Time inversion of Brownian motion

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Discussion Overview

The discussion revolves around the properties of a process defined as X_{t} = tB_{1/t} for t ≠ 0 and X_{0} = 0, with a focus on proving that it is a Brownian Motion without relying on its characterization as a Gaussian process. Participants explore the implications of certain properties of Brownian motion, particularly the relationship between increments and their distributions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to prove that X_{t}-X_{s}=X_{t-s} for 0 ≤ s < t without using the Gaussian process definition.
  • Another participant argues that X_{t}-X_{s} is not equal to X_{t-s}, although they have the same distribution, suggesting that covariance should be calculated to conclude that X is a Brownian Motion.
  • A later reply questions the necessity of avoiding the Gaussian process definition, stating that Brownian motion is inherently a Gaussian process.
  • Confusion arises regarding the distribution of increments, with a participant noting that while B_{t}-B_{s} = B_{t-s} holds for standard Brownian motion, the distributions do not match under certain conditions.
  • Another participant clarifies that the equality of distributions is often misrepresented as equality of the random variables themselves, emphasizing the importance of independence in the context of normal distributions.
  • One participant lists key properties of Brownian motion, including the distribution of increments and their independence, suggesting these can be used to establish the characteristics of the process in question.
  • A participant expresses realization about the importance of the independence of increments in understanding the properties of Brownian motion.

Areas of Agreement / Disagreement

Participants exhibit disagreement regarding the necessity of the Gaussian process definition and the interpretation of distributional equality versus equality of random variables. The discussion remains unresolved as participants explore different aspects of the properties of Brownian motion.

Contextual Notes

There is a noted absence of the Gaussian process concept in the course material, which may limit the discussion's scope. Additionally, the independence of increments and the rules for adding normal distributions are points of contention that remain unresolved.

InvisibleBlue
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Hi,

I'm trying to prove that X=(X_{t})_{t\geq0} is a Brownian Motion, where X_{t} = tB_{1/t} for t\neq0 and X_{0} = 0. I don't want to use the fact that it's a Gaussian process. So far I am stuck in proving:
\[<br /> X_{t}-X_{s}=X_{t-s} \quad \forall \quad 0\leq s&lt;t<br /> \]<br />
Anyone has any ideas?
 
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X_t-X_s is not equal to X_{t-s}, although they do have the same distribution.

Just calculate the distribution of X_t-X_s, and show that X_t-X_s, X_u have zero covariance for u <= s <= t. You should be able to conclude that X is a BM from that.

Edit: missed the bit where you said that you don't want to assume that its a Gaussian process. Why not? BM is defined as a Gaussian process, and you must use something.
 
gel said:
X_t-X_s is not equal to X_{t-s}, although they do have the same distribution.

Just calculate the distribution of X_t-X_s, and show that X_t-X_s, X_u have zero covariance for u <= s <= t. You should be able to conclude that X is a BM from that.

Edit: missed the bit where you said that you don't want to assume that its a Gaussian process. Why not? BM is defined as a Gaussian process, and you must use something.

So You think it's enough to show that those 2 have the same distribution?

Problem is that the notion of Gaussian Process is not introduced or used in this course. I guess I just have to use the properties without saying where it comes from.
 
Actually, I'm now suddenly very confused. We say that B_{t} - B_{s} = B_{t-s} for B = (B_{t})_{t\geq0} brownian motion. So this must mean that they have the same distribution. But if they are standard brownian motion (i.e B_{t} ~ N(0,t)) then we get that

B_{t} - B_{s} ~ N(0, t+s) and B_{t-s} ~ N(0, t-s)
(this is by the rules for adding and subtracting normal distributions)

clearly not equally distributed.

Am I missing something here?
 
InvisibleBlue said:
Problem is that the notion of Gaussian Process is not introduced or used in this course.

doesn't seem very good.

InvisibleBlue said:
Actually, I'm now suddenly very confused. We say that B_{t} - B_{s} = B_{t-s}

Ok, you don't mean that they are equal. Just that they have the same distribution. Sometimes equal distributions are expressed by an = sign with a 'd' above or below it. Just saying B_t-B_s = B_{t-s} says that they are actually equal, which is wrong.

InvisibleBlue said:
B_{t} - B_{s} ~ N(0, t+s) and B_{t-s} ~ N(0, t-s)
(this is by the rules for adding and subtracting normal distributions)

What rules for adding normals are you referring too? There are no general rules - just rules for adding independent normals, which you seem to be using (but B_t,B_s are not independent) and rules for adding joint normals.
 
I think you to need to know at least the following,

1) B_t-B_s has N(0,t-s) distribution, and is independent of {B_u:u <= s}
2) independent normals are joint normal.
3) linear combinations of joint normals are joint normal.
4) joint normals with 0 covariance are independent.

You should be able to show that
(1) uniquely defines all finite distributions of BM
(1)+(2)+(3) => BM is joint normal at different times.

and, with (4), you should be able to answer your question.
 
aahhhh! I completely forgot about the whole independence issue!

Thanks a lot. This really helped!
 

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