Equality cases in a certain case of Jensen's inequality

Question

Suppose that $Y$ is an independent copy of a random variable (r.v.) $X$ with a zero-mean nondegenerate distribution. Is there a non-tautological, preferably simple characterization of the cases when $$E|X-Y|=E|X|\,\text{?} \tag{1}\label{1}$$

This question is a modification of this previous one.

The inequality $E|X-Y|\ge E|X|$ always holds. To get it, condition on $X$ and then apply Jensen's inequality to the zero-mean random variable $Y$.

If $p\in(1,\infty)$ and $E|X|^p<\infty$, then $E|X-Y|^p>E|X|^p$ -- because then the function $|\cdot|^p$ is strictly convex.

In view of the well-known expression of the absolute moments in terms of the characteristic function (c.f.), equality \eqref{1} can be restated as $$\int_0^\infty\frac{dt}{t^2}\,(\Re f(t)-|f(t)|^2)=0.$$

Answer 1

Since the distribution of the r.v. $X$ is zero-mean, $X$ integrable (hence also the copy $Y$).

The equality is attained iff the distribution of $X$ is carried by at most two points.

Indeed, by independence of $X$ and $Y$, $$E\big(|X-Y|\big|X\big) = f(X) \text{ where } f(x) := E|x-Y|.$$ Hence $$f(X) =E\big(|X-Y|\big|X\big) \ge |X| \text{ a.s.}$$ Looking at the expectations, one deduces that
\begin{equation} \begin{aligned} E|X-Y| = E|X| &\iff f(X) = |X| \text{ a.s.} \\ &\iff \text{ for $P_X$ a.e. } x, \quad f(x) = |x| \\ &\iff \text{ for $P_X$ a.e. } x, E|x-Y| = |E(x-Y)| \\ &\iff \big(\text{ for $P_X$ a.e. } x, Y-x \ge 0 \text{ a.s. or } Y-x \le 0 \text{ a.s.}\big) \\ &\iff \big( \text{ for $P_X$ a.e. } x, \mathrm{ess}\inf Y \ge x \text{ or } \mathrm{ess}\sup Y \le x \big). \end{aligned} \end{equation}

Since $X$ and $Y$ have the same distribution, the last condition means that the support of $P_X$ contains at most two points (namely the essential inf and the essential sup).