Say we have following SDE (Vasicek): $$dr(t) =(b-ar_t) dt + \sigma dW_t$$
I am able to reach an integral form of this SDE : $$r(t) = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}] + \sigma e^{-at}\int_0^t e^{as}dW_s$$
From here, I would like to conclude that $r(t)$ is Gaussian but I don't know how to proceed.
I somehow understand that
$$E[r(t)] = r(0) e^{-at} + \frac{b}{a}[1 - e^{-at}]$$
and that
$$Var[r(t)]= \sigma e^{-at}\int_0^t e^{as} dW_s$$
First, note $$\mathbb{E^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=0 $$ and $$\mathbb{Var^Q}\left[\int_0^t e^{-a(t-s)}dW_s\right]=\mathbb{E^Q}\left[\int_{0}^{t} e^{-2a(t-s)}ds\right]=\frac{1}{2a}(1-e^{-2at}) $$ therefore $$\mathbb{E^Q}[r_t]=r_0 e^{-at} + \frac{b}{a}(1 - e^{-at})$$ $$\mathbb{Var^Q}(r_t)=\frac{\sigma^2}{2a}(1-e^{-2at})$$ second
The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $X_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Itô integral of $X_t$ with respect to $W_t$ up to time t is a random variable $$\int_{0}^{t}X_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{X({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ Set $X_s=e^{as}$, $X_s$ is a deterministic function thus $$\int_0^t e^{-a(t-s)}dW_s\sim N\left(0\quad,\quad\frac{1}{2a}(1-e^{-2at})\right) $$
