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The steps to showing that a process $(W_t)_{t \geq 0}$ is a Brownian motion (BM) are as follows:

(1)$W_0 = 0$

(2) $ \forall t ~~~W_t$ is continuous

(3)$W_t \sim N(0,t)$

(4)$W_{t+s}-W_{s} \sim N(0,t)$

(5)$W_{t+s}-W_{s} \bot \mathscr F_s=(W_u)_{u\in(0,s)} $

For example, if we take $X_t = W_{2t}-W{t}$ we can show that if $X$ satisfies these 5 steps then it is also a BM.

My question is that if a Gaussian process is fully characterized by its covariance and mean functions, is it enough to show that $X$ has the same covariance and mean function as $W$ instead of having to go through the 5 steps, or is there a contradictory case in which $X$ has the same mean and covariance function but does not satisfy some other condition such as continuity or independent increments? Answers should hopefully provide that contradiction.

saz
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WeakLearner
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1 Answers1

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If $(X_t)_{t \geq 0}$ is a Gaussian process with the same mean and covariance function as a Brownian motion, then $(X_t)_{t \geq 0}$ satisfies (1),(3)-(5). Moreover, it follows from Kolmogorov's continuity theorem that $(X_t)_{t \geq 0}$ has a modification with (exclusively) continuous sample paths. However, the process $(X_t)_{t \geq 0}$ does not necessarily have continuous sample paths.

saz
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  • Do you know of any examples where this occurs?? – WeakLearner Mar 16 '15 at 08:18
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    @dimebucker91 Suppose $(B_t){t \geq 0}$ is a Brownian motion on the probability space $([0,1],\mathcal{B}[0,1])$ endowed with the Lebesgue measure. For each fixed $\omega \in [0,1]$ we set $$W(t,\omega) := \begin{cases} B(t,\omega), & t \neq \omega, \ 0, & t=\omega \end{cases}$$ Then $(W_t){t \geq 0}$ does not have continuous sample paths, but its still a Gaussian process with the same covariance function as a BM. – saz Mar 16 '15 at 15:16