= Solution
At time zero there has been no survival selection, so the <population hazard of a survival mixture> is the original mean individual hazard:
$$
\boxed{\overline h(0)=\theta\mathbb EU=\theta.}
$$
At very large times, surviving individuals overwhelmingly come from the smallest-frailty end of the distribution. Their hazards approach the lower endpoint $\theta/2$, so
$$
\boxed{\lim_{t\to\infty}\overline h(t)=\frac\theta2.}
$$
For a direct check, differentiating the logarithm of the <uniform frailty survival mixture> gives
$$
\overline h(t)=\frac\theta2+\frac1t-\frac{\theta}{e^{\theta t}-1},\qquad t>0.
$$
Its expansion at zero is $\theta-\theta^2t/12+O(t^3)$, and its large-time limit is $\theta/2$. The changing population hazard does not imply that an individual's hazard declines. The <decreasing population hazard under constant individual hazards> identity makes the selection mechanism precise: with $\Lambda=\theta U$,
$$
\overline h'(t)=-\operatorname{Var}(\Lambda\mid T>t)\leq0.
$$
It follows by differentiating the ratio $\mathbb E[\Lambda e^{-\Lambda t}]/\mathbb E[e^{-\Lambda t}]$. Thus the surviving mixture becomes less frail over time even though every conditional hazard is constant.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-46-uniform-frailty.png]
{title=Uniform frailty survival mixture and declining population hazard despite constant individual hazards}
The figure takes $\theta=1$. The population survivor curve averages conditional exponential curves, while its hazard falls from the initial mean to the minimum individual hazard.
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