze^{-z}}{(w+1)^2} \end{aligned}$$

Relevant Note

  • $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $A^{-1} = \frac{1}{|A|} \times \begin{bmatrix} d & -b \\\ -c & a \end{bmatrix}$, where $|A|=ad-bc$