# Mathematical argument

Convention: N(μ, σ²) with σ² ≥ 0; σ² = 0 is the degenerate normal (a constant), as stated in proposition A.

## Proposition B — TRUE
*If X and Y are independent with X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²), then X + Y is normal.*

Proof sketch. The characteristic function of N(μ, σ²) is φ(t) = exp(iμt − σ²t²/2) (valid also for σ² = 0).
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 ~ N(μ₁+μ₂, σ₁²+σ₂²). ∎
(Independence is used exactly once: to factor the joint expectation. More generally, (X,Y) jointly Gaussian suffices.)

## Proposition A — FALSE
*Marginal normality of X and Y alone does not imply X + Y is normal, even allowing degenerate normals.*

Note: the familiar example Y = −X does **not** refute A here, because X + Y = 0 is a constant, which A counts as normal. We need a non-degenerate, non-normal sum.

**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.

1. *X is normal:* by construction, N(0,1).
2. *Y is normal:* for any Borel set B, P(Y ∈ B) = ½P(X ∈ B) + ½P(−X ∈ B) = P(X ∈ B), since −X ~ X by symmetry of N(0,1). So Y ~ N(0,1).
3. *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) + P(W = 1, X = 0) = 1/2.
   A non-degenerate normal (σ² > 0) is continuous and gives probability 0 to every point, so X + Y is not a non-degenerate normal. A degenerate normal is a point mass, but X + Y is not a.s. constant: P(X + Y > 0) = P(W = 1)P(X > 0) = 1/4 > 0 and P(X + Y = 0) = 1/2 < 1. So X + Y is not normal in either sense. Its law is the mixture ½δ₀ + ½N(0,4), with mean 0 and variance 2. ∎

Remarks. (X, Y) is not jointly Gaussian (the vector (X,Y) lives on the two lines y = ±x). Also Cov(X, Y) = E[W]E[X²] = 0, so X and Y are uncorrelated yet dependent normals whose sum is not normal — showing "uncorrelated normals" is also insufficient; independence (or joint normality) is what B and its generalisation need.

**Why the departure is visible numerically.** Against any continuous CDF, a distribution with an atom of mass p differs in sup-norm by at least p/2 (the CDF jumps by p at the atom while the continuous one does not). With p = 1/2 the KS distance of the sum to *any* normal is ≥ 0.25 in population, far above the ~1/√n ≈ 0.003 sampling noise at n = 100000.

## Numerical illustration (out/counterexample.csv, generated by out/analysis.py)
numpy.random.default_rng(20261029); draw x = standard_normal(n), then w = choice([−1, 1], n); y = w·x; n = 100000.
Summary statistics are in out/results.json under `counterexample_values` (sample KS of x+y vs fitted normal 0.2508; KS of the marginals ≈ 0.002). The simulation illustrates, but does not prove, the claims above.