$E(X(T)\mid X(s),s\leq t)=X(t)\xi(t)$. What is $X$?

Question

Suppose $X(t)$ is some stochastic process. We know that if $X(t)$ is a martingale, then

$$E(X(T)\mid X(s),s\leq t)=X(t)$$

for all $t\geq 0.$

However, suppose $X$ is not a martingale, but it satisfies

$$E(X(T)\mid X(s),s\leq t)=X(t)\xi(t)$$

where $\xi$ is a deterministic function. Is this some kind of a more "general" martingale? I was wondering if anyone knew if these types of processes are important. Thanks!

Answer 1

Let's denote $\mathcal F_t := \sigma(X_s:s\leq t)$. Your hypothesis is $\mathbb E[X_t \mid \mathcal F_s] = X_s\xi(s)$ for some deterministic $\xi$.

However, we know that $$X_r\xi(r)=\mathbb E[X_t\mid\mathcal F_r] = \mathbb E[\mathbb E[X_t\mid\mathcal F_s]\mid\mathcal F_r] = \mathbb E[X_s\xi(s)\mid\mathcal F_r] = X_r\xi(s)\xi(r)$$ for any $r\leq s\leq t$.

Therefore, apart from some trivial cases, $\xi \equiv1$.