relations between expectation conditioned on subsets and conditional expectation

Question

Let $\mathbb{E}[|Y|]< \infty$. Set $\mathbb{E}[Y |B] = \frac{\mathbb{E}[Y \mathbb{1}_B]}{P(B)}$ for $B$ with $P(B) > 0$. Suppose that $\mathbb{E}[Y | X \in [a,b]] \in [a,b]$ for all $a<b$ with $\mathbb{P}(X \in [a,b] ) >0$ holds. My question is whether we can show that this implies that $\mathbb{P} ( \mathbb{E}[Y |X] \in [a,b] | X \in [a,b]) =1$, without first showing that $Y$ and $X$ are a.s. equal.

Answer 1

Yes, it is direct if you abstract a little and call the r.v. $Z_{ab}:=E[Y|X\in [a,b]]$. This r.v. is defined in $\Omega_{ab}\subseteq \Omega$, where $\Omega_{ab}=\{\omega\in\Omega: X(\omega)\in [a,b]\}$.

Your initial statement "it is given that..." can be expressed as $Z_{ab}\in [a,b]$ for all $\omega\in\Omega_{ab}$, which directly implies that the desired probability is one.

Answer 2

1. $$\mathbb{E}[Y|B]=\frac{\mathbb{E}[Y1_B]}{\mathbb{P}(B)}= \int_{\Omega} Y(\omega)\frac{1_B(\omega)\mathbb{P}(d\omega)}{\mathbb{P}(B)} =\int_{\Omega} Y(\omega)\mathbb{P}(d\omega|B).$$

2. Let $X\sim Unif(0,2)$ and $Y\sim Unif(-2,0)$ independent. Then $$ \mathbb{E}[Y|X\in [0,2]]=\mathbb{E}[Y]=-1\notin [0,2].$$