Chapter 6.2, Problem 21E

Let X have a Weibull distribution with parameters α and β, so

$\begin{array}{ccc}E(X)& =& \beta \cdot \Gamma (1\text{ + 1/}\alpha )\\ V(X)& =& {\beta }^2\{\Gamma (1\text{ + 2/}\alpha )\text{ - [}\Gamma \text{(1 + 1/}\alpha {\text{)]}}^2\}\end{array}$

1. a. Based on a random sample X₁,.., X_n, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.
2. b. If n = 20, $\bar{x}$ = 28.0, and ${\displaystyle \sum x_i^2}\text{ = 16,500}$, compute the estimates. [Hint: [Γ(1.2)]²/ Γ(1.4) = .95.]

To determine

Write the equations for the method of moments estimators of $\beta$ and $\alpha$.

Show that the estimate of $\beta$ can be obtained from a table of gamma function after obtaining the estimate of $\alpha$ and the estimate of $\alpha$ is the solution to a complicated equation involving the gamma function.

Answer to Problem 21E

The equations for the method of moments estimators of $\beta$ and $\alpha$ are:

$\underset{\_}{\widehat{\beta }=\frac{\overline{X}}{\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)}}$, and

$\underset{\_}{\frac{\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}}{{\overline{X}}^2}=\frac{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}}$.

Explanation of Solution

Given info:

The random variable X has a Weibull distribution with parameters $\alpha$ and $\beta$, such that $E\left(X\right)=\beta \cdot \Gamma \left(1+\frac{1}{\alpha }\right)$ and $V\left(X\right)={\beta }^2\left\{\Gamma \left(1+\frac{2}{\alpha }\right)-{\left[\Gamma \left(1+\frac{1}{\alpha }\right)\right]}^2\right\}$. A random sample of n observations $X_1,X_2,\mathrm{...},X_n$ is obtained.

Calculation:

First, calculate $E\left(X^2\right)$ for the population having Weibull distribution:

It is known that:

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2\\ E\left(X^2\right)=V\left(X\right)+{\left[E\left(X\right)\right]}^2\end{array}$.

Now, for a Weibull distribution,

$\begin{array}{c}V\left(X\right)={\beta }^2\left\{\Gamma \left(1+\frac{2}{\alpha }\right)-{\left[\Gamma \left(1+\frac{1}{\alpha }\right)\right]}^2\right\}\\ ={\beta }^2\cdot \Gamma \left(1+\frac{2}{\alpha }\right)-{\beta }^2\cdot {\left[\Gamma \left(1+\frac{1}{\alpha }\right)\right]}^2\\ ={\beta }^2\cdot \Gamma \left(1+\frac{2}{\alpha }\right)-{\left[\beta \cdot \Gamma \left(1+\frac{1}{\alpha }\right)\right]}^2\\ ={\beta }^2\cdot \Gamma \left(1+\frac{2}{\alpha }\right)-{\left[E\left(X\right)\right]}^2\left(\text{from given information}\right).\end{array}$

Thus,

$\begin{array}{c}V\left(X\right)+{\left[E\left(X\right)\right]}^2={\beta }^2\cdot \Gamma \left(1+\frac{2}{\alpha }\right)\\ E\left(X^2\right)={\beta }^2\cdot \Gamma \left(1+\frac{2}{\alpha }\right).\end{array}$

For a random sample of size n, the first sample raw moment is the sample mean, that is $\overline{X}$ and the second sample raw moment is $\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}$.

Method of moments:

The method of moments estimator of the m^th population moment is found by equating with the m^th sample moment with the m^th population moment and then solving for the parameters.

Using the method of moments,

$\overline{X}=E\left(X\right)$ and $\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}=E\left(X^2\right)$.

Denote $\widehat{\alpha }$ and $\widehat{\beta }$ as the method of moments estimators as $\alpha$ and $\beta$.

Thus, the method of moments estimators are obtained as follows:

$\begin{array}{c}\overline{X}=E\left(X\right)\\ =\widehat{\beta }\cdot \Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\\ \underset{\_}{\widehat{\beta }=\frac{\overline{X}}{\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)}}.\end{array}$

This equation involves the gamma function.

Again,

$\begin{array}{c}\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}=E\left(X^2\right)\\ ={\widehat{\beta }}^2\cdot \Gamma \left(1+\frac{2}{\widehat{\alpha }}\right).\end{array}$

Substitute the expression for $\widehat{\beta }$ in this equation:

$\begin{array}{c}\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}={\left[\frac{\overline{X}}{\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)}\right]}^2\cdot \Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)\\ \underset{\_}{\frac{\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}}{{\overline{X}}^2}=\frac{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}}.\end{array}$

This equation involves the gamma function.

This equation is independent of $\widehat{\beta }$ and thus, can be solved to obtained $\widehat{\alpha }$. An observation of this equation reveals that it is quite a complicated equation. Thus, the estimate of $\underset{\_}{\alpha }$ is the solution to a complicated equation involving the gamma function.

Once the second equation is solved, the estimated value of $\underset{\_}{\alpha }$, $\underset{\_}{\widehat{\alpha }}$, can be substituted in the first equation to obtain $\underset{\_}{\widehat{\beta }}$, the estimate for $\underset{\_}{\beta }$.

Thus,

To determine

Compute the estimates when $n=20$, $\overline{x}=28.0$ and ${\displaystyle \sum x_i{}^2}=16,500$.

Answer to Problem 21E

The estimate of $\widehat{\alpha }$ is 1.2. The estimate of $\widehat{\beta }$ is $\underset{\_}{\frac{28}{\Gamma \left(1.2\right)}}$.

Explanation of Solution

Calculation:

First, put $n=20$, $\overline{x}=28.0$ and ${\displaystyle \sum x_i{}^2}=16,500$ in:

$\begin{array}{c}\frac{\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i{}^2}}{{\overline{X}}^2}=\frac{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}\\ \frac{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}=\frac{\frac{16,500}{20}}{{\left(28.0\right)}^2}\\ =\frac{825}{784}\\ =\mathrm{1.0523.}\end{array}$

Now, taking reciprocal of both sides of the above equation,

$\begin{array}{c}\frac{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}=\frac{1}{1.0523}\\ =0.9503\\ \approx \mathrm{0.95.}\end{array}$

It is given in hint that:

$\frac{{\left[\Gamma \left(1.2\right)\right]}^2}{\Gamma \left(1.4\right)}=0.95$.

As a result,

$\frac{{\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2}{\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)}=\frac{{\left[\Gamma \left(1.2\right)\right]}^2}{\Gamma \left(1.4\right)}$.

The above equation is an identity, indicating that:

${\left[\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)\right]}^2={\left[\Gamma \left(1.2\right)\right]}^2$, and,

$\Gamma \left(1+\frac{2}{\widehat{\alpha }}\right)=\Gamma \left(1.4\right)$.

From the second equation,

$\begin{array}{c}1+\frac{2}{\widehat{\alpha }}=1.4\\ \frac{2}{\widehat{\alpha }}=1.4-1\\ =0.4\\ \widehat{\alpha }=\frac{2}{0.4}\end{array}$

$=\underset{\_}{5}$.

Substitute $\widehat{\alpha }=5$ and $\overline{x}=28.0$ in the expression for $\widehat{\beta }$:

$\begin{array}{c}\widehat{\beta }=\frac{\overline{X}}{\Gamma \left(1+\frac{1}{\widehat{\alpha }}\right)}\\ =\frac{28}{\Gamma \left(1+\frac{1}{5}\right)}\\ =\frac{28}{\Gamma \left(1+0.2\right)}\\ =\underset{\_}{\frac{28}{\Gamma \left(1.2\right)}}.\end{array}$