I'm trying to prove that the expected value of the stochastic integral (integral w.r.t to the centered, right-continuous with orthogonal increments) is zero. I've come up with a very simple argument that I feel might be wrong.
If I denote the integral as $I(f)$, we've got $I(f_n) \to I(f)$ in norm in the $L_2$ space. From the continuity of the scalar product we get $$EI(f_n) = \langle I(f_n),1 \rangle \to \langle I(f),1 \rangle = EI(f)$$
Since obviously $EI(f_n)=0$ ($Z$ is centered), $EI(f)=0$.
Is this correct? If not, I'd appreciate a nudge in the right direction. Thanks!
edit: Details. $f_n$ are simple functions (piece-wise constant) that converge to $f$. The integral of $f$ is then defined as a limit of integrals of the simple functions. More rigorously:
We have the process $\{Z_\lambda, \lambda \in [a,b] \}$ which is centered, right mean-continuous (in $L_2$), with orthogonal increments ($E(X_{t_4}- EX_{t_3})(\overline{X_{t_2} - EX_{t_1}})=0$ for $t_4>t_3>t_2>t_1$), the interval $[a,b]$ is bounded. Integral of a square-integrable simple (piece-wise constant) function is defined as $$\int^b_a f dZ = \sum^n_{k=1} c_k (Z_{\lambda_k} - Z_{\lambda_{k-1}})$$ For a (nonsimple) function $f$, we take $f_n$ converging to $f$ (simple functions are a dense subset) and define $$I(f) = \text{l.i.m.}_{n\to \infty} I(f_n) := \text{l.i.m.}_{n\to \infty} \int^b_a f_n dZ$$ Where the limit is the limit in the mean.
Hopefully that clears this up, if not, let me know.
