Q. 5.24 - A First Course in Probability

Question

Show that a plot of $\log \left(\log (1-\mathrm{F}(\mathrm{x}){)}^{-1}\right)$ against $\log \left(\mathrm{X}\right)$ will be a straight line with slope $\mathrm{\beta }$ when $\mathrm{F}(-)$ is a Weibull distribution function. Show also that approximately $63.2$ percent of all observations from such a distribution will be less than $\mathrm{\alpha }$. Assume that $\mathrm{v}=0$.

Step-by-Step Solution

Answer

The value of$\mathrm{P}(\mathrm{X}\le \mathrm{\alpha })=\mathrm{F}\left(\mathrm{\alpha }\right)=1-{\mathrm{e}}^{-\left(\frac{\mathrm{\alpha }}{\mathrm{\alpha }}\right)}$is $0.6321$.

Step 1: Weibull distribution

The Weibull distribution ,

$\mathrm{v}=0$

$\mathrm{F}\left(\mathrm{x}\right)=1-{\mathrm{e}}^{-{\left(\frac{\mathrm{x}}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}}$

For $\mathrm{x}>0$, Essentially, it's the same as zero.

So, 

$1-\mathrm{F}\left(\mathrm{x}\right)={\mathrm{e}}^{-{\left(\frac{\mathrm{x}}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}}$

$\log (1-\mathrm{F}(\mathrm{x}\left)\right)=-{\left(\frac{\mathrm{x}}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}$

Step 2: Explanation

Inverse function is,

$\log (1-\mathrm{F}(\mathrm{x}\left)\right)$

 $\mathrm{y}=\log (1-\mathrm{F}(\mathrm{x}\left)\right)$

$\mathrm{y}=-{\left(\frac{\mathrm{x}}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}\Rightarrow -\mathrm{y}={\left(\frac{\mathrm{x}}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}\Rightarrow (-\mathrm{y}{)}^{\frac{1}{\mathrm{\beta }}}$

$=\frac{\mathrm{x}}{\mathrm{\alpha }}\Rightarrow \mathrm{x}=\mathrm{\alpha }·(-\mathrm{y}{)}^{\frac{1}{\mathrm{\beta }}}$

Hence, 

$\log (1-\mathrm{F}(\mathrm{x}){)}^{-1}=\mathrm{\alpha }·(-\mathrm{x}{)}^{\frac{1}{\mathrm{\beta }}}$

Substitute all values,

we get,

$\log (1-\mathrm{F}(\mathrm{x}){)}^{-1}=\mathrm{\alpha }·{\mathrm{x}}^{\frac{1}{\mathrm{\beta }}}$

So,

$\log \left(\log (1-\mathrm{F}(\mathrm{x}){)}^{-1}\right)=\log \left(\mathrm{\alpha }·{\mathrm{x}}^{\frac{1}{\mathrm{\beta }}}\right)$

$=\mathrm{log\alpha }+\frac{1}{\mathrm{\beta }}\mathrm{logx}$

This function can be divided into two parts. Differentiation in terms of $\log \left(\mathrm{x}\right)$ means that the first derivation equals $\frac{1}{\mathrm{\beta }}$, and so the slope equals $\mathrm{\beta }$.

As a result, we've demonstrated the first portion of this task.

To demonstrate that the second assertion is correct, we must compute$\mathrm{P}(\mathrm{X} \le \mathrm{\alpha })$ , where $\mathrm{x}$ has a Weibull distribution with parameters $\mathrm{\alpha }$,$\mathrm{\beta }$and$\mathrm{v}=0$.

Hence,

$\mathrm{P}(\mathrm{X}\le \mathrm{\alpha })=\mathrm{F}\left(\mathrm{\alpha }\right)=1-{\mathrm{e}}^{-{\left(\frac{\mathrm{\alpha }}{\mathrm{\alpha }}\right)}^{\mathrm{\beta }}}$

$=1-{\mathrm{e}}^{-1}=0.6321$