Blumenthal's 0-1 law see theorem 5.8/5.9 tells us that an event in the germ $\sigma-$ algebra has either probability zero or one with respect to a measure induced by a Brownian motion starting in some point $x.$ (Theorem 5.8)

Conversely, one can show that this is also the case for events in the tail $\sigma-$ algebra. This is not the standard Kolmogorov 0-1 law, but I want to call it that way, as it is also name in the used in the link. (Theorem 5.9)

Now, as you can see in the proof of theorem 5.9, that the two sigma algebras (tail and germ) can be mapped onto each other. Thus, the Kolmogorov 0-1 law for Brownian motion is in some sense the same as Blumenthal's 0-1 law.

My question is now the following: I am almost sure, that the probability $P^x(A)$ does not depend on $x$ for events in the tail sigma algebra, but afaik it is in general dependend on $x$ in Blumenthal's 0-1 law. Now, since the two sigma algebras agree (according to the proof given in the link), this does not make sense. Either both should depend on $x$ or they are both independent.

Does anybody know where my reasoning is wrong? I have been thinking about this for quite a time now but since I am not a probability theorist, I thought that somebody who is familiar with these things can easily help me out here.

  • $\begingroup$ "The two $\sigma$-fields agree" - careful with that word "agree". What's constructed is a map from one to the other. It doesn't preserve the measure $P^x$, unless $x=0$. $\endgroup$ Nov 9 '15 at 7:52

The issue, informally, is this. Suppose $W_t$ is a Brownian motion started at some $x \in \mathbb{R}^d$. Let $X_{t} = t W_{1/t}$ as in the linked notes. Then $X_t$ is a Brownian motion, but started at 0, not $x$. (By the strong law of large numbers, we have $\lim_{t \to 0} X_t = \lim_{s \to \infty} \frac{1}{s} W_s = 0$ almost surely.)

To see the issue more formally, it's helpful to work with an explicit sample space $\Omega$. Let's take $\Omega = C((0, \infty), \mathbb{R}^d)$, to be the set of all continuous $\omega : (0, \infty) \to \mathbb{R}^d$, equipped with the Borel $\sigma$-algebra coming from the topology of uniform convergence on compact sets. For each $t > 0$, let $W_t : \Omega \to \mathbb{R}^d$ be simply $W_t(\omega) = \omega(t)$. And for each $x$, let $P^x$ be the unique probability measure on $\Omega$ under which $\{W_t\}$ is a Brownian motion started at $x$.

Now consider the map $\Phi : \Omega \to \Omega$ given by $(\Phi\omega)(t) = t \omega(1/t)$. I agree with your claim that $\Phi$ maps the tail $\sigma$-algebra to the germ $\sigma$-algebra and vice versa. But it sounds like you are thinking that $\Phi$ also preserves all the measures $P^x$, and that is not true. What is true is that $P^x(\Phi^{-1}(A)) = P^0(A)$ for all events $A$.

The Kolmogorov 0-1 law says that for any $x$, the tail $\sigma$-algebra is $P^x$-almost trivial. So a consequence is that the germ $\sigma$-algebra is $P^0$-almost trivial. In other words, Kolmogorov for $P^x$ directly implies Blumenthal for $P^0$ only. If you want to prove Blumenthal for $P^x$, you need one more (easy) step; say, that $P^x$ is the pushforward of $P^0$ under translation by $x$, and this translation preserves the germ $\sigma$-algebra.

  • $\begingroup$ @user82546: The tail and germ $\sigma$-algebras of $\Omega$ do not coincide as sets! What they show is that the tail $\sigma$-algebra of $W_s$ coincides with the germ $\sigma$-algebra of $X_s$. $\endgroup$ Nov 10 '15 at 0:43
  • $\begingroup$ thanks, I got your point and now I understood your answer. thank you very much. $\endgroup$
    – user82546
    Nov 10 '15 at 0:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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