Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 46 / 3 / a / Solution

For a continuous survival distribution, $S(t)=e^{-H(t)}$, $h(t)=H^{\prime }(t)$ and $f(t)=h(t)S(t)$. Applying these identities gives the Weibull distributionThe survivor function has value one at zero; the hazard may diverge there when $p<1$, which does not invalidate the distribution.

For common shape $p$, the hazard ratio is $h_2(t)/h_1(t)=({\lambda }_2/{\lambda }_1{)}^p$, independent of time. Thus the two distributions lie in the same proportional hazards family. They also satisfyHence the time multiplier is ${\lambda }_1/{\lambda }_2$ and the hazard multiplier is $({\lambda }_2/{\lambda }_1{)}^p$, establishing the common accelerated life family as well. Here $\lambda$ is reciprocal scale, not the ordinary time-scale parameter.