# The probability of Levy process staying at a point

Assume $X_{t}$ is a 1-dimensional Levy process on a probability $(\Omega, \mathcal{F}, P)$. For a fixed point $x$ in the state space and fixed $t\neq 0$, what's the value of $P(\omega: X_{t}(\omega)=x)$?

In the case of Brownian motion, $P(\omega: X_{t}(\omega)=x)=0$.

In the case of Pure Jump Levy process, for example $X_{t}$ is a Poisson process, $P(\omega: X_{t}(\omega)=x)\neq 0$.

What confuses me is the case that $X_{t}$ is a Levy process with generating triplets $(\sigma, \gamma, \nu)$. Here $\sigma\neq 0$ is for the Brownian motion part, $\nu\neq 0$ is for the pure jump part. Is $P(\omega: X_{t}(\omega)=x)\neq 0$?

## 1 Answer

As long as $\sigma\ne 0$, we get $P(X_t=x)=0$.

By the Levy-Ito decomposition there is a Brownian motion $Y$ and an independent process $Z$ with $X=Y+Z$. Then for $t>0$, $$P(X_t=x)=P(Y_t=x-Z_t)=0$$ since $Y$ is independent of $Z$.

• Yes, you are right. If the generating triplet of $X_{t}$ is $(\sigma\neq 0, \gamma, \nu)$, $P(X_{t}=x)=0$. If the generating triplet of $X_{t}$ is $(0, 0, \nu)$, $P(X_{t}=x)$ is dependent on the measure $\nu$. If $\nu$ is absolute with respect to Lebesgue measure, $P(X_{t}=x)$ is still equal to 0. – Hengyu Zhou Mar 9 '14 at 1:39
• The motivation of this question is as follows. Assume $X_{t}$ is a Levy process. For any fixed $t$, is $P(X_{t-})=0$? Maybe you already answer my questions. But I am still not sure what's happening when $X_{t}$ is a pure jump process. – Hengyu Zhou Mar 9 '14 at 1:48
• Is that a typo, "$P(X_{t-})=0$"? – Bjørn Kjos-Hanssen Mar 9 '14 at 2:21
• It's a typo. I mean $P(X_{t-}=0)=0$. – Hengyu Zhou Mar 9 '14 at 23:57