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Let $L_t$ be a Levy process that has no positive jumps, but is not strictly decreasing, i.e $$ L_t = \gamma t + \sigma B_t + J_t, $$ where $B_t$ is a Brownian motion, $J_t$ is a pure jump process with only negative jumps.Suppose further, that either $\sigma > 0$ or the Levy measure of $J_t$ is absolutely continuous. In such a case, it is known that the 1-dim distribution of $L_t$ is absolutely continuous.

Let $p(t,x)$ be the probability density function of $L_t$: $$P(L_t \in B) = \int_B p(t,s) ds.$$

Fix $a > 0$.

Now for $t > 0$ and $x < a$, let $$ q(t,x) = \int_0^t \frac{a}{s} p(s,a)p(t-s,x-a)ds. $$ I have reason to believe that $q(t,x) \to p(t,a)$ as $x\nearrow a$, but haven't been able to prove it.

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  • $\begingroup$ If $\gamma=1>0, sigma=0$, and $J_t$ is a Poisson process, the 1-dim distribution of $L_t$ doesn't have a density. $\endgroup$
    – JGWang
    Commented Aug 5, 2020 at 3:06
  • $\begingroup$ Agreed. Thank you. I'll correct it. $\endgroup$
    – bm76
    Commented Aug 10, 2020 at 14:02
  • $\begingroup$ $L_t$ I notice is probably not the best symbol here. It us often used for local times. Anyway, here the definition is clear. $\endgroup$
    – bm76
    Commented Aug 12, 2020 at 5:14

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