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I am reading a number of papers by different authors which are introductions to stochastic differential equations. All of these papers define the Wiener process $W_t$ (Brownian motion) quite simply by a few properties such as

  • $W_0=0$ with probability 1
  • $E(W_t)=0$
  • $Var(W_t-W_s) = t-s$

where $W_t$ is a family of real random variables indexed by the set of nonnegative real numbers $t$.

Most of them go on to employ $W_t$ in a stochastic differential equation in which some probability space $(\Omega,{\mathbb A},P)$ is implicitly assumed in the sense that the SDE is defined pathwise for $\omega \in \Omega$ in a form such as

$$dX_t(\omega)=f(t,X_t(\omega)) dt + g(t,X_t(\omega)) dW_t(\omega)$$

Discussion then proceeds without explicitly constructing the probability space containing paths $\omega$. Searching for "probability space of Wiener process" gives nothing. Searching for "probability space of Brownian motion" gives many resources, such as this one, which, as an exercise, states that "the collection of random variables $(\prod_t)_{t\in{\mathbb R}_+}$ defined on the probability space $(C[0,\infty),{\mathbb B}(C[0,\infty)),\mu)$ is a Brownian motion". It seems like quite a lot of machinery in complex analysis and measure theory is required to construct the space.

Question: What is the simplest and most intuitive, easiest-to-explain construction of the probability space of paths $\omega$ invoked but not defined in the typical presentation of the basic form of an SDE? I am trying to explain this to myself and others. I'm hoping for an explanation which, while not pretending that measure theory and complex analysis don't exist, makes minimal use of measure theoretic constructions to define the space. I'm not looking for proofs, just a construction of the space which would be invoked in a proof.

Possible directions:

  • The MIT notes say "the sample space $\Omega$ is barely mentioned because we can identify $\omega \in \Omega$ with $B_\omega$ a continuous function". Beyond the notational confusion, this would make $\Omega$ a function space.
  • In this question, $\Omega$ is understood to be any uncountable set, such as the real numbers, as an index into that function space, to be constructed, in one of several ways (Fourier series, random walks, and maybe more).
  • This Math StackExchange question also expresses confusion about $\Omega$ and the accepted answer "gets rid of" $\Omega$.
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  • $\begingroup$ You can consider the historical approach of Wiener by random Fourier series (math.mit.edu/~jerison/103/handouts/brownian.13.pdf) $\endgroup$
    – Jean Marie
    Nov 16, 2019 at 8:16
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    $\begingroup$ The MIT handout says "the sample space $\Omega$ is rarely mentioned in probability theory", hence my question. Also the handout goes for $B_\omega(t)$ whereas other notes go for $B_t(\omega)$ so presentation is not standardized at all across the literature. $\endgroup$ Nov 16, 2019 at 13:24

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I'm still in the process of learning this myself, so I may well make mistakes here, but since no one else seems to be interested in answering the question, I'll give it a shot (based in part on the answer you cited yourself). Also, this is anything but rigorous!

Wiener processes are often motivated by taking the limit $$ \lim_{n\to\infty} W_n(t) = \lim_{n\to\infty} \frac{1}{\sqrt{n}} \sum_{1\leq k \leq \lfloor nt \rfloor} \xi_k,$$ where $\xi_k$ are i.i.d. random variables with mean 0 and variance 1. If we impose no further restrictions on $\xi_k$, then they might each be any real number, i.e. $\xi_k \in \mathbb{R}$. Further, we can require each $\xi_k$ to be the identity map on the underlying probability space $\Omega_k$, which induces a probability distribution $P_k$ of $\Omega_k \ni \omega_k$ via $P^{\xi_k}$. Then we can identify the elements $\omega_k$ as precisely the numbers in $\mathbb{R}$, i.e. $\Omega_k \simeq \mathbb{R}$.

Now, since $W_n(t)$ depends on $n$ different $\xi_k$, via the identifications we just made, the underlying probability space of $W_n(t)$ must be $\mathbb{R}^n$. At this point you can note that the underlying probability space of the continuous Wiener process $W(t)$ cannot be $\mathbb{R}$: The cardinality of that set would be far too small.

In the following, I'll draw a heuristic comparison to the way physicists usually treat the transition from finite-dimensional Hilbert spaces to infinite-dimensional Hilbert spaces (see $^\ast$): Intuitively, in taking the limit $n\to\infty$, we must somehow get to "$\mathbb{R}^\infty$". By the fact that the incremental distribution of the Wiener process shall be $$W(t+\delta t) - W(t) \sim \mathcal{N}(0, \delta t),$$ we see that for $\delta t \to 0$, the probability of "large" jumps becomes vanishingly small. Therefore, the probability of discontinuities occurring also becomes vanishingly small, and the correct way of obtaining "$\mathbb{R}^\infty$" in this case should be the set of (a.e.$^{\ast\ast}$) continuous functions. Since $t$ is typically allowed to take values from 0 to infinity, we can identify the probability space of the Wiener process as $\Omega = C[0,\infty)$. In the posts you linked to, this is called the canonical realization of Brownian motion, since, along the way, we implicitly defined $W(t,\cdot)$, considered as a map from the probability space to another measurable space, to be the identity map on $\Omega = C[0,\infty)$.


$^{\ast}$In introductions to quantum physics, infinite-dimensional Hilbert spaces are often introduced in the same way. Then the appropriate infinite-dimensional "limit" turns out to be the space of square-integrable functions $L^2(\mathbb{R})$, or its extension as a rigged Hilbert space.


$^{\ast\ast}$Concerning continuity vs. a.e. continuity (see also this quant. finance post), I think that's largely a matter of definition and construction. The construction I gave above does allow for discontinuities, but the probability of such a discontinuous path occurring is vanishing. Suppose $\omega$ is such a path with a discontinuity at $t_0$. Then $$W(t_0+\delta t) - W(t_0) \overset{!}{\sim} \mathcal{N}(0,\delta t),$$ and therefore the corresponding density function is $$p(\delta x) \propto \exp\left(-\frac{\left(\delta x\right)^2}{2\delta t}\right),$$ but since the discontinuity means that $\left|\delta x\right| > 0$ even as $\delta t \to 0$, it follows that $p(\delta x) \to 0$ and thus the probability of this particular path $\omega$ occurring also vanishes, and we are free to ignore any discontinuous paths.

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  • $\begingroup$ Thanks for working on this. You might like @Oliver Diaz answer to a more recent question I posed. He shows how to get an infinite sequence of any distribution from a single random number and then gives a probability space for any stochastic process. math.stackexchange.com/questions/3760724/… $\endgroup$ Sep 1, 2020 at 20:34

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