Suppose I have two Brownian motions $X$ and $Y$, which are independent. In other words, for any finite set of times $0 < t_1 < t_2 < \cdots < t_n$ the random vectors $(X(t_1),\ldots , X(t_n))$ and $(Y(t_1),\ldots , Y(t_n))$ are independent.
While I know that the increments $X(t)-X(s)$ and $X(u)-X(v)$ are independent for any $v < u \leq s < t$ (by definition), I would like to show that any two increments $X(t) - X(s)$ and $Y(t) - Y(s)$ are independent.
While this seems intuitively obvious, and I have a hand-wavey argument in mind, I'm having trouble making it precise. Is there a nice argument for a Brownian motion novice?
They are functions of the independent vectors $\bigl(X(s),X(t)\bigr)$ and $\bigl(Y(s),Y(t)\bigr)$, QED.