2
$\begingroup$

Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathcal F\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$
  • $X\in\mathcal L^1(\operatorname P)$
  • $A\in\mathcal A$
  • $Y:\Omega\to\mathbb R$ be $\mathcal A$-measurable with $\operatorname E\left[1_A|Y|\right]<\infty$

Are we able to show that $$1_A\operatorname E\left[X\mid\mathcal F\right]=1_AY\text{ almost surely}\Leftrightarrow\forall F\in\mathcal F:\operatorname E\left[1_A1_FX\right]=\operatorname E\left[1_A1_FY\right]\tag1?$$

I'm especially unsure whether we need to impose any stronger measurability assumption on $Y$ and how $(1)$ is related to $$\operatorname E\left[\left.X\right|_A\mid\left.\mathcal F\right|_A\right]=\left.Y\right|_A\;\left.\operatorname P\right|_A\text{-almost surely}\Leftrightarrow\left.Y\right|_A=\left.\operatorname E\left[X\mid\mathcal F\right]\right|_A\;\left.\operatorname P\right|_A\text{-almost surely}\tag2.$$

In order to prove "$\Rightarrow$" in $(1)$ we obviously need to use that, by definition, $\forall F\in\mathcal F:\operatorname E\left[1_FX\right]=\operatorname E\left[1_A\operatorname E\left[X\mid\mathcal F\right]\right]$ (it would be clear, if $A\in\mathcal F$; do we need this?). And for "$\Leftarrow$" we somehow to argue with measure uniqueness. But I struggle to fill out the details in both directions.

$\endgroup$
2
$\begingroup$

In general, (1) fails: let $\mathcal F$ be the $\sigma$-algebra containing $\emptyset$ and $\Omega$. Then (1) reads $$ 1_A\operatorname E\left[X \right]=1_AY\text{ almost surely}\Leftrightarrow \operatorname E\left[1_A X\right]=\operatorname E\left[1_A Y\right], $$ which is not true if $A=\Omega$, $X$ and $Y$ have the same expectation but $Y$ is not almost surely constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.