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

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Let U1,,UnU_1,\ldots,U_n be independent uniform random variables on [1,1][-1,1], and set

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

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

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

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

\originalleft(mX(t)t\aftergroup\originalright)mX(t)1mX(t)2<12n.\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.

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Current status (as of August 2026): The problem appears open, with no recorded activity or verified result.

Sources & referencesView supporting material

Primary source

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

Solutions 1

Proof

The conjecture holds for every n1n\ge1 and every 1<t<1-1<t<1. In fact, there is an exact positive-deficit identity, and the constant 1/(2n)1/(2n) is sharp.

Write

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

and define

V=Var(XX>t),D=E[1ni=1n(UiX)2|X>t].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

m(mt)=1m22n(1+12n)VD2n.(1)\boxed{\displaystyle m(m-t) = \frac{1-m^2}{2n} - \left(1+\frac1{2n}\right)V - \frac D{2n}.} \tag{1}

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

E[Uh(U)]=E[1U22h(U)],\mathbb E[Uh(U)] = \mathbb E\left[ \frac{1-U^2}{2}h'(U) \right],

since the boundary term vanishes at U=±1U=\pm1. Apply this conditionally in coordinate UiU_i to

h(Ui)=(Xt)+,h(Ui)=1n1{X>t}h(U_i)=(X-t)_+, \qquad h'(U_i)=\frac1n\mathbf1_{\{X>t\}}

almost everywhere. Summing over the independent coordinates yields

E[X(Xt)+]=12n2E[i=1n(1Ui2)1{X>t}].(2)\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 P(X>t)>0\mathbb P(X>t)>0. The left side becomes

E[X2X>t]tm=V+m(mt).\mathbb E[X^2\mid X>t]-tm =V+m(m-t).

Meanwhile,

1ni=1nUi2=X2+1ni=1n(UiX)2,\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

1m2VD2n.\frac{1-m^2-V-D}{2n}.

Equating these expressions proves (1).

Every nonempty upper tail has strictly positive conditional variance V>0V>0, while D0D\ge0 and 0<m<10<m<1. Therefore

m(mt)1m2=12n(2n+1)V+D2n(1m2)<12n.\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 12/nt<11-2/n\le t<1 and h=1th=1-t, the coordinate deficits 1Ui1-U_i, conditioned on X>tX>t, are uniformly distributed on

{yi0:i=1nyi<nh}.\left\{y_i\ge0:\sum_{i=1}^ny_i<nh\right\}.

Hence

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

and

m(mt)1m2=mn(1+m)12n(t1).\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 nn.

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.

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