i should prove that there is not a function f(t) such that Z(t)=f(t) * W(t^2-2t+2) is a brownian motion.where W(t) is standard BM with zero mean and t variance.i know that variance of Z(t) should have a form like a*t with a positive and finite.i know that W(t^2-2t+2) is equal in low to W(t^2) - W(2t) + W(2).I think that i cannot find f(t) because the last term, W(2), has variance independent from t so i cannot have a linear form for the variance of Z(t) for any f(t).Can anyone give me some suggestions to formalize this or say me if my solution is uncorrect.thanks very much

You can reason by analysis-synthesis, suppose Zt is a BM :1) You should have E[Zt^2] = t = f(t)^2 * E[W(t^2 - 2t -+2)] = f(t)^2 * (t^2 - 2t -+2)] Then : f(t) = SQRT(t / (t^2 - 2t -+2)).2) You should have for any t, t1 > 0 : E[(Zt - Zt1) * Zt1] = 0 eq. to E[(f(t) * Wt - f(t1) * Wt1) * Wt1 * f(t1)] = 0eq. to t1*f(t)*f(t1) - f(t1)^2 * t1 = 0,which implies f(t) = f(t1) for any t, t1 > 0, and this can't be considering 1)Then, there is no function f such that...

thanks, but if we think about f(t) as a non deterministic function can you also find a condition about non existence of f(t) such that f(t)*W(t^2-2t+2) be a BM?also here it's intuitively clear but i don't know how to formalize it.thanks again