Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Probability functions ▷ Cumulative distribution function of strictly decreasing function

Theorem: Let $X$ be a random variable with possible outcomes $\mathcal{X}$ and let $g(x)$ be a strictly decreasing function on the support of $X$. Then, the cumulative distribution function of $Y = g(X)$ is given by

$\label{eq:cdf-sdfct} F_Y(y) = \left\{ \begin{array}{rl} 1 \; , & \text{if} \; y > \mathrm{max}(\mathcal{Y}) \\ 1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \; , & \text{if} \; y \in \mathcal{Y} \\ 0 \; , & \text{if} \; y < \mathrm{min}(\mathcal{Y}) \end{array} \right.$

where $g^{-1}(y)$ is the inverse function of $g(x)$ 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: The support of $Y$ is determined by $g(x)$ and by the set of possible outcomes of $X$. Moreover, if $g(x)$ is strictly decreasing, then $g^{-1}(y)$ is also strictly decreasing. Therefore, the cumulative distribution function of $Y$ can be derived as follows:

1) If $y$ is higher than the highest value $Y$ can take, then $\mathrm{Pr}(Y \leq y) = 1$, so

$\label{eq:cdf-sdfct-p1} F_Y(y) = 1, \quad \text{if} \quad y > \mathrm{max}(\mathcal{Y}) \; .$

2) If $y$ belongs to the support of $Y$, then $F_Y(y)$ can be derived as follows:

$\label{eq:cdf-sdfct-p2} \begin{split} F_Y(y) &= \mathrm{Pr}(Y \leq y) \\ &= 1 - \mathrm{Pr}(Y > y) \\ &= 1 - \mathrm{Pr}(g(X) > y) \\ &= 1 - \mathrm{Pr}(X < g^{-1}(y)) \\ &= 1 - \mathrm{Pr}(X < g^{-1}(y)) - \mathrm{Pr}(X = g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\ &= 1 - \left[ \mathrm{Pr}(X < g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \right] + \mathrm{Pr}(X = g^{-1}(y)) \\ &= 1 - \mathrm{Pr}(X \leq g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\ &= 1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \; . \end{split}$

3) If $y$ is lower than the lowest value $Y$ can take, then $\mathrm{Pr}(Y \leq y) = 0$, so

$\label{eq:cdf-sdfct-p3} F_Y(y) = 0, \quad \text{if} \quad y < \mathrm{min}(\mathcal{Y}) \; .$

Taking together \eqref{eq:cdf-sdfct-p1}, \eqref{eq:cdf-sdfct-p2}, \eqref{eq:cdf-sdfct-p3}, eventually proves \eqref{eq:cdf-sdfct}.

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Metadata: ID: P186 | shortcut: cdf-sdfct | author: JoramSoch | date: 2020-11-06, 04:12.