Why some process like $\sigma \int_0^t e^{-\alpha(t-s)}dB(s)$ is Gaussian or normal?

Question

Almost all books about stochastic differential equation have the statement says some process is Gaussian directly. For example section 3.5 of "STOCHASTIC DIFFERENTIAL EQUATIONS AND APPLIICATIONS 2nd" by professor Xuerong Mao on page 102 directly says $$\sigma \int_0^t e^{-\alpha(t-s)}dB(s)$$ follows the normal distribution $N(0,\sigma^2(1-e^{-2\alpha t})/2\alpha)$. If it is normal, I know how to calculate the mean and variance. But why this is normal? I saw almost all books state similar conclusion directly.

Another example, in "Stochastic Differential Equations" by Bernt Oksendal on lemma 6.2.3 it says: $$M_t^{(n+1)} = M_0 + \int_0^t H(s)M_s^{(n)}ds + \int_0^t K(s) dB_s,\quad n=0,1,2,\cdots$$ where $H(t), K(t) \in \mathbb{R}^{2 \times 2}$ and $$dM_t = H(t)M_tdt + K(t)dB_t, M_0 = [X_0, 0]^T.$$ Then it wrote $M_t^{(n)} $ is Gaussian directly too.

Are these because of some property of Brownian motion?

Now I have a process $$(t-s)B_s + \int_s^t(t-u)dB_u.$$ Is this a Gaussian?

Answer 1

Indeed, for deterministic $\sigma$ we have $\int \sigma(s)dB_{s}\sim N(0,\int \sigma^{2}(s)ds)$. The proof just follows from the approximation by sums of independent Gaussians.

See here The ito integral is gaussian and that the limit of Gaussians is still Gaussian The limit of a convergent Gaussian random variable sequence is still a Gaussian random variable

But if $\sigma$ is random, then not necessarily eg. Is the Ito integral of a predictable process Gaussian distributed?

No. Let $\sigma_t = 1$ for all $t$ with probability $1/2$ and $2$ with probability $1/2.$ $\sigma_t$ is previsible. Then $W_T$ is a 50-50 mixture of normal with variance $T$ and $4T$, which is not a normal.