Let mathbfX_1,…,mathbfX_n be i.i.d. matrices with common mean matrix EmathbfX_1=mathbfM and pth central moment matrix mathbfV_p:=Ebigl(operatornameabs(mathbfX_1-mathbfM)bigr)^p. For any mathbfA∈S_d^++, Matrix p-Chebyshev inequality. There exists a function f:[1,2]×N→(0,∞) that grows sublinearly in its second argument and is bounded in its first argument, such that Ebigl(operatornameabs(mathbfX_1+…+mathbfX_n-n mathbfM)^p bigr)preceq n,f(p,d),mathbfV_p; and, consequently, for overlinemathbfX_n=frac1n(mathbfX_1+…+mathbfX_n), Pbigl(operatornameabs(overlinemathbfX_n-mathbfM)npreceq mathbfAbigr)leqslant n^1-pf(p,d)operatornametr(mathbfV_p mathbfA^-p). This is proposed as an ideal matrix extension of the scalar and vector p-Chebyshev inequalities. The source provides no resolution, so the conjecture remains open.

Matrix p-Chebyshev inequality

About 2 years old · traced to 2024

Let $\mathbf{X}_1,\dots,\mathbf{X}_n$ be i.i.d. matrices with common mean matrix $\mathbb{E}\mathbf{X}_1=\mathbf{M}$ and $p$ th central moment matrix

\mathbf{V}_p:=\mathbb{E}\bigl(\operatorname{abs}(\mathbf{X}_1-\mathbf{M})\bigr)^p.

For any $\mathbf{A}\in\mathcal{S}_d^{++}$ , Matrix $p$ -Chebyshev inequality. There exists a function $f:[1,2]\times\mathbb{N}\to(0,\infty)$ that grows sublinearly in its second argument and is bounded in its first argument, such that

\mathbb{E}\bigl(\operatorname{abs}(\mathbf{X}_1+\dots+\mathbf{X}_n-n\mathbf{M})^p\bigr)\preceq n\,f(p,d)\,\mathbf{V}_p;

and, consequently, for $\overline{\mathbf{X}}_n=\frac{1}{n}(\mathbf{X}_1+\dots+\mathbf{X}_n)$ ,

\mathbb{P}\bigl(\operatorname{abs}(\overline{\mathbf{X}}_n-\mathbf{M})\npreceq\mathbf{A}\bigr)\leqslant n^{1-p}f(p,d)\operatorname{tr}(\mathbf{V}_p\mathbf{A}^{-p}).

This is proposed as an ideal matrix extension of the scalar and vector $p$ -Chebyshev inequalities. The source provides no resolution, so the conjecture remains open.

References

Primary source

Hongjian Wang and Aaditya Ramdas, “Positive Semidefinite Matrix Supermartingales”, arXiv:2401.15567 (2025).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

Wang and Ramdas proposed a matrix version of Chebyshev’s inequality in 2024, and a posted construction now claims it fails in two dimensions for every subquadratic exponent, but that disproof has not been independently checked.

The conjecture asks for a dimension-sublinear moment bound for sums of independent centered matrices, with a corresponding tail inequality for their empirical mean. Wang and Ramdas introduced this question in 2024 alongside related concentration results.

Known results

Wang and Ramdas (2024): an exchangeable positive-semidefinite matrix $p$ -Chebyshev inequality using raw moments.
Wang and Ramdas (2024): a central-moment trace bound for exchangeable symmetric matrices, rather than the requested Loewner-order sum bound.
A 2024 result gives randomized matrix Chebyshev inequalities for $1\leq p<2$ , but only for a single random matrix.

Posted attempt

A posted construction claims a counterexample in dimension $d=2$ : for every fixed $1\leq p<2$ , four bounded positive-semidefinite observations force $f(p,2)$ to diverge as two rank-one projections become nearly aligned. It therefore claims the conjecture is false in the entire subquadratic range, while $p=2$ remains valid; the calculation has not been independently verified.

Current status (as of August 2026): The proposed inequality has an unverified claimed counterexample for $1\leq p<2$ , so that range is not settled; the $p=2$ case is reported as valid, and no verified resolution of the full statement is recorded.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

Source-version clarification: this statement appears as Conjecture A.4 only in versions 1–3 of Wang and Ramdas, https://arxiv.org/html/2401.15567v3#A1.SS2 . The conjecture was removed beginning with version 4 in January 2025 and is absent from the current version 6. The argument below addresses the withdrawn historical claim still reproduced on this problem page; it does not challenge the distinct valid results in the current paper.

In fact, for every fixed 1≤p<2, there is no finite distribution-independent constant f(p,2), even in dimension d=2 with n=2 bounded positive-semidefinite observations.

Fix 0<θ<π/2, let s=sin θ, and define rank-one orthogonal projections

Q=e₁e₁ᵀ, R=vvᵀ, v=(cos θ,sin θ).

Let Y be uniformly distributed on {Q,−Q,R,−R}, and let X=I+Y. Then 0≼X≼2I, its mean is M=I, and projection idempotence gives

V_p=E|X−M|^p=E|Y|^p=(Q+R)/2.

For independent copies Y₁,Y₂, the four ordered pairs

(Q,−R), (−R,Q), (−Q,R), (R,−Q)

have combined probability 1/4. Since

(Q−R)²=s²I,

each contributes |Y₁+Y₂|^p=s^pI. All other terms are positive semidefinite; therefore

E|Y₁+Y₂|^p≽(s^p/4)I.

Testing the conjectured matrix inequality

E|Y₁+Y₂|^p≼2f(p,2)V_p

against e₂ yields

s^p/4≤f(p,2)s²,

and consequently

f(p,2)≥s^{p−2}/4.

As θ↓0, this lower bound diverges for every 1≤p<2. Hence no finite dimension-two constant exists; allowing arbitrary dimension dependence cannot repair the claim.

There is also a fully rational certificate at p=1. For each integer m≥1 set

v=((m²−1)/(m²+1),2m/(m²+1)), R=vvᵀ,

and retain the same four-point law. Exact enumeration gives

E|Y₁+Y₂|=V₁+[m/(2(m²+1))]I.

For w=(−1,m), one has

wᵀV₁w=1, wᵀE|Y₁+Y₂|w=1+m/2,

so the claimed bound forces

f(1,2)≥1/2+m/4

for every m. At p=2 the usual variance identity does hold with f(2,2)=1; the obstruction concerns precisely the strictly subquadratic range.