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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) sigma algebra agree and thuscan 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. ButNow, 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.

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 $x.$ 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.

Now, as you can see in the proof of theorem 5.9 the two sigma algebras (tail and germ) sigma algebra agree and thus, the Kolmogorov 0-1 law for Brownian motion is 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. But since the two sigma algebras agree (according to the proof given in the link), this does not make sense.

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.

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.

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user82546
user82546
  • 111
  • 4

Blumenthal and Kolmogorov 0-1 law

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 $x.$ 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.

Now, as you can see in the proof of theorem 5.9 the two sigma algebras (tail and germ) sigma algebra agree and thus, the Kolmogorov 0-1 law for Brownian motion is 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. But since the two sigma algebras agree (according to the proof given in the link), this does not make sense.

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.