Let $X_1$ and $X_2$ be i.i.d. random variables from a distribution $D$ on the real numbers with finite variance (and therefore finite mean). Assume that the probability of $X_i = 0$ is $0$. Must it be true that $$ \mathbb{E}\left[ \frac{X_1 X_2}{X_1^2 + X_2^2} \right] \ge 0? $$ If not, what is the infimum over all such distributions of this expectation?


The expectation is always finite. It is possible for the expectation to be $0$, when $D$ is symmetric about $0$. My conjecture is that this expectation is necessarily nonnegative. Of course without the denominator, $\mathbb{E}[X_1 X_2] = \mathbb{E}[X_1] \cdot \mathbb{E}[X_2] = \mu^2 \ge 0$. But with the denominator, it is not so clear.

I imagine this may be very elementary: I am not an expert in most inequalities used in probability theory.

I tried expanding the fraction with partial fractions over the complex numbers, getting $$ \frac{X_1 X_2}{X_1^2 + X_2^2} = \frac{\tfrac12 X_2}{X_1 + i X_2} + \frac{\tfrac12 X_2}{X_1 - i X_2}, $$ but I don't have an idea for how to evaluate these expectations, either.

This question is a result of my previous question. Specifically we can write $$ \mathbb{E}\left[ \frac{(X_1 + X_2)^2}{X_1^2 + X_2^2}\right] = 1 + 2 \cdot \mathbb{E}\left[ \frac{X_1 X_2}{X_1^2 + X_2^2} \right], $$ and in the answer to my previous question it seemed to be the case that the former expectation never goes below $1$. This is equivalent to the present question about the latter expectation.

  • $\begingroup$ What does your second sentence mean? $\endgroup$ – Ted Shifrin Aug 24 '18 at 17:58
  • 1
    $\begingroup$ @TedShifrin Probability of $0$ under distribution $D$ is $0$. (Necessary to prevent division by $0$). What is unclear about it currently? $\endgroup$ – 6005 Aug 24 '18 at 18:02
  • 1
    $\begingroup$ Oh, sorry, I misread it. Thanks. $\endgroup$ – Ted Shifrin Aug 24 '18 at 18:17

One can check that the `kernel' $k(u,v)=uv/(u^2+v^2)$ is positive semidefinite. For instance by noting that $$\tag{1}k(u,v)=\int_0^\infty (ue^{-u^2x})(ve^{-v^2x})\,dx.$$ See this wikipedia article for basic facts about these functions.

The desired inequality is a direct consequence of this: your expectation, $\mathbb Ek(X_1,X_2)$ is one of the quadratic expressions guaranteed to be non-negative by the PSD property of $k$, or is approximated by such expressions.

In greater detail: Since the finitely supported probability measures are dense in the space of all probability measures on $\mathbb R$, in the weak topology, there exists a sequence of finitely supported probability measures $P_n$ converging weakly to the probability distribution of $X_1$. Since $k$ is continuous and bounded, we have $$\mathbb E k(X_1,X_2) = \lim_n \iint k(u,v) P_n(du) P_n(dv).$$ Assume $P_n$ assigns measure $p_i$ to $u_i$, for finitely many values of $i$ (I'm suppressing the notation for the dependence on $n$ here), so $P_n = \sum_i p_i \delta_{u_i}$. Then $\iint k(u,v) P_n(du) P_n(dv)=\sum_{i,j} p_i p_j k(u_i,u_j);$ this latter quantity is known to the non-negative by the positive definiteness of $k$. So $\mathbb E k(X_1,X_2)$ is the limit of non-negative quantities, so is also non-negative.

Another way of using the integral representation (1) above is to notice that $$\mathbb E k(X_1,X_2)= \int_0^\infty \mathbb E(X_1\exp(-tX_1^2))\, \mathbb E(X_1\exp(-tX_2^2))\,dx = \int_0^\infty \left(\mathbb E X_1\exp(-t X_1^2)\right)^2\,dt\ge 0.$$ One needs to use something like the Tonelli theorem to justify the equation here.

  • $\begingroup$ Thanks! So: we define $f_u: [0,\infty) \to \mathbb{R}$ by $f_u(x) = ue^{-u^2 x}$, we note that $f_u \in L^2$ and $k(u,v) = \langle f_u, f_v \rangle$. Then we know that any such-defined kernel function is positive semidefinite by this definition since $\sum_{i,j=1}^n c_i c_j k(u_i,v_j) = \langle \sum_i c_i f_{u_i}, \sum_j c_j f_{u_j} \rangle = \| \sum_i c_i f_{u_i} \|_2^2$. Finally the desired expectation is $\mathbb{E}[k(X_1, X_2)] = \int \int k(x_1, x_2) d \mu \; d \mu$, ... $\endgroup$ – 6005 Aug 26 '18 at 15:32
  • $\begingroup$ ... which is nonnegative. Do you have a reference or name for this last theorem -- i.e. that this is "one of the quadratic expressions guaranteed to be nonnegative by the PSD property of $k$"? $\endgroup$ – 6005 Aug 26 '18 at 15:33
  • 1
    $\begingroup$ Yes, I agree with your detail in your first comment. I have added stuff to my answer, giving one way to fill in the dots. Some huffing and puffing is needed somewhere; I hope this is not excessive. $\endgroup$ – kimchi lover Aug 26 '18 at 16:02
  • $\begingroup$ Thanks a lot for your answer. $\endgroup$ – 6005 Aug 26 '18 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.