Suppose I have two independent Levy processes $X_t$ and $Y_t$, both not continuous.

Is anyone familiar and can refer me to a result(or a counterexample) which states that

${\displaystyle \sum_{0\leq s\leq t}}|\bigtriangleup X_{s}(\omega)\bigtriangleup Y_{s}(\omega)|=0 $ for all $t\in \mathbb{R}$ a.s?

A different yet equivalent formulation of this is

$\bigtriangleup X_{t}=0$ or $\bigtriangleup Y_{t}=0$ a.s. for all $t\in \mathbb{R}$

In words, every two independent Levy processes have no simultaneous jumps a.s. I know it holds for independent Poisson processes and I'm wondering if it generalizes.

  • 2
    $\begingroup$ You should easily find results stating that for any deterministic time $t$, a Lévy process has almost surely no jumps at time $t$. The statement you're looking for is then an immediate corollary. $\endgroup$ Oct 6, 2013 at 20:10
  • $\begingroup$ Thank you. I'm familiar with the fact that for a Levy process $X_t$ and some $s\in \mathbb{R}$ $\bigtriangleup X_{s}=0$ a.s. However, since the number of $s\in \mathbb{R}$ for which a jump occurred might not be measurable I don't see how it answers my question. $\endgroup$
    – Ofer
    Oct 7, 2013 at 11:32
  • 4
    $\begingroup$ use independence and fubini. $\endgroup$ Oct 7, 2013 at 12:03

3 Answers 3


Consider a finite time interval, let $J_i$ denote the set of (times of) jumps of the $i$'th process, and let $J_{i,n}$ denote the set of jumps of the $i$'th process of size $(\frac 1 n,\frac 1{n-1}]$. Note that $J_{i,n}$ is finite a.s. and therefore $J_{i,n}\cap J_{j,m}=\emptyset$ a.s. for all $n,m$ and $i\neq j$. Since there are countably many such intersections, it follows that $J_i\cap J_j = \emptyset$ a.s.

  • $\begingroup$ @ mpr : I am not sure you are answering the question. Regards $\endgroup$
    – The Bridge
    Oct 10, 2013 at 8:02
  • $\begingroup$ The question, as I read it, was to show that independent Levy processes have no simultaneous jumps, which is to say that $J_i\cap J_j$ is empty. Maybe I am missing something. $\endgroup$
    – mpr
    Oct 10, 2013 at 11:25
  • $\begingroup$ $J_{i,n}$ is not finite, it could be countably infinite. You can see this from the Levy-Ito decomposition. $\endgroup$
    – Bati
    Oct 10, 2013 at 12:44
  • $\begingroup$ Countably infinite number of jumps of size greater than $1/n$ in a finite time interval? $\endgroup$
    – mpr
    Oct 10, 2013 at 12:57

The following proposition will be proved:

Proposition Suppose that $X,Y$ are càdlàg adapted processes and $X$ have no fixed time of discontinuity, that is $$ \mathsf{P}(\Delta X_t\ne 0)=0, \qquad \forall t\ge 0.$$ If $X,Y$ are independent, then $$ \mathsf{P}\Bigl(\sum_{s}|\Delta X_s| |\Delta Y_s|>0\Bigr) =\mathsf{P}(\text{$X,Y$ have simultaneous jumps})=0.\tag{1}$$ Remark. If $X,Y$ are two independent Lévy processes, then it satisfy the conditions of above proposition.
Proof. 1 Since $X$ is càdlàg adapted process, then the random set \begin{align}\{\Delta X\ne 0\}&=\bigcup_{m=1}^\infty[\![S_m]\!],\\ \{(\omega,t)\in\Omega\times\mathbb{R}_+: \Delta X_t(\omega)\ne 0\} &=\bigcup_{m=1}^\infty\{(\omega,t)\in\Omega\times\mathbb{R}_+:S_m(\omega)=t\}. \end{align} where $S_m$ is a sequence stopping time (cf. Jacod, J. and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003. I.1.32, p.8). This means \begin{gather} \{\omega:\Delta X_t(\omega)\ne0\}=\bigcup_{m=1}^\infty\{\omega:S_m(\omega)=t\}, \qquad \forall t\in\mathbb{R}_+.\\ \{\omega: \text{the path of $X_s(\omega)$ has jumps}\}=\bigcup_{m=1}^\infty\{\omega:S_m(\omega)<\infty\} \end{gather} The $\{S_m, m\ge 1\}$ is also a sequence stopping time with respect to natural filteration of $X$, that is $$ (S_m\le t)\in \mathcal{F}^X_t\stackrel{\text{def}}{=}\bigcap_{u\ge t} \sigma(X_s,s\le u)\vee \mathcal{N},\qquad \forall t>0.$$ 2 Similarly, for $Y$, $$\{\Delta Y\ne 0\}=\bigcup_{n=1}^\infty[\![T_n]\!], \qquad (Y_n\le t)\in \mathcal{F}^Y_t. $$
For $X$ and $Y$, \begin{align} &\biggl\{\omega: \sum_{s}|\Delta X_s| |\Delta Y_s|>0\biggr\}=\{\omega: \text{$X_s(\omega),Y_t(\omega)$ have simultaneous jumps}\}\\ &\quad =\bigcup_{m=1}^\infty\bigcup_{n=1}^\infty\{\omega:S_m(\omega)=T_n(\omega)<\infty\}. \tag{2} \end{align}

3 Since $X$ has no fixed time of discontinuity, that is $$ \mathsf{P}(\{S_m=s\})\le \mathsf{P}\Bigl(\bigcup_{m\ge 1}\{S_m=s\}\Bigr) =\mathsf{P}(\Delta X_s\ne 0)=0, \quad\forall s>0, m\ge 1.$$ Hence $F_{S_m}(s)=\mathsf{P}(S_m\le s)$, the distribution function of $S_m$, is continuous and $$\Delta F_{S_m}(s)=0,\qquad \forall s\ge 0, m\ge 1.\tag{3}$$

4. Using the independence of $X, Y$ and the measurability of $\{S_m, m\ge 1\}(\{T_n,n\ge 1\})$ with respect to $\mathcal{F}^X_{\infty}(\mathcal{F}^Y_{\infty}$, respectively), we find that $\{S_m, m\ge 1\},\{T_n,n\ge 1\}$ are independent and
\begin{align} &\mathsf{P}(S_m=T_n<\infty)=\int_0^\infty\int_0^\infty1_{\{s=t\}}(s,t)\,dF_{S_m}(s)dF_{T_n}(t)\\ &\quad=\int_0^\infty \Delta F_{S_m}(t)dF_{T_n}(t)=0 \quad \forall m,n\ge 1.\qquad \text{by (3)}. \tag{4} \end{align} Now (1) follows from (2),(4).


Besides the detailed answer you got from the double posting on math.stackexchange, one can even say a bit more:

When you consider the 2-dim Lévy process $(X(t), Y (t))$ without Gaussian part, the components are independent if and only if the support of the corresponding Lévy measure ν is in the set $\{(x, y) : xy = 0\}$. Intuitively, the measure is concentrated on the axis and the two components of the process never jump simultaneously. (see Proposition 5.3. in [1])

As a textbook reference I recommend to have a look at Section 5.4 in [1].


[1] Cont, R., Tankov, P. Financial modelling with jump processes, Chapman Hall/CRC, Boca Raton, FL, 2004, xvi+535


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