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Brownian Motion question

  1. Mar 6, 2012 #1
    I need to show that (Wt)2 is a brownian motion

    So let Vt = (Wt)2

    I need to first show that Vt+s - Vs ~ N(0,t)

    Vt+s - Vs = (Wt+s)2 - (Ws)2 = (Wt+s + Ws)(Wt+s - Ws)

    (Wt+s - Ws) ~ N(0,t)

    But is (Wt+s + Ws) ~ N(0,t)?

    If it is what happens when I multiply two RV's that are normally distributed together? What can I say about the variance of the new distribution?

    Thanks
     
  2. jcsd
  3. Mar 6, 2012 #2

    Ray Vickson

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    You are being asked to show something that is FALSE: W_t^2 is not a Brownian motion, unless you are using an unusual definition of Brownian motion. The square of a normal random variable is essentially a chi-squared random variable with parameter 1; that is N(0,1)^2 = Chi^2(1).

    RGV
     
  4. Mar 7, 2012 #3
    Hi Ray

    Thanks for responding!

    Yeah sorry, the question was show if it is or is not a Brownian Motion, for some odd reason I assumed it was because the rest of the examples were

    How do I show it is not a brownian motion then, which property of a BM does not hold?
    Thanks
     
  5. Mar 7, 2012 #4

    Ray Vickson

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    What properties of BM are you trying to verify or deny? I have seen alternative definitions of BM; which one are you using?

    RGV
     
  6. Mar 7, 2012 #5
    These:

    2q1updw.png

    Also in the question I should note that W is a standard brownian motion
     
  7. Mar 7, 2012 #6

    Ray Vickson

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    Does property (i) hold?

    RGV
     
  8. Mar 7, 2012 #7
    That's what I was attempting in my original post.

    So my brownian motion V is not normally distributed? Not sure how to show that (apart from my earlier attempt)
     
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