# Mathematical argument

# Proposition A — FALSE

"If X and Y are each normally distributed, then X + Y is normally distributed (constants count 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, N(0,1).
2. Y is normal: 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 Y ~ N(0,1).
3. X + Y is not normal: X + Y = (1 + S)X, which is 0 when S = −1 and 2X when S = 1. So P(X + Y = 0) ≥ P(S = −1) = 1/2 (exactly 1/2, since P(X = 0) = 0). A non-degenerate normal law is continuous and gives probability 0 to every point, so X + Y is not a non-degenerate normal. It is not a constant either: P(X + Y > 0) = P(S = 1)P(X > 0) = 1/4 > 0. Its law is the mixture ½δ₀ + ½N(0,4), with mean 0 and variance 2, and it is not normal of any kind (degenerate or otherwise).

Also note Cov(X, Y) = E[S]E[X²] = 0: X and Y are uncorrelated but dependent, each normal, and (X, Y) is not jointly (bivariate) normal. That is why the theorem in B does not apply. The general lesson is that normal marginals do not imply joint normality.

# Proposition B — TRUE

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

Proof sketch. Let X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) be independent, with σᵢ² ≥ 0 (a degenerate case is a constant). 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). The characteristic function determines the distribution uniquely (Lévy uniqueness theorem), so X + Y ~ N(μ₁+μ₂, σ₁²+σ₂²). If both variances are 0, the sum is a constant, and the proposition's convention counts that as normal. ∎

(Equivalently, (X, Y) is bivariate normal with diagonal covariance, and any linear combination of a jointly normal vector is normal.)