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Chapter 4: Q75E (page 182)

Let \({\rm{X}}\) have a Weibull distribution with the pdf from. Verify that \({\rm{\mu = \beta \Gamma (1 + 1/\alpha )}}\). (Hint: In the integral for \({\rm{E(X)}}\), make the change of variable \({\rm{y = (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}\), so that \({\rm{x = \beta }}{{\rm{y}}^{{\rm{1/\alpha }}}}{\rm{.)}}\)

Short Answer

Expert verified

Make following substitutions :

\(\begin{array}{c}{\rm{x = \beta \times }}{{\rm{y}}^{{\rm{1/\alpha }}}}{\rm{dx}}\\{\rm{ = }}\frac{{\rm{\beta }}}{{\rm{\alpha }}}{\rm{ \times }}{{\rm{y}}^{{\rm{(1 - \alpha )/\alpha }}}}{\rm{ \times dy}}\end{array}\)

And use following definition:

\({\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right){\rm{ = }}\int_0^\infty {{{\rm{y}}^{{\rm{1/\alpha }}}}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - y}}}}{\rm{ \times dy}}\)

Step by step solution

01

Definition of Weibull distribution

The Weibull Distribution is a continuous probability distribution that can be used to analyse life statistics, model failure times, and determine product reliability.

02

Verify the equation

If \({\rm{X}}\) has a Weibull distribution, the pdf of \({\rm{X}}\)can be written as follows:

\({\rm{f(x;\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 0}}}\\{\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{\rm{ \times }}{{\rm{x}}^{{\rm{\alpha - 1}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}&{{\rm{x}} \ge {\rm{0}}}\end{array}} \right.\)

Then \({\rm{X}}\)'s predicted value can be stated as follows:

\(\begin{array}{c}{\rm{E(X) = }}\int_{ - \infty }^\infty x {\rm{ \times f(x) \times dx}}\\ = \int_0^\infty x {\rm{ \times }}\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{\rm{ \times }}{{\rm{x}}^{{\rm{\alpha - 1}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}{\rm{ \times dx}}\\{\rm{E(X) = }}\int_0^\infty \alpha {\rm{ \times }}{\left( {\frac{{\rm{x}}}{{\rm{\beta }}}} \right)^{\rm{\alpha }}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}{\rm{ \times dx}}\end{array}\)

Now we make following substitution :

\(\begin{array}{c}{\left( {\frac{{\rm{x}}}{{\rm{\beta }}}} \right)^{\rm{\alpha }}}{\rm{ = y}}\\{\rm{x = \beta \times }}{{\rm{y}}^{{\rm{1/\alpha }}}}\\{\rm{dx = }}\frac{{\rm{\beta }}}{{\rm{\alpha }}}{\rm{ \times }}{{\rm{y}}^{{\rm{(1 - \alpha )/\alpha }}}}{\rm{ \times dy}}\end{array}\)

The integral can then be changed to:

\(\begin{array}{c}{\rm{E(X) = }}\int_0^\infty \alpha {\rm{ \times y \times }}{{\rm{e}}^{{\rm{ - y}}}}{\rm{ \times }}\frac{{\rm{\beta }}}{{\rm{\alpha }}}{\rm{ \times }}{{\rm{y}}^{{\rm{(1 - \alpha )/\alpha }}}}{\rm{ \times dy}}\\{\rm{ = \beta \times }}\int_0^\infty {{{\rm{y}}^{{\rm{1/\alpha }}}}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - y}}}}{\rm{ \times dy}}\end{array}\)

We can now write using the Gamma function definition.

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

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Most popular questions from this chapter

Let Z have a standard normal distribution and define a new rv Y by \[{\text{Y = \sigma Z + \mu }}\]. Show that Y has a normal distribution with parameters \[{\text{\mu }}\] and \[{\text{\sigma }}\]. (Hint: \[{\text{Y\poundsy}}\]if \[{\text{Z\pounds}}\]? Use this to find the cdf of Y and then differentiate it with respect to y.)

Let X 5 the time it takes a read/write head to locate the desired record on a computer disk memory device once the head has been positioned over the correct track. If the disks rotate once every \({\bf{25}}\) milliseconds, a reasonable assumption is that X is uniformly distributed on the interval\(\left( {{\bf{0}},{\rm{ }}{\bf{25}}} \right)\). a. Compute\({\bf{P}}\left( {{\bf{10}} \le {\bf{X}} \le {\bf{20}}} \right)\). b. Compute \({\bf{P}}\left( {{\bf{X}} \le {\bf{10}}} \right)\). c. Obtain the cdf F(X). d. Compute E(X) and \({\sigma _X}\).

An ecologist wishes to mark off a circular sampling region having radius \({\rm{10\;m}}\). However, the radius of the resulting region is actually a random variable \({\rm{R}}\) with pdf

\({\rm{f(r) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{3}}}{{\rm{4}}}\left( {{\rm{1 - (10 - r}}{{\rm{)}}^{\rm{2}}}} \right)}&{{\rm{9}} \le {\rm{r}} \le {\rm{11}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

What is the expected area of the resulting circular region?

The article "A Probabilistic Model of Fracture in Concrete and Size Effects on Fracture Toughness" (Magazine of Concrete Res., \({\rm{1996: 311 - 320}}\)) gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size \({\rm{n = 18}}\) observations on toughness for high strength concrete (consistent with one of the histograms); values of \({{\rm{p}}_{\rm{i}}}{\rm{ = (i - }}{\rm{.5)/18}}\) are also given.

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{.47}}}&{{\rm{.58}}}&{{\rm{.65}}}&{{\rm{.69}}}&{{\rm{.72}}}&{{\rm{.74}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.0278}}}&{{\rm{.0833}}}&{{\rm{.1389}}}&{{\rm{.1944}}}&{{\rm{.2500}}}&{{\rm{.3056}}}\\{{\rm{ Observation }}}&{{\rm{.77}}}&{{\rm{.79}}}&{{\rm{.80}}}&{{\rm{.81}}}&{{\rm{.82}}}&{{\rm{.84}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.3611}}}&{{\rm{.4167}}}&{{\rm{.4722}}}&{{\rm{.5278}}}&{{\rm{.5833}}}&{{\rm{.6389}}}\\{{\rm{ Observation }}}&{{\rm{.86}}}&{{\rm{.89}}}&{{\rm{.91}}}&{{\rm{.95}}}&{{\rm{1}}{\rm{.01}}}&{{\rm{1}}{\rm{.04}}}\\{{{\rm{p}}_{\rm{i}}}}&{{\rm{.6944}}}&{{\rm{.7500}}}&{{\rm{.8056}}}&{{\rm{.8611}}}&{{\rm{.9167}}}&{{\rm{.9722}}}\end{array}\)

Construct a Weibull probability plot and comment.

Let the ordered sample observations be denoted by \({y_1}, {y_2}, \ldots , {y_n}\)(\({y_1}\) being the smallest and \({y_n}\)the largest). Our suggested check for normality is to plot the \({\phi ^{ - 1}}((i - .5)/n),{y_i})\)pairs. Suppose we believe that the observations come from a distribution with mean, and let \({w_1}, \ldots , {w_n}\)be the ordered absolute values of the \({x_i}'s\). A half 颅normal plot is a probability plot of the\({w_i}'s\). More specifically, since \(P\left( {|Z| \le w} \right) = P\left( { - w \le Z \le w} \right) = 2\phi \left( w \right) - 1\), a half-normal plot is a plot of the \(\left( {{\phi ^{ - 1}}/ \left\{ {\left( {\left( {i - .5} \right)/n + 1} \right)/2} \right\},{w_i}} \right)\)pairs. The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: \(23.78, 21.27,\)\(1.44, 2.39, 12.38, \)\(243.40, 1.15, 23.96, 22.34, 30.84.\)
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