= Solution
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 $T_z=e^{a+bz}U^c$,
$$
\boxed{S_z(t)=S_U\left(t^{1/c}e^{-(a+bz)/c}\right),\qquad t>0.}
$$
It follows that $S_z(t)=S_0(te^{-bz})$, proving the <accelerated life family> property with time multiplier $e^{bz}$. In fact the scaling identity $T_z\overset d=e^{bz}T_0$ holds for any fixed real $c$.
The given density integrates to $\mathbb P(X\leq x)=1-e^{-e^x}$, so $U=e^X$ has the unit <exponential distribution>. Its survivor is $e^{-u}$. Consequently
$$
S_z(t)=\exp\left[-e^{-(a+bz)/c}t^{1/c}\right],\qquad
h_z(t)=\frac1c e^{-(a+bz)/c}t^{1/c-1},
$$
and
$$
\boxed{h_z(t)/h_0(t)=e^{-bz/c}.}
$$
These 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 $a=0,c=-1,b=\log2$ 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 $\mathbf1_{\{t<e^{a+bz}\}}$, 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.
Back to article page
