Does expectation inequality imply conditional expectation inequality?

Question

Given a probability space $\left(\Omega\text{, }\mathcal{F}\text{, }\mathbb{P}\right)$ and two random variables defined on it, does it hold true that $$ \mathbb{E}\left(X\right)<\mathbb{E}\left(Y\right)\hspace{0.5cm}\Rightarrow\hspace{0.5cm}\mathbb{E}\left(X|\mathcal{F}\right)<\mathbb{E}\left(Y|\mathcal{F}\right) $$ ?

If not, could you please give me some good counterexample?

Answer 1

Let it be that $X$ only takes values in $\{0,1\}$ with $P(X=1)=p\in(0,0.5)$.

Let $Y=1-X$.

Then: $$\mathbb EX=p<1-p=\mathbb EY$$

If $\mathcal F=\sigma(X)$ then: $$\mathbb E(X\mid\mathcal F)=X\text{ and }\mathbb E(Y\mid\mathcal F)=Y$$

But we do not have $X(\omega)<Y(\omega)$ for every $\omega\in\Omega$ because $P(X=1,Y=0)=p>0$.

So we cannot state that $\mathbb E(X\mid\mathcal F)<\mathbb E(Y\mid\mathcal F)$.

Answer 2

$E(X|\mathcal F)=X$ and $E(Y|\mathcal F)=Y$ because $X$ and $Y$ are already measurable w.r.t. $\mathcal F$. To get a counter example take $X$ with $N(0,1)$ distribution and $Y=1-X$.