I am given the following interest rate model with SDE: $dX_t=\alpha \,dt+\sigma \,dB_t$ where $α$ and $σ$ are constants, and $B_t$ is a Brownian motion. I am supposed to determine $$ E\left[ \int_0^T X_t \,dt\right] \quad\text{ and }\quad V \left[\int_0^T X_t \,dt\right] $$ Any clues? I am lost. I am able to work out SDE from a given function using Ito's formula and then apply stochastic integrals. But I have no idea for this qns.
Any hints appreciated!
Clearly, \begin{align} X_t&=X_0+\alpha\,t+\sigma\,B_t\,,\\ E[X_t]&=X_0+\alpha\,t\,,\\ E[X_uX_v]&=(X_0+\alpha\,u)(X_0+\alpha\,v)+\sigma^2\min(u,v)\,.\tag{1} \end{align} Therefore, $$\tag{2} E\left[\int_0^TX_t\,dt\right]=\int_0^TE[X_t]\,dt=X_0T+\frac{\alpha T^2}{2}\,. $$ To find the variance we can first calculate $$\tag{3} E\left[\left(\int_0^TX_t\,dt\right)^2\right]=\int_0^T\int_0^TE[X_uX_v]\,du\,dv\, $$ and use $$ V\left[\int_0^TX_s\,ds\right]=E\left[\left(\int_0^TX_t\,dt\right)^2\right] -\underbrace{E\left[\int_0^TX_s\,ds\right]^2}_{(X_0T+\frac{\alpha T^2}{2})^2\text{ by (1) }}\,. $$ Using (1) the calculation of (3) is straightforward.