I have some questions concerning the canonical construction of a Brownian motion and would be very happy if someone can help me.

In my probability lecture I have seen the following definitions:

Let $W = \{W_t : t \geq 0\}$ be a Brownian motion starting from $0$ defined on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and let $C = C(\mathbb{R}_+, \mathbb{R})$ be the space of real-valued continuous functions $$x: \mathbb{R}_+ \rightarrow \mathbb{R}, t \mapsto x(t).$$ $C$ is endowed with the topology induced by uniform convergence on compact sets. We denote by $\mathcal{C}$ the associated Borel $\sigma$-field. $\mathcal{C}$ is generated by cylindrical sets $$B = \{x \in C : x(t_1) \in A_1, \ldots, x(t_m) \in A_m\}$$, where $t_1, \ldots, t_m \in [0, \infty)$ and $A_1, \ldots, A_m \in \mathcal{B}(\mathbb{R})$.

My questions are:

  1. Why is the event $\{\omega \in \Omega : \{W_t(\omega) : t \geq 0\} \in B\}$ necessarily an element of $\sigma(W)$ for all $B \in \mathcal{C}$ and why does this imply that $\mathbb{W}(B) = \mathbb{P}(W \in B)$ on $(C, \mathcal{C})$ is a probability measure (called the Wiener measure) ?

  2. Does the definition of the Wiener measure depend on the initial choice of the Brownian motion $W$ ?

  3. I would like to prove that the canonical process $$X : C \rightarrow C : x = \{x(t) : t \geq 0\} \mapsto \{X_t(x) : t \geq 0\}$$, defined by $$X_t(x) = x(t)\ \forall\ x \in \mathbb{C}, \forall\ t \geq 0$$ is a Brownian motion. For this, I have to check the conditions of the definition of a Brownian motion. It is clear that for every fixed $\omega \in \Omega$, $t \mapsto X_t(\omega)$ is continuous. However, how do I show $X$ is Gaussian with zero mean and $$cov(X_t(x), X_s(x)) = \mathbb{E}[x(t)x(s)] = min(s, t) ?$$

Any hint or comment will be much appreciated.

Thanks for your help.

  • $\begingroup$ One small quibble: $m=1,...,t_m$ is confusing with multiple uses of $m$ possibly. $\endgroup$ – jdods May 10 '18 at 11:54
  • 1
    $\begingroup$ Question 1 sounds like $W_t\in B$ for all $t$, which sounds like a null set if $B$ is bounded. $\endgroup$ – jdods May 10 '18 at 11:58

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