I was trying to compute $E[e^{\lambda X_t}|\mathcal{F_s}]$, where $X_t=\int_0^t(W_s-\frac{s}{t}W_t) ds$, $\mathcal{F}$ is associated to $W$. I tried the following.
1) Splitting the integral $\int_0^s$ and $\int_s^t$: $$E[e^{\lambda[\int_0^t(W_s-\frac{s}{t}W_t ds)+\int_s^t (W_s-\frac{s}{t}W_t ds)]}|\mathcal{F_s}]$$ 2)taking out all the measurable elements $$e^{\lambda\int_0^sW_sds}E[e^{\lambda [W_t(-s^2/2t -\frac{1}{t} (\frac{t^2-s^2}{2}) + \int_s^tW_sds]}\mathcal{F_s}]$$ 3)Taking increments, I arrive here $$e^{\lambda\int_0^sW_sds+\lambda W_s(t/2-s)}E[e^{\lambda (W_t-W_s)(t/2-s)-\lambda\int_s^t(W_t-W_s)ds}|\mathcal{F_s}]$$ that expectation should be independent of $\mathcal{F_s}$, hence $$E[e^{\lambda (W_t-W_s)(t/2-s)-\lambda\int_s^t(W_t-W_s)ds}]$$ the mean of the exponent is $0$, my problem is the variance, and in particular
$$E[(W_t-W_s)\int_s^t(W_t-W_s)ds]$$
ps continues from here (but I think it was off-topic there): Determine the distribution of $\int_0^t (W_s-\frac{s}{t}W_t) ds$, where $(W_s)_{s\geq 0}$ is a brownian motion