I learned that Brownian motion is any stochastic process $(W_{t})_{t\geq0}$ that satisfies four well-known properties. But I still don't understand how these four properties uniquely determine Brownian Motion.
The four properties are
$W_0=0$,
for times $ 0 \leq t_1 \leq t_2 <...< t_n
the random variables $W_{t_2} - W_{t_1}$, $W_{t_3}-W_{t_2}$,...,$W_{t_n}-W_{t_{n-1}}$ are independent. For any $0\leq s \leq t
the increment $W_{t} - W_{s}$ has the Gaussian distribution with mean 0 and variance $t-s$. For all $\omega$ in a set of probability one, $B_t (w)$ is a continuous function of $t$.
More precisely,
Let $(W_{t})_{t\geq0}$ and (W_{t}')_{t\geq0} be two stochastic processes that satisfy the four properties of a Brownian motion. Prove that these stochastic processes must then have the same finite dimensional distributions, that is, E[f(W_{t_{1}},...,W_{t_{n}})]=E[f(W_{t_{1}}^{'},...,W_{t_{n}}^{'})] for any $n\geq1$ , $t_{1},...,t_{n}\geq0$ , and bounded measurable $f:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ .
- How do I go about showing this....I am having trouble with multivariate gaussians. 2. Does the Kolmogorov's Extension theorem play some sort of a role here?