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If a Lévy process can be defined as a right continuous process with existing left limits, then according to the following linked theorem (Prove that the number of jump discontinuities is countable for any function) I would say an arbitrary trajectory of a Lévy process can have countable many jumps on a given interval (or compact time set).

Even though, there are Lévy processes that jumps continuum infinitely many times on a given time interval, e.g.: Cauchy process...

As I understand, the Cauchy process jumps in every $t$, since its Lévy triplet is $(0,0,\nu)$, where $\nu (dx)=dx/(\pi x^2)$ according to https://en.wikipedia.org/wiki/Cauchy_process. So there is no continuous part in a Cauchy process, which means it jumps in almost every $t$. Is it even possible for a Lévy process to jump continuum infinitely many times?

It seems a contradiction to me. Where do I go wrong?

Kapes Mate
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    The Cauchy process does only jump a countable number of times and I hope you haven't read the contrary somewhere... Having no continuous part only means that there is no continuous motion. Take the following much easier example of a Lévy process with characteristics $(0,0,\nu)$ with a finite measure $\nu$. Then it has no continuous part, but a.s. only a finite(!) number of jumps! Now, the Cauchy process is more complicated. And the important part is that the set of jumps times is dense(!!) in $[0,+\infty)$, but not uncountable. – Mushu Nrek Jan 05 '23 at 11:52
  • But doesn't it mean the Cauchy process has piecewise constant parts so there is a positive probability, that a given "increment" is zero?...but if the Cauchy process has Cauchy distributed increments/jumps, then taking zero should have zero probability...or because the set of jumps is dense it is not true that the process has piecewise constant parts?...Anyway, you are right about noting that the first Lévy parameter equal to zero and the second parameter sigma=0 just means that there is no trend and diffusion part, but it doesn't claim that there is no continuous part, e.g.: Poisson process – Kapes Mate Jan 07 '23 at 09:32
  • @MushuNrek, if you wrote down your previous comment as an answer I would mark it as a good answer...it already helped me a lot... – Kapes Mate Jan 07 '23 at 09:40
  • I don't exactly understand what you mean with the zero increment. If you detail your thoughts, I might be able to give some insights. If you want to have a look at some very weird behaviour that can happen when you have a countable number of jumps, you might want to look up the Cantor function (https://en.wikipedia.org/wiki/Cantor_function). It is not exactly the same thing, as the Cantor function is continuous (so that it doesn't have jumps!). I think it is only suited to write this down as an answer, when everything is cleared up :) – Mushu Nrek Jan 08 '23 at 10:07
  • Ok...So where do I go wrong with the following train of thoughts? Cauchy process is a pure jump process and the number of jumps are countable (as you write above). I think it implies the following: with positive probability, there are $t$ points, where $X_{t+h}-X_{t}=0$ (other words: $\mathbf{P}\left(X_{t+h}-X_{t}=0\right)>\varepsilon)$. But it sounds like contradiction, since $X_{t+h}-X_{t}$ are Cauchy distributed variables (with zero parameter), and we know that the Cauchy distribution is a continuous distribution, therefore $\mathbf{P}\left(X_{t+h}-X_{t}=0\right)=0$. – Kapes Mate Jan 08 '23 at 14:45
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    It does indeed imply the opposite: for every $t$ and every $h$, $X_{t+h}\neq X_t$ a.s. Imagine for simplicity that the set of jumps is exactly $\mathbb{Q}$. (This is not the case, but the set of jumps behaves in exactly the same way as it is countable and dense!) Then, for any $t$ and any $h$, there is always some $q\in\mathbb{Q}$ such that $q\in (t,t+h)$ [because $\mathbb{Q}$ is dense!]. That means that there will always be a jumps between any two points of time and therefore, the process isn't constant on any time interval. – Mushu Nrek Jan 09 '23 at 15:15
  • Agree, but still don't understand 100%... when you write we can always find a $q$ between $\left(t,t+h\right)$, since $\mathbb{Q}$ is dense, doesn't it mean it jumps in every $t$? You say a $q_{1}\in\left(t,t+h\right)$, where $X$ jumps, but then I can always say a $q_{2}\in\left(t,t+q_{1}\right)$ where $X$ also has jump (i.e. a “closer jump point to $t$”), but we can play this game forever, so I would say it means $X$ is not continuous in $t$, i.e. $X$ jumps in $t$. But the same reasoning hold for every $t$, so the cardinalty of jumps why is not continuum? – Kapes Mate Jan 11 '23 at 11:33
  • I think it is a very similar "problem" to the case of Dirichlet function... – Kapes Mate Jan 11 '23 at 11:35
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    I agree that this is extremely difficult to visualise and I am not sure that I can provide much help here as I struggle just as much with it! Perhaps, the following example helps a bit: instead of a Cauchy process, we simply the situation by considering a pure jump Lévy process $X$ of finite variation, i.e. a Lévy process with characteristics $(0,0,\nu)$ such that $\int_{\mathbb{R}} x; d\nu(x) < +\infty$ (which is much stronger than the usual assumption. In this context, if we write $\Delta X_t := X_t - X_{t-}$ for the jumps, then we can write $X = X_0 + \sum_{s \leq t} \Delta_s$. – Mushu Nrek Jan 11 '23 at 14:00
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    This sum can only make sense if the number of jumps is countable! (And even then, we must make sure that it converges, which is only the case because the process has finite variation!!) If there were an uncountable number of jumps, we couldn't even make sense of this sum. This does not entirely answer your question, but perhaps it gives you a bit more intuition... – Mushu Nrek Jan 11 '23 at 14:01
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    In the general context, you have to think about jumps in the following way: there is a jump at $t$ iff $\Delta X_t = X_t - X_{t-}\neq 0$. Note that this object exists, because $X$ is càdlàg! Then, saying that there the jump times are exactly $\mathbb{Q}$ means exactly that the function is continuous on $\mathbb{R}\setminus\mathbb{Q}$. This might sound very weird, but this is very similar to the behaviour of the function discussed here: https://math.stackexchange.com/questions/960887/show-that-the-function-f-is-continuous-only-at-the-irrational-points. – Mushu Nrek Jan 11 '23 at 14:07
  • Probably it is a dead conversation but let's give it a try. (Recently I've come across with this question again.) Please confirm whether my conclusion is correct: So you are stating that the Cauchy process is continuous outside of a zero-measure set (similarly like in the attached example in question 960887, $f$ is continuous on $\mathbb{R}\setminus\mathbb{Q}$ but not on $\mathbb{Q}$). But a trajectory of a Cauchy process is different in a sense from the Dirichlet function that the Dirichlet function is NOWHERE continuous, i.e. the Dirichlet function is continuous on a set with zero measure. – Kapes Mate Mar 02 '25 at 10:32
  • Or you saying that the Cauchy process rather behaves like the Dirichlet function, i.e. it has countable many (infinite dense) jumps AND it is nowhere continuous (i.e. every point is a discontinuity point)? Sorry for my confusion @MushuNrek , but as you pointed out, it is indeed not an intuitive topic for me – Kapes Mate Mar 02 '25 at 12:03

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Cauchy processes are a special type of Lévy processes, which are notoriously difficult to visualize and are very counterintuitive. Perhaps, the following can help a bit.

Lévy processes come with three flavors: a deterministic drift, a diffusion and a jump part. The deterministic drift is easy to understand, so let's ignore it. The diffusion part is not trivial, but if one understands Brownian motions, then this part is fine, too. The last part is maybe the most confusing one: jumps seem to be rather simple, but as soon as there are infinitely many of them, it becomes hairy. Since Cauchy processes are pure jump processes, we will simply ignore the two other parts, simplifying our lives by a lot.

Assume first that the jump measure $\nu$ is finite. This is the easy case as we can view the corresponding Lévy process is a compound Poisson process, i.e. a sum of iid variables with law given by the renormalization of $\nu$ and indexed up to a Poisson clock at rate $\vert \nu\vert$. In particular, this process is piecewise constant. So far so good.

If we want to extend this definition to infinitely many jumps, we have to assume that $$ \int_{\mathbb R\setminus \{0\}} 1\wedge \vert x\vert \;d\nu(x) < +\infty. $$ That tells us that there are a) finitely many macroscopic jumps and that b) microscopic jumps of size $\vert x\vert \approx 0$ cannot occur too often (loosely speaking). In this situation, we can still view the corresponding Lévy process $X$ from the perspective of being the sum of its jumps: $$ X_t = X_0 + \sum_{s\leq t} \Delta X_s. $$ (Our assumption ensures that this sum converges almost surely.) From a slightly different perspective: we may also view $X$ as the limit of the processes corresponding to taking $\nu_n(dx) := {1}_{\vert x\vert \geq \frac 1 n}\nu(dx)$, i.e. a limit of compound Poisson processes. Although we have this nice description, a visualization becomes harder. When $\nu$ has infinite mass, there are countably many jumps and the set of jump times is dense. That means in particular that in between any two time points, there is always a jump (although microscopically small). An immediate consequence is that the process cannot be piecewise constant, because there is no interval without jumps!

Interestingly, one can still make sense of measures that do not satisfy the above, as long as $$ \int_{\mathbb R\setminus \{0\}} 1\wedge \vert x\vert^2 \;d\nu(x) < +\infty. $$ Now, the construction is much less trivial as the above limiting procedure does not converge and also the sum $\sum_{s\leq t} \Delta X_s$ does not converge. Morally, the problem is that there are too many small jumps so that one has to compensate the jumps through a drift. (This is also why these are usually not referred to as "pure jump", because the pure jump version cannot exist.) I say "morally", because that is only true in the prelimit and cannot be interpreted in this way in the limit. I personally am not aware of an easy visualization of how the Lévy process behaves in this setting. Unfortunately, Cauchy processes are part of this category...

Now that we have this description of Lévy processes, I finally turn to the question you ask: how often do Lévy processes jumps and what does that mean for the continuity of pure jump Lévy processes? The first question was answered above: Lévy processes with infinite jump measure jump countably often and the jump times are dense. In particular, the process does not jump at every moment in time. In some sense even stronger: the probability that the process jumps at some given time $t$ is $0$. In other words: there is no fixed point of discontinuity. More generally, one of the defining features of Lévy processes is that they are stochastically continuous, meaning that if you pick any point in time $t$, then continuous in probability, i.e. $X_s \to X_t$ in probability whenever $s\to t$.

I suspect that each path the Lévy process is continuous outside of the set of jump times, but I do not have a reference for this and don't know a proof. So this is only a vague guess. But the reason for my intuition is the same as for the function defined in this question. In some sense, if $t$ is not a jump time, then every macroscopic jump is a bit away from it and although there are jumps arbitrarily close to it, they also become arbitrarily small.

Mushu Nrek
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