Proof of Simplification of Conditional Expectation of Product of Random Variables Could someone please provide detailed steps to prove or disprove the following?
$E[XY\mid XY>k] = E[XE[Y\mid XY>k]]$
Here, $X,Y$ are random variables that could be discrete or continuos and follow any probability distribution. $E$ is the expectation operator. $k$ is a constant.
As forum members have provided their feedback, a few interesting cases arise and are worth looking into.
CASE 1
If $X$ and $Y$ are dependent. Is the above identity true?
CASE 2
If $X$ and $Y$ are independent, and could be from any distribution. Is the above identity true?
CASE 3
How about the popular normal distribution? If $X$ and $Y$ are normally distributed, is the above identity true?
CASE 4
How about if one of them is normally distributed and the other is log-normal? Say, $X$ is normal and $Y$ and is log-normal, is the above identity true?
STEPS TRIED
A very rough sketch of the proof as given by another forum member:
$$E[XY\mid XY\in B] = \int\int_B xy f_{XY\mid B}\,dx\,dy = \int y \Big(\int x f_{X\mid B}\,dx\Big) f_Y\,dy$$
RELATED QUESTION
The above question is a generalization of a key step required in the simplification of this question: Conditional Expected Value of Product of Normal and Log-Normal Distribution
 A: The equation is not true even when $X$ and $Y$ are independent.

Suppose, for instance, that $X$ and $Y$ are positive variables with an Exponential distribution.  For any $k\gt 0$ and $X\gt 0$ we can easily work out that
$$\mathbb{E}(Y\,|\, XY \gt k)  = \frac{\int_{k/X}^\infty y \exp(-y)dy}{\int_{k/X}^\infty \exp(-y)dy}=\frac{\exp(-k/X)(1+k/X)}{\exp(-k/X)}= 1 + \frac{k}{X}.$$
Therefore
$$\mathbb{E}(X\, \mathbb{E}(Y\,|\, XY \gt k)) = \mathbb{E}\left(X\left( 1 + \frac{k}{X}\right)\right) = \mathbb{E}(X + k) = k+1.$$
On the other hand,
$$\mathbb{E}(XY\,|\, XY \gt k) = \frac{\iint_{xy\gt k}xy \exp(-x-y) dx dy}{\iint_{xy\gt k}\exp(-x-y) dx dy}.$$
After integrating over $y$ (from $k/x$ to $\infty$) and substituting $t = x/\sqrt{k}$, both integrals can be related to the modified Bessel function of the second kind, $K_n$ (for $n=1, 2$):
$$\iint_{xy\gt k}xy \exp(-x-y) dx dy = 2 k \left(\sqrt{k} K_1\left(2 \sqrt{k}\right)+K_2\left(2
   \sqrt{k}\right)\right),$$
$$\iint_{xy\gt k}\exp(-x-y) dx dy = 2 \sqrt{k} K_1\left(2 \sqrt{k}\right).$$
Therefore
$$\mathbb{E}(XY\,|\, XY \gt k) = k+\frac{\sqrt{k} K_2\left(2 \sqrt{k}\right)}{K_1\left(2 \sqrt{k}\right)}.$$
Asymptotically as $k$ grows large,
$$ k+\frac{\sqrt{k} K_2\left(2 \sqrt{k}\right)}{K_1\left(2 \sqrt{k}\right)} = k + \sqrt{k} + \frac{3}{4} + O(k^{-1/2}),$$
showing that it cannot equal $k+1$ for all $k$.  Therefore the equation of the question is not true.
