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mathjax |
author |
affiliation |
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date |
title |
chapter |
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theorem |
sources |
proof_id |
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username |
proof |
true |
Joram Soch |
BCCN Berlin |
joram.soch@bccn-berlin.de |
2020-11-05 20:12:00 -0800 |
Cumulative distribution function of a strictly decreasing function of a random variable |
General Theorems |
Probability theory |
Cumulative distribution function |
Cumulative distribution function of strictly decreasing function |
| authors |
year |
title |
in |
pages |
url |
Taboga, Marco |
2017 |
Functions of random variables and their distribution |
Lectures on probability and mathematical statistics |
retrieved on 2020-11-06 |
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P186 |
cdf-sdfct |
JoramSoch |
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:
- If $y$ is higher than the highest value $Y$ can take, then $\mathrm{Pr}(Y \leq y) = 1$, so
$$ \label{eq:cdf-sdfct-p1}
F_Y(y) = 1, \quad \text{if} \quad y > \mathrm{max}(\mathcal{Y}) ; .
$$
- If $y$ belongs to the support of $Y$, then $F_Y(y)$ can be derived as follows:
$$ \label{eq:cdf-sdfct-p2}
\begin{split}
F_Y(y) &= \mathrm{Pr}(Y \leq y) \\
&= 1 - \mathrm{Pr}(Y > y) \\
&= 1 - \mathrm{Pr}(g(X) > y) \\
&= 1 - \mathrm{Pr}(X < g^{-1}(y)) \\
&= 1 - \mathrm{Pr}(X < g^{-1}(y)) - \mathrm{Pr}(X = g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\
&= 1 - \left[ \mathrm{Pr}(X < g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \right] + \mathrm{Pr}(X = g^{-1}(y)) \\
&= 1 - \mathrm{Pr}(X \leq g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) \\
&= 1 - F_X(g^{-1}(y)) + \mathrm{Pr}(X = g^{-1}(y)) ; .
\end{split}
$$
- 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}.