Is there an easy way to construct, on the same filtered probability space, a Brownian motion $W$ and a Poisson process $N$, such that $W$ and $N$ are not independent ?
[Math] Correlated Brownian motion and Poisson process
probabilitystochastic-processes
Related Solutions
The generator of a sum of (time- and space-) homogeneous independent Markov processes is a sum of generators.
Indeed, let $X$ and $Y$ be these processes. Denote by $T_t$ and $S_t$ their Markov semigroups, $A$ and $B$ their generators, $\mathsf E_{x,y}$ the expectation given $X_0=x, Y_0=y$. Then for any $f\in C_b(\mathbb R)$ (bounded continuous), using the independence and homogeneity, $$ \mathsf E_{x,y}[f(X_t + Y_t)] = \mathsf E_{x,y}[\mathsf E_{x,y}[f(X_t + z)]|_{z=Y_t}]= \mathsf E_{x,y}[\mathsf E_{x+z,y}[f(X_t)]|_{z=Y_t}] \\ = \mathsf E_y[T_t f(x+z)|_{z=Y_t}] = S_t T_tf (x+y). $$ Similarly, it equals to $T_t S_t f (x+y)$, in particular, $T_t$ and $S_t$ commute (and so do $A$ and $B$). Now for $f\in C_b^2(\mathbb R)$ (bounded twice continuously differentiable with bounded derivatives), differentiating this equality in $t$ for $t=0$, we get $$ \frac{d}{dt}\mathsf E_{x,y}[f(X_t + Y_t)]|_{t=0} = (A+B)f(x+y), $$ which means that $A+B$ is the generator of $X+Y$.
There is another way to look at this. A time- and space-homogeneous process is a Lévy process, which can be written as a sum $$ at + b W_t + Z_t, $$ where $W_t$ is a standard Wiener process, and $Z_t$ is a pure jump process. When you add two independent things of such kind, you arrive to $$ (a_1+a_2)t + \sqrt{b_1^2 + b_2^2}\, W_t + (Z_t^1 + Z_t^2). $$ The process $Z_t = Z_t^1 + Z_t^2$ is again a pure jump process, and the jump intensities at a given level just add up. If you translate these things to generator, everything will add up to; for instance, the diffusion part is $\left(\sqrt{b_1^2 + b_2^2}\right)^2 \frac{d^2}{dx^2} = (b_1^2 + b_2^2)\frac{d^2}{dx^2}$.
Generally, the respective solution to $$ {\rm d}S_t=S_t\left(\mu{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right) $$ with $$ {\rm d}W_t^1{\rm d}W_t^2=\rho{\rm d}t $$ would be $$ S_t=S_0\exp\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. $$
According to the SDE, the quadratic variation of $S_t$ is, intuitively, \begin{align} {\rm d}\left<S\right>_t&={\rm d}S_t{\rm d}S_t\\ &=S_t^2\left(\mu{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)^2\\ &=S_t^2\left(\mu^2{\rm d}t^2+\sigma_1^2{\rm d}W_t^1{\rm d}W_t^1+\sigma_2^2{\rm d}W_t^2{\rm d}W_t^2+\right.\\ &\quad\quad\left.2\mu\sigma_1{\rm d}t{\rm d}W_t^1+2\mu\sigma_2{\rm d}t{\rm d}W_t^2+2\sigma_1\sigma_2{\rm d}W_t^1{\rm d}W_t^2\right)\\ &=S_t^2\left(\sigma_1^2{\rm d}t+\sigma_2^2{\rm d}t+2\sigma_1\sigma_2\rho{\rm d}t\right). \end{align}
Thus thanks to Ito's lemma, \begin{align} {\rm d}\log S_t&=\frac{{\rm d}S_t}{S_t}-\frac{1}{2}\frac{{\rm d}\left<S\right>_t}{S_t^2}\\ &=\left(\mu{\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\right)-\frac{1}{2}\left(\sigma_1^2+\sigma_2^2+2\rho\sigma_1\sigma_2\right){\rm d}t\\ &=\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right){\rm d}t+\sigma_1{\rm d}W_t^1+\sigma_2{\rm d}W_t^2\\ &={\rm d}\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. \end{align} This eventually yields $$ S_t=S_0\exp\left[\left(\mu-\frac{1}{2}\sigma_1^2-\frac{1}{2}\sigma_2^2-\rho\sigma_1\sigma_2\right)t+\sigma_1W_t^1+\sigma_2W_t^2\right]. $$
With this solution, as well as the given correlation between $W_t^1$ and $W_t^2$, it is no longer hard to figure out the distribution of $S_t$, and the probability would follow straightforwardly.
Best Answer
In case (W,N) has independent increments, then W and N are also independent, since a Brownian Motion has no common jumps with the Poisson process. This of course doesn't say, there is no probability space, where (W,N) has dependent increments, but it gives you a hint, how it might be constructed.
See e.g. Kallenberg - Foundations of Modern Probability 13.6 for a proof