91Ó°ÊÓ

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

a)

Let random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) have the same distribution as pmf or pdf \(f\left( {x;{\theta _1},{\theta _2}, \ldots ,{\theta _m}} \right),m \^I N\) with unknown parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\).

By equating sample moments to corresponding population moments and solving for unknown parameters\(\widehat {{{\rm{\theta }}_{\rm{1}}}}{\rm{,}}\widehat {{{\rm{\theta }}_{\rm{2}}}}{\rm{, \ldots ,}}\widehat {{{\rm{\theta }}_{\rm{m}}}}\), the moment estimators\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)may be obtained.

The specified distribution in this exercise is Weibull's distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\),forwhichmomentestimatorsmustbedeveloped.

The sample moment of first order is

\({\rm{\bar X = }}\frac{{\rm{1}}}{{\rm{n}}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{X}}_{\rm{n}}}} \right)\)

and the population moment of first order is

\({\rm{E(X) = \beta \times \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)\)

The first equation in the system of equations from which the moment estimators are obtained is

\({\rm{\bar X = E(X)}}\)

The sample moment of second order is:

\(\frac{{\rm{1}}}{{\rm{n}}}\left( {{\rm{X}}_{\rm{1}}^{\rm{2}}{\rm{ + X}}_{\rm{2}}^{\rm{2}}{\rm{ + \ldots + X}}_{\rm{n}}^{\rm{2}}} \right)\)

And the population moment of second order is:

\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = V(X) + (E(X)}}{{\rm{)}}^{\rm{2}}}{\rm{ - }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\rm{\alpha }}}} \right){\rm{ - }}{{\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)} \right)}^{\rm{2}}}} \right\}{\rm{ - }}{\left\{ {{\rm{\beta \times \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)} \right\}^{\rm{2}}}{\rm{ - }}{{\rm{\beta }}^{\rm{2}}}{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\rm{\alpha }}}} \right)\)

The moment estimators are derived from the second equation in the system of equations.

\(\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} {\rm{ - E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\)

As a result, the system of equations that must be solved for\({\rm{\hat \alpha }}\) and \({\rm{\hat \beta }}\) is

\(\begin{array}{*{20}{r}}{{\rm{\bar X - \hat \beta \times \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\hat \alpha }}}}} \right){\rm{,}}}\\{\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} {\rm{ - }}{{{\rm{\hat \beta }}}^{\rm{2}}}{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{{\rm{\hat \alpha }}}}} \right){\rm{.}}}\end{array}\)

Hence, \(\beta \)can be computed from the first equation as

\({\rm{\hat \beta = }}\frac{{{\rm{\bar X}}}}{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\dot \alpha }}}}} \right)}}{\rm{.}}\)

In order to compute\({\rm{\hat \beta }}\), first \({\rm{\hat \alpha }}\) needs to be determined and gamma functioned evaluated.

From first equation, by squaring both sides, the following stands

\({{\rm{\bar X}}^{\rm{2}}}{\rm{ = }}{{\rm{\hat \beta }}^{\rm{2}}}{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\hat \alpha }}}}} \right){\rm{.}}\)

or equally

\({{\rm{\hat \beta }}^{\rm{2}}}{\rm{ = }}\frac{{{{{\rm{\bar X}}}^{\rm{2}}}}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\hat \alpha }}}}} \right)}}{\rm{.}}\)

Now plug in in the second equation:

\(\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} {\rm{ = }}\frac{{{{{\rm{\bar X}}}^{\rm{2}}}}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)}}{\rm{ \times \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{{\rm{\hat \alpha }}}}} \right)\)

or equally

\(\frac{{\rm{1}}}{{\rm{n}}}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} }}{{{{{\rm{\bar X}}}^{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\rm{\alpha }}}} \right)}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)}}{\rm{.}}\)

The only unknown variable in the last equation is \(\hat \alpha \), and the moment estimator alpha may be obtained by solving this equation. The moment estimator \({\rm{\hat \beta }}\)can be obtained once the equation is solved.

Because, as stated in the exercise, solving this equation is difficult, there is no need to solve it. The goal was to figure out how to get the method estimators alpha and \({\rm{\hat \beta }}\).

b)

Consider the given information,

\(\frac{{\rm{1}}}{{\rm{n}}}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} }}{{{{{\rm{\bar X}}}^{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{{\rm{\hat \alpha }}}}} \right)}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)}}\)

For given\({\rm{n = 20,\bar x = 28}}\), and \(\sum {{\rm{x}}_{\rm{i}}^{\rm{2}}} {\rm{ = 16,500,\hat \alpha }}\) needs to be found.

The following is true:

\(\frac{{\rm{1}}}{{{\rm{20}}}}{\rm{ \times }}\left( {\frac{{{\rm{16,500}}}}{{{\rm{2}}{{\rm{8}}^{\rm{2}}}}}} \right){\rm{ = 1}}{\rm{.05}}\)

therefore,

\(\frac{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\frac{{\rm{\alpha }}}{{\rm{\alpha }}}}}} \right)}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)}}{\rm{ = 1}}{\rm{.05}}{\rm{.}}\)

From the hint

\(\frac{{{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}}}{{{\rm{\Gamma (1}}{\rm{.4)}}}}{\rm{ = 0}}{\rm{.95}}\)

or equally

\(\begin{array}{l}\frac{{{\rm{\Gamma (1 + 0}}{\rm{.4)}}}}{{{{\rm{\Gamma }}^{\rm{2}}}{\rm{(1 + 0}}{\rm{.2)}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.95}}}}\\\frac{{{\rm{\Gamma (1 + 0}}{\rm{.4)}}}}{{{{\rm{\Gamma }}^{\rm{2}}}{\rm{(1 + 0}}{\rm{.2)}}}}{\rm{ = 1}}{\rm{.05}}\end{array}\)

which means that

\(\begin{array}{l}\frac{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{{\rm{\bar \alpha }}}}} \right)}}{{{{\rm{\Gamma }}^{\rm{2}}}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\dot \alpha }}}}} \right)}}{\rm{ = 1}}{\rm{.05}}\\{\rm{ = }}\frac{{{\rm{\Gamma (1 + 0}}{\rm{.4)}}}}{{{{\rm{\Gamma }}^{\rm{2}}}{\rm{(1 + 0}}{\rm{.2)}}}}\end{array}\)

Therefore, because of this equality, the following must hold

\(\frac{{\rm{2}}}{{{\rm{\hat \alpha }}}}{\rm{ = 0}}{\rm{.4}}\)

hence,

\({\rm{\hat \alpha = 5}}{\rm{.}}\)

From the estimator

\({\rm{\hat \beta = }}\frac{{{\rm{\bar X}}}}{{{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{\bar \alpha }}}}} \right)}}\)

The estimate is computed as follows:\({\rm{\hat \beta = }}\frac{{{\rm{28}}}}{{{\rm{\Gamma (1}}{\rm{.2)}}}}\)

The average value of a function \(f\left( x,y \right)\) over a rectangle \(R\) is defined to be \({{f}_{ave}}=\frac{1}{A\left( R \right)}\iint\limits_{R}{f\left( x,y \right)dA}\).

Find the average value of \(f\) over the given rectangle, \(f\left( x,y \right)={{e}^{y}}\sqrt{x+{{e}^{y}}}\), \(R=\left( 0,4 \right)\times \left( 0,1 \right)\).

Given: \(f\left( x,y \right)={{e}^{y}}\sqrt{x+{{e}^{y}}}\)

\(R=\left( 0,4 \right)\times \left( 0,1 \right)\)

To find: average value of f.