Conditional expectation and an independent $\sigma$-algebra

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space, $$\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$$ be sub-$$\sigma$$-algebras of $$\mathcal{F}$$ and let $$X$$ be a real-valued random variable that is independent of $$\mathcal{H}$$. Does $$\mathbb{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\mathbb{E}[X\mid\sigma(\mathcal{G})]$$ hold?

Intuitively, I think this should hold, by I can not think of a proof for this. I would be grateful for help.

This is not true. Let $$U,V$$ be i.i.d. $$N(0,1)$$, $$X=U+V, Y=U$$ and $$Z=U-V$$. It is well known (and easy to prove) that $$X$$ and $$Z$$ are independent. But $$E[X|Y,Z]=U+V$$ and $$E[X|Y]=E[U+V|U]=U$$. So $$\mathcal G=\sigma (Y), \mathcal H=\sigma (Z)$$ provides a counter-example.

Here is another counterexample: Let $$Y,Z$$ be independent variables taking the values $$\pm 1$$ with probability $$1/2$$ each. Then $$X=YZ$$ is independent of $$Z$$, yet it satisfies $$E[X|Y,Z]=X$$ and $$E[X|Y]=0$$.

Yuval Peres and geetha290krm already provide counter-examples. It is worth mentioning that the additional assumption that $$\mathcal H$$ is independent of $$\sigma(X)\vee\mathcal G$$ makes equality $$\mathbb{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\mathbb{E}[X\mid\sigma(\mathcal{G})]$$ true, as already shown here.

• That is really helpful, thanks a lot! Sep 21, 2022 at 14:15