Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Quoted from Wikepedia:

The Wiener process can be constructed as the scaling limit of a random walk, or other discrete-time stochastic processes with stationary independent increments. This is known as Donsker's theorem.

I was wondering if the scaling limit of any random walk is always a Wiener process, or just the scaling limits of some special kinds of random walks are, such as Gaussian random walk?

share|cite|improve this question
up vote 4 down vote accepted

In light of your other question, one should say there are a lot of things referred to by the name "random walk". In its simplest form, Donsker's theorem is about a process $X_n$ in $\mathbb{R}^d$ whose increments $X_n - X_{n-1}$ are iid with any distribution that has zero mean and finite variance. In particular, the increments do not have to be Gaussian. It then asserts that the scaling limit of the process $X_n$ is $d$-dimensional Brownian motion (aka Wiener process).

There are lots of other stochastic processes that have Brownian motion as their scaling limits, and still other processes that have a scaling limit that is not Brownian motion. Is there something specific you're interested in?

share|cite|improve this answer
Thanks, Dr Eldredge! So is this Donsker's theorem related to the central limit theorem in some way, since both in some sense do not put restriction on the specific distribution and the limit somehow involves Gaussian distribution? – Tim Nov 6 '10 at 16:41
@Tim: Absolutely. In a sense, Donsker's theorem is the central limit theorem for stochastic processes; you also hear it called a "functional central limit theorem". And it says that Brownian motion is an essential stochastic process to study, for the same reason that the Gaussian distribution is essential in finite dimensions. – Nate Eldredge Nov 6 '10 at 18:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.