Problem 2.2: Vandermonde Moment Matrix is Positive Definite

From Rigollet Chapter 2, Problem 2.2.

Let X = (1, Z, ..., Z^{d-1})ᵀ ∈ ℝᵈ be a random vector where Z is a random variable. Show that the matrix E(XXᵀ) is positive definite if Z admits a probability density with respect to the Lebesgue measure on ℝ.

Key idea

For any nonzero c ∈ ℝᵈ, we have cᵀ E[XXᵀ] c = E[(cᵀX)²] = E[(Σᵢ cᵢ Z^i)²] ≥ 0, with equality only if the polynomial p(z) = Σᵢ cᵢ z^i is zero a.e. w.r.t. μ. Since μ is absolutely continuous with respect to Lebesgue measure, p must be zero Lebesgue-a.e., which means p is the zero polynomial (a nonzero polynomial of degree < d has at most d-1 roots). This contradicts c ≠ 0.

noncomputable def Rigollet.vandermondeVec (d : ℕ) (z : ℝ) : Fin d → ℝ

The Vandermonde vector: given a real number z, produces the vector (1, z, z², ..., z^{d-1}) ∈ ℝᵈ.

noncomputable def Rigollet.vandermondeMomentMatrix (d : ℕ) (μ : MeasureTheory.Measure ℝ) : Matrix (Fin d) (Fin d) ℝ

The moment matrix E[XXᵀ] for the Vandermonde vector X = (1, Z, ..., Z^{d-1})ᵀ. The (i,j) entry is E[Z^{i+j}].

theorem Rigollet.problem_2_2_vandermonde_pos_def (d : ℕ) (hd : 0 < d) (μ : MeasureTheory.Measure ℝ) (hμ : MeasureTheory.IsProbabilityMeasure μ) (habs : μ.AbsolutelyContinuous MeasureTheory.volume) : (vandermondeMomentMatrix d μ).PosDef

Problem 2.2: If Z has a density with respect to Lebesgue measure on ℝ (i.e., the distribution μ is absolutely continuous w.r.t. volume), then the moment matrix E[XXᵀ] for X = (1, Z, ..., Z^{d-1})ᵀ is positive definite.

This holds because cᵀ E[XXᵀ] c = E[(p(Z))²] where p(z) = Σᵢ cᵢ z^i, and a nonzero polynomial cannot vanish on a set of positive Lebesgue measure.