I have been thinking about this problem for a project I have and cannot seem to find a solution.
We are in the following situation: we have a variable $X\sim Exp(\lambda)$ and an independent variable $Y$ that admits a density. We observe $Z=X+Y$. The idea is that $X$ is some Poisson event. Once it occurred, we realize that "the event occurred" only $Y=t$ periods after it did occur. This "stochastic lag" is independent of when the event occurred (the lag $Y$ is independent, of course, the date of observation of $Z$ is not).
My question is the following: suppose there are two variables $Y$ and $Y'$, and corresponding $Z=X+Y$, $Z'=X+Y'$. Can we find conditions on the distribution of $Y$ and $Y'$ such that the hazard rate of $X|Z$ ($X$ conditional on $Z$) is everywhere above $X|Z'$? One guess would be that it works if $Y$ first-order stochastically dominates $Y'$, but I couldn't show it.
Here are some elements. We want to back up the hazard rate of $X$ from $Z$. We can use Bayes rule and a convolution. Letting $f_X,f_Y$ and $f_Z$ being the pdf and $f_{X|Y}$ the conditional pdfs: \begin{equation*} f_{X|Z}(X=t_0|Z=t)=\frac{f_X(t_0)f_Y(t-t_0)}{\int_0^t f_X(s)f_Y(t-s)ds}. \end{equation*} The corresponding survival probability is: \begin{equation*} 1-F_X|Z(X=t_0|Z=t)=1-\frac{\int_0^{t_0}f_X(s)f_Y(t-s)ds}{\int_0^t f_X(s)f_Y(t-s)ds}=\frac{\int_{t_0}^{t}f_X(s)f_Y(t-s)ds}{\int_0^t f_X(s)f_Y(t-s)ds}. \end{equation*}
The hazard rate will be defined as: $h_{X|Z}(t_0)=\frac{f_X|Z(X=t_0|Z=t)}{1-F_X|Z(X=t_0|Z=t)}=\frac{f_X(t_0)f_Y(t-t_0)}{\int_{t_0}^{t}f_X(s)f_Y(t-s)ds}. $
We have $h_{X|Z}(t_0)\geq h_{X|Z'}(t_0)$ iff $\frac{h_{X|Z}(t_0)}{h_{X|Z'}(t_0)}\geq 1$ iff \begin{equation*} \frac{f_{Y}(t-t_0)\int_{t_0}^{t}f_X(s)f_{Y'}(t-s)ds}{f_{Y'}(t-t_0)\int_{t_0}^{t}f_X(s)f_Y(t-s)ds}\geq 1. \end{equation*}
From there, we can see that some form of stochastic dominance will play a role but that we will also probably need to involve the densities, but I wasn't able to go further.
Thank you for your help!