This question is making me crazy for a long time:
Is $\boldsymbol{X}=\left\{(W(t),W(at)); t \ge 0\right\}$ a Lévy process where $W(t)$ is a standard Brownian motion and $a \in \left(0,1\right)$?
My answer is no, since increments are not independent. Indeed if I compute the covariance matrix $\boldsymbol{\Sigma}$ for $s<t$ I get:
\begin{equation*} \boldsymbol{\Sigma} = cov\left(\boldsymbol{X}(t)-\boldsymbol{X}(s),\boldsymbol{X}(s)\right) \end{equation*}
I obtain that:
\begin{equation} \boldsymbol{\Sigma} = \begin{bmatrix} (W(t)-W(s))W(s) & (W(t)-W(s))W(as) \\ (W(at)-W(as))W(s) & (W(at)-W(as))W(as) \end{bmatrix} \end{equation}
If I look at the term $\boldsymbol{\Sigma}_{2,1}$ I have that:
\begin{equation*} \boldsymbol{\Sigma}_{2,1} = \left\{ \begin{array}{ll} at-as & \quad at <s \\ s-as & \quad at>s \end{array} \right. \end{equation*} Since $\boldsymbol{\Sigma}_{1,2}$ is not zero, the increments are not independent and hence $\boldsymbol{X}$ is not a Lévy proess.
Nevertheless, if I compute the characteristic function at $t$ of $\boldsymbol{X}$ I have:
\begin{equation*} \phi_{X(t)}(u_{1},u_{2}) = e^{-\frac{1}{2} t [u_{1},u_{2}] \begin{bmatrix} 1 & a \\ a & a \end{bmatrix} [u_{1},u_{2}]^{T}} \end{equation*} which is the characteristic function of an infinitely divisible law, by the Lévy-Kinchin theorem. Hence the process must be Lévy. And this is in contrast to what we proved above.
Probably I'm missing something but I can't figure out what is wrong in my reasoning.
I could be wrong, but I don't think Levy-Khintchine theorem says what you think it does. Namely, it does not say that "given the characteristic function of the process at time t, if it is infinitely divisible, then the process is Levy". I think it says that "there exists a Levy process whose law matches the law at time t", in which case, it is easy to see what the process should be, since you gave its covariance. Namely, let $W_1(t), W_2(t)$ be iid BMs. Then, $X(t) \stackrel{d}{=} (\sqrt{1-a} W_1(t) + \sqrt{a} W_2(t), \sqrt{a} W_2(t)) $ for each fixed $t$, and this should be a Levy process.