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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?

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  • $\begingroup$ You want to prove that the random variable $X_t$ is Gaussian or that the process $(X_t)_{t\in[0,1)}$ is Gaussian? $\endgroup$ – Davide Giraudo Dec 21 '14 at 22:23
  • $\begingroup$ @DavideGiraudo That the process is Gaussian $\endgroup$ – Slug Pue Dec 21 '14 at 22:26
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    $\begingroup$ "I have actually never been able to find a precise definition of what a Gaussian process is" First paragraph of the obvious. $\endgroup$ – Did Dec 22 '14 at 11:05
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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.

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