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PURE MATH FORMULA: Fisher Distance on Δ₇

Zero English inside formulas. Each equality justified. Verifiable numerically.


THE CHAIN

Given: p, q ∈ Δ₇ (probability simplex, 8 dimensions)

Step 0 — Chentsov's metric:

g_p(u,v) = Σᵢ₌₁⁸ (uᵢ vᵢ / pᵢ)

Justification: Chentsov 1972, Amari 1985. Unique metric respecting sufficient statistics.


Step 1 — The √p embedding:

φ : Δ₇ → S⁷,    φ(p) = (√p₁, √p₂, ..., √p₈)

Lemma: ‖φ(p)‖₂ = 1

‖φ(p)‖₂² = Σᵢ₌₁⁸ (√pᵢ)² = Σᵢ₌₁⁸ pᵢ = 1

Justification: p ∈ Δ₇ ⇒ Σpᵢ = 1 by definition.


Step 2 — Pullback of round metric:

(φ* g_{S⁷})_p(u,v) = ¼ · g_p(u,v)

Proof sketch:

Let c(t) be a curve in Δ₇, c(0) = p, ċ(0) = v.

γ(t) = φ(c(t)) = (√c₁(t), ..., √c₈(t))

γ̇ᵢ(0) = vᵢ / (2√pᵢ)

Round metric on S⁷:

⟨γ̇, γ̇⟩_{S⁷} = Σᵢ γ̇ᵢ² = Σᵢ vᵢ² / (4pᵢ) = ¼ · Σᵢ vᵢ²/pᵢ = ¼ · g_p(v,v)

Justification: Chain rule + direct computation.


Step 3 — Geodesics are great circles:

S⁷ has round metric ⇒ geodesics are great circles.

Great-circle distance:

d_{S⁷}(a,b) = arccos(⟨a,b⟩)

Justification: Standard Riemannian geometry of the sphere.


Step 4 — Inner product on S⁷:

⟨φ(p), φ(q)⟩ = Σᵢ₌₁⁸ √(pᵢ qᵢ)

Justification: Definition of φ + Euclidean inner product.


Step 5 — Fisher distance (THE FORMULA):

d_F(p,q) = 2 · d_{S⁷}(φ(p), φ(q))
         = 2 · arccos(⟨φ(p), φ(q)⟩)
         = 2 · arccos( Σᵢ₌₁⁸ √(pᵢ qᵢ) )

Justification: Steps 2+3+4 combined. The factor 2 comes from Step 2 (g = 4·φ*g_{S⁷}).


THE CLOSED-FORM RESULT

┌─────────────────────────────────────────────────────┐
│                                                     │
│   d_F(p,q) = 2 · arccos( Σᵢ₌₁ⁿ √(pᵢ qᵢ) )         │
│                                                     │
│   Domain: p, q ∈ Δₙ (any dimension n ≥ 2)          │
│   Range: [0, π]                                     │
│   Equality: d_F(p,q) = 0  ⟺  p = q                │
│   Max: d_F(p,q) = π  when p, q are antipodal       │
│        (e.g., p = (1,0,...,0), q = (0,1,0,...,0)) │
│                                                     │
└─────────────────────────────────────────────────────┘

VERIFICATION INSTANCE (n=8)

Inputs:

p = (0.3, 0.1, 0.15, 0.05, 0.2, 0.08, 0.07, 0.05)
q = (0.2, 0.2, 0.1, 0.1, 0.15, 0.1, 0.1, 0.05)

Step A — Compute √(pᵢqᵢ):

√(0.3×0.2) = 0.24494897
√(0.1×0.2) = 0.14142136
√(0.15×0.1) = 0.12247449
√(0.05×0.1) = 0.07071068
√(0.2×0.15) = 0.17320508
√(0.08×0.1) = 0.08944272
√(0.07×0.1) = 0.08366600
√(0.05×0.05) = 0.05000000

Step B — Sum:

S = 0.97586930

Step C — Arccos:

arccos(0.97586930) = 0.22012896

Step D — Multiply by 2:

d_F(p,q) = 2 × 0.22012896 = 0.44025792

OUTPUT: d_F(p,q) ≈ 0.440258


PROPERTIES (all verifiable)

Symmetry:

d_F(p,q) = 2·arccos(Σ√(pᵢqᵢ)) = 2·arccos(Σ√(qᵢpᵢ)) = d_F(q,p)  ✓

Identity:

d_F(p,p) = 2·arccos(Σ√(pᵢpᵢ)) = 2·arccos(Σpᵢ) = 2·arccos(1) = 0  ✓

Triangle inequality:

d_F(p,q) ≤ d_F(p,r) + d_F(r,q)  for all p,q,r ∈ Δ₇

Proof: Great-circle distance on S⁷ satisfies triangle inequality.
       Pullback by isometry preserves triangle inequality.  ✓

Bound:

0 ≤ d_F(p,q) ≤ π

Proof: arccos: [-1,1] → [0,π]. The argument Σ√(pᵢqᵢ) ∈ [0,1]
       by Cauchy-Schwarz: (Σ√(pᵢqᵢ))² ≤ (Σpᵢ)(Σqᵢ) = 1.  ✓

WHY THIS IS THE RIGHT FORMULA

  1. Chentsov's theorem says: any metric respecting sufficient statistics MUST be the Fisher metric (up to constant).

  2. The √p embedding maps Δ₇ → S⁷ isometrically (up to factor 4).

  3. Geodesics on S⁷ are great circles with known distance formula.

  4. Pulling back gives the Fisher distance formula above.

There is no choice in this formula. It is forced by the geometry of the probability simplex combined with Chentsov's uniqueness result.