$X,Y,Z $ s.t. $E[X | Y ] = E[X | Z ] \implies E[X | Y ] = E[X | Z ] = E[X | Y,Z ]$?

Question

Is it true that :

$ \forall X,Y,Z $ $$E[X | Y ] = E[X | Z ] \implies E[X | Y ] = E[X | Z ] = E[X | Y,Z ]$$

I tried to find $X,Y,Z$ with X not independant of $Y,Z$ and $$E[X | Y ] = E[X | Z ] = E[X]$$ but without success.

Has anyone a counter example for that ? It seems blatantly false.

Answer 1

It is false. Let $Y,Z \sim \text{Bernoulli}(\frac{1}{2})$ be independent and $X=\begin{cases}0 & Y=Z \\ 1 & Y \neq Z \end{cases}$. Notice that $X\sim \text{Bernoulli}(\frac{1}{2})$ as well, $X$ is independent of $Y$, $X$ is independent of $Z$, but $X,Y,$ and $Z$ are not mutually independent. In fact, $X$ is measurable with respect to the sigma algebra generated by $Y$ and $Z$. So we have: $$E[X\mid Y] = E[X\mid Z] = \frac{1}{2}$$ $$E[X \mid Y,Z] =X.$$

Answer 2

$$Y:=|X|, Z:=X^2, X\sim N(0,1) \implies E[X\mid Y]=E[X\mid Z]=E[X]$$ Then they are obviously not independent. The issue is, that once you start to package information about the sign into either Y or Z in order to get something different for $E[X\mid Y,Z]$ you run into issues, since that moves the expected value above or below zero. So you need to somehow fix that.

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So what about this: $$\begin{align} &Z\sim U\{-1,1\} \perp X\sim N(0,1), Y=Z\cdot X\\ \implies& E[X\mid Z]=E[X] \end{align}$$ And since $X=Z^2 X = Z Y$ we get $$E[X\mid Y]=E[ZY\mid Y]=YE[Z\mid Y]=0=E[X]$$

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Lastly if we also want Z and X be dependent, we could do something like: $$U\sim U\{-1,1\} \perp X\sim N(0,1)$$ Let $$Y:=UX$$ and $$ Z=\begin{cases} |X| & U=1\\ |X|+1 & U=-1 \end{cases} $$ Then neither none of the variables are independent, but: $$E[X\mid Y]=0=E[X \mid Z]=E[X]$$ And $$E[X\mid Z,Y]=X $$