# Weibull Distribution | Standard | Two Parameter | Mean | Mode

## Weibull Distribution

Weibull distribution is one of the most widely used probability distribution in reliability engineering.

## Definition of Weibull Distribution

A continuous random variable $X$ is said to have a Weibull distribution with three parameters $\mu$, $\alpha$ and $\beta$ if the random variable $Y =\big(\frac{X-\mu}{\beta}\big)^\alpha$ has the exponential distribution with p.d.f.

 $$\begin{equation*} f(y) = e^{-y}, y>0. \end{equation*}$$

The probability density function of $X$ is
 \begin{align*} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & x>\mu, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{align*}

## Two-parameter Weibull Distribution

Let $\mu=0$. Then the pdf of two parameter Weibull distribution is given by

 \begin{align*} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x}{\beta}\big)^\alpha}, & x>0, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{align*}

In two parameter Weibull distribution, if $\alpha=1$, then $X$ has an exponential distribution with parameter $\frac{1}{\beta}$.

 $$\begin{equation*} f(x) = \frac{1}{\beta} e^{-x/\beta}, x>0, \beta>0. \end{equation*}$$

## Standard Weibull Distribution

Let $\mu=0$ and $\beta =1$. Then the pdf of standard Weibull distribution is

 \begin{align*} f(x;\alpha)&= \begin{cases} \alpha x^{\alpha-1}e^{-x^\alpha}, & x>0, \alpha>0; \\ 0, & Otherwise. \end{cases} \end{align*}

In standard Weibull distribution, if $\alpha=1$, then $X$ has an exponential distribution with parameter $1$.
 $$\begin{equation*} f(x) = e^{-x}, x>0. \end{equation*}$$

## Mean and Variance of Two-parameter Weibull Distribution

The $r^{th}$ raw moment of Two-parameter Weibull distribution is

 $$\begin{eqnarray*} \mu_r^\prime &=& E(X^r) \\ &=&\int_0^\infty x^r f(x)\; dx\\ &=& \int_0^\infty x^r\frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha}\; dx \end{eqnarray*}$$

Let $\big(\frac{x}{\beta}\big)^\alpha = y$

$\Rightarrow \alpha\big(\frac{x}{\beta}\big)^{\alpha-1}\frac{1}{\beta}dx =dy$.

$\Rightarrow \frac{\alpha}{\beta}\big(\frac{x}{\beta}\big)^{\alpha-1} dx = dy$.

Also, for $x=0$, $y=0$ and for $x=\infty$, $y=\infty$. Hence,
 $$\begin{eqnarray*} \mu_r^\prime &=& \int_0^\infty \big(\beta y^{1/\alpha}\big)^re^{-y}\; dy\\ &=& \beta^r \int_0^\infty y^{\frac{r}{\alpha}+1-1}e^{-y}\; dy\\ &=& \beta^r \Gamma (\frac{r}{\alpha}+1) \end{eqnarray*}$$

Hence,

 $$\begin{equation*} \text{mean } = \mu_1^\prime = \beta \Gamma (\frac{1}{\alpha}+1). \end{equation*}$$

 $$\begin{equation*} \mu_2^\prime = \beta^2 \Gamma (\frac{2}{\alpha}+1). \end{equation*}$$

Therefore,
 $$\begin{eqnarray*} \text{Variance } = \mu_2 &=& \mu_2^\prime - (\mu_1^\prime)^2 \\ &=& \beta^2 \Gamma (\frac{2}{\alpha}+1) - \bigg(\beta \Gamma (\frac{1}{\alpha}+1) \bigg)^2\\ &=& \beta^2 \bigg(\Gamma (\frac{2}{\alpha}+1) -\bigg(\Gamma (\frac{1}{\alpha}+1) \bigg)^2\bigg). \end{eqnarray*}$$

## Distribution function of Weibull Dist.

Let $X\sim W(\mu,\alpha,\beta)$. Then the pdf of $X$ is

 \begin{align*} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & x>\mu, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{align*}

The distribution function of Weibull distribution is

 $$\begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=& \int_\mu^x f(x)\; dx\\ &=&\int_\mu^x \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}\; dx \end{eqnarray*}$$

Let $Y = -e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}$. Then

 $$\begin{eqnarray*} \frac{dY}{dx} &=& -\bigg[- e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha} \frac{\alpha}{\beta}\big(\frac{x-\mu}{\beta}\big)^{\alpha-1} \bigg]\\ &=& \frac{\alpha}{\beta}\big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}\\ &=& f(x). \end{eqnarray*}$$

Thus $\int f(x) \; dx = Y$. Using this, the distribution function of $X$ is
 $$\begin{eqnarray*} F(x) &=& \bigg[-e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha} \bigg]_\mu^x \\ &=& \bigg[-e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha} + e^{-\big(\frac{\mu-\mu}{\beta}\big)^\alpha} \bigg]\\ &=& 1- e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}. \end{eqnarray*}$$

## Median of Weibull Distribution

Let $M$ be the median of the distribution.

 $$\begin{equation*} F(M) = P(X\leq M) =\frac{1}{2}. \end{equation*}$$

Thus

 $$\begin{eqnarray*} & & F(M) =\frac{1}{2} \\ &\Rightarrow & 1- e^{-\big(\frac{M-\mu}{\beta}\big)^\alpha}=\frac{1}{2}\\ &\Rightarrow & e^{-\big(\frac{M-\mu}{\beta}\big)^\alpha}=\frac{1}{2}\\ &\Rightarrow &-\big(\frac{M-\mu}{\beta}\big)^\alpha=\log_e \frac{1}{2}\\ &\Rightarrow &\big(\frac{M-\mu}{\beta}\big)^\alpha=\log_e 2\\ &\Rightarrow & \big(\frac{M-\mu}{\beta}\big)=(\log_e 2)^{1/\alpha}\\ &\Rightarrow & M=\mu+\beta(\log_e 2)^{1/\alpha}. \end{eqnarray*}$$

Hence, median of the Weibull distribution is $M=\mu+\beta(\log_e 2)^{1/\alpha}$.

## Mode of Weibull Distribution

Let $X\sim W(\mu,\alpha,\beta)$. Then the pdf of $X$ is

 \begin{align*} f(x;\alpha, \beta)&= \begin{cases} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & x>\mu, \alpha, \beta>0; \\ 0, & Otherwise. \end{cases} \end{align*}

Taking $\log_e$, we get

 $$\begin{equation*} \log_e f(x) = c + (\alpha-1)\log_e \big(\frac{x-\mu}{\beta}\big) - \bigg(\frac{x-\mu}{\beta}\bigg)^\alpha \end{equation*}$$

Differentiating w.r.t. $x$ and equating to zero, we get
 $$\begin{eqnarray*} & & \frac{\partial \log_e f(x)}{\partial x}=0\\ \Rightarrow & &0+(\alpha-1) \frac{1}{(x-\mu)/\beta}\frac{1}{\beta}-\alpha \bigg(\frac{x-\mu}{\beta}\bigg)^{\alpha-1}\frac{1}{\beta}=0\\ \Rightarrow & &\frac{\alpha-1}{x-\mu} = \frac{\alpha}{\beta} \bigg(\frac{x-\mu}{\beta}\bigg)^{\alpha-1}\\ \Rightarrow & &\frac{\alpha-1}{\alpha} = \bigg(\frac{x-\mu}{\beta}\bigg)^{\alpha}\\ \Rightarrow & & \frac{x-\mu}{\beta} = \bigg(\frac{\alpha-1}{\alpha}\bigg)^{1/\alpha}\\ \Rightarrow & & x = \mu +\beta \bigg(\frac{\alpha-1}{\alpha}\bigg)^{1/\alpha}. \end{eqnarray*}$$

And
 $$\begin{equation*} \frac{\partial^2 \log_e f(x)}{\partial x^2}=0\bigg|_{x=x_0}<0. \end{equation*}$$

Hence, mode of the Weibull distribution is
 $$\begin{equation*} x_0 =\mu +\beta \bigg(\frac{\alpha-1}{\alpha}\bigg)^{1/\alpha}. \end{equation*}$$

## Characterization of Weibull Distribution

Let $X_i$, $i=1,2,\cdots, n$ be i.i.d. random variables. Then $\min {X_1,X_2,\cdots, X_n}$ has a Weibull distribution if and only if the common distribution of $X_i$'s is a Weibull distribution.

## Moment Generating Function

The moment generating function of Weibull distribution does not exist.

## Conclusion

In this tutorial, you learned about theory of Weibull distribution like the probability density function, stanradr weibull distribution, mean, variance, median, mode and other properties of Laplace distribution.

To read more about the step by step examples and calculator for Weibull distribution refer the link Weibull Distribution Calculator with Examples. This tutorial will help you to understand how to calculate mean, variance of Weibull distribution and you will learn how to calculate probabilities and cumulative probabilities for Weibull distribution with the help of step by step examples.