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If I have a variable Z that is normally distributed, Z~N(0,1), what would be the distribution of √t Z, t>=0? Can I say the process Xt = √t Z is a Brownian Motion?

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – user20160 Feb 21 '18 at 11:28
  • $\begingroup$ You will want to distinguish a single random variable $Z$ from a parameterized set of random variables $(Z_t)$, all having a standard Normal distribution. When $t$ lies within an interval of nonnegative real numbers, the set $(\sqrt{t}\,Z_t)$ enjoys many of the properties of Brownian Motion. (To obtain Brownian Motion you need to make some additional assumptions about the multivariate distributions, such as the distribution of $(Z_t,Z_s)$ for $s\ne t.$) $\endgroup$ – whuber Feb 21 '18 at 16:14
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  • $X_t = \sqrt t Z \sim N(0,t)$
  • $X_t$ is not a Brownian Motion because its increments are not independent. For example $X_t−X_s$ is not independent from $X_s -X_r$, $\forall t>s>r$.
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