Let Y and Z be two random variables on $(\Omega,\mathcal{F},P)$, it is well-known that $\sigma(Z)\subset \sigma(Y)$ is equivalent to that there exists some Borel measurable function such that $Z=f(Y)$.
I am wondering if the analogous is true for stochastic processes. Let ${Y_t}$ and $Z_t$ be two stochastic processes on $(\Omega,\mathcal{F},\mathcal{F}_t,P)$. Suppose that the natural filtration of $Z$ is contained in the natural filtration of $Y$, i.e. $\sigma(Z_s,0\le s\le t)\subset \sigma(Y_s,0\le s\le t)$. Is it true that there exists some Borel measurable function such that $Z_t=f(t,Y_t)$?
No, this is not true. As an example, let $Y$ be a Brownian motion and define $Z$ by $Z_t = 0$ for $t \in [0,2)$ and $Z_t = Y_1$ for $t \in [2,\infty)$. Clearly $\sigma(Z_s, 0 \le s \le t) \subset \sigma(Y_s, 0 \le s \le t)$ for all $t$, but there is no function $f$ such that $Z_t = f(t,Y_t)$ for all $t$. For example, $Z_2 = Y_1$ and $Y_1 \ne f(2,Y_2)$ for any $f$ because $\sigma(Y_1) \not \subset \sigma(Y_2)$.