# Mathematical argument

Notation: N(μ, σ²) with σ² ≥ 0; σ² = 0 is the degenerate (constant) normal. φ_W(t) = E[e^{itW}] is the characteristic function.

## Proposition B — TRUE
*If X and Y are independent and each normal, X + Y is normal.*

**Proof sketch.** Let X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²). Then φ_X(t) = exp(iμ₁t − σ₁²t²/2), φ_Y(t) = exp(iμ₂t − σ₂²t²/2). 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) (degenerate if both variances are 0). ∎

## Proposition A — FALSE
*If X and Y are each normal (no assumption on their joint law), X + Y is normal (constants count).*

Marginal normality does not pin down the joint law, so it does not pin down the law of the sum.

**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. *Y is N(0,1).* For any Borel set B, P(Y ∈ B) = ½P(X ∈ B) + ½P(−X ∈ B) = P(X ∈ B), because −X has the same law as X (N(0,1) is symmetric). So X and Y are both standard normal.
2. *X + Y is not normal.* X + Y = (1 + S)X, which is 0 when S = −1 and 2X when S = 1. Hence P(X + Y = 0) ≥ P(S = −1) = 1/2 (in fact exactly 1/2, because P(X=0)=0).
   - A non-degenerate normal distribution has no atoms, so it gives probability 0 to {0}; X + Y has an atom of mass 1/2, so it is not non-degenerate normal.
   - A degenerate normal is a point mass (probability 1 on one point). But P(X + Y = 0) = 1/2 < 1 and, on {S = 1}, X + Y = 2X has a continuous N(0,4) law, so X + Y is not constant.
   Therefore the law of X + Y, the mixture ½δ₀ + ½N(0,4), is not normal in either sense. ∎

(Also note Cov(X, Y) = E[S]E[X²] = 0: X and Y are uncorrelated yet dependent, so the counterexample also shows "uncorrelated normal marginals" is not enough; joint normality or independence is what is needed.)

Remark on the obvious candidate Y = −X: it gives X + Y ≡ 0, which is a degenerate normal and therefore *not* a counterexample under the statement's convention; this is why the random-sign construction is used.

**Numerical illustration** (not part of the proof): `out/analysis.py` draws n = 100000 pairs with seed 20261029 (design.csv) and writes them to `out/counterexample.csv`. Population value of the KS distance between the law of X + Y and the fitted N(0, 2): the true CDF jumps from 1/4 to 3/4 at 0 whereas the N(0,2) CDF equals 1/2 there, so the distance is 1/4. The sample value is reported as `counterexample_values.ks_sum_vs_normal` in results.json.