= Solution
A <frailty model> represents unobserved heterogeneity using a nonnegative <frailty random variable> $U$ that multiplies an individual's <baseline hazard>. With $H_0(t)=\int_0^th_0(s)\,ds$, the individual and population <survivor functions> are
$$
S(t\mid U=u)=e^{-uH_0(t)},\qquad \overline S(t)=\int_0^\infty e^{-uH_0(t)}g(u)\,du.
$$
Thus the population <survivor function> is the <Laplace transform> of the frailty distribution at $H_0(t)$; at points where differentiation is justified, $\overline h(t)=h_0(t)\mathbb E[U\mid T>t]$. The last identity follows by differentiating the integral and dividing $-\overline S'$ by $\overline S$.
Here $g(u)=e^{-u}$ for $u\geq0$, $\theta>0$, and $H_0(t)=\theta t^2/2$. Direct integration gives
$$
\boxed{\overline S(t)=\frac1{1+\theta t^2/2},\qquad \overline h(t)=\frac{\theta t}{1+\theta t^2/2}.}
$$
This is <population hazard under exponential frailty>. For every fixed $u>0$ the individual <hazard function> $u\theta t$ increases linearly. In contrast,
$$
\overline h'(t)=\frac{\theta(1-\theta t^2/2)}{(1+\theta t^2/2)^2},
$$
so the population hazard increases up to $t=\sqrt{2/\theta}$, then decreases, tending to zero like $2/t$. More vulnerable individuals leave the <risk set> earlier, leaving survivors with smaller frailties. This population decline is selection, not a decline in any fixed individual's hazard. If $\theta=0$, survival is identically one and both hazards vanish; the positive-rate comparison is then inapplicable.
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