Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

It is said that a random variableX has an increasing failure rate if the failure rate h(x) defined in Exercise 18 is an increasing function of xfor x> 0, and it is said that Xhas a decreasing failure rate if h(x) is a decreasing function of x for x > 0. Suppose that X has the Weibull distribution with parameters a and b, as defined in Exercise 19. Show thatX has an increasing failure rate if b > 1, and X has a decreasing failure rate if b < 1.

Short Answer

Expert verified

LetX has an increasing failure rate if b > 1, and X has a decreasing failure rate if b < 1.

Step by step solution

01

Given information

LetX has the Weibull distribution with parameters a and b.One need to prove that X has an increasing failure rate if b > 1, and X has a decreasing failure rate if b < 1.

02

Proof of X has an increasing failure rate if b > 1, and X has a decreasing failure rate if b < 1.

The failure rate of X is

\(h\left( x \right) = \frac{{f\left( x \right)}}{{1 - F\left( x \right)}}\)

Differentiating both sides with respect to x one get

\(\begin{array}{l}h'\left( x \right) = \frac{{f'\left( x \right)\left( {1 - F\left( x \right)} \right) - \left( { - f\left( x \right)} \right)f\left( x \right)}}{{{{\left\{ {1 - F\left( x \right)} \right\}}^2}}}\,\,{\rm{since}}\,F'\left( x \right) = f\left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{f'\left( x \right)\left( {1 - F\left( x \right)} \right) + {{\left( {f\left( x \right)} \right)}^2}}}{{{{\left\{ {1 - F\left( x \right)} \right\}}^2}}}\end{array}\)

Now,\(h'\left( x \right) > 0\,or\, < 0\,iff\,f'\left( x \right) > 0\,or\, < 0\).Now,

\(\begin{array}{l}f\left( x \right) = \frac{b}{{{a^b}}}{x^{b - 1}}{e^{ - {{\left( {\frac{x}{a}} \right)}^b}}}\\f'\left( x \right) = \frac{b}{{{a^b}}}\left( {\left( {b - 1} \right){x^{b - 2}}{e^{ - {{\left( {\frac{x}{a}} \right)}^b}}} + \frac{b}{{{a^b}}}{x^{b - 1}}{e^{ - {{\left( {\frac{x}{a}} \right)}^b}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {b - 1} \right) \times k\,{\rm{where}}\,k > 0\\f'\left( x \right) > 0\,{\rm{if}}\,b > 1\,{\rm{and}}\, < 0\,if\,b < 1\end{array}\)

Hence X has an increasing failure rate if b > 1, and X has a decreasing failure rate if b < 1.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free