I am currently studying infinitely divisible distributions and came across the following exercise:
Suppose the distribution of $X=(X_1,...,X_d)$ is infinitely divisble with the generating triple $(A,\nu,\gamma)$ from the Levy-Khinchin representation theorem.
I want to show that $X_1,...,X_d$ are independent if and only if $A$ is a diagonal matrix and the support of $\nu$ is contained in the union of the coordinate axis of $\mathbb{R}^d$.
By assumption we know that for the characteristic function of $X$ it holds: \begin{align} \psi_X(t) = \exp(-1/2<t,At>+i<\gamma,t> + \int_{\mathbb{R}^d}(e^{i<t,x>}-1-i<t,x>1_D(x))\nu(dx)) \end{align} where D is the closed unit ball. If we now assume that A is diagonal with entries $a_1,...,a_d$ and the support of $\nu$ is contained in the union of the coordinate axis of $\mathbb{R}^d$ we get \begin{align} \psi_X(t) &= \exp(-1/2\sum_{j=1}^d a_jt_j^2+ i\sum_{j=1}^d\gamma_jt_j + \sum_{j=1}^d \int_{\mathbb{R}}(e^{it_jx}-1-it_jx1_D(x))\nu(dx))\\ &= \prod_{j=1}^d \exp(-1/2a_jt_j^2+ i\gamma_jt_j + \int_{\mathbb{R}}(e^{it_jx}-1-it_jx1_D(x))\nu(dx)) \end{align}
So i want to show if this holds, that $X_1,...,X_d$ are independent. If i knew that every $X_i$ had the characteristic function $\exp(-1/2a_jt_j^2+ i\gamma_jt_j + \int_{\mathbb{R}}(e^{it_jx}-1-it_jx1_D(x))\nu(dx))$ then i would be finished since then the characteristic function of $X$ would be the product of the characteristic functions of the $X_i$. But i dont know if i can get any information about the distribution of the $X_i$. I think if the $X_i$ were also identically distributed it would immediately follow from a property of infinitely divisible distributions. However, without this additional assumption i don't see that. Anybody how can help me with that? Thanks in advance
Just take $t_j=0$ for $j \neq i$ in your formula for $\psi_X(t)$. This gives the characteristic function of $X_i$.