Is it true that $\mathbb{E}\{X\}=\mathbb{E}\{Y\}$ $\Rightarrow$ $\mathbb{E}\{X|\mathcal{F}\}=\mathbb{E}\{Y|\mathcal{F}\}$?

Question

Given $(\Omega$, $\mathcal{F}$, $\mathbb{P})$ and two r.v.'s $X$ and $Y$ defined on it, does it hold true that: $$\mathbb{E}\{X\}=\mathbb{E}\{Y\}\Rightarrow\mathbb{E}\{X|\mathcal{F}\}=\mathbb{E}\{Y|\mathcal{F}\}$$ ? If so, how can one show that is true?

Answer 1

Suppose that it is true.

Now let $A\in\mathcal F$ such that $0<P(A)<1$.

Then we can find a constant $c$ with $P(A)=c(1-P(A))$.

Let $X:=\mathsf{1}_A$ and $Y:=c\mathsf{1}_{A^\complement}$ so that $\mathbb EX=\mathbb EY$.

However then: $$P(A)=\mathbb EX\mathsf1_A=\mathbb E[\mathbb E[X\mid\mathcal F]\mathsf{1}_A]=\mathbb E[\mathbb E[Y\mid\mathcal F]\mathsf{1}_A]=\mathbb EY\mathsf1_{A}=0$$

This contradicts $0<P(A)$ and we conclude that for every $A\in\mathcal F$ we must have $P(A)\in\{0,1\}$.

That means that every random variable defined on the probability space is degenerated.

In that (uninteresting) situation the statement indeed holds.