Every continuous Markov process $X$ adapted to a Brownian filtration is $X_t = f(t,B_t)$ for some function $f$?

Question

Let $X$ be continuous Markov process adapted to a filtration generated by Brownian motion $B$. Does there exist a function $f$ such that $X_t = f(t,B_t)$?

My guess is that it should have such representation. However, I am not able to produce any proof.

Answer 1

No. Try $X_t=S_t-B_t$, where $S_t=\sup\limits_{0\leqslant s\leqslant t}B_s$.

Another class of examples: $X_t=\displaystyle\int_0^tg(s)\,\mathrm dB_s$ with $g$ not identically $0$, not identically $1$, and not identically $-1$.