The coordinates of a vector-valued <Compound Poisson process> are independent exactly when its <Lévy measure> is supported on the union of the coordinate axes. For marks $(g_1(U),g_2(U))$ with $U$ uniform on $[0,1]$ and continuous $g_i$, this is equivalent to
$$
g_1(y)g_2(y)=0
$$
for every $y\in[0,1]$.

Independent coordinates of a compound Poisson process

The joint process is a two-dimensional <Compound Poisson process> whose <Lévy measure> is
$$
\nu(B)=\lambda\,\operatorname{Leb}
\{y\in[0,1]:(g_1(y),g_2(y))\in B\}.
$$
Two coordinates of a Lévy process are independent exactly when its Lévy measure charges only the coordinate axes and its Gaussian covariance has no cross term. Here there is no Gaussian part, so independence is equivalent to
$$
g_1(y)g_2(y)=0
\quad\text{for almost every }y.
$$
Since $g_1g_2$ is continuous, this is equivalent to pointwise vanishing. Conversely, when the product vanishes, the mark sets where $g_1$ and $g_2$ are nonzero are disjoint; independent thinning of the <Poisson random measure> gives independent coordinate processes. Hence
$$
\boxed{(X_t^{g_1})_{t\geq0}\text{ and }(X_t^{g_2})_{t\geq0}
\text{ are independent}
\iff g_1(y)g_2(y)=0\ \text{for every }y\in[0,1].}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 6 / d / Solution

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A compound Poisson process has the form
$$
X_t=\sum_{n=1}^{N_t}Y_n,
$$
where $N$ is a <Poisson process> and the independent identically distributed marks $Y_n$ are independent of $N$. It is a <Lévy process> with a finite Lévy measure.

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 Incoming links: Compound Poisson process