The motivation of this question is to show that two probabilities on $C_{0}^{n}(0,1)$ (the space of continuous $\mathbb R^{n}$ valued process on $[0,1]$ starting from zero) induced by two non-degenerate $n$ dimensional Brownian motions $W_{1}$ and $W_{2}$ are equivalent to each other.
Intuitively, it may be correct, if we think of these two probability measures by Gaussian measures on the space $C_{0}^{n}(0,1)$ in terms of abstract Wiener space. However, due to the lack of knowledge in this area, I could not find a rigorous proof.
I appreciate if one can provide a detailed proof. Thanks.
In this below, I will given one example to make the question clear: Suppose $W_1$ is 2-D standard BM, $W_2$ is given by $W_2 (t) = [1, 0; \rho, \sqrt{1 - \rho^2}] W_1(t)$ for some $\rho\in (0,1)$. Are $W_1$ and $W_2$ are singular or equivalent?
Motivated from the comments below, $W_1$ and $W_2$ seems to be mutually singular. The argument is that $W_1$ is concentrated to the set $A_1 = \{x\in C_0^2[0,1]: [x](t) = I t\}$, but $W_2$ is concentrated on another disjoint set $A_2 = \{x \in C_0^2[0,1]: [x](t) = [1, \rho; \rho, 1] t\}$, where $[x](t)$ is quadratic variation of $x$ on $[0,t]$.
However, consider a set $B= \{x\in C_0^2: x^1(1)>0\}$, then $P(W_1\in B) = P(W_2\in B) = 1/2$, which contradicts to singularity? It's not contradiction to singularity, since $B\cap A_1$ and $B\cap A_2$ are both not empty.
