I have a question concerning the definiton of Brownian motion. Usually (e.g. on Wikipdia) one demands a brownian motion $\lbrace B_t\rbrace_{t\in[0,\infty)}$ to satisfy the following condition:
$\forall 0\leq t_0<t_1<...<t_m: B_{t_1}-B_{t_0},...,B_{t_m}-B_{t_{m-1}}$ are independent random variables.
But today I come across with formally another definition of Brownian motion! Instead of the above condition the book demands the following property:
$\forall t,h\geq 0$, $B_{t+h}-B_t$ is independent of $\lbrace B_u : 0\leq u\leq t \rbrace$.
Now I'm wondering whether both properties are indeed equivalent or not?! Hopefully someone can help me.
Yes, they are indeed equivalent.
Proof: Let $X_1,\ldots,X_n$ arbitrary random variables and denote by
$$S_j := \sum_{k=1}^j X_k$$
the $j$-th partial sum. Since $$X_j = S_j - S_{j-1} \qquad \quad S_j = \sum_{k=1}^j X_k$$
it is not difficult to see that
$$\sigma(X_1,\ldots,X_n) = \sigma(S_1,\ldots,S_n).$$
If we set $X_j := B_{t_j}-B_{t_{j-1}}$ for a Brownian motion $(B_t)_{t \geq 0}$, then this shows
$$\sigma(B_{t_1},\ldots,B_{t_n}) = \sigma(B_{t_1}-B_{t_0},\ldots,B_{t_n}-B_{t_{n-1}}). \tag{1}$$
Consequently,
$$\begin{align*} \sigma\{B_u; u \leq t\} &\stackrel{\text{Def}}{=} \sigma \left( \bigcup_{\substack{t_1<\ldots<t_n \leq t \\n \in \mathbb{N}}} \sigma(B_{t_1},\ldots,B_{t_n}) \right) \\ &\stackrel{(1)}{=} \sigma \bigg( \underbrace{\bigcup_{\substack{t_1<\ldots<t_n \leq t \\n \in \mathbb{N}}} \sigma(B_{t_1}-B_{t_0},\ldots,B_{t_n}-B_{t_{n-1}})}_{=:\mathcal{G}} \bigg). \end{align*}$$
Since $\mathcal{G}$ is a $\cap$-stable generator of $\sigma(B_u; u \leq t)$, we conclude that $B_{t+u}-B_t$ is independent from $\sigma(B_u; u \leq t)$ if, and only if, $B_{t+h} -B_t$ is independent from $\mathcal{G}$. And this is equivalent to $B_{t+h}-B_t$ being independent from $B_{t_n}-B_{t_{n-1}},\ldots,B_{t_1}-B_{t_0}$ for all $t_1 < \ldots < t_n \leq t$.
Literature: See e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes (Lemma 2.14).