Question

Let \(X\) denote time until failure of a device and let \(r(x)\) denote the hazard function of \(X\). (a) If \(r(x)=c x^{b}\); where \(c\) and \(b\) are positive constants, show that \(X\) has a Weibull distribution; i.e., $$ f(x)=\left\\{\begin{array}{ll} c x^{b} \exp \left\\{-\frac{c x^{b+1}}{b+1}\right\\} & 0

Short answer

The exercise consists of demonstrating the form of a Weibull distribution when the hazard function is a power of 'x', the form of a Gompertz distribution when the hazard function is exponential in 'x', and the form of a Rayleigh distribution when the hazard function is linear in 'x'. The solutions are obtained by finding the cumulative distribution functions corresponding to each hazard functions and differentiating them if necessary to find the probability distribution functions.

Step-by-step solution

Weibull Distribution

Start with the given hazard function \(r(x)=c x^{b}\). The cumulative distribution function (CDF) of a random variable with that hazard function is given by \(R(x) = \exp(-\int_0^x r(t) dt)\). Calculate the integral and exponentiate to get the cumulative density function \(R(x)= \exp\left(-\frac{c x^{b+1}}{b+1}\right)\). The Probability Density Function (PDF) is obtained by differentiating the CDF: \(f(x) = R'(x) = c x^{b} \exp \left(-\frac{c x^{b+1}}{b+1}\right)\), which is the PDF of a Weibull distribution.

Gompertz Distribution

Start with the given hazard function \(r(x)=c e^{b x}\). As before, calculate the cumulative distribution function \(R(x) = \exp(-\int_0^x r(t) dt) = \exp\left(-\frac{c}{b}\left(1-e^{b x}\right)\right)\). Now, the cumulative distribution function (CDF) is \(F(x) = 1 - R(x) = 1 - \exp \left\{-\frac{c}{b}\left(1-e^{b x}\right)\right\}\). The hazard function defined this way leads to the CDF of a Gompertz distribution.

Rayleigh Distribution

Start with the given hazard function \(r(x)=b x\). Calculate the cumulative distribution function as before to obtain \(R(x) = \exp\left(-\frac{b x^{2}}{2}\right)\). Differentiate to find the PDF: \(f(x) = R'(x) = b x e^{-b x^{2} / 2}\). This is the PDF of a Rayleigh distribution.