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Time inversion of Brownian motion

  1. Dec 22, 2008 #1
    Hi,

    I'm trying to prove that [tex]X=(X_{t})_{t\geq0}[/tex] is a Brownian Motion, where [tex]X_{t} = tB_{1/t}[/tex] for [tex]t\neq0[/tex] and [tex]X_{0} = 0[/tex]. I don't want to use the fact that it's a Gaussian process. So far I am stuck in proving:
    [tex]\[
    X_{t}-X_{s}=X_{t-s} \quad \forall \quad 0\leq s<t
    \]
    [/tex]
    Anyone has any ideas?
     
  2. jcsd
  3. Dec 23, 2008 #2

    gel

    User Avatar

    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.
     
  4. Jan 4, 2009 #3
    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.
     
  5. Jan 4, 2009 #4
    Actually, I'm now suddenly very confused. We say that [tex]B_{t} - B_{s} = B_{t-s}[/tex] for [tex]B = (B_{t})_{t\geq0}[/tex] brownian motion. So this must mean that they have the same distribution. But if they are standard brownian motion (i.e [tex]B_{t}[/tex] ~ [tex]N(0,t)[/tex]) then we get that

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

    clearly not equally distributed.

    Am I missing something here?
     
  6. Jan 4, 2009 #5

    gel

    User Avatar

    doesn't seem very good.

    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.

    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.
     
  7. Jan 4, 2009 #6

    gel

    User Avatar

    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.
     
  8. Jan 8, 2009 #7
    aahhhh! I completely forgot about the whole independence issue!

    Thanks a lot. This really helped!
     
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