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Atlas.HighDimensionalStatistics.code.Chapter1.Problem_1_4

Problem 1.4: Sub-Gaussian random matrices #

noncomputable def Problem_1_4.matrixOpNorm {n m : } (A : Matrix (Fin n) (Fin m) ) :
Instances For
    theorem Problem_1_4.problem_1_4a_matrix_subgaussian {Ω : Type u_1} {x✝ : MeasurableSpace Ω} {μ : MeasureTheory.Measure Ω} (x✝¹ : MeasureTheory.IsProbabilityMeasure μ) {n m : } {A : ΩMatrix (Fin n) (Fin m) } {σsq : } (hSG : ∀ (i : Fin n) (j : Fin m), IsSubGaussian (fun (ω : Ω) => A ω i j) σsq μ) (hIndep : ProbabilityTheory.iIndepFun (fun (p : Fin n × Fin m) (ω : Ω) => A ω p.1 p.2) μ) (x : EuclideanSpace (Fin m)) (y : EuclideanSpace (Fin n)) (hx : x = 1) (hy : y = 1) :
    IsSubGaussian (fun (ω : Ω) => inner ((Matrix.toEuclideanLin (A ω)) x) y) σsq μ

    Problem 1.4 (a)

    theorem Problem_1_4.integral_exp_sub_one_sub {Ω : Type u_1} [MeasurableSpace Ω] {μ : MeasureTheory.Measure Ω} [MeasureTheory.IsProbabilityMeasure μ] {f : Ω} (hf : MeasureTheory.Integrable f μ) (hef : MeasureTheory.Integrable (fun (ω : Ω) => Real.exp (f ω)) μ) (hmean : (ω : Ω), f ω μ = 0) :
    (ω : Ω), Real.exp (f ω) - 1 - f ω μ = (ω : Ω), Real.exp (f ω) μ - 1

    Helper: integral of exp(f) - 1 - f equals E[exp(f)] - 1 when E[f] = 0.

    If X is sub-Gaussian with σ² = 0, then X = 0 a.e. Proof: E[exp(sX)] ≤ 1 for all s. By Jensen, E[exp(X)] ≥ exp(E[X]) = 1. So E[exp(X)] = 1. Since exp(x) ≥ 1 + x with equality iff x = 0, the function g(ω) = exp(X(ω)) - 1 - X(ω) is nonneg with integral 0, hence 0 a.e. This forces X = 0 a.e.

    theorem Problem_1_4.problem_1_4b_operator_norm_bound {Ω : Type u_1} {x✝ : MeasurableSpace Ω} {μ : MeasureTheory.Measure Ω} (x✝¹ : MeasureTheory.IsProbabilityMeasure μ) {n m : } {A : ΩMatrix (Fin n) (Fin m) } {σsq : } (hσsq : 0 σsq) (hSG : ∀ (i : Fin n) (j : Fin m), IsSubGaussian (fun (ω : Ω) => A ω i j) σsq μ) (hIndep : ProbabilityTheory.iIndepFun (fun (p : Fin n × Fin m) (ω : Ω) => A ω p.1 p.2) μ) (hMeas : ∀ (i : Fin n) (j : Fin m), Measurable fun (ω : Ω) => A ω i j) :
    ∃ (C : ), 0 < C (ω : Ω), matrixOpNorm (A ω) μ C * σsq * (m + n)

    Problem 1.4 (b)