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Background Information:

The process $W = (W_t:t\geq 0)$ is a $\mathbb{P}$-Brownian motion if and only if

i) $W_t$ is continuous, and $W_0 = 0$

ii) the value of $W_t$ is distributed, under $\mathbb{P}$, as a normal random variable $N(0,t)$,

iii) the increment $W_{s+t} - W_{s}$ is distributed as a normal $N(0,t)$, under $\mathbb{P}$, and is independent of $\mathcal{F}_s$, the history of what the process did up to time $s$.

Question:

If $Z$ is a normal $N(0,1)$, then the process $X_t = \sqrt{t}Z$ is continuous and is marginally distributed as a normal $N(0,t)$. Is $X$ a Brownian motion?

From the definition Brownian motion above it seems that we directly satisfy the 2 conditions. Although I believe we need to show the third condition to indeed conclude that $X$ is of Brownian motion. I am just not sure how to provide a formal solution to this question.

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    $\begingroup$ Can you please check whether $X_{s+t}-X_s$ has the distribution of $N(0, t)$? $\endgroup$ – Gordon Nov 22 '16 at 15:20
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Aside from the independence requirement for the increments, that is, the independence of $X_{s+t}-X_s$ and $\mathcal{F}_s$, you can check whether the increment $X_{s+t}-X_s$ has the distribution of $N(0, t)$. In fact, note that \begin{align*} X_{s+t}-X_s &= (\sqrt{s+t}-\sqrt{s}) Z\\ &\sim N\left(0,\, (\sqrt{s+t}-\sqrt{s})^2\right), \end{align*} which obviously does not have the distribution of $N(0, t)$. That is, the process $\{X_t, \, t\ge 0\}$ is not a Brownian motion.

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    $\begingroup$ Who could have guessed! :) $\endgroup$ – Quantuple Nov 22 '16 at 16:09

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