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?
Best Answer
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.