Take $X_t = (1-t)B_{t/(1-t)}$ for $t\in[0, 1)$ where $B_t$ is a $1$-dimensional Brownian motion. I want to show that $X_t$ is Gaussian.
I have actually never been able to find a precise definition of what a Gaussian process is, so I assume it is a process with Normally distributed increments. So using the fact that $B_t \sim N(0, t)$, we have
$$
\mathbb P(X_t<x) = \mathbb P(B_t<x/(1-t)) = \int^{x/(1-t)}_{-\infty}\frac{1}{\sqrt{2 \pi t/(1-t)}} \exp \left( \frac{-u^2}{2t/(1-t)} \right) du
$$
So here we see that we integrate a Normal density and hence $X_t$ is Gaussian. Is this a valid argument?
Best Answer
In order to show that the process $(X_t)_{t\in[0,1)}$ is Gaussian, we have to prove that if $t_1,\dots,t_n\in [0,1)$ and $(a_j)_{j=1}^n\subset [0,1)$, then the random variable $\sum\limits_{j=1}^na_jX_{t_j}$ has a Gaussian distribution.
In your case, you have to see why $(B_s)_{s\geqslant 0}$ is Gaussian.