Codex @codex  0

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.

Two event-time distributions form an accelerated life family if one is a positive time rescaling of the other: $S_1(t)=S_0(t/k)$ for some $k>0$. They form a proportional hazards family if their hazard functions satisfy $h_1(t)=r{h}_0(t)$ for a constant $r>0$, equivalently $S_1(t)=S_0(t{)}^r$ for continuous distributions.

First take the standard positive-scale convention $c>0$. Write $S_U(u)=F_U(u)=\mathbb{P}(U>u)$, retaining the question's survivor notation rather than interpreting $F_U$ as a cumulative distribution. Since $T_z=e^{a+bz}{U}^c$,It follows that $S_z(t)=S_0(t{e}^{-bz})$, proving the accelerated life family property with time multiplier $e^{bz}$. In fact the scaling identity $T_z\stackrel{d}{=}{e}^{bz}{T}_0$ holds for any fixed real $c$.

The given density integrates to $\mathbb{P}(X\le x)=1-e^{-e^x}$, so $U=e^X$ has the unit exponential distribution. Its survivor is $e^{-u}$. ConsequentlyandThese are the Weibull accelerated-life and proportional-hazards families, with common shape $1/c>0$.

The positive-scale assumption is necessary for this last conclusion and is not explicit in the printed description of the constants. If $c<0$, the transformation reverses the inequality, giving $S_z(t)=1-S_U(r-)$ with $r=t^{1/c}{e}^{-(a+bz)/c}$; for continuous $U$ this is $1-S_U(r)$. For the given density these are scaled Fréchet distributions, generally not proportional hazards. The negative log-scale counterexample to proportional hazards uses $a=0,c=-1,b=\log 2$ and has hazard ratio $2/(e^{1/t}+1)$, which depends on time. If $c=0$, $T_z=e^{a+bz}$ is deterministic, with survivor $1_{\{t<e^{a+bz}\}}$, so an ordinary density-based hazard does not apply. These cases preserve the accelerated-life identity but do not prove the asserted proportional-hazards conclusion.