Chapt 4e - 4.12 The Hazard Function 151 Recall that in...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4.12 The Hazard Function 151 Recall that in Chapter 1 we defined the reliability function R ( t ) of a compo- nent as the probability that the component has not failed by time t . Thus, we can relate the reliability function R X ( x ) of X to its CDF as follows: R X ( x ) = P [ X > x ] = 1- P [ X x ] = 1- F X ( x ) Thus, the hazard function can be defined in terms of the reliability function as follows: h X ( x ) = f X ( x ) R X ( x ) We now show that by specifying the hazard function, we uniquely specify the reliability function and, hence, the CDF of a random variable. Since F X ( x ) = 1- R X ( x ) , we have that f X ( x ) = d dx F X ( x ) = d dx { 1- R X ( x ) } = - d dx R X ( x ) Thus, h X ( x ) = d dx F X ( x ) R X ( x ) =- d dx R X ( x ) R X ( x ) = - d dx ln R X ( x ) Integrating both sides from 0 to x we obtain [ ln R X ( t ) ] x = - integraldisplay x h X ( t ) dt Since R X ( ) = 1, we have that ln R X ( x ) = - integraldisplay x h X ( t ) dt R X ( x ) = 1- F X ( x ) = exp braceleftbigg- integraldisplay x h X ( t ) dt bracerightbigg Example 4.26 The time until a component fails is exponentially distributed with a mean of 200 hours. Given that the component has not failed after operating for 150 hours, calculate the hazard function. Solution Let T denote the time until the component fails. Then the PDF and CDF of T are given by f T ( t ) = e- t , = 1 / 200 , t F T ( t ) = integraldisplay t f T ( x ) dx = 1- e- t 152 Chapter 4 Special Probability Distributions Thus, the hazard function is given by h T ( t ) = f T ( t ) 1- F T ( t ) = e- t e- t = Since is the failure rate, we see that the hazard function in this case is a constant that is equal to the failure rate. trianglesolid Example 4.27 Determine the hazard function of a component whose time to failure X is the so-called Weibull random variable with parameters and and whose PDF and CDF are given by f X ( x ) = x - 1 e- x , x ; , F X ( x ) = 1- e- x Solution The Weibull distribution is widely used in reliability modeling. When = 1, we obtain the exponential distribution; and when = 2, we obtain the Rayleigh distribution that is popularly used to model different types of interfer- ence in communication systems. The hazard function of the Weibull distribution is given by h X ( x ) = f X ( x ) 1- F X ( x ) = x - 1 e- x e- x = x - 1 trianglesolid Example 4.28 The hazard function of a certain random variable Y is given by h Y ( y ) = . 5 y , y 0. What is the PDF of Y ? Solution The reliability function of Y is given by R Y ( y ) = exp braceleftbigg- integraldisplay y h Y ( t ) dt bracerightbigg = exp braceleftbigg- integraldisplay y . 5 tdt bracerightbigg = e- . 25 y 2 = 1- F Y ( y ) Therefore, the CDF of Y is given by F Y ( y ) = 1- R Y ( y ) = 1- e- . 25 y 2 Finally, the PDF of Y is given by f Y ( y ) = d dy F y ( y ) = . 5 ye- . 25 y 2 y trianglesolid 4.13 Chapter Summary 153...
View Full Document

This note was uploaded on 01/05/2010 for the course STAT 350 taught by Professor Carlton during the Fall '07 term at Cal Poly.

Page1 / 10

Chapt 4e - 4.12 The Hazard Function 151 Recall that in...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online