How to derive the semigroup and infinitesimal generator for the compound Poisson process? Let $(N_t: t\geq0)$ be a Poisson process and $(Y_i)_{i\in\mathbb N}$ a series of i.i.d. integrable random variables on $\mathbb Z$. Put
$X_t = \sum_{i=1}^{N_t} Y_i$ for $N_t > 0$ and $X_t$ for $N_t=0$.

I want to formulate the forward and backward equations for the process $(X_t: t\geq0)$. But I'm stuck at finding its semigroup and generator (Q-matrix).

So I'm looking for
$$P'(t)=Q P(t),\ t\geq0, \qquad P(0)=\mathrm I \qquad \mathrm{(backward\ equation)}$$
and
$$P'(t)=P(t)Q,\ t\geq0, \qquad P(0)=\mathrm I \qquad \mathrm{(forward\ equation)}$$
where
$\left(P(t)\right)_{t\geq0}$ is the semigroup and $Q$ the Q-matrix of $(X_t)$.
I know the the Q-matrix and semigroup for the Poisson process $(N_t)$ are
$Q = \lambda(P-\mathrm I)$
and
$$P(t)=\mathrm e^{-\lambda t} \sum_{k=0}^\infty \frac{(\lambda t)^k}{k!}P^k$$
where $P$ is the matrix of transition probabilities (on a countable state space $S$).
For $(N_t)$ we have the backward equation given as
$$P'_{ij}(t)
=\sum_{k\in S}q_{ik}P_{kj}(t)
=\lambda P_{i+1,j}(t)-\lambda P_{ij}(t), \qquad t>0 \qquad \tag{1}$$
and the forward equation as
$$P'_{ij}(t)
=\sum_{k\in S}P_{ik}(t)q_{kj}
=\lambda P_{i,j-1}(t)-\lambda P_{ij}(t), \qquad t>0. \qquad \tag{2}$$
If I understand the definition correctly, then $(X_t)$ is similar to a Poisson process, but after its Exp($\lambda$) waiting time it doesn't jump deterministically $+1$, but according to the random variables $Y_i$. However, I don't know how to compute the semigroup and Q-matrix for $(X_t)$, nor can I find any suitable reference.
Can someone help me get there?
 A: The miracle of infinitesimal generator is that it often tells you literally what is happening to you process. Here the generator is 
$$
Qf(x) = \lambda\int_{\mathbb R} \big(f(y+x) - f(x)\big)F(dy).
$$
Let me decipher this for you. $\color{blue}\lambda$ before the integral is the jump intensity. The integrand tells you what to do: $\color{blue}{-f(x)}$ means removing the moving particle from its current location $x$. $\color{blue}{+f(x+y)}$ means placing the removed article to the new location $x+y$. $\color{blue}{F(dy)}$ means that the length $y$ of jump to new location should be chosen according to the distribution $F$. 
In your particular case, where the jumps are integer valued (so that e.g. $P(Y_i = n) = a_n$, $n\in \mathbb{Z}\setminus \{0\}$), you'll get 
$$
Qf(m) = \lambda\sum_{n\in\mathbb{Z}} a_n\big(f(m+n) - f(m)\big).
$$
The corresponding "matrix" is then equal to $[Q] = (q_{ij})_{i,j\in\mathbb Z}$ with 
$$
q_{ij} = \begin{cases}
\lambda a_{i-j}, i\neq j,\\
-\lambda, i=j.
\end{cases}
$$
To find the semigroup, you need to identify $e^{tQ}$. This does not always have a nice form. In the case of usual Poisson process, the generator can be decomposed as $Q = -\lambda I + \lambda S$, where $S$ is the right shift. Since these two operators commute, $$
e^{tQ} = e^{-t\lambda I} e^{-t\lambda S} = e^{-\lambda t}\sum_{n=0}^\infty \frac{(\lambda t)^n}{n!}S^n;
$$
here you can clearly see that the increments of your process have Poisson distribution (as expected).
For a general compound Poisson process you may write your answer as some mixture of convolutions of $Y$ with Poissonian weights, but this is not very exciting.
