How can variance and mean of brownian motion be stated

Question

I have seen in the construction of brownian motion that the Its mean is 0 and variance t. i.e $E(B(t))=0$ , $Var(B(t))=t$ Why in a problem sheet does it state $B_{j}(t)$ for $j=1,...,n$ are brownian motions with variances $\sigma^{2}_{j}$. How can these be brownian motion with these variances... is it saying $Var(B_{j}(t)) \neq t$

Answer 1

If $B(t)$ is a standard Brownian motion, then indeed $\mathbb E[B(t)]=0$ and $\mathsf{Var}(B(t))=t$ for $t\geqslant0$. Let $X(t) = \sigma B(t) + \mu t$ for some $\mu\in\mathbb R$ and $\sigma>0$. Then $X(t)\sim\mathcal N(\mu t, \sigma^2 t)$, and is called a Brownian motion with drift $\mu$ and volatility (or variance) $\sigma^2$.

In your example, it seems that each $B_j(t)$ is a Brownian motion with zero drift and volatility $\sigma^2_j$.

Answer 2

We usually denote by $\sigma^2$ the volatility of a brownian motion, not its variance, which is $\sigma^2 \times t$. It might help you to understand.