Let us define the following cumulative distribution: \begin{align} \Pr (Y(t)\geq a)=\sum_{n=0}^\infty \frac{e^{-kt}(kt)^n }{n!}\int_a^\infty[\circledast_{f}\circ H_a ]^n\delta (x) dx \end{align} where the operator $[\circledast_{f_J}\circ H_a ]$ denotes the multiplication by a Heaviside function with a cut off at $a$ followed by the convolution by a pdf $f$. This is a cumulative distribution function in the sense that $\Pr (Y(t)> 0)=0$ and $\Pr (Y(t)\geq -\infty)=1$ (or we could say that $1-\Pr (Y(t)\geq a)$ is the CDF).

A global question would be: what can I know about this distribution, given the pdf $f$?

An intuitive property would be $\langle Y(t) \rangle\xrightarrow[t\rightarrow +\infty]{}-\infty \quad \forall f \;|\; \mu_f<0$

A more specific one is: how to compute $\langle Y(+\infty) \rangle$ given $f$?

My first idea was to notice that for $t\rightarrow + \infty$, we can just consider terms with a large $n$ in the series. Then I wanted to take inspiration from the central limit theorem. Given that the pdf $f$ has definite first and second moments: \begin{align} [\circledast_{f}]^n\delta (x)=f*f*...*f\sim\mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)\quad \text{as }\quad n\rightarrow \infty \end{align} However we have here \begin{align} \int_a^\infty[\circledast_{f}\circ H_a]^n\delta (x)dx\sim \quad ? \quad <\int_a^\infty \mathcal{N}\Big(\frac{x-\mu_f n}{\sigma_f \sqrt{n}}\Big)dx\quad \text{as }\quad n\rightarrow \infty \end{align}

Here are other complementary informations:

For the special case $f(x)=p_-\delta(x+1)+p_+\delta(x-1)$, I have found, with a random walk approach, that $\langle Y(+\infty) \rangle = \frac{1}{1-\frac{p_+}{p_-}}$ for $p_+>p_-$.

following this kind of approximation along with the previous result, we end up with the following approximation: $\langle Y(+\infty) \rangle \approx \frac{1}{2}\Big(\sqrt{\mu_f^2+\sigma_f^2}-\frac{\mu_f^2+\sigma_f^2}{\mu_f}\Big)$ for $\mu_f>0$, which is not exact according to numerical simulations but which gives the order of magnitude.

This post is related to this one.