I have a stochastic process $dX(t)=k[A-Y(t)]f(t)dt$, where $Y(t)$ is the Ornstein-Uhlenbeck process, and satisfies: $dY(t)=k[A-Y(t)]dt+\sigma dW(t)$ and $Y(0)=0$. $f(t)=0$ for $t<\tau$ and $f(t)=1$ for $t\geq\tau$. Now I want to solve for $X(t)$ and $\mathbb{E}(X(t))$, $\mathbb{V}(X(t))$.
I could plug in $Y(t)=A(1-e^{-kt})+\sigma\int_0^t e^{-k(t-s)}dW_s$, but how shall I proceed? I am new to SDE, and any help is highly appreciated! (I find this question related to How to solve system of stochastic differential equations?)
EDIT:
I found this question: https://mathoverflow.net/questions/143245/distribution-of-integral-of-exponential-of-wiener-process. Why does this hold? $\sigma e^{as}\int_0^s e^{-as}dW_s=\beta W_s$ where $\beta^2=\frac{\sigma^2}{2a}(1-e^{-2as})$? Actually I cannot obtain the right formula of variance for the OU process based on this equation.
Assuming this is correct, my attempt for $\tau=0$: (Let me know if anything is wrong)
$Y(t)=A(1-e^{-kt})+\beta(t)W(t)$ where $(\beta(t))^2=\frac{\sigma^2}{2k}(1-e^{-2kt})$. So $dX(t)=(kAe^{-kt}-k\beta(t) W_t)dt$. Thus $\mathbb{E}(X(T))=\int_0^T kAe^{-kt}dt$.
I am still confused in obtaining $\mathbb{V}(X(t))$, and thanks for any help in advance!
