Suppose $X(t)$ is an O-U process. We define $f(t,X(t))=\pi(t,X(t))(X(t)+a)$, where a is a constant. Now I need to calculate the variance of $\int_{s}^{T}\pi[(X(t)+a)dt+bdW(t)]$. Here a and b are both constants, W(t) is a standard wiener process.
I calculate $Var(\int_{s}^{T}\pi[(X(t)+a)dt+bdW(t)])=Var(\int_{s}^{T}f(t,X(t))dt+b\pi(t,X(t)) dW(t))$.
Firstly, I calculate $E[\int_{s}^{T}f(t,X(t))dt+b\pi(t,X(t)) dW(t)]=E[\int_{s}^{T}f(t,X(t))dt]$.
Then I calculate $E[(\int_{s}^{T}f(t,X(t))dt+b\pi(t,X(t)) dW(t))^{2}]=E[\int_{s}^{T}(f(t,X(t))dt+b\pi(t,X(t)) dW(t))^{2}]=E[\int_{s}^{T}b^{2}\pi^{2}dt]$
Therefore, I think the variance of $\int_{s}^{T}\pi[(X(t)+a)dt+bdW(t)]$ is $E[\int_{s}^{T}b^{2}\pi^{2}dt]-(E[\int_{s}^{T}f(t,X(t))dt])^{2}$. But the answer is given as $Var[\int_{s}^{T}b\pi(t,X(t)) dW(t)]$.
Could you give me any hint where i am wrong? It seems the difference is comes from $(E[\int_{s}^{T}f(t,X(t))dt])^{2}$. The answer seems to say that this part is equal to 0, but I don't get it.