I have the following in my notes and I'm not sure if it's true or not. Any help would be highly appreciated.
If $\{W_t\}_{t\geq0}$ is a standard Brownian motion stochastic process, $\Delta>0$ and $n\in\mathbb{N}$ then $W_{n\Delta}-W_{(n-1)\Delta}$ is mean zero and variance $\Delta$ normal random variable.
I am using the following definition of Brownian motion, as taken from Harrison's Stochastic flow systems:
A stochastic process $\{W_t\}_{t\geq0}$ has independent increments if the random variables $X_{t_1}-X_{t_0},\ldots,X_{t_n}-X_{t_{n-1}}$ are independent for all $n\geq 1$ and all $0\leq t_0\leq\ldots\leq t_n<\infty$. It has stationary independent increments if the distribution of $X_t-X_s$ depends only on $t-s$. A Standard Brownian motion is a stochastic process that has continuous sample paths, stationary independent increments and $X_t\sim N(0,t)$.
That the difference is mean zero, I can see. What I'm not convinced about is that variance wil be $\Delta$. Any thoughts?