= Solution
Let population 1 have baseline $\theta t$ and population 2 baseline $\lambda\theta t$, with $\theta>0$ and $\lambda>1$, and identical unit-rate exponential frailty distributions initially. Their population <hazard functions> have ratio, for $t>0$,
$$
\boxed{R(t)=\frac{\overline h_2(t)}{\overline h_1(t)}=\lambda\frac{1+\theta t^2/2}{1+\lambda\theta t^2/2}.}
$$
At exactly $t=0$, both population hazards are zero, so their literal ratio is $0/0$ and is undefined. The intended initial comparison is the right-hand limit,
$$
\boxed{R(0+)=\lambda,\qquad \lim_{t\to\infty}R(t)=1.}
$$
With $z=\theta t^2/2$, $dR/dz=\lambda(1-\lambda)/(1+\lambda z)^2<0$, so the marginal <hazard ratio> decreases from $\lambda$ toward one. The faster-hazard population loses highly frail individuals sooner; its survivors become less frail than the first population's survivors.
Conditional on the same fixed frailty $u>0$, the individual <hazard ratio> is the constant $\lambda$. Marginalizing over frailty therefore destroys <proportional hazards>, even though they hold at the individual level. A population-level <Cox proportional-hazards model> that ignores frailty would impose an incorrect constant <hazard ratio> here. Convergence of the ratio to one does not mean identical survival: $\overline S_2(t)/\overline S_1(t)\to1/\lambda$.
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