Question

Show that the failure rate (hazard function) of the Pareto distribution is $$ \frac{h(x)}{1-H(x)}=\frac{\alpha}{\beta^{-1}+x} $$ Find the failure rate (hazard function) of the Burr distribution with cdf $$ G(y)=1-\left(\frac{1}{1+\beta y^{\tau}}\right)^{\alpha}, \quad 0 \leq y<\infty $$ In each of these two failure rates, note what happens as the value of the variable increases.

Short answer

The hazard function for the Pareto distribution is \(\frac{\alpha}{\beta^{-1}+x}\) and for the Burr distribution it can be found by differentiating its CDF and substituting appropriately into the hazard function. As for the behavior of the hazard function as the variables increase, it must be observed by substituting increasing values of the variables into the respective functions.

Step-by-step solution

Prove the hazard function of Pareto Distribution

The failure rate or hazard function is given by \(h(x) = \frac{f(x)}{1 - H(x)}\) where \(f(x)\) is the pdf and \(H(x)\) is the cdf. For the Pareto distribution, \(f(x) = \frac{\alpha \beta^\alpha}{x^{\alpha + 1}}\) for \(x \geq \beta\) and \(H(x) = 1 - (\frac{\beta}{x})^\alpha\). We substitute these functions into the hazard function, simplify it, and if we do the math correctly, it should result in \(\frac{\alpha}{\beta^{-1}+x}\).

Find the hazard function of Burr Distribution

We start with the CDF of the Burr distribution, \(G(y) = 1 - \left(\frac{1}{1 + \beta y^\tau}\right)^\alpha\). To get the hazard function we first have to get the pdf which we get by differentiating the CDF, \(g(y) = G'(y)\). We then substitute the pdf and \(1-G(y)\) into the hazard function.

Analyze the behaviour as variable increases

To analyze the behavior of the hazard functions as the variable increases, one has to observe how the functions change as \(x\) increases for the Pareto distribution and as \(y\) increases for the Burr distribution.