Results 1 to 6 of 6

Math Help - brownian motion

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    136

    brownian motion

    also for a brownian motion process

    why is 2B(3) - B(2) - 2B(1) = N(0,1) + N(0,1) + N(0,4)

    thanks alot
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,117
    Thanks
    383
    Awards
    1
    Quote Originally Posted by chogo View Post
    also for a brownian motion process

    why is 2B(3) - B(2) - 2B(1) = N(0,1) + N(0,1) + N(0,4)

    thanks alot
    What???

    Please define your terms!

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    Posts
    136
    sorry

    if X(t) = -1 for 0 <= t <= 1
    X(t) = 1 for 1 < t <= 2
    X(t) = 2 for 2 < t <= 3

    then the ito integral int_0^3 X(t)dB(t) = 2B(3) - B(2) - 2B(1)

    that i understand. But why does 2B(3) - B(2) - 2B(1) have a distribution Normal(0,6) or if put as sums N(0,1) + N(0,1) + N(0,4).

    I know from the definition of wiener process that Bt - Bs = N(0,t-s) for 0<s<t
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    Posts
    136
    ive learnt on this forum if no one responds its cos they dont understand the question. Is this the case or does no one know this?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by chogo View Post
    sorry

    if X(t) = -1 for 0 <= t <= 1
    X(t) = 1 for 1 < t <= 2
    X(t) = 2 for 2 < t <= 3

    then the ito integral int_0^3 X(t)dB(t) = 2B(3) - B(2) - 2B(1)

    that i understand. But why does 2B(3) - B(2) - 2B(1) have a distribution Normal(0,6) or if put as sums N(0,1) + N(0,1) + N(0,4).

    I know from the definition of wiener process that Bt - Bs = N(0,t-s) for 0<s<t
    I have only a casual knowledge of Brownian motion, but I think it goes like this. The integral is

    int_0^3 X(t)dB(t) = 2(B(3)-B(2)) + (B(2)-B(1)) - (B(1)-B(0)).

    That (B(t) - B(t-1)) ~ N(0,1) for t = 1,2,3 follows from (B(t) - B(s)) ~ N(0,t-s) when s = t-1 and B(0) = 0. Next, after the multiplication by 2, 2(B(3)-B(2)) ~ N(0,2^2) = N(0,4). Finally, use the sum of independent normal variables has a normal distribution with variance equal to the sum of the variances to get

    2(B(3)-B(2)) + (B(2)-B(1)) - (B(1)-B(0)) ~ N(0,4+1+1) = N(0,6).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2006
    Posts
    136
    that was a complete world of help THANK YOU SO MUCH
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. brownian motion
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 24th 2011, 02:13 AM
  2. Brownian Motion
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 1st 2010, 10:32 AM
  3. Brownian motion
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 10th 2009, 08:35 AM
  4. More Brownian motion
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 10th 2009, 01:40 AM
  5. Brownian Motion
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 8th 2007, 09:18 PM

Search Tags


/mathhelpforum @mathhelpforum