# 44 Weibull Distribution

## 44.1 Probability Distribution Function

A random variable $$X$$ is said to have a Weibull Distribution with parameters $$\alpha$$ and $$\beta$$ if its probability density function is:

$f(x)=\left\{ \begin{array}{ll} \alpha\beta x^{\beta-1}e^{-\alpha x^{\beta}},&0<x,\ 0<\alpha,\ 0<\beta\\ 0 & otherwise \end{array}\right.$

## 44.2 Cumulative Distribution Function

\begin{aligned} \int\limits_{0}^{x}\alpha\beta t^{\beta-1}e^{-\alpha t^\beta}dt &= \alpha\beta\int\limits_{0}^{x}t^{\beta-1}e^{-\alpha t^\beta}dt \\ &= \alpha\beta\Big[\frac{-1}{\alpha\beta}e^{-\alpha t^\beta}\Big]_0^x \\ &= \alpha\beta\Big[\frac{-1}{\alpha\beta}e^{-\alpha x^\beta} + \frac{1}{\alpha\beta}\Big] \\ &= \frac{\alpha\beta}{\alpha\beta}\big(-e^{-\alpha x^\beta}+1\big) \\ &= 1-e^{-\alpha x^\beta} \end{aligned}

Using this result, we can write the Cumulative Distribution Function as

$F(x)=\left\{ \begin{array}{ll} 1-e^{-\alpha x^\beta},& 0<x,\ 0<\alpha,\ 0<\beta\\ 0 & otherwise \end{array}\right.$

## 44.3 Expected Values

\begin{aligned} E(X) &= \int\limits_{0}^{\infty}x\alpha\beta x^{\beta-1}e^{-\alpha x^{\beta}}dx \\ &= \alpha\beta\int\limits_{0}^{\infty}x x^{\beta-1}e^{-\alpha x^{\beta}}dx \\ ^{[1]} &= \alpha\beta\int\limits_{0}^{\infty} \Big(\big(\frac{y}{\alpha}\big)^\frac{1}{\beta}\Big)^\beta e^{-y}\frac{1}{\alpha\beta} \Big(\frac{y}{\alpha}\Big)^{\frac{1}{\beta}-1}dy \\ &= \frac{\alpha\beta}{\alpha\beta}\int\limits_{0}^{\infty} \Big(\frac{y}{\alpha}\Big)^\frac{\beta+1}{\beta} \Big(\frac{y}{\alpha}\Big)^{\frac{1}{\beta}-1}e^{-y}dy \\ &= \int\limits_{0}^{\infty}\Big(\frac{y}{\alpha}\Big) ^{\frac{\beta+1}{\beta}-\frac{1}{\beta}-1}e^{-y}dy \\ &= \int\limits_{0}^{\infty}\Big(\frac{y}{\alpha}\Big) ^{\frac{\beta+1}{\beta}-1}e^{-y}dy \\ &= \frac{1}{\alpha^{\frac{\beta+1}{\beta}-\frac{\beta}{\beta}}} \int\limits_{0}^{\infty}y^{\frac{\beta+1}{\beta}-1}e^{-y} \\ &= \alpha^{-\frac{1}{\beta}} \int\limits_{0}^{\infty}y^{\frac{\beta+1}{\beta}-1}e^{-y} \\ ^{[2]} &= \alpha^{-\frac{1}{\beta}}\Gamma\Big(\frac{\beta+1}{\beta}\Big) \end{aligned}

1. $$y=\alpha x^\beta\ \Rightarrow x=(\frac{y}{\alpha})^\frac{1}{\beta}\ \Rightarrow dx=\frac{1}{\alpha\beta}(\frac{y}{\alpha})^{\frac{1}{\beta}-1}$$
2. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx =\Gamma(\alpha)$$

\begin{aligned} E(X^2) &= \alpha\beta\int\limits_{0}^{\infty}x^2x^{\beta-1}e^{-\alpha x^\beta}dx \\ &= \alpha\beta\int\limits_{0}^{\infty}x^{\beta+1}e^{-\alpha x^\beta}dx \\ ^{[1]} &= \alpha\beta\int\limits_{0}^{\infty}\Big(\big(\frac{y}{\alpha}\big)^\frac{1}{\beta}\Big)^{\beta+1} e^{-y}\frac{1}{\alpha\beta}\big(\frac{y}{\alpha}\big)^{\frac{1}{\beta}-1}dy \\ &= \frac{\alpha\beta}{\alpha\beta}\int\limits_{0}^{\infty}\bigg(\frac{y}{\alpha}\bigg)^{\frac{\beta+1}{\beta}} \bigg(\frac{y}{\alpha}\bigg)^{\frac{1}{\beta}-1}e^{-y}dy \\ &= \int\limits_{0}^{\infty}\bigg(\frac{y}{\alpha}\bigg)^{\frac{\beta+1}{\beta}+\frac{1}{\beta}-1}e^{-y}dy \\ &= \frac{1}{\alpha^{\frac{\beta+2}{\beta}-\frac{\beta}{\beta}}} \int\limits_{0}^{\infty}y^{\frac{\beta+2}{\beta}-1}e^{-y}dy \\ &= \alpha^{-\frac{2}{\beta}}\int\limits_{0}^{\infty}y^{\frac{\beta+2}{\beta}-1}e^{-y}dy \\ ^{[2]} = \alpha^{-\frac{2}{\beta}}\Gamma\Big(\frac{\beta+2}{\beta}\Big) \end{aligned}

1. $$y=\alpha x^\beta\ \Rightarrow x=(\frac{y}{\alpha})^\frac{1}{\beta}\ \Rightarrow dx=\frac{1}{\alpha\beta}(\frac{y}{\alpha})^{\frac{1}{\beta}-1}$$
2. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx =\Gamma(\alpha)$$

\begin{aligned} \mu &= E(X) \\ &= \alpha^{-\frac{1}{\beta}}\Gamma\Big(\frac{\beta+1}{\beta}\Big)\\ \\ \\ \sigma^2 &= E(X^2) - E(X)^2 \\ &= \alpha^{-\frac{2}{\beta}}\Gamma\Big(\frac{\beta+2}{\beta}\Big) - \alpha^{-\frac{2}{\beta}}\Gamma\Big(\frac{\beta+1}{\beta}\Big)^2 \\ &= \alpha^{-\frac{2}{\beta}}\Big[\Gamma\Big(\frac{\beta+2}{\beta}\Big) - \Gamma\Big(\frac{\beta+1}{\beta}\Big)^2\Big] \end{aligned}

## 44.4 Theorems for the Weibull Distribution

### 44.4.1 Validity of the Distribution

$\int\limits_{0}^{\infty}\alpha\beta x^{\beta-1}e^{-\alpha x^\beta}dx = 1$

Proof:

\begin{aligned} \int\limits_{0}^{\infty}\alpha\beta x^{\beta-1}e^{-\alpha x^\beta}dx &= \alpha\beta\int\limits_{0}^{\infty}x^{\beta-1}e^{-\alpha x^\beta}dx \\ ^{[1]} &= \alpha\beta\int\limits_{0}^{\infty}\Big(\big(\frac{y}{\alpha}\big)^ \frac{1}{\beta}\Big)^{\beta-1} e^{-y}\big(\frac{y}{\alpha}\big)^{\frac{1}{\beta}-1}\frac{1}{\alpha\beta}dy &= \frac{\alpha\beta}{\alpha\beta}\int\limits_{0}^{\infty}\big(\frac{y}{\alpha}\big)^\frac{\beta}{-1} \big(\frac{y}{\alpha}\big)^{\frac{1}{\beta}-1}e^{-y}dy \\ &= \int\limits_{0}^{\infty}\big(\frac{y}{\alpha}\big)^{\frac{\beta-1}{\beta}+\frac{1-\beta}{\beta}}e^{-y}dy \\ &= \int\limits_{0}^{\infty}\frac{y^0}{\alpha^0}e^{-y}dy \\ &= \int\limits_{0}^{\infty}y^{1-1}e^{-y}dy \\ ^{[2]} &= \Gamma(1)=1 \end{aligned}

1. $$y=\alpha x^\beta\ \Rightarrow x = (\frac{y}{\alpha})^\frac{1}{\beta}\ \Rightarrow dx = \frac{1}{\alpha\beta}(\frac{y}{\alpha})^{\frac{1}{\beta}-1}$$
2. $$\int\limits_{0}^{\infty}x^{\alpha-1}e^{-x}dx = \Gamma(\alpha)$$