Problem 1.4: Sub-Gaussian random matrices #
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 : ℕ}
{A : Ω → Matrix (Fin n) (Fin m) ℝ}
{σsq : ℝ}
(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 ℝ (Fin m))
(y : EuclideanSpace ℝ (Fin n))
(hx : ‖x‖ = 1)
(hy : ‖y‖ = 1)
:
IsSubGaussian (fun (ω : Ω) => inner ℝ ((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 → ℝ}
(hf : MeasureTheory.Integrable f μ)
(hef : MeasureTheory.Integrable (fun (ω : Ω) => Real.exp (f ω)) μ)
(hmean : ∫ (ω : Ω), f ω ∂μ = 0)
:
Helper: integral of exp(f) - 1 - f equals E[exp(f)] - 1 when E[f] = 0.
theorem
Problem_1_4.subgaussian_zero_ae
{Ω : Type u_1}
[MeasurableSpace Ω]
{μ : MeasureTheory.Measure Ω}
[MeasureTheory.IsProbabilityMeasure μ]
{X : Ω → ℝ}
(hSG : IsSubGaussian X 0 μ)
:
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 : ℕ}
{A : Ω → Matrix (Fin n) (Fin m) ℝ}
{σsq : ℝ}
(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)
:
Problem 1.4 (b)