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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

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    $\begingroup$ Suppose $X$ is non-constant and has a finite lower bound $L$, so that $\Pr(X\lt L)=0$. Let $Y=X$. When $k\lt 0$, $XY=X^2\gt 0\gt k$, whence $$\mathbb{E}(XY|XY\gt k)=\mathbb{E}(XY)=\mathbb{E}(X^2)$$ while $$\mathbb{E}(X\,\mathbb{E}(Y|XY\gt k))=\mathbb{E}(X\,\mathbb{E}(Y))=\mathbb{E}(X)\mathbb{E}(X)=\mathbb{E}(X)^2.$$ If these two expressions were equal, $$\text{var}(X)=\mathbb{E}(X^2)-\mathbb{E}(X)^2=0,$$ implying $X$ is constant. This contradiction shows the assertion is not generally true when $Y$ and $X$ are dependent. It could hold by accident for particular $X,Y$ and $k$ however. $\endgroup$
    – whuber ♦
    Commented Jul 22, 2015 at 15:36
  • $\begingroup$ @Whuber. Much appreciated thanks ... 1) To confirm if $X$ and $Y$ are independent the identity in the question holds for all values of $X$ and $Y$. 2) How do you get E(XY|XY>k)=E(XY) and E(Y|XY>k)=E(Y) and E(XE(Y))=E(X)E(X)? Could you please elaborate? $\endgroup$
    – texmex
    Commented Jul 23, 2015 at 4:23
  • $\begingroup$ For $k\lt 0$ the event $XY\gt k$ is the universal event of probability $1$, whence conditioning on it changes nothing. $\endgroup$
    – whuber ♦
    Commented Jul 23, 2015 at 11:54
  • $\begingroup$ @whuber. Thanks for the clarification. To confirm, if $X$ and $Y$ are independent does the identity in the question hold for all values of $X$ and $Y$. $\endgroup$
    – texmex
    Commented Jul 24, 2015 at 6:53

1 Answer 1

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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.

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  • $\begingroup$ I think there is a mistake on the first line. $\infty>E[Y] = E[E[Y\mid A]] = E[1 + \frac{k}{X}] = \infty$. $\endgroup$
    – Hunaphu
    Commented Sep 7, 2015 at 15:42
  • $\begingroup$ @Hunaphu Please explain what "$A$" is in your assertion, because I cannot match what you have written with anything that appears in this answer. $\endgroup$
    – whuber ♦
    Commented Sep 7, 2015 at 16:11
  • $\begingroup$ @whuber Please note, I added a few cases in the question (some unresolved) since I think these might be of interest to anyone following this topic, also for completeness, and to link it back to the question where this came up from. Hence, please add anything as you see fit. $\endgroup$
    – texmex
    Commented Sep 18, 2015 at 12:15
  • $\begingroup$ @Hunaphu Could you please clarify whether the assertion at the top (The main question here) holds in all the four cases (as framed in the question). I am not sure if you remember, but this is from the steps you had out lined for the simplification of the following question: stats.stackexchange.com/questions/157954/… $\endgroup$
    – texmex
    Commented Sep 18, 2015 at 12:16

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