关于连续型随机变量的分布转换，除了使用反CDF函数将均匀分布函数转换为其他分布函数以外，存在一种通用办法实现不同分布之间的转换，不只局限于均匀分布。

假定存在随机变量$X$，其累积分布函数为$F_X(x)$。若存在单调函数$g$，使得$Y=g(X)$（因为是单调函数，因此$X$与$Y$的关系为单射），则为了计算$Y$的累积分布函数$F_Y(y)$，

设$g(x)$此时为严格单调递增

$\begin{array}{ccc}{F}_Y\left(y\right)& =& Pr\{Y⩽y\}\\ & =& Pr\left\{g\right(X)⩽y\}\\ & =& Pr\{X⩽g^{-1}(y\left)\right\}\\ & =& F_X\left(g^{-1}\right(y\left)\right)\end{array}$
\begin{array}{l}
F_Y(y) & = & Pr\{ Y\leqslant y \} \\
& = & Pr\{g(X)\leqslant y\} \\
& = & Pr\{ X\leqslant g^{-1}(y) \} \\
& = & F_X(g^{-1}(y))
\end{array}

此时对终式左右对x求导(累积分布函数导数为概率密度函数)

$\begin{array}{c}{f}_Y\left(y\right)\frac{dy}{dx}=f_X\left(g^{-1}\right(y\left)\right)\\ f_Y\left(y\right)=\left(\frac{dy}{dx}{)}^{-1}{f}_X\right(g^{-1}\left(y\right))\end{array}$
\begin{array}{l}
f_Y(y)\frac{dy}{dx}=f_X(g^{-1}(y)) \\
f_Y(y) = (\frac{dy}{dx})^{-1}f_X(g^{-1}(y)) \\
\end{array}

再设$g(x)$此时为严格单调递减

$\begin{array}{ccc}{F}_Y\left(y\right)& =& Pr\{Y⩽y\}\\ & =& Pr\left\{g\right(X)⩽y\}\\ & =& 1-Pr\{X⩽g^{-1}(y\left)\right\}\\ & =& 1-F_X\left(g^{-1}\right(y\left)\right)\end{array}$
\begin{array}{l}
F_Y(y) & = & Pr\{ Y\leqslant y \} \\
& = & Pr\{g(X)\leqslant y\} \\
& = & 1 - Pr\{ X\leqslant g^{-1}(y) \} \\
& = & 1 - F_X(g^{-1}(y))
\end{array}

同样的有
$f_Y(y) = -(\frac{dy}{dx})^{-1}f_X(g^{-1}(y))$

则有$f_Y(y) = \vert (\frac{dy}{dx})^{-1} \vert f_X(g^{-1}(y))$。

PBRT中讲到，当我们有一概率密度为$p_x(x)$的随机变量$X$，想要转换为满足已知分布为$p_y(y)$的随机变量$Y$，那么即有下式成立(此过程与上式正好相反)

$\begin{array}{ccc}{F}_X\left(x\right)& =& Pr\{X⩽x\}\\ & =& Pr\left\{g\right(X)⩽g(x\left)\right\}\\ & =& Pr\{Y⩽g(x\left)\right\}\\ & =& F_Y\left(g\right(x\left)\right)\end{array}$
\begin{array}{l}
F_X(x) & = & Pr\{ X\leqslant x \} \\
& = & Pr\{g(X)\leqslant g(x)\} \\
& = & Pr\{ Y\leqslant g(x) \} \\
& = & F_Y(g(x))
\end{array}

$g(x) = F_Y^{-1}(F_X(x))$或当$g(X)$为单调减函数时$g(x) = F_Y^{-1}(1 - F_X(x))$，PBRT中未对此处进行详细说明。

因此当$X$为满足$[0, 1]$区间均匀分布的随机变量时，$F_X(x)=x$，那么上式中$F_X(x))$或是$1 - F_X(x)$都满足$[0, 1]$区间均匀分布，则$g(x)=F_Y^{-1}(x)$，即Inverse Transform Sampling得证。

多维随机变量的转换$T(x)$为双射变换，则满足$f_Y(y)=f_Y(T(x))=\frac{f_X(x)}{\vert J_T(x)\vert }$，且$T(x)=(T_1(x),\cdots T_n(x))$

$J_T(x)=\left[$
\begin{matrix}
\frac{\partial T_1}{\partial x_1} & \cdots & \frac{\partial T_1}{\partial x_n} \\
\vdots & \ddots & \vdots \\
\frac{\partial T_n}{\partial x_1} & \cdots & \frac{\partial T_n}{\partial x_n} \\
\end{matrix}
\right]