Brownian motion or not?

Question

Suppose that $(X_t , t\in [0;1])$ are independent normal r.v with mean 0 and variance $\sigma^2 _{t}$. Is this process brownian motion?

Answer 1

You first need $\sigma_t=t$, since the Brownian motion should have variance $t$ at time $t$.

A condition for $X$ to be a Brownian motion is that for any $s<t$, $$X_t-X_s\sim N(0,t-s)\qquad (1).$$

From the independence of $X_t$ and $X_s$, you get $$X_t-X_s\sim N(0,t+s),$$ which is a contradiction to the condition $(1)$.

Answer 2

The answer is never, because, if $(W_t)_{t\geqslant0}$ is a Brownian motion, then $W_{1/2}$ and $W_1$ are not independent while you assumed that $X_{1/2}$ and $X_1$ are independent.