0
$\begingroup$

In M\"orters and Peres' book Brownian Motion, in constructing Brownian motions, they wrote:

We have thus constructed a continuous process $B: [0, 1] → \mathbb R$ with the same finite dimensional distributions as Brownian motion. Take a sequence $B_0,B_1, . . .$ of independent $C[0, 1]$-valued random variables with the distribution of this process, and define $\{B(t): t \geq 0\}$ by gluing together the parts.

My question is: Why does the sequence $B_0,B_1,\dots$ exist? I only know that we can construct an i.i.d sequence of real valued random variables, but I don't know that can be done with $C[0,1]$-valued random variables.

$\endgroup$
1
  • $\begingroup$ The $B_i$ are defined as the uniform limit of piecewise linear functions. See e.g. "Fig. 1.2. The first three steps in the construction of Brownian motion" in the text. $\endgroup$
    – Math1000
    Commented Apr 8, 2018 at 8:52

2 Answers 2

Reset to default
1
$\begingroup$

If $\mu$ is the Borel measure on $C[0,1]$ induced by $B:[0,1] \to \mathbb R$ then the coordinate maps on the infinite product $(C[0,1],\mu) \times (C[0,1],\mu) \cdots $ will serve as the sequence $\{B_n\}$. There is no need to define $B_n$'s on the same space as the one on which $B$ is defined.

$\endgroup$
2
  • $\begingroup$ Thanks! Now I have another question. Exercise 1.2 of the book asks us to show that $B(t)$ is jointly measurable w.r.t. $t$ and $\omega$. This is easy if we just look at the original definition of $B:[0,1]\to \mathbb R$. But how could we show joint measurability for $B:[0,\infty)\to \mathbb R$? $\endgroup$
    – Tongou Yang
    Commented Apr 8, 2018 at 19:55
  • $\begingroup$ Thanks! I have found my answer here almostsure.wordpress.com/2009/11/03/… $\endgroup$
    – Tongou Yang
    Commented Apr 9, 2018 at 0:55
0
$\begingroup$

Actually, there is another, arguably more concrete way to think about it. It shows that there is Brownian motion on $[0,1]$ with Lebesgue measure.

The Lévy construction can be carried out on $[0,1]$ with Lebesgue measure. By the Lévy construction, you obtain a Brownian motion with time index set $[0,1]$. In other words, there is a measurable map $f: [0,1] \rightarrow \mathbb R^{[0,1]}$ such that $f$, seen as random variable, is a Brownian motion. But then, for any uniformly distributed random variable $U$ on $[0,1]$, $f(U)$ is a Brownian motion with time index set $[0,1]$.

Now, it is a well-known fact that on a given prob. space there is a uniform iff there is an i.i.d. sequence of Bernoullis, and each infinite countable set admits a countable partition of countable subsets. Hence, on $[0,1]$ with Lebesgue measure there is an infinitely countable i.i.d. sequence of uniformly distributed random variables, say $(U_n)_{n\in\mathbb N}$. Hence, $B_n := f(U_n)$, $n\in\mathbb N$, defines the required sequence of i.i.d. Brownian motions on a compact time interval.

Finally, for $n\in\mathbb N$, and $t\in [n,n+1)$, set $$B(t):= \sum_{k=0}^n B_k(\min(t-k,1)) = B_0(1) + \dots + B_{n-1}(1) + B_n(t-n).$$ $B$ is a Brownian motion with time index set $\mathbb R_+$ on $[0,1]$ with Lebesgue measure.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .