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Gaussian mixture lower bound for weakly sparse covariance

Statement

Assume AC. Fix τ,M>0, β>1, and 0≤q<1. There are v,b>0 and P0<∞, depending only on these fixed parameters, such that the following holds. Let p≥P0, n≥1, p≥nβ and 0≤c≤Mn(1−q)/2(log⁡p)−(3−q)/2. Put a=log⁡(p)/n, s=τ/4, ϵ=va and k=⌊c2(sϵ)q⌋. If c/[2(sϵ)q]≥2, then Rn(Pq(τ,c))≥bc2a1−q. For q=0, (sϵ)q=1. The proof gives all Gaussian mixture and conditional remainder estimates locally; no external mixture-affinity lemma is assumed.

Facts & Assumptions

Given: The class, fixed constants, scales and nontrivial sparse-row regime in the statement.

[F1]

Under AC, a positive-definite covariance defines a Gaussian law with the specified covariance and normal unit-direction projections. Multivariate normal law, including singular covariance.

[F2]

Strong off-diagonal sums ∑i≠j∣σij∣q≤c imply the weak-ℓq condition; for q=0 the condition counts nonzero entries. Weak-lq covariance class with directional sub-Gaussian tails.

[F3]

Minimax risk is the infimum of maximal expected squared operator loss over measurable estimators. Minimax squared operator risk.

[F5]

Finite-sum Cauchy–Schwarz applies to Euclidean vectors and finite weighted sums, and the operator norm is the maximum Euclidean output norm over unit inputs. Euclidean spheres and closed balls as subspaces of Rn The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| The Cauchy-Schwarz inequality for finite sums.

[F6]

The independent standard-normal coordinates in the Gaussian construction have their product joint law; their scalar density is (2π)−1/2e−x2/2, product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets, and positive operators have nonnegative square roots. Multivariate normal law, including singular covariance Standard normal and normal laws Independent random elements have product joint law On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n} A non-negative operator has a unique non-negative square root.

[F7]

Under Countable Choice, an invertible linear change of variables satisfies the Borel Lebesgue substitution formula; the stated AC assumption supplies Countable Choice. Borel change of variables from the compact-support formula and Radon uniqueness AC supplies countable selections and prescribed serial paths.

[F8]

Tonelli's theorem factors the nonnegative cross-moment integrand across the finite product sample space. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product.

[F9]

Total variation distance is sup⁡E∣P(E)−Q(E)∣ for probability laws on a common measurable space. Total variation distance for probability laws.

[F10]

Cauchy–Schwarz for square-integrable random variables controls the density-ratio integral under a probability law. Cauchy-Schwarz for random variables.

Proof

1.1givenF1F2F4F5algebraconstruct

Choose v2<(β−1)/(4β) and put r=⌊p/2⌋. Write x=c/[2(sϵ)q], so x≥2 and k=⌊x⌋≥2 and k≥x/2. The radius condition gives x≤Cn/(log⁡p)3/2, hence k2≤Cn/(log⁡p)3, kϵ≤C/log⁡p, and nϵ4k2≤C/log⁡p. Also k2p2v2/r≤Cp1/β+2v2−1/(log⁡p)3→0, uniformly over admitted n,c. All constants depend only on fixed parameters. Increase P0 so r≥16k, 2kϵ≤1/3, and the small bounds below hold. Let Λ be all r×r zero-one matrices with every row sum k and every column sum at most 2k; cyclic shifts of a k-set show it is nonempty. For γ∈{0,1}r, embed Dγλ as the upper-right block of a symmetric off-diagonal matrix A(γ,λ), using two disjoint r-coordinate blocks and an extra zero coordinate if p is odd, and put Σγ,λ=s(Ip+ϵA(γ,λ)). Every absolute row sum of A is at most 2k. For any z, ∣zTAz∣≤12∑j,l∣Ajl∣(zj2+zl2)≤2k∥z∥22; applying this to unit eigenvectors and using [F4] bounds the eigenvalues of A in [−2k,2k]. Thus Σγ,λ is positive definite with eigenvalues in [2s/3,4s/3]⊂(0,τ]. Each column has at most 2k off-diagonal entries of magnitude sϵ. If 0<q<1, its weak-ℓq quantity is at most 2k(sϵ)q≤c; if q=0, its nonzero count is at most 2k≤c. Hence [F2] puts every covariance in the radius-c class. The standard-normal density gives P(∣Z∣>t)=e−t2/22/π∫0∞e−w2/2−tw dw≤e−t2/2, so [F1] verifies the directional-tail condition. For any rectangular matrix M and vector z, [F5] gives ∣(Mz)i∣2≤(∑j∣Mij∣)(∑j∣Mij∣zj2). Summing over rows yields ∥Mz∥22≤(max⁡i∑j∣Mij∣)(max⁡j∑i∣Mij∣)∥z∥22, hence ∥M∥op2≤(max⁡i∑j∣Mij∣)(max⁡j∑i∣Mij∣). Therefore each cross-block matrix C=ϵDγ−1L below has ∥C∥op≤2kϵ≤2kϵ.

2.1givenstep 1.1algebra

Give (γ,λ) the uniform prior on {0,1}r×Λ. Fix bit i, say i=1 by row symmetry, and condition on the remaining bit vector γ−1 and nuisance rows L=(λ2,…,λr). The number of possible first-row supports is N(L)=(d(L)k), where d(L) counts columns whose nuisance sum is less than 2k. Since there are (r−1)k nuisance ones, d(L)≥r/2. The marginal prior of L is π(L)=N(L)/∣Λ∣, independent of γ−1; conditional on L, a support A is uniform among its N(L) available k-sets. For each fixed γ−1, when γ1=0 let Qγ−1,L be the Gaussian n-sample law, which does not depend on A; when γ1=1, let Pγ−1,A,L be the n-sample law and P‾γ−1,L its uniform support mixture. In the chi-square expansion draw two independent available supports A,B given L. Their joint mass with L is proportional to 1/N(L), so given A,B,γ−1 the nuisance rows are not asserted uniform.

3.1givenF1F4F6F7F8step 1.1step 2.1algebra

Fix γ−1 and let C∈Rr×r be the cross-block matrix with first row zero and remaining rows ϵγlλl. Under Qγ−1,L, after canceling s, the covariance is B0=(IrCCTIr), with an extra identity coordinate if p is odd. Put R=(Ir−CTC)1/2; it is positive definite because ∥C∥op<1. The block matrix T=(Ir0CTR) satisfies TTT=B0. Whiten by T−1, extended by the identity on any extra coordinate. If e=e1 is the first coordinate in the left block and vA=1A is the right-block support vector, then T−1e=e and T−1vA=R−1vA. Thus the alternative covariances become I+ΔA and I+ΔB, where ΔA=euT+ueT, ΔB=ewT+weT, u=ϵR−1vA, w=ϵR−1vB, and u,w⊥e. With H=(Ir−CTC)−1, uTw=t=ϵ21ATH1B. The operator norms of ΔA,ΔB equal ∥u∥2,∥w∥2, respectively. Since ∥R−1∥op≤(1−∥C∥op2)−1/2, these norms are at most ϵk/1−∥C∥op2<1/4 for large P0; here ϵk=(kϵ)/k≤C/(2log⁡p). For positive-definite S, [F1], [F6] and [F7] give the Lebesgue density fS(x)=(2π)−p/2(det⁡S)−1/2exp⁡(−xTS−1x/2): the independent standard-normal coordinates have product density, and the invertible linear map S1/2 changes variables with determinant (det⁡S)1/2. For the one-observation laws, write fA,fB,f0 for the positive densities of the two alternatives and the whitened baseline. Put M=(I+ΔA)−1+(I+ΔB)−1−I. Each inverse has smallest eigenvalue at least (1+1/4)−1=4/5, hence M⪰(3/5)I. The product density fAfB/f0 has exponent matrix M and normalizing factor (2π)−p/2[det⁡(I+ΔA)det⁡(I+ΔB)]−1/2. Diagonalizing M by [F4], applying an orthogonal change of variables and scaling each one-dimensional standard-normal density [F6, F7] gives ∫fA(x)fB(x)f0(x) dx=[det⁡(I+ΔA)det⁡(I+ΔB)det⁡M]−1/2. The identity (I+ΔA)M(I+ΔB)=I−ΔAΔB and determinant multiplicativity For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B) reduce it to det⁡(I−ΔAΔB)−1/2. Since u,w⊥e, ΔAΔB=teeT+uwT. On span⁡{e}, I−ΔAΔB contributes determinant 1−t. On e⊥, if u=0 the rank-one determinant is 1; otherwise choose an orthonormal basis whose first vector is u/∥u∥2, so I−uwT has only its first row changed and determinant 1−wTu=1−t. The full determinant is (1−t)2, and the one-observation cross integral is (1−t)−1. The n-sample laws have product densities by the product joint-law supplier in [F6]; repeated Tonelli [F8] therefore makes their cross integral the nth power, (1−t)−n.

4.1givenF4F5step 1.1step 3.1algebra

Write J=∣A∩B∣, Xl=∣A∩supp⁡λl∣ and Yl=∣B∩supp⁡λl∣. The identity H−I=CT(I−CCT)−1C follows from H−I=(I−CTC)−1CTC and (I−CTC)CT=CT(I−CCT). The Neumann series (I−CCT)−1=∑m≥0(CCT)m converges because ∥C∥op<1; since C has nonnegative entries, H−I is entrywise nonnegative and t−Jϵ2≥0. The eigenvalues of CCT lie in [0,∥C∥op2], so [F4] gives ∥(I−CCT)−1∥op≤(1−∥C∥op2)−1. Also (C1A)l=ϵγlXl and (C1B)l=ϵγlYl. Cauchy--Schwarz [F5] and 2ab≤a2+b2 yield 0≤t−Jϵ2≤ϵ42(1−∥C∥op2)∑l=2r(Xl2+Yl2). Since ∥H∥op≤(1−∥C∥op2)−1 and ∥1A∥2=∥1B∥2=k, [F5] also gives t≤ϵ2k/(1−∥C∥op2). Since k≥2 and kϵ≤C/log⁡p, ϵ2k=(kϵ)2/k=o(1), so t=o(1). The derivative of −log⁡(1−x) is at most 2 for 0≤x≤t≤1/2; since 1−∥C∥op2≥8/9, the preceding bound gives (1−t)−n≤(1−Jϵ2)−nexp⁡{θ∑l=2r(Xl2+Yl2)},θ=2nϵ4.

5.1givenstep 1.1step 2.1step 4.1algebra

Fix A,B,γ−1 and all nuisance rows except row l, and let aj be the sum in column j over those fixed rows. A support S for row l is feasible exactly when ∣S∣=k, aj+1{j∈S}≤2k for every column, and aj+1{j∈S}<2k for every j∈A∪B (so both first-row supports remain available). Thus before prior weighting row l is uniform over the k-subsets of the fixed set V={j∈A∪B:aj≤2k−2}∪{j∉A∪B:aj≤2k−1}. At most (r−2)k/(2k)≤r/2 columns outside A∪B have aj=2k, so deleting these and A∪B leaves at least r/2−2k≥r/4 available columns; hence d0=∣V∣≥r/4. The first-row availability count is d(L)=#{j:aj+1{j∈S}<2k}; as S varies, only columns outside A∪B with aj=2k−1 can change this count, so its range has width at most k≤2k. Since d(L)≥r/2 and (d+1k)/(dk)=(d+1)/(d+1−k), the ratio of any two weights 1/N(L) is at most exp⁡(Ck2/r)≤2 after increasing P0. Thus the actual conditional density of the row support is at most twice the uniform-subset density on V. If m=∣A∩V∣, singleton and ordered-pair inclusion probabilities give EXl=mk/d0≤Ck2/r and E[Xl(Xl−1)]=m(m−1)k(k−1)/(d0(d0−1))≤Ck4/r2. Hence EXl2≤C(k2/r+k4/r2); the same bound holds for Yl and under the weighted law with twice the constant. Since θk2=o(1), ez−1≤2z for 0≤z≤1 yields E[exp⁡{θ(Xl2+Yl2)}∣all other rows,A,B,γ−1]≤1+Cθ(k2/r+k4/r2).

6.1givenstep 1.1step 5.1algebra

Put fl=exp⁡{θ(Xl2+Yl2)} and B0=1+Cθ(k2/r+k4/r2). Every factor except fl is measurable when all rows except l are fixed. Thus for any subset S containing l, conditional expectation gives E[∏j∈Sfj∣A,B,γ−1]≤B0E[∏j∈S∖{l}fj∣A,B,γ−1]. Iterating over the finitely many nuisance rows yields E[exp⁡{θ∑l=2r(Xl2+Yl2)}∣A,B,γ−1]≤B0r−1≤exp⁡{Cnϵ4(k2+k4/r)}=1+o(1) uniformly in A,B,γ−1. No row independence is used.

7.1givenF5F9F10step 1.1step 2.1step 3.1step 4.1step 6.1algebra

Given L, A,B are independent uniform k-subsets of a set of size d≥r/2. Expanding zJ=(1+(z−1))J and using E(Jj)=(kj)(k)j(d)j≤[k2/(d−k+1)]jj! gives E[zJ∣L]≤exp⁡{k2(z−1)/(d−k+1)}. Since J≤k and ϵ2k=o(1), (1−Jϵ2)−n≤exp⁡(2nϵ2J)=p2v2J. Define χ2(P‾γ−1,L,Qγ−1,L)=∫(dP‾γ−1,L/dQγ−1,L−1)2 dQγ−1,L. Finite expansion of the mixture density gives 1+χ2(P‾γ−1,L,Qγ−1,L)=EA,B∣L,γ−1(1−t)−n. The posterior of L given A,B,γ−1 has weights proportional to 1/N(L); step 6.1 bounds the conditional remainder factor by 1+o(1) uniformly in A,B,γ−1. Since d−k+1≥r/3, the overlap generating-function bound is uniform in L,γ−1 and yields Eγ−1,L[1+χ2(P‾γ−1,L,Qγ−1,L)]≤(1+o(1))exp⁡(Ck2p2v2/r)=1+o(1), using step 1.1. For each fixed (γ−1,L), if R=dP‾/dQ, then EQ(R−1)=0, so the positive and negative parts of R−1 have equal integrals, and the event {R≥1} attains the supremum in [F9]; hence TV⁡(P‾,Q)=12EQ∣R−1∣; [F10] bounds this by 12χ2(P‾,Q). Finite Cauchy--Schwarz [F5] over the common prior weights on (γ−1,L) bounds the average conditional total variation by 12Eγ−1,Lχ2, at most 1/2 after increasing P0. The same mixing weights are used under bit i=0 and bit i=1, so the full sample-law mixtures have total variation at most 1/2 by the triangle inequality for the density integral. Their density affinity satisfies ∫min⁡(dPi,0,dPi,1)=1−TV⁡(Pi,0,Pi,1)≥1/2.

8.1givenF3F4F5step 1.1step 7.1algebra∎

For parameters with h different bits, apply their covariance difference to the vector that is one on the right block and zero elsewhere. Each changed bit contributes a distinct left-block coordinate of magnitude sϵk, independently of support choices, so the squared output norm is at least h(sϵk)2 while the input squared norm is r. Hence ∥Σθ−Σθ′∥op2≥h(sϵk)2/r. Given any measurable matrix estimator, choose a nearest finite covariance target with a fixed tie order. The finite distance comparisons are measurable: [F5] bounds the operator norm by the Frobenius norm, and its triangle inequality follows by applying the Euclidean triangle inequality to every unit vector, so distance to each fixed target is continuous. A wrong target decision forces squared estimation loss at least one quarter of the squared target separation. For each bit, the sum of its two conditional testing errors is at least the density affinity, since pointwise the test contributes at least the minimum of the two densities. By step 7.1, under the uniform prior each bit error probability is at least 1/4, so the average number of bit errors is at least r/4. The maximal risk is at least the average risk and therefore at least (sϵk)2/16. Finally, k≥c/[4(sϵ)q] implies this is at least bc2a1−q for a positive constant depending only on the fixed parameters.

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