# Mathematical argument — F5-conf-03

# Proposition A (FALSE)

"If X and Y are each normally distributed, then X + Y is normally distributed (a constant counts as normal)."

Counterexample. Let X ~ N(0,1) and let S be independent of X with P(S = 1) = P(S = −1) = 1/2. Define Y = S·X.

1. X is normal: by construction.
2. Y is normal: for any t, P(Y ≤ t) = ½P(X ≤ t) + ½P(−X ≤ t) = ½Φ(t) + ½Φ(t) = Φ(t), using the symmetry of N(0,1) (−X ~ N(0,1)). So Y ~ N(0,1).
3. X + Y is not normal: X + Y = (1 + S)X, which equals 0 when S = −1 and 2X when S = 1. Hence its law is the mixture ½δ₀ + ½N(0,4).
   • It has an atom: P(X + Y = 0) = ½ (P(X = 0) = 0 so the S = 1 branch adds nothing). A non-degenerate normal law has no atoms.
   • It is not constant: Var(X + Y) = E[(1+S)²X²] = E[(1+S)²]·1 = ½·4 = 2 > 0.
   • Therefore X + Y is neither a non-degenerate nor a degenerate normal. ∎

(X and Y are uncorrelated — Cov(X,Y) = E[S]E[X²] = 0 — but not independent, and (X, Y) is not jointly normal; this is exactly the gap the proposition overlooks.)

Quantitative separation. The closest normal by moments is N(0, 2), whose CDF at 0 is ½, while the CDF of X + Y jumps from ¼ to ¾ at 0. Hence the Kolmogorov–Smirnov distance between the law of X + Y and N(0,2) is at least ¼ (and in fact any normal is at distance ≥ ¼ from it, since the jump of size ½ cannot be matched by a continuous CDF closer than ½·½). The simulation (out/analysis.py, out/counterexample.csv, seed and n from data/design.csv) reproduces this: counterexample_values.ks_sum_vs_normal ≈ 0.251, frac_sum_exactly_zero ≈ 0.500.

# Proposition B (TRUE)

"If X and Y are independent and each normal, then X + Y is normal."

Proof sketch. Let X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²) independent (σᵢ² ≥ 0, allowing degenerate cases). The characteristic function of N(μ, σ²) is φ(t) = exp(iμt − σ²t²/2). By independence, φ_{X+Y}(t) = φ_X(t)φ_Y(t) = exp(i(μ₁+μ₂)t − (σ₁²+σ₂²)t²/2), which is the characteristic function of N(μ₁+μ₂, σ₁²+σ₂²). By the uniqueness theorem for characteristic functions, X + Y ~ N(μ₁+μ₂, σ₁²+σ₂²) (a constant if both variances are 0). ∎

The simulation baseline (baseline_independent_sum.ks_sum_vs_normal ≈ 0.002) is consistent with this but is not the proof.