# Prove that $E\left(\frac{XY}{X^2+Y^2}\right) \geqslant 0$ for i.i.d. $X$ and $Y$

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.

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

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 $k(t, x) = x e^{-x^2 t} \mathbf{1}_{\{t \geq 0\}}$.

• 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
– πr8
Apr 22, 2017 at 22:52
• @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 $+\infty$)? Apr 23, 2017 at 6:39
• @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. Apr 23, 2017 at 6:51
• @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. Apr 23, 2017 at 7:16
• 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 $X$. Apr 23, 2017 at 7:20