I am trying to solve the following exercise in Oksendal's book:
Let $B_t$ be Brownian motion and fix $t_0\ge 0$. Prove that $$\bar{B_t}:=B_{t_0+t}-B_{t_0};\quad t\ge 0$$ is a Brownian motion.
I try to prove it by definition. In the book, Brownian motion is defined as
$$P^x(B_{t_1}\in F_1, \cdots, B_{t_k}\in F_k) = \\ \int\limits_{F_1 \times \cdots \times F_k}p(t_1, x, x_1)\cdots p(t_k-t_{k-1}, x_{k-1}, x_k)dx_1 \ldots dx_k, $$ where $$p(t,x,y) = (2\pi t)^{-n/2}\cdot \exp(-\frac{|x-y|^2}{2t})$$
Hence for simplicity, we only need to show $$P^0(\bar{B}_{t_1}\in F_1, \bar{B}_{t_2}\in F_2) = \int\limits_{F_1 \times F_2}p(t_1, 0, x_2) p(t_2-t_1, x_3, x_2)dx_2 dx_3$$ Suppse $B_t$ starts at $x_0$. By definition of $\bar{B_t}$
$P^0(\bar{B}_{t_1}\in F_1,\bar{B}_{t_2}\in F_2)=P^0(B_{t_0+t_1}-B_{t_0}\in F_1, B_{t_0+t_2}-B_{t_0}\in F_2)\tag{1}$
Then I don't know how to handle it. I know I should use $B_t$'s finite dimensional distribution. So if I can write $(1)$ as
$$P^{x_0}[\bigcup_{x_1\in\mathbb{R}^n} \left(B_{t_0}=x_1,B_{t_0+t_1}\in F_1+x_1, B_{t_0+t_2}\in F_2+x_1\right)]\tag{2}\\=\int\limits_{\mathbb{R}^n\times (F_1+x_1) \times (F_2+x_1)}p(t_0, x_0, x_1)p(t_1, x_1, x_2) p(t_2-t_1, x_2, x_3)dx_1 dx_2 dx_3 $$
Then apply change of variable formula and Fubini's theorem, I can get the result.
But I don't know how to justify $(2)$ and the last equality. I think it should be true because it's the conditional probability. But the book didn't introduce this concept and since $\mathbb{R}^n$ is uncountable, I don't know how this is valid in terms of measure theory.
Why aren't you applying the characterization: "Any continuous real-valued process $(X_t)$ that is zero-mean Gaussian process with covariance $\mathrm{cov} (X_t,X_s)=t\wedge s$ is a Brownian motion" ? (Rogers and Williams, page 4).
I write $B'_t$ for your new process. It is a Gaussian process.
Then $B'_t$ has continuous sample paths (since $B_t$ does). Also
$$E(B'_t) = E(B_{t_o+t}) - E(B_{t_0}) = 0. $$
Finally for $s < t$
\begin{align} \mathrm{cov}(B'_t,B'_s) &= E((B_{t_o+t}-B_{t_o})(B_{t_o+s}-B_{t_o}))\\ &= \min(t_0+t, t_o+s) + \min(t_0,t_0) - \min(t_0+s,t_0) + \min(t_0+t,t_0)\\ &= (t_0 +s) + t_0 - t_0 -t_0 = s = \min(s,t). \end{align}
In the second line I have expanded the brackets, used linearity of expectation, and the characterization applied to $(B_t)$.