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I need to prove this (seemingly simple) property of Brownian motion

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]B_t[/itex] is a Brownian motion. I want to show that if you fix [itex]t_0 \geq 0[/itex], then the process [itex]W_t = B_{t_0+t} - B_{t_0}[/itex] is also a Brownian motion.

    2. Relevant equations
    Apparently, a stochastic process [itex]X_t[/itex] is a Brownian motion on [itex]\mathbb R^d[/itex] beginning at [itex]x\in \mathbb R^d[/itex] if it has finite-dimensional distributions given by the following mess (for [itex]0 \leq t_1 \leq \ldots \leq t_k[/itex]):

    P(X_{t_1}\in F_1,\ldots,X_{t_k}\in F_k) = \int_{F_1 \times \cdots \times F_k} p(t_1,x,x_1)p(t_2-t_1,x_1,x_2)\ldots p(t_k - t_{k-1},x_{k-1},x_k)dx_1\ldots dx_k,

    where the [itex]F_j[/itex] are Borel subsets of [itex]\mathbb R^d[/itex] and [itex]p[/itex] is given by

    p(t,x,y) = (2\pi t)^{-(d/2)} \exp \left( -\frac{|x-y|^2}{2t} \right).

    3. The attempt at a solution
    It makes sense that [itex]W_0 = 0[/itex], so I basically am trying to show that

    P(W_{t_1}\in F_1,\ldots,W_{t_k}\in F_k) = \int_{F_1 \times \cdots \times F_k} p(t_1,0,y_1)p(t_2-t_1,y_1,y_2)\ldots p(t_k - t_{k-1},y_{k-1},y_k)dy_1\ldots dy_k.

    I'd like to do this by using the fact that [itex]B_t[/itex] has distributions that look like this. So I tried

    P(W_{t_1}\in F_1,\ldots,W_{t_k}\in F_k) = P(B_{t_0 + t_1} - B_{t_0}\in F_1, \ldots B_{t_0 + t_k} - B_{t_0}\in F_k) = P(B_{t_0 + t_1} \in F_1 + B_{t_0}, \ldots B_{t_0 + t_k} \in F_k + B_{t_0}),

    where [itex]F_j + B_{t_0}[/itex] is just the translate of the Borel set [itex]F_j[/itex] by the vector [itex]B_{t_0}[/itex]. But I'm not sure I can do this, first of all; and second of all, I'm not sure it helps me, because I then would like to make a change of variables to massage the integral I then get. By the observation I just made above, I have

    P(W_{t_1}\in F_1,\ldots,W_{t_k}\in F_k) = \int_{(F_1 + B_{t_0}) \times \cdots (F_k + B_{t_0})} p(t_0 + t_1,x,x_1) \cdots p(t_{k} - t_{k-1},x_{k-1},x_k)dx_1\ldots dx_k.

    So I want to change variables and set [itex]y_j = x_j - B_{t_0}[/itex] to get those pesky [itex]B_{t_0}[/itex]'s out of the integral. But that doesn't help me with the term [itex]p(t_0 + t_1,x,x_1)[/itex]. I need that to read something like [itex]p(t_1,y,y_1)[/itex], but I really do not see how to get rid of the [itex]t_0[/itex]...
    Last edited: Sep 5, 2011
  2. jcsd
  3. Sep 5, 2011 #2
    Hmmm...well, at least I don't feel so bad about not having gotten this now.
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