I'm reading Brownian Motion: An Introduction to Stochastic Processes by Schilling. In the first chapter they give the following definition for some random variable $X_t$ to be considered a Brownian motion:
- $B_0(\omega)=0$ for almost all $\omega$
- $B_{t_n}-B_{t_{n-1}},...,B_{t_1}-B_{t_0}$ are independent for $0 = t_0\leq t_1\leq ... \leq t_n$ for $n\geq 0$
- $B_t-B_s\sim B_{t+h}-B_{s+h}$ for $0\leq s< t$ and $h\geq -s$
- $B_t-B_s\sim \mathcal{N}(0,t-s)$
- $t\mapsto B_t$ is continuous
My confusion is that it seems like condition 2 and 4 are redundant, in that:
$$\begin{align}Cov(B_t-B_s,B_a-B_b)&=\mathbb{E}((B_t-B_s)(B_a-B_b))\\ &=\mathbb{E}(B_tB_a-B_sB_a-B_tB_b+B_sB_b)\\ &=(t\wedge a -s\wedge a-t\wedge b+s\wedge b)\end{align}$$
For nonoverlapping intervals $[t,s]$ and $[a,b]$, such as those described in condition 2, the covariance will always evaluate to 0. Meaning that condition 4 implies condition 2.
I'm wondering if there's some flaw in my reasoning with this, or if I made some mistake up to this point.