If one defines a Levy process as a stochastic process $(X_t)_{t\geq0}$ that has independent and stationary increments with (a.s.) cadlag paths (hence a def. withouth stochastic continuity). How can I conclude that this process has a.s. no jumps at fixed times (i.e. show that the proba of a positive jump at time $t$ for $t$ fixed cannot be positive)?

  • $\begingroup$ What exactly do you mean by "at fixed times"? $\endgroup$ – Chris Janjigian Apr 4 '14 at 23:58
  • $\begingroup$ sry for the vague formulation, i edited my question. again, i ment that for any fixed t (e.g. t=0.5) the probability of a jump at this time is 0. $\endgroup$ – Mr. Barrrington Apr 5 '14 at 6:58

If $(X_t)_{t \geq 0}$ is a process with independent and stationary increments and cadlag paths, then it is stochastically continuous. Indeed: Let $s<t$, then by the stationarity of the increments

$$\mathbb{P}(|X_t-X_s|> \varepsilon) = \mathbb{P}(|X_{t-s}|>\varepsilon)$$

for any $\varepsilon>0$. Since $(X_t)_{t \geq 0}$ has cadlag sample paths and $X_0=0$, we know that $X_r \to 0$ a.s. as $r \downarrow 0$, hence $X_r \to 0$ in probability. Therefore,

$$\mathbb{P}(|X_{t}-X_s| > \varepsilon) \to 0, \qquad s \uparrow t. \tag{1}$$

Now we are ready to prove that $(X_t)_{t \geq 0}$ has no jumps at a fixed time. In $(1)$ we have shown that $X_s \to X_t$ in probability as $s \uparrow t$. On the other hand, as $r \mapsto X_r$ is cadlag, we know that $X_s \to X_{t-}$ as $s \uparrow t$. The uniqueness of the limit yields $X_t = X_{t-}$ a.s.

Alternative solution: Let $t>0$ and denote by $\Delta X_t := X_t-X_{t-}$ the jump height. Then, by Fatou's lemma,

$$\begin{align*} \mathbb{P}(|\Delta X_t|>\varepsilon) &\leq \mathbb{P} \bigg( \bigcup_{n \in \mathbb{N}} \bigcap_{k \geq n} \{|X_t-X_{t-1/k}|>\varepsilon/2\} \bigg) \\ &\leq \liminf_{n \to \infty} \mathbb{P}(|X_t-X_{t-1/n}|>\varepsilon/2) = 0 \end{align*}$$

where we used in the last step the stochastic continuity. Note that we do even have to assume that $(X_t)_{t \geq 0}$ has cadlag sample paths.


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