Let $W_{t}$ be a Wiener process on a standard filtered probability space $(\Omega ,{\mathcal {F}},{\mathcal {F}}_{t},P)$ and let ${\mathcal {G}}_{t}$ be the augmented filtration generated by $B$. If $X$ is a square integrable random variable measurable with respect to ${\mathcal {G}}_{\infty }$, then there exists a predictable process $\Theta$ which is adapted with respect to ${\displaystyle {\mathcal {G}}_{t}}$, such that
$$X = EX + \int_0^T\Theta_s dW_s$$
where $W$ is a Wiener process
I want to prove that Ito representation of $X = \int_0^T W(s) ds$ is in form for $\Theta_s = (T-s)$.
My work so far
Let's calculate the integral $$\int_0^T (T-s)dW_s = \int_0^T TdW_s - \int_0^TsdW_s = T\int_0^TdW_s - \int_0^TsdW_s=$$ $$= TW_T - \int_0^TsdW_s$$
And now I have two problems:
$(1)$ Why $EX = E[\int_0^T W(s)ds] = 0 $ ?
$(2)$ I'm not exactly sure how to calculate this integral $\int_0^TsdW_s$. I read that partial integration should be used but I don't see how.
Could you please give me a hand in justifying $(1)$ and $(2)$ ?
Well for (1) it's only Fubini nothing more (as $E[W_s]=0$), for the (2) you have by stochastic integration by part (look for example Theorem 3) that :
$$TW_T-0.W_0=\int_0^TsdW_s+\int_0^TW_sds+\int_0^Td\langle s,W_s\rangle=\int_0^TsdW_s+\int_0^TW_sds+0$$
So $\int_0^T (T-s)dWs =\int_0^TW_sds$
QED