I get a problem that comes up in the construction of the Lévy-Itõ decomposition. For a Lévy-process $X$ there is a independently scattered poisson random measure $N$, such that for each t, and for each set $A$ such that $0 \ne \bar{A}$, we have that $N(t,A)$, the number of jumps in $(0,t]$ such that the jump size is in $A$ is a poisson distribution with parameter $E[N(1,A)]=\nu(A)$.
I need to show that if I define $Y_a(t)=X(t)-\int_{|x|\ge a}xN(t,dx)$. I have that $E[Y_a(t)]=tE[Y_a(1)]$.
We can show through simple functions that $E[\int_{|x|\ge a}xN(t,dx)]=t\int_{|x| \ge a}xd\nu=tE[\int_{|x| \ge a}xN(t,dx)]$.
But this is only the last part of $Y_a$, what do I do to get the result? Do I have for any Lévy-process that $E[X(t)]=tE[X(1)]$?
Yes, this results holds for any Lévy process which is integrable.
It follows from the independence and stationarity of the increments that
$$\mathbb{E}X_{k/n} = \mathbb{E} \left( \sum_{j=1}^k (X_{j/n}-X_{(j-1)/n}) \right) = k \mathbb{E}(X_{1/n})$$
and, similarly,
$$\mathbb{E}(X_1) = \mathbb{E} \left( \sum_{j=1}^n (X_{j/n}-X_{(j-1)/n}) \right) = n \mathbb{E}(X_{1/n}).$$
Hence,
$$\mathbb{E}X_{k/n} = \frac{k}{n} \mathbb{E}(X_1),$$
i.e.
$$\mathbb{E}X_q = q \mathbb{E}(X_1)$$
for all rationals $q \in \mathbb{Q} \cap [0,\infty)$. Using the continuity of the mapping $t \mapsto \mathbb{E}(X_t)$, this identity extends to all $t \geq 0$: $$\mathbb{E}(X_t) = t \mathbb{E}(X_1).$$