A real valued stochastic process $\left\{ X_{t}:\ t\in\mathbb{R}^{+}\right\} $ is termed additive if $\forall n\in\mathbb{N}$, $0\leq t_{0}<t_{1}<...<t_{n}<+\infty$, $X_{t_{0}},X_{t_{1}}-X_{t_{0}},...,X_{t_{n}}-X_{t_{n-1}}$are independent random variables. Given this definition, it can be easily shown that the probabilty distribution of $X_{t}$ is infinitely divisible (ID) for every $t\geq0$, that is, if $\mu_{t}$ is the probability measure associated to $X_{t}$, then $\forall n\in\mathbb{N}$ there exists a probability measure $\mu_{n,t}$ such that $\mu_{t}=\mu_{n,t}^{*n}$, where ''$*n$'' denotes the $n$-fold iteration of the convolution of a measure with itself. That said, it follows that$\left\{ X_{t}:\ t\in\mathbb{R}^{+}\right\} $ admits the Lévy-Khintchine representation for every $t\geq0$, that is:

$$ \mathbb{E}\left[\exp\left(i\lambda X_{t}\right)\right]=\exp\left\{ i\gamma_{t}\lambda-\frac{1}{2}\sigma_{t}\lambda^{2}+\int_{\mathbb{R}}\left[e^{i\lambda x}-1-i\lambda x\chi_{(0,1)}(x)\right]\nu_{t}(dx)\right\} $$ for a characteristic triple $\left(\gamma_{t},\sigma_{t},\nu_{t}\right)$. Yet, the increments of this processes are not necessarily stationary, so we may not have also the usual Lévy-Ito decomposition which is available for Lévy processes. Now, assuming that the process is also a.s. increasing with $\int_{0}^{1}x\nu_{t}(dx)<\infty,$ $t\geq0$, what we can have is instead the following representation of the Laplace exponent of the process after centralization: $\mathbb{E}\left[\exp\left(\lambda X_{t}\right)\right]=\exp\left\{ -\int_{0}^{+\infty}\left(1-e^{-\lambda v}\right)\nu_{t}(dx)\right\} $ so that $X_{t}=\int_{(0,t]\times\mathbb{R}^{+}}vJ(ds,dv)$, where $J$ is a Poisson random measure on $\mathcal{B}\left(\mathbb{R}^{+}\times\mathbb{R}^{+}\right)$ with intensity measure $\tilde{\nu}$ defined via $\tilde{\nu}\left((0,t]\times C\right)=\nu_{t}(C)$, $\forall C\in\mathcal{B}\left(\mathbb{R}^{+}\right)$, $t\geq0$. At this point my question is: how do we see that such increasing additive process may still be not Lévy even if we end up with a very similar representation? Does everyting lie on the fact that in case of a Lévy increasing additive processe the family of measures $\left\{ \nu_{t}\right\} _{t\geq0}$ would be of the kind: $\nu_{t}=t\nu$, for a Lévy measure $\nu$? And also: it is well known that for every fixed $\varepsilon>0$, the number of jumps $>\epsilon$ of a Lévy process in a finite time interval is a.s. finite and, eventually, this is due to the fact that from the Lévy-Khintchine representation follows: $\int_{\left|x\right|>\varepsilon}\nu(dx)<\infty,\ t\geq0.$ So my feeling is that if we allow also very small jumps (i.e. in a neighbourhood of the origin) and consider a Lévy measure with infinite mass, i.e. $\nu(\mathbb{R})=+\infty$, the probability of having an infinite number of jumps - of whatever maginitude - in a finite time interval is equal to one. If instead the Lévy measure has finite mass, i.e. $\nu(\mathbb{R})<+\infty$, then the probability of this event is zero. On this ground, my question is: in what the jump behaviour of an increasing additive process can be different from the one of a real-valued increasing Lévy process (usually termed subordinator)?

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    $\begingroup$ "it is well known that since Lévy processes have a càdlàg modification they cannot jump infinitely many times in a finite time interval." This statement is actually not correct. A Lévy process can jump infinitely many times in a finite time interval; this does not contradict the cadlag property. (cadlag implies that there are only finitely many jumps of size $>\epsilon$ for any $\epsilon>0$.) $\endgroup$ – saz Jun 21 '16 at 17:18
  • $\begingroup$ @saz: perfectly right, I was going to be more precise on that. Indeed I re-edited the question. $\endgroup$ – Jack London Jun 21 '16 at 17:22

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