Is it OK $E(XY|Y)\overset?=YE(X|Y)$

Question

When it is allowed to pull a r.v. out of the expectation, i.e.

$E(XY|Y)\overset?=YE(X|Y)\tag1$.

or even

$E(Y^2|Y)\overset?=Y^2\tag2$

in the computation rules it is written that if,

$XY\ge0$ or both r.v. are in $L^1(\star)$

it is allowed. But what if other conditions were given instead of $(\star)$, for example

$E(X^2|Y)=Y^2$ and $E(X|Y)=Y$

Is then $(1)\ \&\ (2)$ still valid ?

Answer 1

$$\mathbb E[X\mid \mathcal G]=X$$ when $X$ is $\mathcal G-$measurable. So obviously, for any Borel function, $$\mathbb E[f(Y)\mid Y]=f(Y).$$

Answer 2

Let us make it precise. Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathcal{G}\subseteq\mathcal{F}$ be a sub $\sigma$-algebra. Let $X,Y:\Omega\rightarrow\mathbb{R}$ be random variables such that $Y$ is $\mathcal{G}$-measurable. We have that $E\left[XY\mid\mathcal{G}\right]=YE\left[X\mid\mathcal{G}\right]$, provided that $XY$ and $X$ are integrable. Note that we do not require that $Y$ is integrable.

Firstly, requiring that $XY$ and $Y$ are integrable makes sense because conditional expectation is only defined for integrable random variables (at least, this is true for almost all textbooks). Firstly, we prove that the proposition is true if $Y$ is a bounded. (I skip the proof).

Let $A_{n}=[|Y|\leq n]$. Note that $1_{A_{n}}\rightarrow1$ pointwisely. Now, $E\left[XY1_{A_{n}}\mid\mathcal{G}\right]=E\left[X(Y1_{A_{n}})\mid\mathcal{G}\right]=(Y1_{A_{n}})E\left[X\mid\mathcal{G}\right]$ because $Y1_{A_{n}}$is bounded and $\mathcal{G}$-measurable. Observe that $|XY1_{A_{n}}|\leq|XY|$, $XY$ is integrable, and $XY1_{A_{n}}\rightarrow XY$. By the conditional version of dominated convergence theorem, we have $E[XY1_{A_{n}}\mid\mathcal{G}]\rightarrow E[XY\mid\mathcal{G}]$ (a.e.). Therefore, letting $n\rightarrow\infty$ on both sides, we are done.