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Let a stochastic process $X$ be defined by $X_t=\sqrt{t}\,Z$, where $Z$ is a standard normal random variable. Is $\{X_t,t\ge 0\}$ a standard Brownian Motion?

Fimpellizzeri
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Userabc
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  • What have you tried? What is the definition of standard Brownian motion? Where are you having trouble? – David Feb 09 '17 at 05:22

1 Answers1

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$\{X_t\}$ is not a standard Brownian motion because for $s<t$ $$X_t-X_s=(\sqrt t-\sqrt s)Z$$ which is normally distributed with mean $0$ and variance $(\sqrt t-\sqrt s)$. But not variance $t-s$.

user123
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