Random variable $X$ is symmetric iff $\mathbb{E}\left(X / \left(1 + r^2X^2\right)\right) = 0$ for any $r \in \mathbb{R}$. Can you prove/disprove the following claim?
Let $X$ be a random variable, which takes values in $\mathbb{R}$. Assume that $\mathbb{E}\left(X / \left(1 + r^2X^2\right)\right)$ is defined and finite for any $r \in \mathbb{R}$. The density of $X$ is symmetric about $0$ iff $$\mathbb{E}\left(\frac{X}{1 + r^2X^2}\right) = 0$$ for any $r \in \mathbb{R}$.
A couple of observations:

*

*If $X$ is symmetric about $0$ then the expectation is $0$ for any $r \in \mathbb{R}$ because $x/(1+r^2x^2)$ is an odd function.

*If the expectation condition was instead that $\mathbb{E}(g(X)) = 0$ for any odd function $g$ then it would be clear that $X$ must be symmetric, since we could just choose functions of the form
$$g_s(x) = \begin{cases}
-1 & \text{if } x \in (-s - \epsilon, -s),\\
1 & \text{if } x \in (s, s + \epsilon),\\
0 & \text{otherwise}
\end{cases}$$
for any $s > 0$ and arbitrarily small $\epsilon$'s and just check that the density of $X$ is symmetric.


Context: I was watching a conference talk about e-values (more info about e-values here) and the speaker claimed (unless I misunderstood) that the following hypotheses are equivalent:
$$X \textrm{ is symmetric},$$
and
$$\mathbb{E}\left(1 + \frac{rX}{1 + r^2X^2}\right) \leq 1$$
for any $r \in \mathbb{R}$. The speaker said that the claim can be shown but I couldn't find the proof in any of his citations.
 A: Writing $\beta=1/r^2$, the hypothesis amounts to
$$
0=E\left({X\over \beta+X^2}\right),\qquad\forall \beta>0.
$$
But $1/(\beta+X^2)=\int_0^\infty e^{-t(\beta+X^2)} dt$,
and so we  have (by Fubini)
$$
0=\int_0^\infty e^{-\beta t} E\left[Xe^{-tX^2}\right] dt,\qquad\forall \beta>0.
$$
The function $t\mapsto E[X\exp(-tX^2)]$ is continuous on $(0,\infty)$ (dominated convergence), so by Laplace inversion,
$$
E[Xe^{-tX^2}]=0,\qquad\forall t>0.\qquad\qquad(1)
$$
By Stone-Weierstrass
$$
0=E[Xg(X^2)e^{-tX^2}],\qquad\forall t>0,
$$
for each continuous $g:[0,\infty)\to\Bbb R$ with $\lim_{x\to\infty} g(x) =0$.
In particular, for each continuous $g$ of compact support, upon letting $t\downarrow 0$ we see that
$$
0=E[Xg(X^2)].
$$
Consequently, for each $C^1$ odd function $h$ of compact support
$$
0=E[h(X)],
$$
which means that $-X$ has the same distribution as $X$.
(And if $X$ has a density, the it's symmetric about $0$.)
A: Would it be okay to assume that $r^nE[|X|^n] \to 0$ for some $r>0$, i.e. the moments of $X$ do not grow faster than geometrically? If so, here would be a proof. In particular the mgf of $X$ exists for all $t$ small enough. It's moment generating function $t \mapsto E[e^{tX}] = \sum_n \frac{t^n}{n!}E[X^n]$  is determined by the moments of $X$. Same goes for $-X$ which has mgf  $t \mapsto E[e^{t(-X)}] = E[e^{-tX}] =\sum_n \frac{(-t)^n}{n!}E[X^n]$. Thus in this case to show that $X$ is symmetric, it suffices to show that all its odd moments are zero, since that would imply $X$ and $-X$ have the same mgf and hence the same distribution.
Now let's show your assumption implies all odd moments are zero. Note that by Taylor series $$\frac{rx}{1+(rx)^2} = rx - (rx)^3 + (rx)^5 - \cdots,$$ which is valid for all values of $x \in \mathbb{R}$. Plugging in $X$, taking expectations and using your assumption gives that
$$
E[rX - (rX)^3 + (rX)^5 - \cdots] = 0
$$
for all $r \in \mathbb{R}$.
Since $r_0^nE[|X|^n] \to 0$ for some $r_0>0$ small enough (our initial assumption), we have for $|r| < r_0$
$$
0 = E[rX - (rX)^3 + (rX)^5 - \cdots] = r E[X] - r^3 E[X^3] + r^5 E[X^5] - \cdots.
$$
The right hand side is a convergent power series in $r$ for $|r|<r_0$ that is identically zero, and hence its coefficients are all zero, giving that $E[X^n] = 0$ for all odd $n$, completing the proof.
