A strengthened truncated-mean inequality for sums of uniform random variables

About 4 years old · traced to 2022

Let $U_1,\ldots,U_n$ be independent uniform random variables on $[-1,1]$ , and set

X=\frac{1}{n}\sum_{i=1}^n U_i.

For $-1<t<1$ , define the truncated mean

m_X(t)=\mathbb{E}[X\mid X>t].

Strengthened truncated-mean inequality. For every $-1<t<1$ ,

\mathopen{}\mathclose\bgroup\originalleft(m_X(t)-t\aftergroup\egroup\originalright)\frac{m_X(t)}{1-m_X(t)^2}<\frac{1}{2n}.

This is presented as a stronger inequality motivated by numerical experiments; the supplied text gives no resolution status.

References

Primary source

Shoni Gilboa, Pazit Haim-Kislev and Boaz Slomka, “Isobarycentric Inequalities”, arXiv:2202.07527 (2022).

inequalities probability

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-posted argument claims a complete proof of the inequality and sharpness of its constant, but no independent verification was found.

The strengthened inequality is presented as a conjectural truncated-mean statement arising from the probabilistic work of Shoni Gilboa, Pazit Haim-Kislev, and Boaz Slomka (2022). The cited discussion identifies it as Conjecture 1.5 of that paper.

Posted attempt

An unverified reader-posted proof claims the result for every $n \ge 1$ and $-1<t<1$ , via the identity $m(m-t)=\frac{1-m^2}{2n}-\left(1+\frac{1}{2n}\right)V-\frac{D}{2n}$ , where $V$ is the conditional variance and $D\ge0$ is a conditional dispersion term. It concludes strictness from $V>0$ and claims the constant is sharp as $t\uparrow1$ . The argument has not been independently verified.

Current status (as of August 2026): A complete proof and sharpness claim have been posted, but they remain unverified; absent confirmation, the conjecture is not settled.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

The conjecture holds for every $n\ge1$ and every $-1<t<1$ . In fact, there is an exact positive-deficit identity, and the constant $1/(2n)$ is sharp.

Write

X=\frac1n\sum_{i=1}^nU_i,\qquad m=\mathbb E[X\mid X>t],

and define

V=\operatorname{Var}(X\mid X>t), \qquad D= \mathbb E\left[ \frac1n\sum_{i=1}^n(U_i-X)^2 \,\middle|\,X>t \right].

We prove

\boxed{\displaystyle m(m-t) = \frac{1-m^2}{2n} - \left(1+\frac1{2n}\right)V - \frac D{2n}.} \tag{1}

For $U\sim\operatorname{Unif}[-1,1]$ , integration by parts gives

\mathbb E[Uh(U)] = \mathbb E\left[ \frac{1-U^2}{2}h'(U) \right],

since the boundary term vanishes at $U=\pm1$ . Apply this conditionally in coordinate $U_i$ to

h(U_i)=(X-t)_+, \qquad h'(U_i)=\frac1n\mathbf1_{\{X>t\}}

almost everywhere. Summing over the independent coordinates yields

\mathbb E[X(X-t)_+] = \frac1{2n^2} \mathbb E\left[ \sum_{i=1}^n(1-U_i^2) \mathbf1_{\{X>t\}} \right]. \tag{2}

Divide by $\mathbb P(X>t)>0$ . The left side becomes

\mathbb E[X^2\mid X>t]-tm =V+m(m-t).

Meanwhile,

\frac1n\sum_{i=1}^nU_i^2 = X^2+\frac1n\sum_{i=1}^n(U_i-X)^2,

so the right side of (2), after conditioning, becomes

\frac{1-m^2-V-D}{2n}.

Equating these expressions proves (1).

Every nonempty upper tail has strictly positive conditional variance $V>0$ , while $D\ge0$ and $0<m<1$ . Therefore

\boxed{\displaystyle \frac{m(m-t)}{1-m^2} = \frac1{2n} - \frac{(2n+1)V+D}{2n(1-m^2)} < \frac1{2n}.}

This is precisely the conjecture, with an explicit strictness gap.

Finally, if $1-2/n\le t<1$ and $h=1-t$ , the coordinate deficits $1-U_i$ , conditioned on $X>t$ , are uniformly distributed on

\left\{y_i\ge0:\sum_{i=1}^ny_i<nh\right\}.

Hence

m=1-\frac{nh}{n+1}

and

\frac{m(m-t)}{1-m^2} = \frac{m}{n(1+m)} \longrightarrow\frac1{2n} \qquad(t\uparrow1).

Thus the conjectured constant is best possible for every $n$ .

Source: S. Gilboa, P. Haim-Kislev and B. Slomka, Isobarycentric Inequalities, International Mathematics Research Notices 2023(14), 12298–12323, doi:10.1093/imrn/rnac191; arXiv:2202.07527, Conjecture 1.5.