Brownian Motion 3

Question

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?

Answer 1

$\{X_t\}$ 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 $t-s$.