 Solution

A proportional hazards family has a common baseline hazard $h_0$ and positive constants $a_z$ such that $h_z(t)=a_z{h}_0(t)$. Thus its hazard ratios are constant in time, and its survivor functions satisfy $S_z(t)=S_0(t{)}^{a_z}$. An accelerated life family instead has positive constants $c_z$ such that $T_z$ has the distribution of $c_z{T}_0$. Equivalently, $S_z(t)=S_0(t/c_z)$: all time quantiles are multiplied by the same factor. These two properties are generally different.

Integrating the printed Weibull distribution density, or differentiating the following expression and checking its initial value, givesHere $\lambda$ is an inverse-time parameter: the coefficient of $t^p$ in the cumulative hazard is ${\lambda }^p$, not $\lambda$. This matters when comparing different parameterizations of a Weibull distribution.

With common shape $p$, the hazard ratio isIt does not depend on $t$, proving membership in one proportional hazards family. For the other property,Consequently they also belong to one accelerated life family. A larger inverse-time parameter gives shorter survival times and larger hazards, consistently in both descriptions. This is the common-shape case of Weibull accelerated-life and proportional-hazards families.