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In Wiener's construction of Brownian Motion, it is assumed that there exists a probability space $(\Omega,\mathcal F,\mathbb P)$ and random variables $X_n:\Omega\rightarrow\mathbb R$ for $n\in\mathbb N,$ which are independent and normally distributed. Does anyone know what this probability space actually is and why it exists? Is it unique? Can Brownian Motion be constructed on any measurable space? What role plays then the probability measure $\mathbb P$?

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2 Answers 2

up vote 3 down vote accepted

Let $(\Omega,\Sigma,\mu)$ be an atomless probability space and $P$ be a Polish space, endowed with the Borel $\sigma$-algebra. Let $\nu$ be any measure on $P$. Then there exists a random variable $f:\Omega\to P$ such that $\mu\circ f^{-1}=\nu$. Since $\mathbb{R}^{\mathbb{N}}$ is Polish, the result follows.

The result mentioned above is well-known, but I don't know a simple proof. It suffices to show it for $P=\mathbb{R}$, since by the Borel isomorphism theorem, all uncountable Polish spaces are isomorphic as measurable spaces. There is a standard construction in many probability books for obtaining any measure on $\mathbb{R}$ as the distribution of a random variable on $[0,1]$ from its cumulative distribution function. To show that one can get the uniform distribution on $[0,1]$ as the distribution of a random variable from any atomless probability space, one can use the fact that one can partition an atomless probability space always into two measurable set of equal measure, and iterate this to get a "tree" of measurable sets. From this one can construct an appropriate random variable, but the details are a bit messy I think.

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For simplicity consider the case of all the variables independent and following $\mathcal{N}(0,1)$. You can consider the space $(\Omega, \mathcal{F}, P) = (\mathbb{R}^\mathbb{N}, \mathcal{B}(\mathbb{R}^\mathbb{N}), \mu)$, where $\mu$ is defined by the finite dimensional distributions $$\mu_{i_1, \ldots, i_k}(A) = \int_A \frac{1}{\sqrt{2\pi}^k} e^{\frac{-\sum_{j=0}^k x_j^2}{2}} dx_1 \ldots dx_j \quad A \in \mathbb{R}^k$$

That is, the finite dimensional laws are multivariate normal random variables with independent components each following $\mathcal{N}(0,1)$.

The existence of the measure $\mu$ is a consequence of the Kolmogorov's consistency theorem which assures that given a set of consistent finite dimensional laws the measure $\mu$ will exist in the infinite product space and it will be consistent with the specified finite dimensional laws.

Explicitly being consistent is that on the product space the measure of a cilinder $$ C = A_{i_1} \times A_{i_2} \cdots \times A_{i_k} \times \mathbb{R}^{\mathbb{N} - \{i_1 \, \ldots, i_k\}}$$ will be given by $$ \mu(C) = \mu_{i_1, \ldots, i_k}(A_{i_1} \times \cdots \times A_{i_k})$$ In plain english, every finite dimensional component will have the specified measure, in this case independent normal elements since $$ P(X_{i_1} \in A_{i_1}, \ldots, X_{i_k} \in A_{i_k}) = \mu(C)$$

You can even use this theorem directly to construct the Brownian Motion in $(\mathbb{R}^{[0, T]}, \mathcal{B}(\mathbb{R}^{[0, T]}))$ by specifying the finite dimensional distributions as normal with mean zero and the covariance structure of the BM. But in my opinion is one of the unfunnier ways to construct it ;).

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