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

1. Sep 5, 2011

AxiomOfChoice

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

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

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

where the $F_j$ are Borel subsets of $\mathbb R^d$ and $p$ 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 $W_0 = 0$, 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 $B_t$ 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 $F_j + B_{t_0}$ is just the translate of the Borel set $F_j$ by the vector $B_{t_0}$. 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 $y_j = x_j - B_{t_0}$ to get those pesky $B_{t_0}$'s out of the integral. But that doesn't help me with the term $p(t_0 + t_1,x,x_1)$. I need that to read something like $p(t_1,y,y_1)$, but I really do not see how to get rid of the $t_0$...

Last edited: Sep 5, 2011
2. Sep 5, 2011

AxiomOfChoice

Hmmm...well, at least I don't feel so bad about not having gotten this now.