4.3 Transformations

Suppose that $X$ has pdf $f_{X}(x)$. What is the pdf of $Y = 3X$? Or if $X=(X_1, X_2, X_3)^{\top}$, what is the pdf of

\begin{displaymath}Y = \left( \begin{array}{c} 3X_1 \\ X_1-4X_2 \\ X_3 \end{array} \right) ? \end{displaymath}

This is a special case of asking for the pdf of $Y$ when
\begin{displaymath} X = u(Y) \end{displaymath} (4.43)

for a one-to-one transformation $u$: $\mathbb{R}^p \rightarrow \mathbb{R}^p$. Define the Jacobian of $u$ as

\begin{displaymath}{\cal J} = \left( \frac{\partial x_i}{\partial y_j} \right) = \left( \frac{\partial u_i(y)}{\partial y_j} \right)\end{displaymath}

and let $\abs(\vert{\cal J}\vert)$ be the absolute value of the determinant of this Jacobian. The pdf of $Y$ is given by
\begin{displaymath} f_Y(y) = \abs(\vert{\cal J}\vert) \cdot f_X\{u(y)\}. \end{displaymath} (4.44)

Using this we can answer the introductory questions, namely

\begin{displaymath}(x_1, \ldots, x_p)^{\top} = u(y_1, \ldots, y_p) = \frac{1}{3} (y_1, \ldots, y_p)^{\top} \end{displaymath}

with

\begin{displaymath}{\cal J} = \left( \begin{array}{ccc} \frac{1}{3} & & 0 \\ & \ddots & \\ 0 & & \frac{1}{3} \end{array} \right) \end{displaymath}

and hence $ \abs(\vert{\cal J}\vert) = \left( \frac{1}{3} \right) ^p $. So the pdf of $Y$ is ${\displaystyle \frac{1}{3^p} f_{X} \left( \frac{y}{3} \right)}$.

This introductory example is a special case of

\begin{displaymath}Y = \data{A}X + b, \textrm{ where } \data{A} \quad \textrm{is nonsingular.} \end{displaymath}

The inverse transformation is

\begin{displaymath}X = \data{A}^{-1}(Y-b). \end{displaymath}

Therefore

\begin{displaymath}{\cal J} = \data{A}^{-1}, \end{displaymath}

and hence
\begin{displaymath} f_Y(y) = \abs(\vert\data{A}\vert^{-1})f_X\{\data{A}^{-1}(y-b)\}. \end{displaymath} (4.45)

EXAMPLE 4.12   Consider $X=(X_{1},X_{2})\in\mathbb{R}^2$ with density $f_X(x)=f_{X}(x_{1},x_{2})$,

\begin{displaymath}\data{A} = \left( \begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array} \right), \quad b = \left( 0 \atop 0 \right).\end{displaymath}

Then

\begin{displaymath}Y = \data{A}X + b = \left( \begin{array}{c} X_1+X_2 \\ X_1-X_2 \end{array} \right) \end{displaymath}

and

\begin{displaymath}\vert\data{A}\vert = -2, \quad \abs(\vert\data{A}\vert^{-1}) ... ...eft( \begin{array}{rr} -1 & -1 \\ -1 & 1 \end{array} \right).\end{displaymath}

Hence
$\displaystyle f_Y(y)$ $\textstyle =$ $\displaystyle \abs(\vert\data{A}\vert^{-1}) \cdot f_{X}(\data{A}^{-1}y)$  
  $\textstyle =$ $\displaystyle \frac{1}{2} f_{X} \left\{ \frac{1}{2} \left( \begin{array}{rr} 1 ... ...array} \right) \left( \begin{array}{c} y_1 \\ y_2 \end{array} \right) \right\}$  
  $\textstyle =$ $\displaystyle \frac{1}{2} f_{X} \left\{ \frac{1}{2}(y_1+y_2), \frac{1}{2}(y_1-y_2) \right\}.$ (4.46)

EXAMPLE 4.13   Consider $X\in\mathbb{R}^1$ with density $f_X(x)$ and $Y=\exp(X)$. According to (4.43) $x=u(y)=\log(y)$ and hence the Jacobian is

\begin{displaymath} {\data{J}}=\frac{dx}{dy}=\frac{1}{y}. \end{displaymath}

The pdf of $Y$ is therefore:

\begin{displaymath} f_Y(y)=\frac{1}{y}f_X\{\log(y)\}. \end{displaymath}

Summary
$\ast$
If $X$ has pdf $f_{X}(x)$, then a transformed random vector $Y$, i.e., $X=u(Y)$, has pdf $f_Y(y) = \abs(\vert{\cal J}\vert) \cdot f_X\{u(y)\}$, where ${\cal J}$ denotes the Jacobian ${\cal J}=\left( \frac{\partial u(y_i)} {\partial y_j} \right)$.
$\ast$
In the case of a linear relation $Y=\data{A}X+b$ the pdf's of $X$ and $Y$ are related via $f_Y(y) = \abs(\vert\data{A}\vert^{-1})f_X\{\data{A}^{-1}(y-b)\}$.