I'm tryig to prove the next proposition but I'm lost:

Let $X=(X^{1},\ldots,X^{d})$ be a $d-$dimensional Lévy process with Gaussian coefficient $Q$ and Lévy measure $\Pi.$ Then the Levy processes $X^1,\ldots,X^{d}$ are independent if and only if $Q$ is diagonal matrix and the support of $\Pi$ is contained in the coordinate axes.

I've been working with the definition of characterístic exponent: $$E(\exp(i\langle\lambda,X\rangle))=\exp\{-t\Psi(\lambda)\},\space\lambda\in\mathbb{R}^d.$$

Here $$\Psi(\lambda)=i\langle a,\lambda\rangle+\frac{1}{2}Q(\lambda)+\int_{\mathbb{R}^{d}}(1-e^{ix\lambda}-i\langle\lambda x\rangle1_{\{|x|<1\}})\Pi(dx),\space a\in\mathbb{R}^{d}.$$

Utilizing the above, and independence of Lévy processes we get $$\displaystyle\sum_{j=1}^{d}\Psi(\lambda_{j})=\Psi(\lambda)$$ and


So we have that that $Q$ is matrix diagonable and the support of $\Pi$ it's on the coordinate axes if the equality between integrals on the behind equality holds, but I don't get the last. Because of independence and Fubini theorem we will have $$\displaystyle\prod_{j=1}^{d}\int_{\mathbb{R}}e^{i\lambda_{j}x_{j}}\Pi(dx_{j})=\displaystyle\sum_{j=1}^{d}\int_{\mathbb{R}}e^{i\lambda_{j}x_{j}}\Pi(dx_{j})$$ isn't it? I'm not sure of the behind. I have doubts of this because of the integration domain (the indicator functions.) but I don't know if this is eough to have such equality.

For the other implication I guess the same steps as above works or this purpose, but I'm not sure.

Any kind of help is thanked in advanced.

Edit:I've proved the first implication; for the second I have troubles.

I'd like to prove that

$$\int_{\mathbb{R}^{d}}(1-e^{ix\lambda}-i\langle\lambda x\rangle1_{\{|x|<1\}})\Pi(dx)=\displaystyle\sum_{j=1}^{d}\int_{\mathbb{R}}(1-e^{i\lambda_{j}x_{j}}+i\lambda_{j}x_{j}1_{\{|x_{j}|<1\}})\Pi(dx_{j}))$$ utilizing the fact that $\Pi$ has support contained in coordinate axes. I've been worked with the first term of the above equality, but I don't get a sum of integrals as I'd like.

Is there another way to prove this?

Again, thanks for the help in advanced.

  • $\begingroup$ Do you know that the Lévy triplet is unique? $\endgroup$ – saz Apr 19 '18 at 17:03
  • $\begingroup$ Thanks @ saz! I had forgotten about that. So the other implication is like this, right? $\endgroup$ – Squird37 Apr 19 '18 at 21:44
  • $\begingroup$ yes, it is..... $\endgroup$ – saz Apr 20 '18 at 7:04
  • $\begingroup$ But I don't get the equality between the integral over $\mathbb{R}^{d}$ and the sum of integrals on $\mathbb{R}.$ Could you help me with this? @saz $\endgroup$ – Squird37 Apr 20 '18 at 7:22
  • $\begingroup$ Sorry, I totally forgot to answer you. Note that we can write an integral of the form $$\int_{\mathbb{R}} f(x_j) \, \Pi(dx_j)$$ as an integral with respect to an image measure: $$\int_{\mathbb{R}^d} f(x) \nu_j(dx)$$ where $\nu_j := \Pi \circ \pi_j$ and $\pi_j: \mathbb{R}^d \to \mathbb{R}, x \mapsto x_j$ denotes the projection onto the $j$-th coordinate. Using this, you can write the sum of the integrals $$\sum_{j=1}^d \int_{\mathbb{R}} \dots \Pi(dx_j)$$ as $$\int_{\mathbb{R}^d} \dots \left( \sum_{j=1}^d \nu_j \right)(dx)$$ and because of uniqueness it follows that the latter measure is ... $\endgroup$ – saz May 8 '18 at 12:01

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