Prove $\mathbb{E}[Y\mid \mathcal{A}]=X\Rightarrow\mathbb{E}[Y\mid X]=X$ and $\mathbb{E}[Y\mid \mathcal{A}]=X \not\Leftarrow\mathbb{E}[Y\mid X]=X$ Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let $X$ be a random variable. Let $\mathcal{A}\subseteq\mathcal{F}$ be a sub $\sigma$-algebra of $\mathcal{F}$. Further suppose that $X$ is $\mathcal{A}$-measurable.
I want to prove the following:


*

*$\mathbb{E}[Y\mid \mathcal{A}]=X\Rightarrow\mathbb{E}[Y\mid X]=X$

*$\mathbb{E}[Y\mid \mathcal{A}]=X \not\Leftarrow\mathbb{E}[Y\mid X]=X$
One idea I had for 1. was as follows: $X=\mathbb{E}[Y\mid \mathcal{A}]=\mathbb{E}[\mathbb{E}[Y\mid \mathcal{A}]\mid X]=\mathbb{E}[Y\mid X]$ since $\sigma(X)$ is the smallest $\sigma$-algebra generated by $X$ and thus $\sigma(X)\subseteq\mathcal{A}$ and the smallest $\sigma$-algebra wins. Is this reasoning correct? 
As for 2., I am uncertain as to how to proceed. 
Thank you! 
 A: For (1) there is a standard argument of showing that. Define the events:
$$ E^+:=\Big\{  \mathbb{E}[Y\mid \mathcal{A}]>X \Big\}, \quad E^-:= \Big\{  \mathbb{E}[Y\mid \mathcal{A}]>X \Big\}$$
And we can write $E^+=\cup_{n=1}^\infty E_n^+$ where:
$$ E^+_n:=\Big\{  \mathbb{E}[Y\mid \mathcal{A}]>X+\frac{1}{n} \Big\} $$
This is $\mathcal{A}$-measurable set, and one can show that:
$$ \mathbb{E}\Big[ \Big\vert\mathbb{E}[Y\mid \mathcal{A}]-X \Big\vert \cdot \mathbf{1}_{E_n^+} \Big]= \mathbb{E}\Big[ (\mathbb{E}[Y\mid \mathcal{A}]-X ) \cdot \mathbf{1}_{E_n^+} \Big] \geq \frac{1}{n}\cdot \mathbb{P}(E_n^+) $$
By definition of conditional expectation, this should be $0$, hence $\mathbb{P}(E_n^+)=0$ for all $n$. And by sub-additivity $\mathbb{P}(E^+)=0$. Likewise $\mathbb{P}(E^-)=0$. Hence $\mathbb{E}[Y\mid \mathcal{A}]=X $  almost surely.
This is the brute force approach, but there could be an elegant solution for it.
For (2), if $X$ is constant then it is always measurable and $\sigma(X)=\{\Omega,\emptyset \}$, see this thread. Then $\mathbb{E}[Y\vert X]=\mathbb{E}[Y]$, and you just need an example where $\mathbb{E}[Y\mid\mathcal{A}]$ is not constant.
