# Characteristics of the Weibull distribution

Published on May 17, 2017

One of the most famous inventions of Waloddi Weibull is his continious distribution. It is usefull in many cases like survival analysis or nuclear physics. It is strange, but there is no on the Internet (or I had badly searched) detailed findings of characteristics of the Weibull distribution with three parameters. I want to fix it.

So, this is the probability density (PDF) of the Weibull distribution with three parameters:

$$f(x) =\frac{\gamma (x-a)^{\gamma-1}}{\beta^\gamma} \exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)}$$

The distribution function (CDF) of a random variable:

$$F(x) = P(\xi < x) = \int_{\alpha}^{x} \frac{\gamma (x-a)^{\gamma-1}}{\beta^\gamma} \exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} dx = \Bigg| \begin{matrix} y=\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \\ dy=\frac{\gamma (x-\alpha)^{\gamma - 1}}{\beta ^\gamma} dx \end{matrix} \Bigg| =$$
$$= \int_{0}^{(\frac{x-\alpha}{\beta})^\gamma} e^{-y} dy =1-e^{-(\frac{x-\alpha}{\beta})^\gamma}$$

Mathematical expectation of the Weibull distribution with three parameters:

$$E\xi = \int_{\alpha}^{\infty} x \frac{\gamma (x-a)^{\gamma-1}}{\beta^\gamma} \exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} dx = \Bigg| \begin{matrix} y=\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \\ dy=\frac{\gamma (x-\alpha)^{\gamma - 1}}{\beta ^\gamma} dx\\ x=\sqrt[\gamma]{y}\beta + \alpha \end{matrix} \Bigg| =$$
$$= \int_{0}^{\infty} (\sqrt[\gamma]{y}\beta + \alpha) e^{-y} dy = \alpha \int_{0}^{\infty} e^{-y} dy + \beta \int_{0}^{\infty} \sqrt[\gamma]{y} e^{-y} dy = \alpha + \beta \Gamma(1+\frac{1}{\gamma})$$

Dispersion of the Weibull distribution with three parameters:

$$D\xi = E\xi^2 - (E\xi)^2$$
$$E\xi^2 = \int_{\alpha}^{\infty} x^2 \frac{\gamma (x-a)^{\gamma-1}}{\beta^\gamma} \exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} dx = \Bigg| \begin{matrix} y=\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \\ dy=\frac{\gamma (x-\alpha)^{\gamma - 1}}{\beta ^\gamma} dx\\ x=\sqrt[\gamma]{y}\beta + \alpha \end{matrix} \Bigg| =$$
$$= \int_{0}^{\infty} (\sqrt[\gamma]{y}\beta + \alpha)^2 e^{-y} dy = \beta^2 \int_{0}^{\infty} y^{\frac{2}{\gamma}} e^{-y} dy + 2 \alpha \beta \int_{0}^{\infty} y^{\frac{1}{\gamma}} e^{-y} dy + \alpha^2 \int_{0}^{\infty} e^{-y} dy =$$
$$= \beta^2 \Gamma(1+\frac{2}{\gamma}) + 2 \beta \alpha \Gamma(1+\frac{1}{\gamma}) + \alpha^2$$
$$D\xi = Ex^2 - (Ex)^2 = \beta^2 \Gamma(1+\frac{2}{\gamma}) + 2 \beta \alpha \Gamma(1+\frac{1}{\gamma}) + \alpha^2 - (\alpha + \beta \Gamma(1+\frac{1}{\gamma}))^2 =$$
$$= \beta^2 \Gamma(1+\frac{2}{\gamma}) - \beta^2 \Gamma^2(1+\frac{1}{\gamma}) = \beta^2 (\Gamma(1+\frac{2}{\gamma}) - \Gamma^2(1+\frac{1}{\gamma}))$$

k-th moment of the Weibull distribution with three parameters:

$$E\xi^k = \int_{\alpha}^{\infty} x^k \frac{\gamma (x-a)^{\gamma-1}}{\beta^\gamma} \exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} dx = \Bigg| \begin{matrix} y=\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \\ dy=\frac{\gamma (x-\alpha)^{\gamma - 1}}{\beta ^\gamma} dx\\ x=\sqrt[\gamma]{y}\beta + \alpha \end{matrix} \Bigg| =$$

$$= \int_{0}^{\infty} (\sqrt[\gamma]{y}\beta + \alpha)^k e^{-y} dy = \int_{0}^{\infty} \sum_{i=0}^{k} C^{i}_{k} (\sqrt[\gamma]{y}\beta)^i \alpha^{k-i} e^{-y} dy =$$

$$= \sum_{i=0}^{k} C^{i}_{k} \alpha^{k-i} \beta^i \int_{0}^{\infty} y^{\frac{i}{\gamma}} e^{-y} dy = \sum_{i=0}^{k} C^{i}_{k} \alpha^{k-i} \beta^i \Gamma(1+\frac{i}{\gamma})$$

Characteristic function of the Weibull distribution with three parameters:

$$\varphi (t) = E(exp(itx)) = \int_{\alpha}^{\infty} exp(itx)f(x) dx = \int_{\alpha}^{\infty} \cos(tx) f(x) dx +$$
$$+ i \int_{\alpha}^{\infty} \sin(tx) f(x) dx$$

$$A = \int_{\alpha}^{\infty} \cos(tx) f(x) dx = cos(t\alpha) - \int_{\alpha}^{\infty} t \sin(tx) exp\Big( -\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \Big) dx$$

$$B = i \int_{\alpha}^{\infty} \sin(tx) f(x) dx = i \sin (t\alpha) - \int_{\alpha}^{\infty} i t \cos(tx) exp\Big( -\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \Big) dx$$

$$A+B = \int_{\alpha}^{\infty}[ \cos(tx) + i\sin(tx)] f(x) dx = exp(it\alpha) -$$

$$- \int_{\alpha}^{\infty} t(\sin(tx) - i \cos(tx)) f(x) dx = exp(it \alpha) - \int_{\alpha}^{\infty} t exp(itx) (-i) f(x) dx =$$

$$= exp(it\alpha) - \int_{\alpha}^{\infty} \sum_{i=0}^{\infty} \frac{(it)^i x^i}{i!} (-it) f(x) dx =$$

$$= exp(it\alpha) + \sum_{i=0}^{\infty} \frac{(it)^{i+1}}{i!} \int_{\alpha}^{\infty} x^i exp\Big( -\Big( \frac{x-\alpha}{\beta} \Big)^\gamma \Big) dx =$$

$$= exp(it\alpha) + \sum_{i=0}^{\infty} \frac{(it)^{i+1}}{i!} \sum_{j=0}^{i}C^{j}_{i} \alpha^{i-j} \beta^j \Gamma(1+\frac{j}{\gamma})$$

(0,1) level quantile of the Weibull distribution with three parameters:

$$F(x_\eta)=\eta$$

$$1-e^{-(\frac{x_\eta-\alpha}{\beta})^\gamma} = \eta$$

$$\ln(1-\eta) = -(\frac{x_\eta-\alpha}{\beta})^\gamma$$

$$(-\ln(1-\eta))^{\frac{1}{\gamma}} = \frac{x_\eta-\alpha}{\beta}$$

$$x_\eta = (-\ln(1-\eta))^{\frac{1}{\gamma}} \beta + \alpha$$, where $$\eta\in(0,1)$$

Median of the Weibull distribution with three parameters is the solution of equation $$F(x) = 0,5$$

$$x = (-\ln(1-0,5))^{\frac{1}{\gamma}} \beta + \alpha = (-\ln(0,5))^{\frac{1}{\gamma}} \beta + \alpha = (\ln(2))^{\frac{1}{\gamma}} \beta + \alpha$$

Mode of the Weibull distribution with three parameters is the maximum point of the density distribution:

$$f'(x) = \frac{(\gamma - 1) \gamma (x-\alpha)^{\gamma - 2} exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)}}{\beta ^\gamma} - \frac{\gamma ^2 exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} (x-\alpha)^{\gamma - 1} (\frac{x-\alpha}{\beta})^{\gamma - 1}}{\beta ^{\gamma + 1}}$$

Assume that $$f'(x) = 0$$

$$\frac{(\gamma - 1) \gamma (x-\alpha)^{\gamma - 2} exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)}}{\beta ^\gamma} = \frac{\gamma ^2 exp{(-\Big(\frac{x-\alpha}{\beta}\Big)^\gamma)} (x-\alpha)^{\gamma - 1} (\frac{x-\alpha}{\beta})^{\gamma - 1}}{\beta ^{\gamma + 1}}$$

$$\beta (\gamma - 1) (x-\alpha)^{\gamma - 2} = \gamma (x-\alpha)^{\gamma - 1} (\frac{x-\alpha}{\beta})^{\gamma - 1}$$

$$\beta^\gamma (\gamma - 1) = \gamma (x-\alpha)^{\gamma}$$

$$(x-\alpha)^{\gamma} = \beta^\gamma \frac{\gamma - 1}{\gamma}$$

$$x = \beta \Big(\frac{\gamma - 1}{\gamma}\Big)^{\frac{1}{\gamma}} + \alpha$$

As $$\gamma \leq 1: Mod(\xi)=\alpha$$

As $$\gamma > 1: Mod(\xi)= \beta \Big(\frac{\gamma - 1}{\gamma}\Big)^{\frac{1}{\gamma}} + \alpha$$

I hope that this material will be useful for students who are just looking for a hint in case of writing a term paper :)

Created by Sergey Migalin. © 2013-2019

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