I'm trying to solve the following problem, and I'm not sure my solution path is correct.
Let $(\Omega,F,P)$ be a complete probability space. Let T>0. Let $(W_t)_{t\geq0}$ be a standard $\mathbb{R}$-valued $(F_t)_{t\geq0^-}$-Brownian motion, where $(F_t)_{t\geq0}$ is a filtration fulfilling the usual conditions.
Show the existence of $\alpha \in \mathbb{R}$, a r.v. Z independent of $W_2$ such that $W_1 = \alpha W_2 + Z$. Determine its law (for Z).
I'm not sure how to proceed, I tried writing $W_1 = W_2 + (\alpha - 1)W_2 + Z$, and then $W_2 - W_1 + (\alpha - 1)W_2 + Z = 0$, so we would have $W_2 - W_1 \sim N(0,1)$ and $(\alpha - 1)W_2 \sim N(0,2(\alpha - 1)^2)$. This would lead to $Z \sim N(0, 1 + 2(\alpha - 1)^2)$.
In that sense, you can have any $\alpha \in \mathbb{R}$ and we see the distribution of Z. I'm not sure if what I did is right and rigorously acceptable (for instance I don't think I used the independence between $Z$ and $W_2$).
Thank you in advance.
Your argument fails because $W_2-W_1$ and $W_@ are not independent.
Just take $Z=W_1-\alpha W_2$ where $\alpha$ is to be determined. Then $(W_1,W_2,Z)$ are jointly normal so independence of $Z$ and $W_2$ is equivalent to $EZW_2=0$. This gives $EW_1W_2=\alpha EW_2^{2}$ or $1=(\alpha) (2)$. Thus, $\alpha =\frac 1 2$.