A compound Poisson process is a jump process with two parameters, the rate of the jumps $\lambda$ and the distribution of the jumps $\mu$ ($\mu$ is a probability measure on $\mathbb{R}$). The infinitesimal generator of this process is given by:
$$A_{\lambda,\mu}f(x) = \lambda\int_\mathbb{R} (f(x+z)-f(x))\mu(dz)$$
What happen if I take a brownian motion with generator $\frac{1}{2}f''$ and construct the process with generator $\frac{1}{2}f''+A_{\lambda,\mu}$.
Is this process the sum of a brownian motion with an a independent compound poisson process with parameters $\lambda, \mu$? If the answer is yes, How we can generalize this notion? for example, If we take a general process with generator $L$ and we construct the process with generator $L+A$ is this process the sum of a $L$-process and a independent $A$-process.
I think that total generality it is not true because the $L$-process can have jumps with rate dependent on the state.
Any help will be appreciated!
Best Answer
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 $\left(\sqrt{b_1^2 + b_2^2}\right)^2 \frac{d^2}{dx^2} = (b_1^2 + b_2^2)\frac{d^2}{dx^2}$.