Unfortunately I could not make it to the last probability theory lecture and now I am reading through the notes and have some troubles understanding what is going on.

So we defined Brownian motion as a stochastic process on $[0,\infty)$ such that

1.) $t \mapsto X_t(\omega)$ is a.s. continuous 2.) $X_t$ has stationary $X_{t_i}-X_{t_{i-1}} \sim N(0,t_i-t_{i-1})$ and independent increments where $X_t \sim N(0,t).$

Now, we are following the book "Continuous Time Markov Processes" and I refer to the beginning of Chapter 1.7.

It says: It is most convenient in the case of Brownian motion to take the probability space $\Omega$ to be the space $C[0,\infty)$ of all continuous functions $\omega(.)$ on $[0,\infty).$ This choice is natural because Brownian paths are continuous. The process is defined by $X(t,\omega) = \omega(t).$

The $\sigma$- algebra $F$ is taken to be the smallest one for which the projection $\omega \mapsto \omega(t)$ is measurable for each $t$. Rather than having one probability measue on $(\Omega,F)$ we now have a family $(P^x)$ of probability measures indexed by $x \in \mathbb{R}$. The probability measure is the distribution of $x+B(.),$ where $B$ is standard Brownian motion.

This is all a little bit confusing to me. So far I regarded Brownian motion as a map $X: \Omega \rightarrow \big(C[0,\infty), B(C[0,\infty))\big)$ where $B$ is the Borel sigma algebra on $C[0,\infty)$ and $\Omega$ was a space that I did not really care about.

But now they seem to be changing $\Omega$ which is fairly bizarre to me. Does anybody understand what they want to do and from where to where (in particular equipped with which sigma algebra and measures) they want the brownian motion to go? As it could still be that they want to denote something different with omega, I wanted to ask the experts here.

• The pair of comments there is related. – Did Oct 27 '15 at 9:43
• @Did but the first comment also says that $\Omega$ is left untouched. This actually confuses me even more, as they are talking about $\Omega$ in this book. The only thing I woud actually need is to see from where to where my Brownian motion is supposed to map according to this book. Do you think you could help me with that? – user167575 Oct 27 '15 at 9:52
• As you say, one possible choice, which is the one your source is advocating, is to define the Brownian motion $X$ as the identity map on $C[0,\infty)$ with the suitable sigma-algebra and the suitable probability measure. That is, $X=(X_t)$ with $X_t(\omega)=\omega(t)$ for every $t$. This is called the canonical realization of Brownian motion and $C[0,\infty)$ with the suitable sigma-algebra and the suitable probability measure is the canonical space for Brownian motion. Thus, $X(\omega)=\omega$ and each $X_t$ maps $C[0,\infty)$ to $\mathbb R$. Is this your question? – Did Oct 27 '15 at 10:01
• (All modesty aside, I was referring to my comments on the other page.) – Did Oct 27 '15 at 10:02
• yes, this was really confusing me, thank you. – user167575 Oct 27 '15 at 10:24

There are at least two concepts that you might want to consider separately, before you look at this definition again: the canonical process, and time continuous Markov process. They are changing $\Omega$ because they want the process to be canonical, and they define a whole bunch of probability measures, because this is necessary to define a Markov process.

Just go step by step through the definition:

• $C[0,\infty)$ is just a non-empty set, which is a perfectly fine carrier set for a measurable space.
• For each $t$, the projection $X_t: C[0,\infty) \to \mathbb{R}$, defined by $X_t(f) = f(t)$ is a well-defined function. You also know that $(\mathbb{R}, \mathfrak{B}(\mathbb{R}))$ is a measurable space. Thus, you can define a $\sigma$-algebra $\mathcal{F}$ on $C[0,\infty)$ as the initial $\sigma$-algebra induced by the family of functions $\{X_t\}_t$. Now you have a measurable space $(C[0,\infty), \mathcal{F})$.
• Now, for each starting point $x\in\mathbb{R}$, you define a probability measure $P^x$ as the law of $t\mapsto x + B(t)$, that is, the law of a Brownian motion started at point $x$. So now, you have a measurable space $(C[0,\infty), \mathcal{F})$ and a bunch of probability measures $\{P^x\}_x$ on this space.
• By construction, each projection $X_t: C[0,\infty) \to \mathbb{R}$ is measurable, thus $(X_t)_t$ is a stochastic process on $(C[0,\infty), \mathcal{F}, P^x)$ for each starting point $x\in\mathbb{R}$. The process $(X_t)_t$ is called the canonical process, because it's built from the canonical projections. By construction, $P^x[X_0 = x] = 1$, this is the first property in the definition of a Markov process that one has to verify.
• Depending on how exactly one constructed paths of a Brownian motion, one can more or less easily show that the whole construction indeed constitutes a Markov process $(X_t)_t$ with distributions $(P^x)_{x\in\mathbb{R}}$ on the space $(C[0,\infty), \mathcal{F})$.

The process $(X_t)_t$ with distributions $(P^x)_{x\in\mathbb{R}}$ on the space $(C[0,\infty), \mathcal{F})$ is the Brownian motion explicitly written down as a time-continuous Markov process.

Here is a simpler construction that might help to understand why we can get rid of $\Omega$ and make everything canonical: Suppose you have a probability space $(\Omega, \mathcal{A}, P)$, a measurable space $(\Psi, \mathcal{B})$, and an $\mathcal{A}-\mathcal{B}$-measurable random variable $X$. This variable induces a measure on $(\Psi, \mathcal{B})$ as follows: $B \in\mathcal{B} \mapsto P(X^{-1}(B))$. Thus, $(\Psi, \mathcal{B}, P\circ X^{-1})$ also becomes a probability space. The identity function $Id: \Psi \to \Psi$ is trivially measurable, thus we can interpret the identity function $Id$ as a $\Psi$-valued random variable. Furthermore, notice that $Id$ and $X$ have the same law, because $P[X\in B] = P[X^{-1}(B)] = (P\circ X^{-1})[B] = (P\circ X^{-1})[Id^{-1}(B)] = (P\circ X^{-1})[Id \in B]$. Therefore, we can actually get rid of the original $\Omega$, forget $X$, forget $P$, and instead build a new probability measure on $(\Psi, \mathcal{B})$ and use the identity function on $\Psi$ instead of $X$.

• very interesting and very helpful! So the Brownian motion comes via the $(P^x)_x$ into the game. $P^x[X_0=x] = P(X_0(x+B(t))=x) = P(x+B(0)=x)=1$ as Brownian motion starts at zero, right? Now talking about this sigma algebra $\mathcal{F},$ is this the same sigma algebra as the canonical Borel sigma algebra induced by the supremum norm or is this sigma-algebra different? – user167575 Oct 27 '15 at 10:23
• Yes, it comes in through the $(P^x)_x$. I assume that it has already been constructed previously (e.g. by Levy's construction), the quote in your question is concerned with "re-packaging" the already constructed paths of Brownian motion so that it fits into the time-continuous Markov-processes framework. I added another paragraph to show how one can "forget annoying implementation details" in a simpler case. What about the sigma-algebra: see this question: math.stackexchange.com/questions/1455735/… – Andrey Tyukin Oct 27 '15 at 11:27