Let X(t) be a continuous time random walk, with exponentially distributed waiting times of pdf $f\_T(t)= k e^{-k t}\; t\geq 0$ and a jump sizes pdf $f\_J(x)$. Suppose the initial distribution $\rho\_{X(0)}(x)$.
I would like to prove (or correct) the following intuitive expression for the survival probability above a certain threshold $a$:
\begin{align}
\Pr (X(t')\geq a \; \forall t'\leq t \; | \; \rho\_{X(0)}) &\equiv S\_a(t \; | \; \rho\_{X(0)}) \\
&= \int\_a^\infty \Big(\sum\_{n=0}^\infty \frac{e^{-kt}(kt)^n \mathcal{O}^n}{n!}\Big) \rho\_{X(0)} (x) dx\\
&= \sum\_{n=0}^\infty \frac{e^{-kt}(kt)^n }{n!}\int\_a^\infty\mathcal{O}^n\rho\_{X(0)} (x) dx
\end{align}
where $\mathcal{O}$ is an operator that I believe to be $\circledast\_{f\_J}\circ H\_a $ the multiplication by a Heaviside function with a cut off at $a$ followed by the convolution by the jump size pdf. 


The first step would be to write a renewal equation for $S\_a$, here is my attempt, first for $a<x\_0$ and $\rho\_{X(0)}=\delta\_{x\_0}$:
\begin{align}
S\_{a}(t|\delta\_{x\_0})&=\int\_t^{\infty} f\_T(t')dt'+\int\_0^t \int\_{a}^{\infty}S\_{a}(t-t'|\delta\_{x})f\_J(x-x\_0)f\_T(t')dxdt'
\end{align}
The probability to stay above $a$ up to time $t$ starting from $x\_0$ is the probability to stay at $x\_0$ (no jump) plus the probability to make any jump to $x\geq a$ at time $t'<t$ and to survive above $a$ up to time $t$.
We could then define $S\_a(t \; | \; \rho\_{X(0)})=\int\_a^\infty S\_a(t \; | \; \delta\_{x})\rho\_{X(0)}(x)dx$
\begin{align}
\int\_a^\infty S\_{a}(t|\delta\_{x\_0})\rho\_{X(0)}(x\_0)dx\_0 &= \int\_a^\infty \rho\_{X(0)}(x\_0)dx\_0\int\_t^\infty f\_T(t')dt'\\
&+ \int\_a^\infty \rho\_{X(0)}(x\_0)\int\_0^t \int\_a^\infty S\_a(t-t'|\delta\_{x})f\_J(x-x\_0)f\_T(t')dxdt'dx\_0
\end{align}


We can easily simplify the time part of this integral equation by taking its Laplace transform with respect to $t$. And I see how to get the Poisson distribution.
However, the space part is more complicated. We can write explicitly the Heaviside's functions and take the Fourier transform, but this does not lead me to the expected result. The renewal equation is probably not correctly set.


Do you see anything wrong here? A resolution using a Green function may be another way.



CTRW: solve a renewal equation
Using my previous remark (namely the correction of the operator $ \mathcal{O} $), an expression would just involve the "skeleton" of your continuous time random walk, namely the walk with initial law $ \rho\_{X\_0} \equiv \rho\_0 $ and jump law $ f\_J $, or, $ Y\_k = R\_0 + \sum\_{m = 1}^k J\_m $. Then, this survival probability would be $ \mathbb{P}(R\_0 \geqslant a, Y\_1 \geqslant a, \dots, Y\_{[t]}\geqslant a) = \int\_a^{+\infty} \mathcal{O}^{[t]}\rho\_0(x) dx $ (with the corrected operator $ \mathcal{O} $). Here, $ [t] $ is the integer part of $t$.


Your walk is just $ X\_t = Y\_{N\_t} $ where $ (N\_t) $ is an independent Poisson process of parameter $k$ (i.e. $ \mathbb{P}(N\_t = m) = e^{-k t} (kt)^m/m! $). Thus, what you have written is more of the form $ \mathbb{P}\_{\rho\_0}( Z\_s \geqslant a, \, \forall s \geqslant 0 ) $ where $Z = Y\circ N^{(kt)}$ and $ N^{(kt)} $ is an independent Poisson process of parameter $kt$ (but you need to write $ e^{-kt} $ in place of $ e^{-knt} $).


