Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Probability density function ▷ Probability density function of invertible function

Theorem: Let $X$ be an $n \times 1$ random vector of continuous random variables with possible outcomes $\mathcal{X} \subseteq \mathbb{R}^n$ and let $g: \; \mathbb{R}^n \rightarrow \mathbb{R}^n$ be an invertible and differentiable function on the support of $X$. Then, the probability density function of $Y = g(X)$ is given by

$\label{eq:pdf-invfct} f_Y(y) = \left\{ \begin{array}{rl} f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \; , & \text{if} \; y \in \mathcal{Y} \\ 0 \; , & \text{if} \; y \notin \mathcal{Y} \end{array} \right. \; ,$

if the Jacobian determinant satisfies

$\label{eq:jac-det} \left| J_{g^{-1}}(y) \right| \neq 0 \quad \text{for all} \quad y \in \mathcal{Y}$

where $g^{-1}(y)$ is the inverse function of $g(x)$, $J_{g^{-1}}(y)$ is the Jacobian matrix of $g^{-1}(y)$

$\label{eq:jac} J_{g^{-1}}(y) = \left[ \begin{matrix} \frac{\mathrm{d}x_1}{\mathrm{d}y_1} & \ldots & \frac{\mathrm{d}x_1}{\mathrm{d}y_n} \\ \vdots & \ddots & \vdots \\ \frac{\mathrm{d}x_n}{\mathrm{d}y_1} & \ldots & \frac{\mathrm{d}x_n}{\mathrm{d}y_n} \end{matrix} \right] \; ,$

$\lvert J \rvert$ is the determinant of $J$ and $\mathcal{Y}$ is the set of possible outcomes of $Y$:

$\label{eq:Y-range} \mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace \; .$

Proof:

1) First, we obtain the cumulative distribution function of $Y = g(X)$. The joint CDF is given by

$\label{eq:Y-cdf-s1} \begin{split} F_Y(y) &= \mathrm{Pr}(Y_1 \leq y_1, \ldots, Y_n \leq y_n) \\ &= \mathrm{Pr}(g_1(X) \leq y_1, \ldots, g_n(X) \leq y_n) \\ &= \int_{A(y)} f_X(x) \, \mathrm{d}x \end{split}$

where $A(y)$ is the following subset of the $n$-dimensional Euclidean space:

$\label{eq:A-y} A(y) = \left\lbrace x \in \mathbb{R}^n: g_j(x) \leq y_j \; \text{for all} \; j = 1, \ldots, n \right\rbrace$

and $g_j(X)$ is the function which returns the $j$-th element of $Y$, given a vector $X$.

2) Next, we substitute $x = g^{-1}(y)$ into the integral which gives us

$\label{eq:Y-cdf-s2} \begin{split} F_Y(z) &= \int_{B(z)} f_X(g^{-1}(y)) \, \mathrm{d}g^{-1}(y) \\ &= \int_{-\infty}^{z_n} \ldots \int_{-\infty}^{z_1} f_X(g^{-1}(y)) \, \mathrm{d}g^{-1}(y) \; . \end{split}$

where we have the modified the integration regime $B(z)$ which reads

$\label{eq:B-z} B(z) = \left\lbrace y \in \mathbb{R}^n: y \leq z_j \; \text{for all} \; j = 1, \ldots, n \right\rbrace \; .$

3) The formula for change of variables in multivariable calculus states that

$\label{eq:cov-multi} y = f(x) \quad \Rightarrow \quad \mathrm{d}y = \left| J_f(x) \right| \, \mathrm{d}x \; .$

Applied to equation \eqref{eq:Y-cdf-s2}, this yields

$\label{eq:Y-cdf-s3} \begin{split} F_Y(z) &= \int_{-\infty}^{z_n} \ldots \int_{-\infty}^{z_1} f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \, \mathrm{d}y \\ &= \int_{-\infty}^{z_n} \ldots \int_{-\infty}^{z_1} f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \, \mathrm{d}y_1 \ldots \mathrm{d}y_n \; . \end{split}$

4) Finally, we obtain the probability density function of $Y = g(X)$. Because the PDF is the derivative of the CDF, we can differentiate the joint CDF to get

$\label{eq:Y-cdf-s4} \begin{split} f_Y(z) &= \frac{\mathrm{d}^n}{\mathrm{d}z_1 \ldots \mathrm{d}z_n} \, F_Y(z) \\ &= \frac{\mathrm{d}^n}{\mathrm{d}z_1 \ldots \mathrm{d}z_n} \int_{-\infty}^{z_n} \ldots \int_{-\infty}^{z_1} f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \, \mathrm{d}y_1 \ldots \mathrm{d}y_n \\ &= f_X(g^{-1}(z)) \, \left| J_{g^{-1}}(z) \right| \end{split}$

which can also be written as

$\label{eq:pdf-invfct-qed} f_Y(y) = f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \; .$
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Metadata: ID: P254 | shortcut: pdf-invfct | author: JoramSoch | date: 2021-08-30, 07:05.