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?
- $X_t = \sqrt t Z \sim N(0,t)$
- $X_t$ is not a Brownian Motion because its increments are not independent. For example $X_t−X_s$ is not independent from $X_s -X_r$, $\forall t>s>r$.