Find explicitly the solution of $$dX_t=a(b-X_t)\:dt+\sigma \:dW_t$$What behaviour do you expect if the parameters are:
(i) $a=1,\:b=1,\:\sigma=1,\: \Delta t=1$ and $X_0=1$
(ii) $a=-1,\:b=1,\:\sigma=1,\: \Delta t=1$ and $X_0=1$
(iii) $a=1,\:b=2,\:\sigma=0,\: \Delta t=1$ and $X_0=1$
I get the solution, $$X_T=b(1-e^{-aT})+X_0e^{-aT}+\sigma e^{-aT} \int_0^Te^{as}\:dW_s$$
Using the given parameters, I get,
(i) $X_T=(1-e^{-T})+e^{-T}+e^{-T}\int_0^Te^{s}\:dW_s$
(ii) $X_T=(1-e^{T})+e^{T}+e^{T}\int_0^Te^{-s}\:dW_s$
(iii) $X_T=2(1-e^{T})+e^{T}$
Now, I didn't get what they mean by behaviour. And it seems I face trouble to solve the deterministic integral $\left(\int_0^Te^{s}\:dW_s\text{ or }\int_0^Te^{-s}\:dW_s\right)$.
Any help will be appreicated. Thanks in advance.
The given SDE represents an Ornstein–Uhlenbeck process, which has the solution you provided. The main characteristics are:
$$ \mathbb{E}[X_T] = X_0 e^{-aT}+b(1-e^{-aT}) \\ \text{Var}[X_T] = \frac{\sigma^2}{2a}(1-e^{-2aT}) \\ \text{Cov}(X_t,X_s) = \frac{\sigma^2}{2a} e^{-a(t+s)}(e^{2as}-1) $$
Usually, when we deal with mean-reverting processes, as in this case, we want to know the behaviour of mean and variance in the limit $T \to \infty$.
N.B. You have made some mistakes when you substitute the parameters (I see some minus signs missing from the exponentials)