# Equivalent statements in the definition of Brownian Motion

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:

1. $$B_0(\omega)=0$$ for almost all $$\omega$$
2. $$B_{t_n}-B_{t_{n-1}},...,B_{t_1}-B_{t_0}$$ are independent for $$0 = t_0\leq t_1\leq ... \leq t_n$$ for $$n\geq 0$$
3. $$B_t-B_s\sim B_{t+h}-B_{s+h}$$ for $$0\leq s< t$$ and $$h\geq -s$$
4. $$B_t-B_s\sim \mathcal{N}(0,t-s)$$
5. $$t\mapsto B_t$$ 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.

• I think conditions $(3)$ and $(4)$ are to be taken 'together', since only in $(3)$ we have the specification about $s$ Jun 22, 2022 at 16:59

Condition 2 ensures the increments $$B_{t_n}-B_{t_{n-1}},...,B_{t_1}-B_{t_0}$$ are jointly Gaussian, which is needed for you to use that having $$0$$ covariance implies independence. I think that Condition 3 is redundant with Condition 4, though, since Condition 4 clearly implies Condition 3.