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Let $X$ and $Y$ be two independent identically distributed random variables. Prove that $$E\left(\frac{XY}{X^2+Y^2}\right) \geqslant 0.$$

I tried to manipulate with the expression $\dfrac{XY}{X^2+Y^2}$, e.g. tried to rewrite it in the form $$\dfrac{XY}{X^2+Y^2} = \dfrac{(X+Y)^2}{4(X^2+Y^2)} - \dfrac{(X-Y)^2}{4(X^2+Y^2)}$$ or even in the form: $$\dfrac{XY}{X^2+Y^2} = \dfrac{(X+Y)^2 - (X-Y)^2}{2(X+Y)^2 + 2(X-Y)^2}$$

but it didn't help much.

I am interested to see a proof of this fact or some favorable ideas that may help here.

Reference: The problem was proposed in the Kolmogorov Students' Competition in Probability Theory, 2017.

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  • $\begingroup$ $\mathbb{E}\left(XY\right) \geqslant 0$ is not difficult to show since it equals $\left(\mathbb{E}\left(X\right)\right)^2$ $\endgroup$
    – Henry
    Apr 22, 2017 at 19:16
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    $\begingroup$ If $X$ and $Y$ are independent then $\mathbb{E}\left(XY\right) = \mathbb{E}\left(X\right) \mathbb{E}\left(Y\right)$ while if $X$ and $Y$ are identically distributed then $\mathbb{E}\left(X\right)= \mathbb{E}\left(Y\right)$ providing that all these are finite $\endgroup$
    – Henry
    Apr 22, 2017 at 19:24
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    $\begingroup$ The random variables $\frac{X}{\sqrt{X^2+Y^2}}$ and $\frac{Y}{\sqrt{X^2+Y^2}}$ are not necessarily independent $\endgroup$
    – clark
    Apr 22, 2017 at 19:37
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    $\begingroup$ Perhaps one should study the r.v. $\Theta := \arccos \left ( \frac{X}{\sqrt{X^2+Y^2}} \right )$. $\endgroup$
    – Ian
    Apr 22, 2017 at 20:16
  • $\begingroup$ @Ramil are you sure this statement is true? $\endgroup$
    – amakelov
    Apr 22, 2017 at 21:21

1 Answer 1

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Here is a proof:

$$ \Bbb{E}\left[\frac{XY}{X^2+Y^2}\right] = \Bbb{E}\left[ \int_{0}^{\infty} XY e^{-(X^2+Y^2)t} \, dt \right] = \int_{0}^{\infty} \Bbb{E}[X e^{-X^2 t}]\Bbb{E}[Y e^{-Y^2 t}] \, dt \geq 0. $$

Addendum. My observation is that the expectation can be thought as a quadratic form on the space of finite measures. Indeed, if $\mu$ is the common law of $X$ and $Y$, then

$$ \Bbb{E}\left[ \frac{XY}{X^2+Y^2}\right] = \int_{\Bbb{R}^2} \frac{xy}{x^2+y^2} \, \mu(dx)\mu(dy). $$

So the problem boils down to showing that this form is positive semi-definite. One such way is to find a unitary transformation under which the quadratic form reduces to the usual $2$-norm. Such a transform are often given as integral transformation:

$$ L\mu(t) = \int_{\Bbb{R}} k(t, x) \, \mu(dx). $$

Assuming that $L$ provides such a transformation, it should satisfy

$$ \int_{\Bbb{R}^2} \frac{xy}{x^2+y^2} \, \mu(dx)\mu(dy) = \int_{\Bbb{R}} (L\mu(t))^2 \, dt = \int_{\Bbb{R}^2} \left(\int_{\Bbb{R}} k(t, x)k(t, y) \, dt \right) \mu(dx)\mu(dy). $$

When the dust settles down, it becomes clear what we should hunt: a function $k$ such that

$$ \frac{xy}{x^2+y^2} = \int_{\Bbb{R}} k(t, x)k(t, y) \, dt. $$

It is not hard to guess such a function, and indeed my proof uses $k(t, x) = x e^{-x^2 t} \mathbf{1}_{\{t \geq 0\}}$.

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    $\begingroup$ This is really nice! I just want to comment quickly (because this took me a moment to think through) that $\Bbb{E}[X e^{-X^2 t}]$ isn't necessarily nonnegative, but because it's equal to $\Bbb{E}[Y e^{-Y^2 t}]$, the integrand in the final integrand is nonnegative.+1 $\endgroup$
    – πr8
    Apr 22, 2017 at 22:52
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    $\begingroup$ @SangchulLee this is a relly nice approach! Looks like a magic :) But could you please argue why it is legal to move expectation inside the integral? I.e. why for a random variable $X$ and some function $f(x, t)$ the following should hold $$\operatorname{\mathbb{E}}\left(\int\limits_{a}^b f(X, t)dt\right) = \int\limits_{a}^b \operatorname{\mathbb{E}}(f(X, t))dt$$ (here $a$ and $b$ both can be $-\infty$ or $+\infty$)? $\endgroup$
    – Ramil
    Apr 23, 2017 at 6:39
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    $\begingroup$ @Ramil, That's a good point. Assuming that $\Bbb{P}(X = 0) = 0$ (so that the expectation is well-defined), we have $$\Bbb{E}\left[ \int_{0}^{\infty} \left|XYe^{-(X^2+Y^2)t}\right| \, dt \right] = \Bbb{E}\left[ \frac{|XY|}{X^2+Y^2} \right] \leq \frac{1}{2} $$ and by Fubini's theorem you can still switch the order of integration. $\endgroup$
    – Sangchul Lee
    Apr 23, 2017 at 6:51
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    $\begingroup$ @Ramil, Actually, in the probability theory, expectation is simply an integral over the probability space: $$ \Bbb{E}[X] = \int_{\Omega} X(\omega) \Bbb{P}(d\omega). $$ Note that the notion of types (discrete/continuous/singular continuous) arise only when we begin to compare the law $\Bbb{P}(X \in \cdot)$ of $X$ with the Lebesgue measure on $\Bbb{R}$. In our cases, however, we are only appealing to the Fubini's theorem and we need not appeal to such notion. Everything can be equally carried out with this abstract integral. $\endgroup$
    – Sangchul Lee
    Apr 23, 2017 at 7:16
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    $\begingroup$ But if this abstract concept makes you uncomfortable, you can always rely on the representation $$ \Bbb{E}[X] = \int_{-\infty}^{\infty} x \, dF_X(x) $$ which is again true for any real-valued random variable $X$ having expectation and $F_X(x) = \Bbb{P}(X \leq x)$ is the cdf of $X$. $\endgroup$
    – Sangchul Lee
    Apr 23, 2017 at 7:20

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