Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time inversion of Brownian motion

  1. Dec 22, 2008 #1

    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:
    X_{t}-X_{s}=X_{t-s} \quad \forall \quad 0\leq s<t
    Anyone has any ideas?
  2. jcsd
  3. Dec 23, 2008 #2


    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


    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


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook