# Mathematical argument

# Proposition B (TRUE): X, Y independent, X ~ N(μ1, σ1²), Y ~ N(μ2, σ2²) ⇒ X + Y ~ N(μ1+μ2, σ1²+σ2²).

Proof sketch. The characteristic function of N(μ, σ²) is φ(t) = exp(iμt − σ²t²/2) (σ = 0 allowed: constant). By independence, φ_{X+Y}(t) = E[e^{itX}] E[e^{itY}] = 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 has that law. ∎ (More generally, the conclusion holds whenever (X, Y) is jointly Gaussian, since X+Y is a linear functional of a Gaussian vector.)

# Proposition A (FALSE): marginal normality of X and Y does not imply normality of X + Y.

Note that Y = −X is not a counterexample: X+Y ≡ 0 is a degenerate normal, which the proposition explicitly allows.

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

Y is N(0,1). For any Borel set B, P(Y ∈ B) = ½P(X ∈ B) + ½P(−X ∈ B) = P(X ∈ B), because −X ~ X by symmetry of N(0,1). So X and Y are both standard normal. (They are uncorrelated, E[XY] = E[W]E[X²] = 0, but not independent, since |Y| = |X|, and (X, Y) is not jointly Gaussian.)

X + Y is not normal. X + Y = (1+W)X, which equals 0 when W = −1 and 2X when W = 1. Hence P(X + Y = 0) ≥ P(W = −1) = 1/2, indeed the law of X+Y is the mixture ½·δ₀ + ½·N(0,4). A normal distribution is either continuous (σ > 0, every point has probability 0) or a point mass (σ = 0, P(X+Y = c) = 1 for some c). X+Y has an atom of mass exactly 1/2 at 0 (P(2X = 0) = 0), so it is neither: X+Y is not normal, not even degenerately. Equivalently, E[X+Y] = 0 and Var(X+Y) = 2, yet P(X+Y = 0) = 1/2, impossible for N(0,2). ∎

The KS distance between the law of X+Y and N(0, 2) (matching mean and variance) is exactly 1/4: at 0 the CDF of X+Y jumps from 1/4 to 3/4 while Φ(0) = 1/2, and the gap shrinks away from 0.

# Numerical illustration

out/analysis.py draws n = 100000 pairs from this construction with numpy.random.default_rng(20261029) (design.csv) and writes them to out/counterexample.csv. See out/results.json → counterexample_values and out/report.md. The simulation illustrates, but does not prove, the claims above.