$\mathbb E[X(t)]=0$, $\mathbb E[(X(t))^2]=t$. Does there exist a $\mathcal F_s$ measurable random variable $Y$ such that $\mathbb E[(X(t)-Y)^2]<t-s$

Question

Let $0\le s \le t$ and $X(t)$ a stochastic process with $\mathbb E[X(t)]=0$, $\mathbb E[(X(t))^2]=t$.

Let $\mathcal F_t$ be the natural filtration of $X$.

Does there exist a $\mathcal F_s$ measurable random variable $Y$ such that $\mathbb E[(X(t)-Y)^2]<t-s$ ?

Using linearity of the expected value the LHS is equal to $\mathbb E[X(t)^2]-2\mathbb E[X(t)Y]+\mathbb E[Y^2]=t-2\mathbb E[X(t)Y]+\mathbb E[Y^2]$

Answer 1

If there is no further assumption on the process then this is false. First note that among all $\mathcal F_s$ measurable random variables with finite variance the one that minimizes $E(X_t-Y)^{2}$ is $Y=E(X_t|\mathcal F_s)$. Hence the statement is equivalent to $E(X_t-E(X_t|\mathcal F_s))^{2} <t-s$. Suppose the entire collection $\{X_t\}$ is independent. Then the inequality becomes $t <t-s$ which is false.

Note: in the case of BM we have $E(X_t|\mathcal F_s)=X_s$ so the inequality becomes $t-s<t-s$!