# Mathematical argument

## Proposition A — FALSE
*"If X and Y are each normally distributed, then X + Y is normally distributed (a constant counts as a degenerate normal)."*

Note: the classic counterexample Y = −X is **not** valid here, because X + Y = 0 is a constant, which the proposition counts as normal. We need a sum that is neither normal nor constant.

**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, 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 ~ N(0,1) by symmetry of the standard normal density. 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 ½·δ₀ + ½·N(0,4).
   - P(X+Y = 0) ≥ ½ > 0, so it is not a non-degenerate normal (which assigns probability 0 to every point).
   - P(X+Y = 0) = ½ exactly (P(X=0)=0), so it is not almost surely constant, i.e. not a degenerate normal either.
   - Equivalently, its characteristic function ½ + ½e^{−2t²} is not of the form e^{iμt − σ²t²/2}.
   
   Therefore X + Y is not normal, and A is false. (Remark: Cov(X,Y) = E[S]E[X²] = 0, so X and Y are uncorrelated normals that are not jointly normal — showing joint normality, not marginal normality, is what matters.)

The population Kolmogorov distance between the law of X+Y and N(0,2) (the normal with the same mean 0 and variance 2) is 0.25, attained at the jump at 0 (F(0⁻) = ¼ and F(0) = ¾, versus ½ for N(0,2)). This is far above sampling noise at n = 100 000 (≈0.004), so the departure is clearly visible in the simulation (`out/counterexample.csv`, generated by `out/analysis.py`).

## Proposition B — TRUE
*"If X and Y are independent and each normally distributed, then X + Y is normally distributed."*

**Proof sketch.** Let X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²) (σ² = 0 allowed, the degenerate case). 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), the characteristic function of N(μ₁+μ₂, σ₁²+σ₂²). By the uniqueness theorem for characteristic functions, X + Y ~ N(μ₁+μ₂, σ₁²+σ₂²). (If both variances are 0 the sum is a constant, which is a degenerate normal.) ∎

(Equivalently: independence with normal marginals makes (X,Y) jointly normal, and any linear combination of a jointly normal vector is normal.)