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For $\log T_z=bz-X$, with $e^X$ unit exponential distribution and $b=\log 2$, the survivor functions are $S_z(t)=1-e^{-2^z/t}$ for $t>0$. These are time-scaled Fréchet distributions, so they form an accelerated life family. However their hazard ratio is $2/(e^{1/t}+1)$, tending to zero as $t\downarrow 0$ and to one as $t\to \infty$. It is not constant. Thus a positive-scale assumption matters when deriving a proportional hazards family from the usual log-location-scale model.

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.

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.

The increasing time change $H_0$ preserves the ordering of event times and transforms each censoring time by the same rule. Applying part b to $U_1,U_2$ therefore givesThis is the pairwise race probability for a proportional hazards family.

Writing the two functions in the question as survivor functions, $S_2(t)=S_1(t{)}^{\lambda }$. Since $H_k(t)=-\log S_k(t)$,Differentiating at times where the hazard functions exist gives $h_2(t)=\lambda h_1(t)$. Their hazard ratio is therefore the constant $\lambda$, so they form a proportional hazards family.

A proportional hazards family has hazard functions related bywhere the hazard ratio $c(z)$ is positive and independent of time. Equivalently, its cumulative hazard functions satisfy $H(t\mid z)=c(z)H_0(t)$ and its survivor functions satisfy $S(t\mid z)=S_0(t{)}^{c(z)}$.

Weibull distributions with a common shape $k>0$ form both an accelerated life family and a proportional hazards family. Their hazard functions are $h_z(t)=k{\lambda }_z{t}^{k-1}$, whose ratio is ${\lambda }_z/{\lambda }_0$. Their time multiplier is $({\lambda }_0/{\lambda }_z{)}^{1/k}$. In the representation $\log T_z=a+bz+cX$ with $c>0$ and density $e^{x-e^x}$ for $X$, the parameters are $k=1/c$ and ${\lambda }_z=e^{-(a+bz)/c}$, giving time multiplier $e^{bz}$ and hazard multiplier $e^{-bz/c}$.