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 Weibull random variable $X$ is
$$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
- $\alpha$ is the shape parameter
- $\beta$ is the scale parameter
- $\mu$ is the location parameter.
Two-parameter Weibull Distribution
Let $\mu=0$. Then the pdf of two parameter Weibull distribution is given by
$$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x}{\beta}\big)^\alpha}, & \hbox{$x>0$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
In two parameter Weibull distribution, if $\alpha=1$, then $X$ has an exponential distribution with parameter $\frac{1}{\beta}$.
$$ \begin{equation*} f(x) = \frac{1}{\beta} e^{-x/\beta}, x>0, \beta>0. \end{equation*} $$
Graph of Weibull Distribution
Standard Weibull Distribution
If we let $\mu=0$ and $\beta =1$, then the distribution of $X$ is called standard Weibull distribution. Then the pdf of standard Weibull distribution is
$$ \begin{equation*} f(x;\beta)=\left\{ \begin{array}{ll} \alpha x^{\alpha-1}e^{-x^\alpha}, & \hbox{$x>0$, $\beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
In standard Weibull distribution, if we let $\alpha=1$, then the random variable $X$ has an exponential distribution with parameter $1$.
$$ \begin{equation*} f(x) = e^{-x}, x>0. \end{equation*} $$
$r^{th}$ raw moment of Two-parameter Weibull Distribution
The $r^{th}$ raw moment of Two-parameter Weibull distribution is
$$ \begin{equation*} % \nonumber to remove numbering (before each equation) \mu_r^\prime = \beta^r \Gamma (\frac{r}{\alpha}+1) \end{equation*} $$
Proof
The $r^{th}$ raw moment of Two-parameter Weibull distribution is
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \mu_r^\prime &=& E(X^r) \\ &=&\int_0^\infty x^r f(x)\; dx\\ &=& \int_0^\infty x^r\frac{\alpha}{\beta} \big(\frac{x}{\beta}\big)^{\alpha-1}e^{-\big(\dfrac{x}{\beta}\big)^\alpha}\; dx \end{eqnarray*} $$
Let $\big(\dfrac{x}{\beta}\big)^\alpha = y$
$\Rightarrow \alpha\big(\dfrac{x}{\beta}\big)^{\alpha-1}\dfrac{1}{\beta}dx =dy$.
$\Rightarrow \dfrac{\alpha}{\beta}\big(\dfrac{x}{\beta}\big)^{\alpha-1} dx = dy$.
Also, for $x=0$, $y=0$ and for $x=\infty$, $y=\infty$. Hence,
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \mu_r^\prime &=& \int_0^\infty \big(\beta y^{1/\alpha}\big)^re^{-y}\; dy\\ &=& \beta^r \int_0^\infty y^{\frac{r}{\alpha}+1-1}e^{-y}\; dy\\ &=& \beta^r \Gamma (\frac{r}{\alpha}+1) \end{eqnarray*} $$
Mean and Variance of Two-parameter Weibull Distribution
The mean and variance of Two-parameter Weibull distribution are $E(X) = \beta \Gamma (\dfrac{1}{\alpha}+1)$ and $V(X) = \beta^2 \bigg(\Gamma (\dfrac{2}{\alpha}+1) -\bigg(\Gamma (\dfrac{1}{\alpha}+1) \bigg)^2\bigg)$ respectively.
Proof
Letting $r=1$ in the $r^{th}$ raw moments, we have
$$ \begin{equation*} E(X) = \mu_1^\prime = \beta \Gamma (\frac{1}{\alpha}+1). \end{equation*} $$
Letting $r=2$ in the $r^{th}$ raw moments, we have
$$ \begin{equation*} E(X^2) = \mu_2^\prime = \beta^2 \Gamma (\frac{2}{\alpha}+1). \end{equation*} $$
Then the variance of $X$ is
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \text{Variance } & = & E(X^2) - [E(X)]^2\\ &=& \beta^2 \Gamma (\frac{2}{\alpha}+1) - \bigg(\beta \Gamma (\frac{1}{\alpha}+1) \bigg)^2\\ &=& \beta^2 \bigg(\Gamma (\frac{2}{\alpha}+1) -\bigg(\Gamma (\frac{1}{\alpha}+1) \bigg)^2\bigg). \end{eqnarray*} $$
Distribution function of Weibull Distribution
The distribution function of Three-parameter Weibull distribution is
$$ \begin{equation*} % \nonumber to remove numbering (before each equation) F(x) = 1- e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}. \end{equation*} $$
Proof
Let $X\sim W(\mu,\alpha,\beta)$. Then the pdf of $X$ is
$$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
The distribution function of Weibull distribution is
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) F(x) &=& P(X\leq x) \\ &=& \int_\mu^x f(x)\; dx\\ &=&\int_\mu^x \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}\; dx \end{eqnarray*} $$
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
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) F(x) &=& \bigg[-e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha} \bigg]_\mu^x \\ &=& \bigg[-e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha} + e^{-\big(\frac{\mu-\mu}{\beta}\big)^\alpha} \bigg]\\ &=& 1- e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}. \end{eqnarray*} $$
Median of Weibull Distribution
The median of Three parameter Weibull distribution is $M=\mu+\beta(\log_e 2)^{1/\alpha}$
Proof
Let $M$ be the median of the distribution.
$$ \begin{equation*} F(M) = P(X\leq M) =\frac{1}{2}. \end{equation*} $$
Thus
$$ \begin{eqnarray*} % \nonumber to remove numbering (before each equation) & & F(M) =\frac{1}{2} \\ &\Rightarrow & 1- e^{-\big(\frac{M-\mu}{\beta}\big)^\alpha}=\frac{1}{2}\\ &\Rightarrow & e^{-\big(\frac{M-\mu}{\beta}\big)^\alpha}=\frac{1}{2}\\ &\Rightarrow &-\big(\frac{M-\mu}{\beta}\big)^\alpha=\log_e \frac{1}{2}\\ &\Rightarrow &\big(\frac{M-\mu}{\beta}\big)^\alpha=\log_e 2\\ &\Rightarrow & \big(\frac{M-\mu}{\beta}\big)=(\log_e 2)^{1/\alpha}\\ &\Rightarrow & M=\mu+\beta(\log_e 2)^{1/\alpha}. \end{eqnarray*} $$
Hence, median of the Weibull distribution is $M=\mu+\beta(\log_e 2)^{1/\alpha}$.
Mode of Weibull Distribution
The mode of three parameter Weibull distribution is
$$ \begin{equation*} x_0 =\mu +\beta \bigg(\frac{\alpha-1}{\alpha}\bigg)^{1/\alpha}. \end{equation*} $$
Proof
Let $X\sim W(\mu,\alpha,\beta)$. Then the pdf of $X$ is
$$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$
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.
Hope this article helps you understand Weibull distribution and various proof related to Weibull distributions.
