The exponential distribution is known to have a constant hazard rate, $\lambda = \frac{f(t)}{S(t)} = \frac{\lambda e^{-\lambda t}}{e^{-\lambda t}}$ where $f(t)$ and $S(t)$ are its PDF and survival functions respectively. If we assume the $\lambda$ itself comes from a Gamma distribution and integrate over it, the resulting exponential mixture becomes a Pareto distribution, which is much heavier tailed and has decreasing hazard rate. This and other mixtures are discussed in this paper. All of those mixtures result in distributions with decreasing hazard rates.
Here is an example of such a mixture: let's say the $\lambda$ of the exponential comes from a Gamma distribution with PDF: $g(\lambda) = \theta e^{-\theta \lambda}\frac{(\theta \lambda)^{m-1}}{(m-1)!}$.
The PDF of the resulting exponential-mixture distribution becomes:
$$f_T(t) = \int\limits_0^\infty \lambda e^{-\lambda t} g(\lambda) d\lambda$$
If you work out the integral, this becomes:
$$f_T(t) = \frac{m\theta^m}{(t+\theta)^{m+1}}$$
This is a Pareto distribution and has a decreasing hazard rate.
Conjecture-1: I conjecture that mixing the exponential distribution in this way can't possibly lead to a distribution with increasing hazard rate.
The reason I suspect this is because :
Conjecture-2: Decreasing hazard rate distributions lead to over-dispersed point processes (variance of number of events in any interval of time is greater than the mean). While a time to event distribution with increasing hazard rate will result in an under-dispersed distribution (variance of number of events less than mean). I don't have a proof for this, but have verified it extensively with simulations.
If you believe conjecture-2, then conjecture-1 is equivalent to saying that mixing the exponential distribution can't possibly convert it into an under-dispersed process since you are adding a factor that can only increase the uncertainty.
I'm looking for either a proof or counter-example to conjecture-1, but don't know where to start.
Best Answer
Since $T\mid \Lambda=\lambda \sim \rm{Exp}(\lambda)$, and the pdf of $\Lambda$ is $g$, we have
$$ f_T(t) = \int_0^{+\infty} \lambda e^{-\lambda t}g(\lambda)d\lambda $$
$$ S_T(t) = \int_0^{+\infty} e^{-\lambda t}g(\lambda)d\lambda $$
$$ f_T'(t) = -\int_0^{+\infty} \lambda^2 e^{-\lambda t}g(\lambda)d\lambda $$
$$ S_T'(t) = -f(t)= -\int_0^{+\infty} \lambda e^{-\lambda t}g(\lambda)d\lambda $$
The derivative of the harzard rate is $$ h_T'(t) = \frac {S_T(t)f_T'(t) - f_T(t)S_T'(t)} {S_T(t)^2} = \frac {S_T(t)f_T'(t) + f_T(t)^2} {S_T(t)^2}$$
Since we are only concern about the sign of $h'(t)$, and its denominator is always positive, we just consider the numerator:
$$ S_T(t)f_T'(t) + f_T(t)^2 = -\left[\int_0^{+\infty} e^{-\lambda t}g(\lambda)d\lambda\right] \left[\int_0^{+\infty} \lambda^2 e^{-\lambda t}g(\lambda)d\lambda\right] + \left[\int_0^{+\infty} \lambda e^{-\lambda t}g(\lambda)d\lambda\right]^2 $$
By Hölder's inequality, $$\begin{align} &\int_0^{+\infty} \left| \left[\lambda \sqrt{e^{-\lambda t}g(\lambda)}\right] \left[\sqrt{e^{-\lambda t}g(\lambda)}\right] \right| d\lambda\\ \leq & \left\{\int_0^{+\infty} \left[\lambda \sqrt{e^{-\lambda t}g(\lambda)}\right]^2 d\lambda \right\}^{\frac {1} {2}} \left\{\int_0^{+\infty} \left[\sqrt{e^{-\lambda t}g(\lambda)}\right]^2 d\lambda \right\}^{\frac {1} {2}} \end{align} $$ which implies $h'(t) \leq 0 $ for any positive $t$, i.e. The hazard rate is monotonic decreasing. You may further argue that the inequality is strict.