Manuscript Oai119-A@1 · Oai119-A.tex · proposition · fdff7292

Proposition: A pointwise Hellinger certificate implies information

The Courtade–Kumar implication as a corollary · lines 1085–1146 · proof included or linked

Oai119-A@1#prop:HtoI, Oai119-A@1#eq:quantbridge

Fix one Boolean function and one symmetric-noise experiment on a uniform cube. Write

d = T_{ρ} f, m = E f, a = H (m), b = E H (d), s = H (ρ) .

a - b \leq 1 - s,

then

I_{2} (f (X); Y) \leq 1 - h_{2} (\frac{1 - ρ}{2}) .

More quantitatively, if

η = (1 - s) - (a - b) \geq 0,

then

1 - h_{2} (\frac{1 - ρ}{2}) - I_{2} (f (X); Y) \geq φ (s + η) - φ (s) .

Proof. Concavity of $H$ gives $0 \leq b \leq a \leq 1$ . Equation (eq:MI), (eq:conversionidentity), and Jensen's inequality imply

I_{2} (f (X); Y) = φ (a) - E φ (H (d)) \leq φ (a) - φ (b) .

Put $δ = a - b$ . Convexity implies that an increment of length $δ$ increases when its interval is shifted to the right. Hence

φ (a) - φ (b) \leq φ (1) - φ (1 - δ) = 1 - φ (s + η) .

Here $s + η = 1 - δ \in [0, 1]$ . Since

φ (s) = h_{2} (\frac{1 - ∣ ρ ∣}{2}) = h_{2} (\frac{1 - ρ}{2}),

the assertions follow.

Displayed formulas in this passage

Oai119-A@1#eq:quantbridge · line 1107
$1 - h_{2} (\frac{1 - ρ}{2}) - I_{2} (f (X); Y) \geq φ (s + η) - φ (s) .$
unlabeled · line 1089
$d = T_{ρ} f, m = E f, a = H (m), b = E H (d), s = H (ρ) .$
unlabeled · line 1094
$a - b \leq 1 - s,$
unlabeled · line 1098
$I_{2} (f (X); Y) \leq 1 - h_{2} (\frac{1 - ρ}{2}) .$
unlabeled · line 1103
$η = (1 - s) - (a - b) \geq 0,$
unlabeled · line 1120
$I_{2} (f (X); Y) = φ (a) - E φ (H (d)) \leq φ (a) - φ (b) .$
unlabeled · line 1129
$φ (a) - φ (b) \leq φ (1) - φ (1 - δ) = 1 - φ (s + η) .$
unlabeled · line 1138
$φ (s) = h_{2} (\frac{1 - ∣ ρ ∣}{2}) = h_{2} (\frac{1 - ρ}{2}),$