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

  1. $W_0=0$,

  2. for times $ 0 \leq t_1 \leq t_2 <...< t_n <T$ the random variables $W_{t_2} - W_{t_1}$, $W_{t_3}-W_{t_2}$,...,$W_{t_n}-W_{t_{n-1}}$ are independent.

  3. For any $0\leq s \leq t <T$ the increment $W_{t} - W_{s}$ has the Gaussian distribution with mean 0 and variance $t-s$.

  4. 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 $f:\,\mathbb{R}^{n}\rightarrow\mathbb{R}$ .

  1. 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?
  • $\begingroup$ There are several very different ways to construct Brownian motion. For example, Paul Lévy viewed it through a sequence of dyadic continuous gaussian linear interpolating approximations. Hence you might want to write down the four properties you have in mind. $\endgroup$ – Did Feb 22 '12 at 6:10
  • $\begingroup$ I understand. But, my primary difficulty here is computational: how does one even show E[f(Wt1,...,Wtn)]=E[f(W′t1,...,W′tn)] $\endgroup$ – Raghavan Feb 22 '12 at 6:11
  • $\begingroup$ In some constructions, the value of these marginals is one of the defining properties. $\endgroup$ – Did Feb 22 '12 at 6:12
  • $\begingroup$ Finite dimensional distributions are defined as $$ \nu_{t_1,\dots,t_n}(B_1,\dots,B_n) = \mathsf P\{X_{t_1}\in B_1,\dots,X_{t_n}\in B_n\} $$ and you may find it easier to work with such formulations. However it depends on four properties you have in mind - please write them in your question. $\endgroup$ – Ilya Feb 22 '12 at 6:19
  • $\begingroup$ Ok i made those properties more explicit $\endgroup$ – Raghavan Feb 22 '12 at 6:41

Let me give you some hints. To recover finite-dimensional distributions of the real-value process $X$ it is sufficient (and necessary) to find all the probabilities of the form $$ \mathsf P\{X_{t_1}\in B_1,\dots,X_{t_n}\in B_n\} $$ for any sequence of non-negative reals $t_k$ and Borel subsets of real line $B_k$.

In fact, only 3 first properties are needed to do it. Indeed, let us define $$ p(t,x,y) = \frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(y-x)^2}{2t}\right) $$ for $t>0, x,y\in\mathbb R$ and $p(0,x,y) =\delta_x(y)$. As a result, for $W$ first three properties imply that $p$ is the density function of $W$: $W_t\sim p(t,0,y)$. Clearly, it holds as well for $W'$ since only these 3 properties were used.

Take any sequence of non-negative $t_1,\dots, t_k$ and arrange it in the order of ascending: $0\leq t_{\sigma(1)}\leq t_{\sigma(2)}\leq\dots t_{\sigma(n)}$ where $\sigma$ is an appropriate permutation of $\{1,\dots,n\}$.

Then $$ \mathsf P\{W_{t_1}\in B_1,\dots,W_{t_n}\in B_n\} = \mathsf P\{W_{t_{\sigma(1)}}\in B_{\sigma(1)},\dots,W_{t_{\sigma(n)}}\in B_{\sigma(n)}\} = $$

$$ = \int\limits_{B_{\sigma(1)}}\dots\int\limits_{B_{\sigma(n)}}p(t_{\sigma(1)},0,x_1)p(t_{\sigma(2)}-t_{\sigma(1)},x_1,x_2)\dots p(t_{\sigma(n)} - t_{\sigma(n-1)},x_{n-1},x_n)dx_1\dots dx_n. $$

Clearly, the same expression you obtain for $$ \mathsf P\{W'_{t_1}\in B_1,\dots,W'_{t_n}\in B_n\} $$ as well, since we used only 3 properties of a Brownian motion which are shared by both $W$ and $W'$. Hence the finite-dimensional distributions are the same as needed. If you still want to show that expectations are the same just note that if the probability measure $\nu$ on $\mathbb R^n$ is the same as probability measure $\nu'$ on the same space if and only if for any bounded and meausrable $f:\mathbb R^n\to\mathbb R$ it holds that $$ \int\limits_{\mathbb R^n}f(x)\nu(dx) =\int\limits_{\mathbb R^n}f(x)\nu'(dx). $$


Thank to Il y a, if we consider the Wiener space: $$ \mathbb{W}^{\otimes d}=\left( \mathcal{C} ( [0,T] , \mathbb{R}^d ), (\mathcal{B}_t)_{0 \leq t \leq T}, \mathbb{P} \right) $$ where:

  1. $\mathcal{C} ( [0,T] , \mathbb{R}^d)$ is the space of continuous functions $ [0,T] \rightarrow \mathbb{R}^d$;

  2. $\left( \omega_{t}\right) _{t\geq 0}$ is the coordinate process defined by $\omega_{t}(f)=f\left( t\right) $, $f \in \mathcal{C} ( [0,T] , \mathbb{R}^d )$;

  3. $\mathbb{P}$ is the Wiener measure as shown above;

  4. $(\mathcal{B}_t)_{0 \leq t \leq T}$ is the ($\mathbb{P}$-completed) natural filtration of $\left( \omega_{t}\right)_{0 \leq t \leq T}$.

Then the cordinate process $\left( \omega_{t}\right) _{t\geq 0}$ define a standard Brownian motion.


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