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What is the “true” definition of a Lévy process?

I notice that definitions vary in non-equivalent ways:

1) Wikipedia states that a Lévy process is one that satisfies four particular properties, but these properties do not include the right-continuous property.

2) These notes require a Lévy process to be “right continuous with left-limits”.

3) These notes require a Lévy process to be “right continuous” (without the “with left limits”, why is that missing?)

I observe:

  1. These definitions are not equivalent: In another SE question, I give a simple example of a process that satisfies the 4 properties of wikipedia but is surely not right-continuous: Are nonnegative Lévy processes almost always nondecreasing?

  2. Both wikipedia and the first set of above notes mention that the 4 properties imply a “version” of $X(t)$ is right-continuous (without explanation of what that means). After some further web-searching I find that $Y(t)$ is a “version” of $X(t)$ if $P[Y(t)=X(t)]=1$ for all $t\geq 0$ (which is not the same as $P[Y(t) = X(t) \quad \forall t \geq 0]=1$). This fact does not seem strong enough to justify the wikipedia definition in comparison to the other definitions.

  3. This stackexchange link incorrectly suggests the definitions are all equivalent (the answer is actually a “good” answer but makes an understandable mistake because one would assume the definitions should be equivalent): Definition of Lévy process

My gut reaction is to like the definition in the second set of notes the best (those notes are the most detailed) and to reject the wikipedia definition. It would be useful for someone to give thoughtful and experienced perspective on these distinctions, also to explain why the “left limits” is missing in the third set of notes (i.e., can that be proven back, or what?)

saz
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Michael
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1 Answers1

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Typically one distinguishes between "Lévy processes" and "Lévy processes in law".

Definition: Let $(X_t)_{t \geq 0}$ be a stochastic process such that $X_0 = 0$ almost surely and $(X_t)_{t \geq 0}$ has stationary and independent increments. If $(X_t)_{t \geq 0}$ is right-continuous in probability, i.e. $$\lim_{s \downarrow t} \mathbb{P}(|X_s-X_t|>\delta)=0, \qquad \delta>0, t \geq 0, \tag{1}$$ then $(X_t)_{t \geq 0}$ is a Lévy process in law. If $(X_t)_{t \geq 0}$ has cadlag sample paths with probability 1, then $(X_t)_{t \geq 0}$ is called a Lévy process.

By the stationarity of the increments, $(1)$ is equivalent to right-continuity in probability at $t=0$, i.e. $$\lim_{h \downarrow 0} \mathbb{P}(|X_h|>\delta)=0, \qquad \delta>0. \tag{2}$$

You can find these definitions e.g. in the monograph by Sato and some other books on this topic. Consequently, the definition on wikipedia is actually a "Lévy process in law" and the definition in No. 2 is about "Lévy processes". No idea why No. 3 does only assume right-continuity but you might want to notice that these are "only" slides of a talk and therefore perhaps not the best source to get a rigorous definition. In any cas, right-continuity (almost surely) gives right-continuity in probability, and hence a Lévy process in law.

Any Lévy process is also a Lévy process in law. The converse is not true but we can always pick a Lévy process which equals in distribution a given Lévy process in law.

Theorem: Any Lévy process in law $(X_t)_{t \geq 0}$ has a modification which is a Lévy process, i.e. there exists a Lévy process $(Y_t)_{t \geq 0}$ such that $$\mathbb{P}(X_t=Y_t)=1 \quad \text{for all $t \geq 0$.} \tag{3}$$

Note that $(3)$ implies $$\mathbb{P}(X_{t_1}=Y_{t_1}, \ldots,X_{t_n}=Y_{t_n})=1$$ for any choice of $t_1 \leq \ldots \leq t_n$ and $n \in \mathbb{N}$; in particular $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ have the same finite-dimensional distributions. Consequently, all statements which are only concerned with finite-dimensional distributions (e.g. the Lévy-Khintchine representation) hold for Lévy process iff they hold for Lévy processes in law. As soon as we are interested in properties which depend on the whole path, it is often somewhat tedious to work with Lévy processes in law; e.g. it is not even clear that the supremum $M_t = \sup_{s \leq t} X_s$ is measurable whereas this is immediate for Lévy processes.

saz
  • 123,507