Home page for accesible maths 4 Univariate Transformations 4.1 Introduction 4.3 The Probability Integral Transform

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4.2 Distribution Function Method

The distribution function (cdf) method for evaluating the distribution of a transformation arises from the observation that

\displaystyle F_{Y}(y)=\operatorname{\mathsf{P}}\left({Y\leq y}\right)=% \operatorname{\mathsf{P}}\left({g(X)\leq y}\right).

It proceeds as follows.

1.

find the values of $X$ which correspond to the event $g(X)\leq y$ , let this correspond to the event $X\in A_{y}$ say,
2.

evaluate the probability $\operatorname{\mathsf{P}}\left({Y\leq y}\right)=\operatorname{\mathsf{P}}\left% ({X\in A_{y}}\right)$ ,
3.

differentiate to obtain the pdf of $Y$ .

The figures illustrate the sets $A_{y}$ for various transformations $Y=g(X)$ . When $g(\cdot)$ is monotonically increasing or decreasing $A_{y}$ will always be an interval of the form $(-\infty,x]$ or $[x,\infty)$ and the method is particularly easy to apply in these cases. The method, however, holds whatever the properties of the transformation $g(\cdot)$ . It is best explained through examples.

Figure 4.1: First Link, Second Link, Caption: The set

A_{y}

for two different transformations

g(\cdot)

(the curves).

Figure 4.2: First Link, Second Link, Caption: The set

A_{y}

for two more different transformations

g(\cdot)

(the curves).

Example 4.2.1.

Let $X\sim{\operatorname{\mathsf{Unif}}}(0,2)$ ; what are the densities of

(a)

$Y=X^{2}$ , and
(b)

$V=(X-1)^{2}$ ?

Solution.

\displaystyle\operatorname{\mathsf{P}}\left({X\leq x}\right)=F_{X}(x)=\left\{% \begin{array}[]{ll}0&\quad x\leq 0\\ x/2&\quad 0<x\leq 2\\ 1&\quad x>2\end{array}\right.

(a)

Clearly $0<Y\leq 4$ . For $y$ in this range (and, hence, $y^{1/2}$ between $0$ and $2$ ),

$\displaystyle F_{Y}(y)$ $\displaystyle=\operatorname{\mathsf{P}}\left({Y\leq y}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({X^{2}\leq y}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X\leq y% ^{1/2}}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}y^{1/2}/2.}$

So

$\displaystyle f_{Y}(y)=\left\{\begin{array}[]{ll}{\color[rgb]{0.76,0.01,0}0}&% \quad y\leq 0\\ {\color[rgb]{0.76,0.01,0}\frac{1}{4\sqrt{y}}}&\quad 0<y\leq 4\\ {\color[rgb]{0.76,0.01,0}0}&\quad 0<y\leq 4\end{array}\right.$
(b)

$0\leq V\leq 1$ . For $v$ in this range (hence, both $1-v^{1/2}$ and $1+v^{1/2}$ between $0$ and $2$ ),

$\displaystyle F_{V}(v)$ $\displaystyle=\operatorname{\mathsf{P}}\left({V\leq v}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({(X-1)^{2}\leq v}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({-v^{1/2% }\leq X-1\leq v^{1/2}}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({1-v^{1/% 2}\leq X\leq 1+v^{1/2}}\right)}$

$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1+v^{1/2}}{2}-\frac{1-v^{1/2}}{2}}$

So

$\displaystyle f_{V}(v)=\left\{\begin{array}[]{ll}{\color[rgb]{0.76,0.01,0}0}&% \quad v<0\\ {\color[rgb]{0.76,0.01,0}\frac{1}{2\sqrt{v}}}&\quad 0<v\leq 1\\ {\color[rgb]{0.76,0.01,0}0}&\quad v>1\end{array}\right.$

Example 4.2.2.

A classic: $X\sim\operatorname{Uniform}(0,1)$ . Show that

\displaystyle Y=-\frac{1}{\beta}\log(1-X)

has an $\operatorname{\mathsf{Exp}}(\beta)$ distribution, where $\beta>0$ .

Solution.

	$\displaystyle F_{Y}(y)=\operatorname{\mathsf{P}}\left({Y\leq y}\right)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({-\beta^% {-1}\log(1-X)\leq y}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({\log(1-% X)\geq-\beta y}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X\leq 1% -\exp(-\beta y)}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\exp(-\beta y)}$

for ${\color[rgb]{0.76,0.01,0}y>0.}$ This is the cdf of an $\operatorname{\mathsf{Exp}}(\beta)$ random variable.

Example 4.2.3.

If $X\sim\operatorname{\mathsf{Exp}}(\beta)$ , show that $Y=\beta X$ has an $\operatorname{Exponential}(1)$ distribution.

Solution.

	$\displaystyle F_{Y}(y)$	$\displaystyle=\operatorname{\mathsf{P}}\left({Y\leq y}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({\beta X\leq y}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X\leq y% /\beta}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\exp\{-\beta(y/\beta)\}}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\exp(-y)}$

for ${\color[rgb]{0.76,0.01,0}y>0.}$ This is the cdf of an $\operatorname{Exponential}(1)$ random variable.

Example 4.2.4.

$X\sim N(0,1)$ . Show that $Y=X^{2}$ has a $\operatorname{\mathsf{Gam}}(1/2,1/2)$ distribution. Note: this is also a $\chi^{2}_{1}$ distribution – see Section 11.1.

Solution.

	$\displaystyle F_{Y}(y)$	$\displaystyle=\operatorname{\mathsf{P}}\left({Y\leq y}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({X^{2}\leq y}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({-\sqrt{% y}\leq X\leq\sqrt{y}}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}F_{X}(\sqrt{y})-F_{X}(-\sqrt{y}).}$

To obtain the pdf we differentiate and use the chain rule:

	$\displaystyle f_{Y}(y)$	$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{2}y^{-1/2}f_{X}(\sqrt{y})+% \frac{1}{2}y^{-1/2}f_{X}(-\sqrt{y})}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1}{2}y^{-1/2}\frac{1}{\sqrt{2\pi}% }\exp(-y/2)+\frac{1}{2}y^{-1/2}\frac{1}{\sqrt{2\pi}}\exp(-y/2)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}y^{-1/2}\frac{1}{\sqrt{2\pi}}\exp(-y/2)}$

for $y>0$ , which matches the Gamma pdf equation (3.1) when $\alpha=1/2$ and $\beta=1/2$ since $\Gamma(1/2)=\sqrt{\pi}$ .

Example 4.2.5.

When $X\sim\operatorname{\mathsf{Exp}}(1)$ show that for $\alpha>0$ ,

\displaystyle Y=X^{1/\alpha}

has a $\operatorname{\mathsf{Weib}}(\alpha,1)$ distribution. Hence use Example 4.2.2 to find the transformation of a ${\operatorname{\mathsf{Unif}}}(0,1)$ random variable required to achieve a $\operatorname{\mathsf{Weib}}(\alpha,1)$ random variable.

Solution. $X>0$ so $Y=X^{1/\alpha}$ is real and so is a random variable, with $0<Y<\infty$ ; also, for $x>0,F_{X}(x)=1-\exp(-x)$ .

	$\displaystyle F_{Y}(y)$	$\displaystyle=\operatorname{\mathsf{P}}\left({Y\leq y}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({X^{1/\alpha}\leq y}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({X\leq y% ^{\alpha}}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}1-\exp(-y^{\alpha}),}$

which is the cdf of an $\operatorname{\mathsf{Weib}}(\alpha,1)$ random variable. From Example 4.2.2, $X$ can be written as $X=-\log(1-U)$ , where $U\sim{\operatorname{\mathsf{Unif}}}(0,1)$ , hence $Y=(-\log(1-U))^{1/\alpha}$ .

	$\displaystyle F_{V}(v)$	$\displaystyle=\operatorname{\mathsf{P}}\left({V\leq v}\right)={\color[rgb]{% 0.76,0.01,0}\operatorname{\mathsf{P}}\left({(X-1)^{2}\leq v}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({-v^{1/2% }\leq X-1\leq v^{1/2}}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\operatorname{\mathsf{P}}\left({1-v^{1/% 2}\leq X\leq 1+v^{1/2}}\right)}$
		$\displaystyle={\color[rgb]{0.76,0.01,0}\frac{1+v^{1/2}}{2}-\frac{1-v^{1/2}}{2}}$