Time inversion of Brownian motion

[InvisibleBlue]

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:
$$\[ X_{t}-X_{s}=X_{t-s} \quad \forall \quad 0\leq s<t \]$$
Anyone has any ideas?

 

[gel]

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.

 

[InvisibleBlue]

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.

 

[InvisibleBlue]

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?

 

[gel]

Problem is that the notion of Gaussian Process is not introduced or used in this course.

doesn't seem very good.

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.

$$B_{t} - B_{s}$$ ~ $$N(0, t+s)$$ 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.

 

[gel]

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.

 

[InvisibleBlue]

aahhhh! I completely forgot about the whole independence issue!

Thanks a lot. This really helped!