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Is it possible for a stochastic process $X_t$ to have law $\mathcal N(0,t)$, continuous paths a.s and not be a Brownian motion ? If it is not possible, how to prove it ?

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    $\begingroup$ Let $Z\sim N(0,1)$ and let $X_t=\sqrt t Z$ for all $t$. As a stochastic process it's boring, but it has the right marginal distributions. $\endgroup$ Commented Aug 30, 2020 at 15:17

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It is possible, like pointed out in the comments. In fact, we can characterize Brownian Motion as a stochastic process $X_{t}$ with a.s. continuous sample paths such that

  1. For any $0\le t_{1}<...<t_{n}<\infty$, $(X_{t_{1}},...,X_{t_{n}})$ follows a multivariate normal distribution

  2. For any $t\ge 0$, we have $E[X_{t}]=0$

  3. For any $s,t \ge 0$, we have $Cov(X_{t},X_{s})=\min(s,t)$

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