Probability of exiting on the boundary for a monotone Lévy-type process

Question

Let the continuous function $\ell:\mathbb R \times(0,\infty)\to[0,\infty)$ be a Lévy-type kernel, such that $$ \sup_{x}\int_0^\infty \min\{1,y\}\ell( x, y)\,dy<\infty, $$ and suppose that $\mathcal Lg(x):=\int_0^\infty (g(x-y)-g(x))\ell(x,y)\,dy$ generates a decreasing Lévy-type process $s\mapsto L^{x}_s$, where $x\in\mathbb R$ denotes the starting point. Define the first exit time from $(0,\infty)$ as $\tau_0(x):=\inf\{s>0: L_s^{x}\le 0 \}$. Questions:

(i) Is this enough to prove that $\mathbf P[L_{\tau_0(x)}^{x} =0]=0$ ?

(ii) If (i) is not true, does it become true if we additionally assume that $L^{x}_s$ allows a density for every $s,x>0$ ?

iii) If (i) is not true, is (i) true for the Lévy case $\ell(x,y)=\ell(y)$ ? (I know $\mathbf P[L_{\tau_0(x)}^{x} =0]=0$ is true if $\ell(y) dy=\ell(dy)$ is an infinite measure)

Answer 1

(i,ii) Not quite. A trivial counterexample is $\ell(x, y) \equiv 0$. A simple non-trivial counterexample would be something like $$\ell(x,y) = \begin{cases} 1 / (x^{1/2} y^{3/2}) & \text{if $0 < x \leqslant 1$, $y < x$} \\ 0 & \text{if $0 < x \leqslant 1$, $y \geqslant x$} \\ e^{-y} & \text{otherwise.} \end{cases} $$ Then $L_t$ started at $L_0 \in (0, 1]$ will only have jumps $\Delta L_t$ of size less than $L_t$, until it reaches zero (which happens in finite time with probability one). Afterwards, it will follow a negative subordinator with exponentially distributed jumps.

One can even require $\ell(x,y) > 0$ everywhere, as long as $\ell(x,y)$ is not too large when $0 < x \leqslant 1$ and $y \geqslant x$.

(iii) For Lévy processes, the property we are discussing here is called creeping. As you pointed out, for a driftless subordinator (as in your case), there is no creeping if the Lévy measure is infinite (see e.g. Theorem 5.9 in Kyprianou's book).

Similarly, there is not creeping if $L^x_t$ is a driftless subordinator with a finite Lévy measure with no atoms. Indeed: in this case $P(L^x_{\tau_0(x)} = 0) > 0$ would mean that the amount of time spent by $L^x_t$ at $0$ is postive, and hence $P(L^x_t = 0) > 0$ for some $t$. However, the distribution of $L^x_t$ only contains an atom at $x$.

Finally, if the Lévy measure has an atom at $y < 0$, then of course the process $L^{-k y}_t$ will hit $0$ with positive probability for every $k = 1, 2, \ldots$