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Chapter 4: Q122SE (page 196)

Let X denote the lifetime of a component, with f(x) and F(x) the pdf and cdf of X. The probability that the component fails in the interval \({\rm{(x,x + \Delta x)}}\) is approximately f(x) . The conditional probability that it fails in \({\rm{(x,x + \Delta x)}}\) given that it has lasted at least x is f(x)? \({\rm{f(x) \times \Delta x/(1 - F(x))}}\). Dividing this by \({\rm{\Delta x}}\) produces the failure rate function:

\({\rm{r(x) = }}\frac{{{\rm{f(x)}}}}{{{\rm{1 - F(x)}}}}\)

An increasing failure rate function indicates that older components are increasingly likely to wear out, whereas a decreasing failure rate is evidence of increasing reliability with age. In practice, a 鈥渂athtub-shaped鈥� failure is often assumed.

a. If X is exponentially distributed, what is r(x)?

b. If X has a Weibull distribution with parameters a and b, what is r(x)? For what parameter values will r(x) be increasing? For what parameter values will r(x) decrease with x?

c. Since \({\rm{r(x) = - (d/dx)ln(1 - F(x))}}\), \({\rm{ln(1 - F(x)) = - }}\int {{\rm{r(x)dx}}} \). Suppose

\({\rm{r(x) = }}\left\{ {\begin{aligned}{}{{\rm{\alpha }}\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right)}&{{\rm{0}} \le {\rm{x}} \le {\rm{\beta }}}\\{\rm{0}}&{\rm{\;}}\end{aligned}} \right.\)

so that if a component lasts b hours, it will last forever (while seemingly unreasonable, this model can be used to study just 鈥渋nitial wear out鈥�). What are the cdf and pdf of X?

Short Answer

Expert verified

(a) \({\rm{r(x) = \lambda }}\), if X is exponentially distributed.

(b) \({\rm{r(x) = }}\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}\) , if X has a Weibull distribution with parameters a and b.

Decreasing: \({\rm{0 < \alpha < 1}}\)

Increasing: \({\rm{\alpha > 1}}\)

(c) Cdf is \({\rm{F(x) = 1 - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

Pdf is \({\rm{f(x) = \alpha }}\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right){{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

Step by step solution

01

Definition of Probability

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

02

Calculating r(x) if X is exponentially distributed

Given:

\({\rm{r(x) = }}\frac{{{\rm{f(x)}}}}{{{\rm{1 - F(x)}}}}\)

(a) The exponential distribution's probability density function and cumulative distribution function with parameter \({\rm{\lambda }}\) are:

\({\rm{f(x) = \lambda }}{{\rm{e}}^{{\rm{ - \lambda x}}}}\)

\({\rm{F(x) = 1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}\)

Fill in the following expressions in the given formula:

\({\rm{r(x) = }}\frac{{{\rm{f(x)}}}}{{{\rm{1 - F(x)}}}}{\rm{ = }}\frac{{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda x}}}}}}{{{\rm{1 - }}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \lambda x}}}}} \right)}}{\rm{ = }}\frac{{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda x}}}}}}{{{{\rm{e}}^{{\rm{ - \lambda x}}}}}}{\rm{ = \lambda }}\)

03

Calculating r(x) if X has a Weibull distribution with parameter a and b

(b) The Weibull distribution's probability density function and cumulative distribution function with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\) are:

\({\rm{f(x) = }}\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}\)

\(F(x) = \left\{ {\begin{array}{*{20}{c}}0&{x < 0} \\{1 - {e^{ - {{(x/\beta )}^\alpha }}}}&{x0} \end{array}} \right.\)

Fill in the following expressions in the given formula:

\({\rm{r(x) = }}\frac{{{\rm{f(x)}}}}{{{\rm{1 - F(x)}}}}{\rm{ = }}\frac{{\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}}{{{\rm{1 - }}\left( {{\rm{1 - }}{{\rm{e}}^{\left. {{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}} \right)}}} \right.}}{\rm{ = }}\frac{{\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}}{{{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}}{\rm{ = }}\frac{{\rm{\alpha }}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 1}}}}\)

04

Calculating parameter values

Calculate the derivative:

\({\rm{r'(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{(r(x)) = }}\frac{{{\rm{\alpha (\alpha - 1)}}}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 2}}}}\)

Determine the source:

\(\frac{{{\rm{\alpha (\alpha - 1)}}}}{{{{\rm{\beta }}^{\rm{\alpha }}}}}{{\rm{x}}^{{\rm{\alpha - 2}}}}{\rm{ = 0}}\)

The values of x,\({\rm{\alpha }}\), and \({\rm{\beta }}\) are all greater than zero:

\({\rm{\alpha - 1 = 0}}\)

To each side, add 1:

\({\rm{\alpha = 1}}\)

If the derivative \({\rm{r'(x)}}\) is negative, as it is in the case of \({\rm{\alpha < 1}}\), \({\rm{r(x)}}\) decreases.

If the derivative \({\rm{r'(x)}}\) is positive, as it is for \({\rm{\alpha > 1}}\), \({\rm{r(x)}}\) is growing.

05

Calculating cdf and pdf of X 

\({\rm{r(x) = \alpha }}\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right)\)

Calculate the integral:

\(\int {\rm{\alpha }} {\rm{r(x)dx = }}\int {\rm{\alpha }} \left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right){\rm{dx = \alpha }}\left( {{\rm{x - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{2\beta }}}}} \right)\)

The negative integral of \({\rm{1 - F(x)}}\) is the natural logarithm:

\({\rm{ln(1 - F(x)) = - }}\smallint {\rm{r(x)dx = - \alpha }}\left( {{\rm{x - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{2\beta }}}}} \right)\)

Take each side's exponential:

\({\rm{1 - F(x) = }}{{\rm{e}}^{{\rm{ln(1 - F(x))}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

Remove one from both sides of the equation:

\({\rm{ - F(x) = - 1 + }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

Multiply \({\rm{ - 1}}\) to each side of the equation:

\({\rm{F(x) = 1 - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

06

Calculating cdf and pdf of X

The cumulative distribution function's derivative is the probability density function:

\({\rm{f(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{F(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}} \right)\)

\({\rm{ = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}} \right)\)

\({\rm{ = - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{ - \alpha }}\left( {{\rm{x - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{{\rm{2\beta }}}}} \right)} \right)\)

\({\rm{ = - }}{{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\left( {{\rm{ - \alpha }}\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right)} \right)\)

\({\rm{ = \alpha }}\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\beta }}}} \right){{\rm{e}}^{{\rm{ - \alpha }}\left( {{\rm{x - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\beta }}} \right)}}\)

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b. What is the probability that at most one error is made on both exams combined?

c. Obtain a general expression for the probability that the total number of errors in the two exams is m (where \({\rm{m}}\) is a nonnegative integer). (Hint: \({\rm{A = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:x + y = m}}} \right\}{\rm{ = }}\left\{ {\left( {{\rm{m,0}}} \right)\left( {{\rm{m - 1,1}}} \right){\rm{,}}.....{\rm{(1,m - 1),(0,m)}}} \right\}\)Now sum the joint pmf over \({\rm{(x,y)}} \in {\rm{A}}\)and use the binomial theorem, which says that

\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)
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