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

[AxiomOfChoice]

Homework Statement

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.

Homework 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):

<br /> 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,<br />

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

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

The Attempt at a Solution

It makes sense that W_0 = 0, so I basically am trying to show that

<br /> 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.<br />

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

<br /> 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}),<br />

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

<br /> 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.<br />

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...

 

[AxiomOfChoice]

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