I am trying to do some calculations related to the Vacisek model, but I think I am mixing up concepts and I'm not getting to any solution. Let me explain what the problem is.
The Vacisek model follows the SDE $$dR_t = -a(R_t-b)dt + \sigma dW_t$$ with $R_0 = r_0$, with $a, b, \sigma > 0$ and $(W_t : t \in [0,T])$ is a Brownian Motion with filtration $\{\mathscr{F}_{t}\}_{t\geq0}$. It can be shown that the solution to this SDE is $$ R_t = b +(r_0 - b)e^{-at} + \sigma \int_0^te^{-a(t-s)}dW_s $$
Now, assuming that $a = 0$, I would like to compute $$ \mathbb E\left[\left(\int_t^T R_sds - M_t \right)^2 \middle| \mathscr{F}_t\right] \mbox{ for $t \leq T$} $$
where $M_t = \mathbb E\left[\int_t^T R_sds \middle| \mathscr{F}_t\right]$.
I have started by doing the following: $$ \mathbb E\left[\int_t^T R_sds \middle| \mathscr{F}_t\right] = \mathbb E\left[\int_t^T r_0 + \sigma W_sds \middle| \mathscr{F}_t\right] $$
My problem is how to move forward from here, the conditional expectation is making me doubt whether I should have taken a different approach. Any hints are welcome.
If $a=0$, then $R_s=r_0+\sigma W_s$ for every $s$ hence $$ K_T=\int_t^TR_s\mathrm ds=(T-t)(r_0+\sigma W_t)+\sigma\int_0^{T-t}V_s\mathrm ds, $$ where $V=(W_{t+s}-W_t)_{s\geqslant0}$ is independent of $\mathcal F_t$. Thus, $M_T=E(K_T\mid\mathcal F_t)$ is $$ M_T=(T-t)(r_0+\sigma W_t)+\sigma\int_0^{T-t}E(V_s)\mathrm ds= (T-t)(r_0+\sigma W_t), $$ and $$ K_T-M_T=\sigma\int_0^{T-t}V_s\mathrm ds, $$ is independent of $\mathcal F_t$. Thus, $$ E((K_T-M_T)^2\mid\mathcal F_t)=E((K_T-M_T)^2)=u(T-t), $$ where, for every nonnegative $\tau$, $$ u(\tau)=\sigma^2\iint_{[0,\tau]^2}E(V_sV_u)\mathrm ds\mathrm du=\sigma^2\int_0^{\tau}\int_0^u2s\mathrm ds\mathrm du=\frac13\sigma^2\tau^3.$$