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10 <=, >= replaced by \leq, \geq; [made Community Wiki]

A (standard, real-valued) Brownian motion $W = \{W(t): t >=0\}$\geq 0\}$is commonly defined by the following properties: 1)$W(0) = 0$a.s., 2) the process has independent increments, 3) for all$s,t>=0$s,t \geq 0$ with $s<t$, the increment $W(t) – W(s)$ is normally distributed with mean zero and variance $t-s$, and 4) almost surely, the function $t \mapsto W(t)$ is continuous.

As is well known, the above set of conditions can be reduced to 2), 3') for all $t>=0$, t \geq 0$,$W(t)$has mean zero and variance$t$, and 4). [Note that, in 3'),$W(t)$is not assumed to be normally distributed.] But what about omitting condition 2)? Can you find an example of a process$W$satisfying conditions 1), 3), and 4), but not 2)? [Note that such$W$must have (the Brownian motion) covariance$E[W(s)W(t)] = s$,$0 <= \leq s <=t$; \leq t$; hence, it cannot be a Gaussian process, for otherwise it would be a Brownian motion.]

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A (standard, real-valued) Brownian motion $W = \{W(t): t>=0\}$ is commonly defined by the following properties: 1) $W(0) = 0$ a.s., 2) the process has independent increments, 3) for all $s,t>=0$ with $s<t$, the increment $W(t) – W(s)$ is normally distributed with mean zero and variance $t-s$, and 4) almost surely, the function $t \mapsto W(t)$ is continuous.

As is well known, the above set of conditions can be reduced to 2), 3') for all $t>=0$, $W(t)$ has mean zero and variance $t$, and 4). [Note that, in 3'), $W(t)$ is not assumed to be normally distributed.] But what about omitting condition 2)? Can you find an example of a process $W$ satisfying conditions 1), 3), and 4), but not 2)? [Note that such $W$ must have (the Brownian motion) covariance $E[W(s)W(t)] = s$, $0<= s<=t$; hence, it cannot be a Gaussian process, for otherwise it would be a Brownian motion.]

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