A stochastic process $X(t)$ by definition is gaussian iff all its finite-dimensional joint probability density functions are multivariate gaussian. Namely iff given the times $(t_1,t_2,...,t_n)$ , the random variables $(X(t_1),X(t_2),...,X(t_n))$ are jointly gaussian (n is arbitrary).
My intent is understanding why the solution of the following simple stochastic differential equation is a gaussian process:
$$dX(t)=a(t)dt+b(t)dW(t)$$
where W(t) is a Wiener process with diffusion coefficient D and the initial condition is $X(t_0)=X_0$, where $X_0$ is a random variable that is independent of W(t). a(t) and b(t) are deterministic "regular" functions.
The explicit solution is:
$$X(t)=X_0+ \int_{t_0}^{t}a(s)ds+\int_{t_0}^{t}b(s)dW(s)$$
If the integral $\int_{t_0}^{t}b(t)dW(t)$ were a gaussian process (I'm not sure about this) and the initial condition were a gaussian random variable, than at fixed t, the random variable $X(t)$ would be the sum of 3 independent gaussian random variables ( $\int_{t_0}^{t}a(t)dt$ is in fact degenerate) and would be therefore gaussian. But what about the multiple-times joint densities of $(X(t_1),X(t_2),...,X(t_n))$?