Jumps of independent Lévy processes

Suppose I have two independent Lévy processes $(X_t)_t$ and $(Y_t)_t$, both not continuous.

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

${\displaystyle \sum_{0\leq s\leq t}}|\Delta X_{s}\cdot \Delta Y_{s}|=0$ for all $t\in \mathbb{R}$ a.s?

A different yet equivalent formulation of this is

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

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

3 Answers

Let $(X_t)_{t \geq 0}$, $(Y_t)_{t \geq0}$ be independent Lévy processes with Lévy triplet $(\ell_1,q_1,\nu_1)$ and $(\ell_2,q_2,\nu_2)$, respectively. By the independence of the processes, we know that $(X_t,Y_t)_{t \geq 0}$ is a Lévy process and using Lévy-Khinchine's formula we find that its Lévy measure equals

$$\nu(dx,dy) := \nu_1(dx) \otimes \delta_{\{0\}}(dy)+ \delta_{\{0\}}(dx)\otimes \nu_2(dy)$$

where $\delta_{\{0\}}$ denotes the Dirac measure centered at $0$ (for a proof see e.g. this question). Roughly speaking, the jumps of the process $(X,Y)$ are concentrated on the coordinate axis. Indeed: Let

$$A(\varepsilon) := \mathbb{R}^2 \backslash \bigg(B[0,\varepsilon] \cup \{(x,y) \in \mathbb{R}^2; x = 0 \vee y=0\} \bigg)$$

for $\varepsilon>0$. Obviously, $A(\varepsilon)$ is open and satisfies $\nu(A(\varepsilon)) = 0$. Consequently, there exists $\Omega(\varepsilon)$, $\mathbb{P}(\Omega(\varepsilon))=1$, such that

$$\forall t \geq 0: \Delta (X_t,Y_t)(\omega) = (\Delta X_t(\omega),\Delta Y_t(\omega)) \notin A(\varepsilon)$$

for all $\omega \in \Omega(\varepsilon)$. Set $\Omega' := \bigcap_{k \in \mathbb{N}} \Omega(1/k)$. Then $\mathbb{P}(\Omega')=1$ and

$$\forall t \geq 0: (\Delta X_t(\omega),\Delta Y_t(\omega)) \notin \mathbb{R}^2 \backslash \{(x,y); x = 0 \vee y=0\}$$

for any $\omega \in \Omega'$. This shows that $X$ and $Y$ do not jump simultaneously a.s.

Let $X$, $Y$ be independent real-valued Lévy processes and $\varepsilon>0$. Without loss of generality set $t=1$. Since the processes have cadlag paths, we find

\begin{align*} & \quad \mathbb{P}(\exists s \in [0,1]: \Delta X_s > \varepsilon, \Delta Y_s>\varepsilon) \\ &\leq \mathbb{P} \bigg( \bigcup_{i=0}^{\infty} \bigcap_{j \geq i} \underbrace{\left\{ \exists k \in \{0,\ldots,j\}: |X_{(k+1)/j}-X_{k/j}| > \frac{\varepsilon}{2}, |Y_{(k+1)/j}-Y_{k/j}| > \frac{\varepsilon}{2} \right\}}_{=: A_j} \bigg) \\ &= \mathbb{P} \left( \liminf_{j \to \infty} A_j \right) \leq \liminf_{j \to \infty} \mathbb{P}(A_j) \end{align*}

Using the stationarity of the increments as well as the independence of the processes, we obtain

\begin{align*} \mathbb{P}(A_j) &\leq \sum_{k=0}^j \mathbb{P}\left(|X_{(k+1)/j}-X_{k/j}| > \frac{\varepsilon}{2} \right) \cdot \mathbb{P}\left(|Y_{(k+1)/j}-Y_{k/j}|> \frac{\varepsilon}{2} \right) \\ &= j \cdot \mathbb{P}\left(|X_{1/j}|> \frac{\varepsilon}{2} \right) \cdot \mathbb{P}\left(|Y_{1/j}|> \frac{\varepsilon}{2} \right) \end{align*}

Applying this lemma yields that

$$j \cdot \mathbb{P} \left(|X_j| > \frac{\varepsilon}{2} \right) \leq C, \qquad j \in \mathbb{N}$$

for some constant $C>0$. On the other hand, by the stochastic continuity, $\mathbb{P}(|Y_{1/j}|>\varepsilon/2) \to 0$ as $j \to \infty$. Consequently,

$$\mathbb{P}(\exists s \in [0,1]: \Delta X_s > \varepsilon, \Delta Y_s>\varepsilon) \leq \liminf_{j \to \infty} \mathbb{P}(A_j) = 0$$

Since $\varepsilon>0$ is arbitrary, this finishes the proof.

Note that the jumps of a Lévy process $X$ cannot occur at fixed times. Indeed,

\begin{align*}\DeclareMathOperator \re {Re} \DeclareMathOperator \im {Im} \mathbb{P}(|\Delta X_t|>\varepsilon) &= \mathbb{P} \left( \bigcup_{j \geq 1} \bigcap_{k \geq j} |X_t-X_{t-1/k}| \geq \varepsilon \right) \\ &\leq \liminf_{k \to \infty} \mathbb{P}(|X_t-X_{t-1/k}| \geq \varepsilon) = 0 \end{align*}

by the stochastic continuity of $X$. Thus, $\Delta X_t = 0$ a.s. Obviously, this implies

$$\Delta X_t \cdot \Delta Y_t = 0$$

for any two Lévy processes $X$, $Y$.

Note that $\Delta X_t \cdot \Delta Y_t = 0$ a.s., $t \geq 0$, does in general not imply

$$\sum_{s \leq t} \Delta X_s \cdot \Delta Y_s = 0,$$

see Did's comment below.

• Careful, you just proved the result when X=Y (a case where it does not hold). – Did Oct 7 '13 at 17:33
• @Did I don't see your point. Obviously, $\Delta X_t \cdot \Delta X_t = 0$ a.s for fixed $t$. I did not claim that $\Delta X_t \cdot \Delta Y_t = 0$ implies $\sum_{s \leq t} \Delta X_s \cdot \Delta Y_s$ for any two Lévy processes $X$ and $Y$. – saz Oct 7 '13 at 17:45
• Sorry if I caused any confusion in the formulation of my question. I probably should have put it $\bigtriangleup X_{t}(\omega)=0$ or $\bigtriangleup Y_{t}(\omega)=0$ for all t a.s. This is like the difference between the definition of two processes being a version of one another and being indistinguishable. I suspected it would lead to confusion but I saw it in Jean Bertoin paper on subordinators and I thought it might help. – Ofer Oct 11 '13 at 8:05
• My point is that you state (correctly) that "Note that (1) does in general not imply (2)" but that you only show (1) while the question is about (2). Your add-on makes clear the difference between (1) and (2) (thanks for that)--but now I wonder why you do not answer (2)... – Did Oct 13 '13 at 11:58
• @Did It sounded to me as ofer was aware of a proof of the equivalence $(1) \Leftrightarrow (2)$. Later it became clear that this is not the case. Unfortunately, I recognized belatedly that my own proof doesn't work out either. So if you (or someone else) answer the question, I'll remove my answer. – saz Oct 13 '13 at 13:41

by Fubini's theorem we know that $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|$ is measurable. We can write it this way $\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}$. Take expectation, and by independence,the MCT and the continuity in probability of the Levy processes we have

$E\Bigl(\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}|\Bigr)E\Bigl(|\triangle Y_{s}|\Bigr)1_{\{i-1\leq s\leq i\}}=0$

so $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|=0$ a.s. Since the trajectories of the summand are increasing it implies for all $t>0$.

• Probably, I've missed your point, but: Why should $\sum_{0 \leq s \leq t} |\Delta X_s \cdot \Delta Y_s|$ equal $\sum_{i=1}^{\infty} |\Delta X_s \cdot \Delta Y_s| 1_{(i-1) \wedge t \leq s \leq i \wedge t}$? In my opinion, the latter equals $|\Delta X_s \cdot \Delta Y_a| \cdot 1_{[0,t]}(s)$. – saz Oct 28 '13 at 16:39
• You are right. I fixed it. Thanks. – Ofer Nov 4 '13 at 8:00
• That's not what I meant. Fix $s \in [i-1,i]$ (note that you do not sum over $s$, so it is fixed!), then $$\sum_{i=1}^{\infty} |\Delta X_s \cdot \Delta Y_s| \cdot 1_{i-1 \leq s \leq i} = |\Delta X_s \cdot \Delta Y_s|$$, i.e. $\sum_{i=1}^{\infty} |\Delta X_s \cdot \Delta Y_s|=0$ holds trivially since $\Delta X_s = \Delta Y_s = 0$ a.s. In particular, your calculation does not imply $$\sum_{0 \leq s \leq \infty} |\Delta X_s \cdot \Delta Y_s|=0$$ – saz Nov 4 '13 at 15:55
• Moreover, your "proof" (if it was correct) would also work for $X=Y$: Obviously, we have $$\mathbb{E}(|\Delta X_s \cdot \Delta Y_s|) = \mathbb{E}(|\Delta X_s|^2) = 0$$ and therefore $$\sum_{i=1}^{\infty} |\Delta X_s \cdot \Delta Y_s| 1_{i-1 \leq s \leq i} = 0$$ And as @did already mentioned, the statement does not hold for $X=Y$. – saz Nov 4 '13 at 15:58
• you are right. it's wrong. – Ofer Nov 4 '13 at 17:20