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If a stochastic process, say a Levy process, is continuous in probability, i.e., $$ \lim_{t \to s} P[|X_t-X_s| > \epsilon] = 0 $$

does this imply that $P[|X_t| > N] \to 0 $ as $N \to \infty $ for $ 0 \le t \le c $ ?

Sometimes, Levy processes are defined to incorporate the cadlag property in the definition, and in that case, I think that is clearly true. But, if the existence of a cadlag modification is to be proved then one at some point will need to prove something like what I have above.

I have noticed that K. Ito in his Stochastic Processes (Lemma 3, p. 48) has a result that pertains to additive processes that are continuous in probability and states that "since $X_t $ is continuous in probability, the probability law $\mu_t $ of $X_t $ is continuous in $t. $ Therefore $\{\mu_t\}_{t \in [0, c]} $ is compact ... " and then the condition is clearly true.

Unfortunately, I do not think I understand what it is meant by "the probability law ... is continuous in $t. $

I would be grateful if anyone could help me clarify this technical point.

Thank you, Maurice

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Ito means that $\lim_{s\to t}\mu_s=\mu_t$ in the sense of weak convergence of probability measures. Thus $\{\mu_t\}_{t\in[0,c]}$ (as a subset of the class of probability measures on $\Bbb R$ endowed with the topology of weak convergence) is the continuous image of a compact set, and is therefore itself compact.

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EDIT: @JohnDawkins points out that my answer assumes path continuity, which is different from continuous in probability.

Yes. Suppose there exists a set $C$ with $P(C)=1$ so that for $\omega\in C$, $X_t(\omega)$ is continuous (in $t$). So for any $c\ge0$, $X_t(\omega)$ attains its maximum and minimum at some points in $[0,c]$ (continuous functions attain their max and min on compact sets).

Let $A_n=\{\omega:|X_t(\omega)|>n\ \forall t\in[0,c]\}$. Note $A_{n+1}\subset A_n$. Let $A=\cap_{n=1}^\infty A_n$. Pick $\omega\in C$. Then $\max_{t\in[0,c]}|X_t(\omega)|<\infty$, so there exists some integer $M(\omega)$ so that $M(\omega)>\max_{t\in[0,c]}|X_t(\omega)|$, i.e. $\omega\notin A_{M(\omega)}$, so $\omega\notin A$, i.e. $C\subset A^c$, or equivalently, $A\subset C^c$. So $P(A)\le P(C^c)=0$.

Note $P(A)=\lim_{n\to\infty} P(|X_t(\omega)|>n\ \forall t\in[0,c])$, the quantity of interest.

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    $\begingroup$ "continuous in probability" is not the same as "path continuous, a.s." Indeed, every Lévy process is continuous in probability, but the only path-continuous Lévy processes are Brownian motions. $\endgroup$ – John Dawkins Mar 24 '17 at 19:11
  • $\begingroup$ Whoops, thanks for the note @JohnDawkins. $\endgroup$ – manofbear Mar 24 '17 at 19:16

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