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‎P/cdf-sdfct.md‎

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---
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layout: proof
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mathjax: true
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2020-11-06 04:12:00
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title: "Cumulative distribution function of a strictly decreasing function of a random variable"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Probability functions"
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theorem: "Cumulative distribution function of strictly decreasing function"
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sources:
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- authors: "Taboga, Marco"
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year: 2017
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title: "Functions of random variables and their distribution"
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in: "Lectures on probability and mathematical statistics"
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pages: "retrieved on 2020-11-06"
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url: "https://www.statlect.com/fundamentals-of-probability/functions-of-random-variables-and-their-distribution#hid5"
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proof_id: "P186"
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shortcut: "cdf-sdfct"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [random variable](/D/rvar) with possible outcomes $\mathcal{X}$ and let $g(x)$ be a strictly decreasing function on the support of $X$. Then, the [cumulative distribution function](/D/cdf) of $Y = g(X)$ is given by
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$$ \label{eq:cdf-sdfct}
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F_Y(y) = \left\{
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\begin{array}{rl}
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1 \; , & \text{if} \; y > \mathrm{max}(\mathcal{Y}) \\
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1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \; , & \text{if} \; y \in \mathcal{Y} \\
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0 \; , & \text{if} \; y < \mathrm{min}(\mathcal{Y})
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\end{array}
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\right.
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$$
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where $g^{-1}(y)$ is the inverse function of $g(x)$ and $\mathcal{Y}$ is the set of possible outcomes of $Y$:
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$$ \label{eq:Y-range}
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\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace \; .
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$$
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**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](/D/cdf) of $Y$ can be derived as follows:
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1) If $y$ is higher than the [highest value](/D/max) $Y$ can take, then $\mathrm{Pr}(Y \leq y) = 1$, so
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$$ \label{eq:cdf-sdfct-p1}
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F_Y(y) = 1, \quad \text{if} \quad y > \mathrm{max}(\mathcal{Y}) \; .
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$$
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2) If $y$ belongs to the support of $Y$, then $F_Y(y)$ can be derived as follows:
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$$ \label{eq:cdf-sdfct-p2}
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\begin{split}
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F_Y(y) &= \mathrm{Pr}(Y \leq y) \\
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&= 1 - \mathrm{Pr}(Y > y) \\
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&= 1 - \mathrm{Pr}(g(X) > y) \\
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&= 1 - \mathrm{Pr}(X < g^{-1}(y)) \\
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&= 1 - \mathrm{Pr}(X < g^{-1}(y)) - \mathrm{Pr}(X = g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\
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&= 1 - \left[ \mathrm{Pr}(X < g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \right] + \mathrm{Pr}(X = g^{-1}(y)) \\
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&= 1 - \mathrm{Pr}(X \leq g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\
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&= 1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \; .
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\end{split}
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$$
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3) If $y$ is lower than the [lowest value](/D/min) $Y$ can take, then $\mathrm{Pr}(Y \leq y) = 0$, so
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$$ \label{eq:cdf-sdfct-p3}
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F_Y(y) = 0, \quad \text{if} \quad y < \mathrm{min}(\mathcal{Y}) \; .
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$$
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Taking together \eqref{eq:cdf-sdfct-p1}, \eqref{eq:cdf-sdfct-p2}, \eqref{eq:cdf-sdfct-p3}, eventually proves \eqref{eq:cdf-sdfct}.

‎P/pdf-sdfct.md‎

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---
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layout: proof
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mathjax: true
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2020-11-06 05:30:00
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title: "Probability density function of a strictly decreasing function of a continuous random variable"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Probability functions"
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theorem: "Probability density function of strictly decreasing function"
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sources:
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- authors: "Taboga, Marco"
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year: 2017
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title: "Functions of random variables and their distribution"
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in: "Lectures on probability and mathematical statistics"
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pages: "retrieved on 2020-11-06"
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url: "https://www.statlect.com/fundamentals-of-probability/functions-of-random-variables-and-their-distribution#hid7"
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proof_id: "P188"
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shortcut: "pdf-sdfct"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [continuous](/D/rvar-disc) [random variable](/D/rvar) with possible outcomes $\mathcal{X}$ and let $g(x)$ be a strictly decreasing function on the support of $X$. Then, the [probability density function](/D/pdf) of $Y = g(X)$ is given by
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$$ \label{eq:pdf-sdfct}
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f_Y(y) = \left\{
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\begin{array}{rl}
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-f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; , & \text{if} \; y \in \mathcal{Y} \\
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0 \; , & \text{if} \; y \notin \mathcal{Y}
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\end{array}
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\right.
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$$
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where $g^{-1}(y)$ is the inverse function of $g(x)$ and $\mathcal{Y}$ is the set of possible outcomes of $Y$:
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$$ \label{eq:Y-range}
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\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace \; .
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$$
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**Proof:** The [cumulative distribution function of a strictly decreasing function](/P/cdf-sifct) is
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$$ \label{eq:cdf-sdfct}
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F_Y(y) = \left\{
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\begin{array}{rl}
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1 \; , & \text{if} \; y > \mathrm{max}(\mathcal{Y}) \\
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1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \; , & \text{if} \; y \in \mathcal{Y} \\
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0 \; , & \text{if} \; y < \mathrm{min}(\mathcal{Y})
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\end{array}
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\right.
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$$
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Note that [for continuous random variables, the probability](/D/pdf) of point events is
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$$ \label{eq:pdf-cont}
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\mathrm{Pr}(X = a) = \int_a^a f_X(x) \, \mathrm{d}x = 0 \; .
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$$
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Because the [probability density function is the first derivative of the cumulative distribution function](/P/pdf-cdf)
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$$ \label{eq:pdf-cdf}
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f_X(x) = \frac{\mathrm{d}F_X(x)}{\mathrm{d}x} \; ,
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$$
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the [probability density function](/D/pdf) of $Y$ can be derived as follows:
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1) If $y$ does not belong to the support of $Y$, $F_Y(y)$ is constant, such that
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$$ \label{eq:pdf-sdfct-p1}
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f_Y(y) = 0, \quad \text{if} \quad y \notin \mathcal{Y} \; .
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$$
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2) If $y$ belongs to the support of $Y$, then $f_Y(y)$ can be derived using the chain rule:
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$$ \label{eq:pdf-sdfct-p2}
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\begin{split}
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f_Y(y) &\overset{\eqref{eq:pdf-cdf}}{=} \frac{\mathrm{d}}{\mathrm{d}y} F_Y(y) \\
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&\overset{\eqref{eq:cdf-sdfct}}{=} \frac{\mathrm{d}}{\mathrm{d}y} \left[ 1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \right] \\
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&\overset{\eqref{eq:pdf-cont}}{=} \frac{\mathrm{d}}{\mathrm{d}y} \left[ 1 - F_X(g^{-1}(y)) \right] \\
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&= -\frac{\mathrm{d}}{\mathrm{d}y} F_X(g^{-1}(y)) \\
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&= - f_X(g^{-1}(y)) \, \frac{\mathrm{d}g^{-1}(y)}{\mathrm{d}y} \; .
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\end{split}
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$$
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Taking together \eqref{eq:pdf-sdfct-p1} and \eqref{eq:pdf-sdfct-p2}, eventually proves \eqref{eq:pdf-sdfct}.

‎P/pmf-sdfct.md‎

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---
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layout: proof
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mathjax: true
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author: "Joram Soch"
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affiliation: "BCCN Berlin"
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e_mail: "joram.soch@bccn-berlin.de"
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date: 2020-11-06 04:21:00
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title: "Probability mass function of a strictly decreasing function of a discrete random variable"
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chapter: "General Theorems"
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section: "Probability theory"
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topic: "Probability functions"
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theorem: "Probability mass function of strictly decreasing function"
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sources:
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- authors: "Taboga, Marco"
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year: 2017
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title: "Functions of random variables and their distribution"
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in: "Lectures on probability and mathematical statistics"
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pages: "retrieved on 2020-11-06"
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url: "https://www.statlect.com/fundamentals-of-probability/functions-of-random-variables-and-their-distribution#hid6"
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proof_id: "P187"
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shortcut: "pmf-sdfct"
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username: "JoramSoch"
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---
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**Theorem:** Let $X$ be a [discrete](/D/rvar-disc) [random variable](/D/rvar) with possible outcomes $\mathcal{X}$ and let $g(x)$ be a strictly decreasing function on the support of $X$. Then, the [probability mass function](/D/pmf) of $Y = g(X)$ is given by
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$$ \label{eq:pmf-sdfct}
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f_Y(y) = \left\{
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\begin{array}{rl}
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f_X(g^{-1}(y)) \; , & \text{if} \; y \in \mathcal{Y} \\
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0 \; , & \text{if} \; y \notin \mathcal{Y}
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\end{array}
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\right.
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$$
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where $g^{-1}(y)$ is the inverse function of $g(x)$ and $\mathcal{Y}$ is the set of possible outcomes of $Y$:
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$$ \label{eq:Y-range}
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\mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace \; .
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$$
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**Proof:** Because a strictly decreasing function is invertible, the [probability mass function](/D/pmf) of $Y$ can be derived as follows:
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$$ \label{eq:pmf-sdfct-qed}
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\begin{split}
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f_Y(y) &= \mathrm{Pr}(Y = y) \\
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&= \mathrm{Pr}(g(X) = y) \\
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&= \mathrm{Pr}(X = g^{-1}(y)) \\
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&= f_X(g^{-1}(y)) \; .
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\end{split}
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$$

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