It's the first time I'm working with conditional expectation, so I need to ensure my reasoning is correct:
Let $\Omega = [0, \pi]$ with the Borel $\sigma$-algebra and probability equal to normalized Lebesgue measure. Let $X(\omega) = \sin{\omega}, Y(\omega) = \cos{\omega}$. Calculate $E((X+Y)^2|X)$ and $E((X+Y)^2|XY)$.
First of all, since X and Y are trigonometric functions, $X^2 + Y^2 = 1$, which leads to $(X+Y)^2 = 2XY + 1$. Therefore:
$E((X+Y)^2|X) = E(2XY+1|X) = 2 E(XY|X) + E(1|X)$, because conditional expected value is linear. Now:
$E(1|X)=1$, since the function is constant.
$E(XY|X) = Y E(X|X) = XY$, because $Y$ is $X$-measurable and bounded.
Thus the final result is $E((X+Y)^2|X) = 2XY +1$. I've got the same result in the second conditional expectation due to similar reasoning.
How wrong am I? Should I extend this answer by calculating $\int \limits_{A} (2XY +1)dP$ for some Borel $A$ from this $\sigma$-algebra?
Just a little bit.
Now $\mathsf E((X+Y)^2\mid XY) ~=~ 1+2XY$ is okay.
However, $\mathsf E((X+Y)^2\mid X) ~$$=~ 1$ because $\mathsf E(XY\mid X) = 0$
Since $\mathsf E(XY\mid X) ~=~ \mathsf E(X\cos(\arcsin X)\mid X) ~=~0$
(For each value of $X$ there exists two equally possible values of $Y$, one positive and one negative.)