(Weibull distribution) Let X be a continuous random variable with
$f(x)= \begin{cases} \mathrm{e}^{-x}, \quad x \gt 0, \\ 0, \qquad x \leq 0. \end{cases}$
Obtain the density of the random variable
$Y=\alpha X^{\beta}.$
where 𝛼 > 0 and 𝛽 > 0 are arbitrary constants.

Question

(Weibull distribution) Let X be a continuous random variable with

[latex]f(x)= \begin{cases} \mathrm{e}^{-x}, \quad x \gt 0, \ 0, \qquad x \leq 0. \end{cases}[/latex]

Obtain the density of the random variable

[latex]Y=\alpha X^{\beta}.[/latex]

where 𝛼 > 0 and 𝛽 > 0 are arbitrary constants.

Accepted Answer

We apply the formula

[latex]f_{Y}(\mathrm y)=|(\mathrm g^{-1})^{\prime}(\mathrm y)|f(\mathrm g^{-1}(\mathrm y))\qquad\qquad\qquad\qquad(7.19)[/latex]

from Proposition 6.2, where the function [latex]\mathrm g[/latex] is defined by [latex]\mathrm g(x)=\alpha x^{\beta}.[/latex]

The derivative of this function is

[latex]\mathrm g^{\prime}(x)=\alpha\beta x^{\beta-1},[/latex]

which is positive for x > 0, and so [latex]\mathrm g[/latex] is a strictly increasing function. In order to find the inverse function, [latex]\mathrm g^{-1}[/latex], we solve the equation [latex]\mathrm g(x) = \mathrm y[/latex]. This gives

[latex]\alpha x^{\beta}=\mathrm y\iff x^{\beta}={\frac{\mathrm y}{\alpha}}\iff x=\left({\frac{\mathrm y}{\alpha}}\right)^{1/\beta}.[/latex]

Therefore,

[latex](\mathrm g^{-1})(\mathrm y)=\left(\frac{\mathrm y}{\alpha}\right)^{1/\beta}.[/latex]

The derivative of this function, with respect to y, is

[latex](\mathrm g^{-1})^{\prime}(\mathrm y)={\frac{1}{\beta}}\cdot{\frac{\mathrm y^{1/\beta-1}}{\alpha^{1/\beta}}}.[/latex]

We thus see from (7.19) that the density of Y is

[latex]f_{Y}(\mathrm y)=\left|{\frac{1}{\beta}}\cdot{\frac{\mathrm y^{1/\beta-1}}{\alpha^{1/\beta}}}\right|\exp\left\{-\left({\frac{\mathrm y}{\alpha}}\right)^{1/\beta}\right\}=\frac{1}{\beta\alpha^{1/\beta}}\cdot \mathrm y^{1/\beta-1}\cdot\exp\left\{-\left(\frac{\mathrm y}{\alpha}\right)^{1/\beta}\right\},\quad \mathrm y\gt 0.[/latex]

This distribution is known as the Weibull distribution, after the famous Swedish mathematician Waloddi Weibull.

Question 7.13: (Weibull distribution) Let X be a continuous random variable......

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