Mathematical argument
Convention: a "normal" law is N(μ, σ²) with σ² ≥ 0; σ² = 0 is the point mass at μ (a constant).
Proposition A — FALSE
Claim: "If X and Y are each normal, then X + Y is normal."
Counterexample. Let X ~ N(0,1) and let W be independent of X with P(W = 1) = P(W = −1) = 1/2. Set Y = W·X.
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X is normal: N(0,1) by construction.
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Y is normal: 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 Y ~ N(0,1).
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X + Y is not normal: X + Y = (1 + W)X, which equals 0 when W = −1 and 2X when W = 1. So P(X + Y = 0) = P(W = −1) + P(W = 1, X = 0) = 1/2.
- A non-degenerate normal law (σ² > 0) has a density and gives probability 0 to every single point. So it cannot have an atom of mass 1/2.
- A degenerate normal is a constant, but X + Y is not constant: P(X + Y > 0) = P(W = 1)P(X > 0) = 1/4 > 0, and in the same way P(X + Y < 0) = 1/4.
So X + Y is not normal under either convention. (Its law is the mixture ½δ₀ + ½N(0,4). It has mean 0 and variance E[(1+W)²]E[X²] = 2, but it is not N(0,2).)
So proposition A is false: normal marginals do not make a sum normal. What is missing is joint normality of (X, Y). Here (X, Y) is not bivariate normal, because Cov(X, Y) = E[W]E[X²] = 0 even though X and Y are not independent (|Y| = |X|).
Numerical illustration (seed 20261029, n = 100000, data in out/counterexample.csv):
- KS distance of x from N(0,1): 0.0028. KS distance of y from N(0,1): 0.0016. Both are below the approximate 5% critical value of 0.0043.
- Exactly 50.05% of the simulated x + y are 0.
- KS distance between x + y and the normal fitted to it (mean −0.0036, sd 1.406): 0.251. This is roughly 58 times the critical value.
Proposition B — TRUE
Claim: "If X and Y are independent and each normal, then X + Y is normal."
Proof sketch (characteristic functions). Let X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) with σᵢ² ≥ 0. The characteristic function of N(μ, σ²) is φ(t) = exp(iμt − σ²t²/2); this formula also holds for σ² = 0. Independence gives 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). Characteristic functions determine the distribution uniquely (Lévy's uniqueness theorem). So X + Y ~ N(μ₁+μ₂, σ₁²+σ₂²), which is degenerate only if both summands are constants. ∎
(Equivalently: independence of normal X and Y implies that (X, Y) is bivariate normal, and every linear combination of a jointly normal vector is normal.)