1# Mathematical argument (F5-conf-03)23## Proposition A — FALSE4*"If X and Y are each normally distributed, then X + Y is normally distributed (constants count as normal)."*56**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.78*Marginals are normal.* X ~ N(0,1) by construction. For any Borel set B,9P(Y ∈ B) = ½P(X ∈ B) + ½P(−X ∈ B) = P(X ∈ B), because X and −X have the same law (the N(0,1) density is symmetric). Hence Y ~ N(0,1).1011*The sum is not normal.* X + Y = (1 + W)X, which equals 0 when W = −1 and 2X when W = 1. So12P(X + Y = 0) ≥ P(W = −1) = 1/2 (exactly 1/2, since P(X = 0) = 0), while Var(X + Y) = E[(1+W)²X²] = E[(1+W)²]·E[X²] = 2 > 0.13Every normal law is either non-degenerate (absolutely continuous, so no atoms) or degenerate (a point mass, variance 0). X + Y has an atom of mass 1/2 but positive variance, so it is neither: X + Y is not normally distributed. ∎1415(Note that X and Y are uncorrelated — Cov(X,Y) = E[W]E[X²] = 0 — yet (X, Y) is not bivariate normal; joint normality is what fails.)1617## Proposition B — TRUE18*"If X and Y are independent and each normal, then X + Y is normal."*1920**Proof sketch.** Let X ~ N(μ₁, σ₁²), Y ~ N(μ₂, σ₂²) (σ² = 0 allowed). 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). By the uniqueness theorem for characteristic functions, X + Y ~ N(μ₁+μ₂, σ₁²+σ₂²) (degenerate if both variances are 0). ∎2122Equivalently: independence of normal marginals makes (X, Y) jointly (bivariate) normal, and any linear combination of a jointly normal vector is normal. This is precisely the hypothesis missing in A.23