Mathematical argument (F5-conf-03)
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.
- Y is N(0,1). For any Borel set B, P(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). So both marginals are standard normal.
- X + Y is not normal. X + Y = (1 + S)X, which equals 0 when S = −1 and 2X when S = 1. Hence P(X + Y = 0) ≥ P(S = −1) = 1/2 (in fact exactly 1/2, since P(X = 0) = 0).
- A non-degenerate normal law is absolutely continuous, so it puts probability 0 on any single point; X + Y puts 1/2 on {0}, so it is not a non-degenerate normal.
- It is not degenerate (constant) either: Var(X + Y) = E[(1+S)²X²] = E[(1+S)²]·E[X²] = 2·1 = 2 > 0.
- The law of X + Y is the mixture ½δ₀ + ½N(0,4), which is neither normal nor constant. ∎
Remarks: Cov(X,Y) = E[S]E[X²] = 0, so X and Y are uncorrelated but not independent (|Y| = |X|). (X, Y) is not jointly (bivariate) normal: a jointly normal pair would make X+Y normal. The proposition fails because marginal normality does 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), Y ~ N(μ₂, σ₂²) independent (σᵢ² ≥ 0, allowing degenerate cases). 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). (If both variances are 0, the sum is a constant, which counts as degenerate normal.) ∎