Integral w.r.t. Brownian motion is gaussian process

Question

Consider the stochastic term from the solution of the SDE of an Ornstein-Uhlenbeck process:

$$X_t=\sigma \int_0^te^{-\lambda(t-s)} \, dBs, $$ where $\sigma>0$.

How can be seen that $X_t$ is a Gaussian process?

Answer 1

If you write out the Riemann sums which converge to the stochastic integral defining $X_t$, each is a linear combination of the jointly Gaussian random variables $B_{s_i}$ for some finitely many values of $s_i$, with deterministic coefficients. So the Riemann sum defines a Gaussian random variable. Then the integral in $X_t$ is a limit i.p. of Gaussian random variables, which must therefore be Gaussian (this is easy to check using characteristic functions, for instance).

A similar argument would show that any linear combination $\sum a_i X_{t_i}$ is likewise Gaussian, so $X_t$ is a Gaussian process.