Invariance of Brownian motion under orthogonal transformations

Question

Let $\left(B_t\right)_{t \in [0,\infty)}$ be an $n$-dimensional Brownian motion with start at $x \in \mathbb{R}^n$, and let $A$ be an orthogonal $n \times n$ real matrix. I'm trying to show that $AB$ is again an $n$-dimensional Brownian motion. I've succeeded in showing that each component is a one-dimensional Brownian motion, but I'm having a hard time showing that the components are independent. Any help would be appreciated.

Answer 1

Set $W_t := A \cdot B_t$ where $A$ is an orthogonal matrix.

1. Since $W_0=A \cdot x$ we see that $(W_t)_{t \geq 0}$ starts at $A \cdot x$.
2. As $t \mapsto B_t$ is continuous, we find that $t \mapsto A \cdot B_t$ is continuous.
3. Let $0 \leq s \leq t$. Then, $$\begin{align*} \mathbb{E}e^{\imath \, \xi^T \cdot (W_t-W_s)} = \mathbb{E}e^{\imath \, \xi^T (A \cdot (B_t-B_s))} = \mathbb{E}e^{\imath \, (A^T\xi)^T \cdot B_{t-s}} \end{align*}$$ for any $\xi \in \mathbb{R}^n$. Here we used the stationarity of the increments of $(B_t)_{t \geq 0}$. As $B_{t-s} \sim N(0,(t-s) I_n)$, we get $$\mathbb{E}e^{\imath \, \xi^T \cdot (W_t-W_s)} = \exp \left(- \frac{1}{2} (t-s) |A^T \cdot \xi|^2 \right).$$ It follows straight from the orthogonality of $A$ that $|A^T \cdot \xi|=|\xi|$. Hence, $$\mathbb{E}e^{\imath \, \xi^T \cdot (W_t-W_s)} = \exp \left(- \frac{1}{2} (t-s) |\xi|^2 \right). \tag{1}$$ This shows that $W_{t-s} \sim W_t-W_s \sim N(0,(t-s) I_n)$.
4. Since $(B_t)_{t \geq 0}$ is a Brownian motion, the increments $$B_{t_1}-B_{t_0}, \ldots,B_{t_n}-B_{t_{n-1}}$$ are independent. Obviously, this implies that $$A (B{t_1}-B_{t_0}),\ldots,A(B_{t_n}-B_{t_{n-1}})$$ are independent. Hence, $(W_t)_{t \geq 0}$ has independent increments.