= Solution
For $\sigma^2>0$, a <gamma distribution> with unit mean and <variance> $\sigma^2$ has shape and rate $\psi=1/\sigma^2$. Substituting its <Laplace transform> into the preceding result gives
$$
\overline F^{(z)}(t)
=\left(1+\sigma^2e^{\beta z}H_0(t)\right)^{-1/\sigma^2}.
$$
Its negative logarithmic derivative is the population <hazard function>:
$$
\overline h^{(z)}(t)
=\frac{e^{\beta z}h_0(t)}{1+\sigma^2 e^{\beta z}H_0(t)}.
$$
Consequently, at times with $h_0(t)>0$, the <gamma frailty hazard ratio> is
$$
\boxed{r(t)=e^\beta\frac{1+\sigma^2H_0(t)}{1+\sigma^2 e^\beta H_0(t)}.}
$$
At a time when both hazards vanish this formula is a continuous extension; the literal ratio $0/0$ is undefined.
Write $a=e^\beta>0$ and $x=\sigma^2H_0(t)\geq0$. Two useful identities are
$$
r-1=\frac{a-1}{1+ax},\qquad
\frac{dr}{dx}=\frac{a(1-a)}{(1+ax)^2}.
$$
For small cumulative exposure $x$, $r=a\{1+(1-a)x+O(x^2)\}$, so it starts at the conditional <hazard ratio> $e^\beta$. If $\beta>0$, it decreases towards one; if $\beta<0$, it increases towards one; and if $\beta=0$, it equals one throughout. It stays between $e^\beta$ and one and never crosses one.
For fixed $\sigma^2>0$, \b[the late-time limit is one provided $H_0(t)\to\infty$]. If instead the <integrated hazard> tends to a finite value $K$, the limiting <hazard ratio> is $a(1+\sigma^2K)/(1+a\sigma^2K)$, which need not be one. Elapsed time alone does not imply unbounded exposure.
For small $\sigma^2$ at a fixed time, $r\to e^\beta$. At $\sigma^2=0$ the frailty degenerates to $U=1$, giving an ordinary <proportional hazards model>. For large $\sigma^2$ at a fixed time with $H_0(t)>0$, $r\to1$; at time zero, however, it remains $e^\beta$. Thus that variance limit is not uniform near zero. Larger heterogeneity accelerates the movement of the population <hazard ratio> towards one. The mechanism is <survival selection>: frailer individuals leave the population sooner, particularly in the higher-hazard group, while each individual's conditional multiplier remains fixed.
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