# Mathematical argument

Convention: "normal" means N(mu, sigma^2) with sigma^2 >= 0. sigma^2 = 0 is the degenerate case, a constant.

# Proposition B (TRUE): if X and Y are independent with X ~ N(mu1, s1^2) and Y ~ N(mu2, s2^2), then X+Y ~ N(mu1+mu2, s1^2+s2^2).

Proof sketch. The characteristic function of N(mu, s^2) is phi(t) = exp(i mu t - s^2 t^2 / 2). Because X and Y are independent, phi_{X+Y}(t) = phi_X(t) phi_Y(t) = exp(i(mu1+mu2)t - (s1^2+s2^2)t^2/2). This is the characteristic function of N(mu1+mu2, s1^2+s2^2). By the uniqueness theorem for characteristic functions, X+Y has that law. The argument still holds when s1 = 0 or s2 = 0 (degenerate). QED.

# Proposition A (FALSE): normal marginals alone do not make X+Y normal.

Counterexample. Let X ~ N(0,1). Let S be independent of X with P(S=1) = P(S=-1) = 1/2. Set Y = S X.

• X is normal, by construction.
• Y is N(0,1). For any Borel set B, P(Y in B) = 1/2 P(X in B) + 1/2 P(-X in B) = P(X in B), because -X has the same law as X (N(0,1) is symmetric). The first equality uses the independence of S and X.
• X+Y is not normal. X+Y = (1+S)X. This equals 0 when S = -1 and equals 2X when S = 1. So P(X+Y = 0) >= P(S=-1) = 1/2: the law has an atom at 0. A non-degenerate normal law has a continuous distribution function, so it has no atoms. X+Y is also not a constant: Var(X+Y) = E[(1+S)^2 X^2] = E[(1+S)^2] E[X^2] = 2, which is positive. Equivalently, P(X+Y > 1) = 1/2 P(2X > 1) > 0. Hence X+Y is neither a non-degenerate normal nor a degenerate (constant) normal. QED.

Remarks. (X, Y) is not jointly (bivariate) normal. If it were, every linear combination, including X+Y, would be normal. X and Y are uncorrelated, since E[XY] = E[S] E[X^2] = 0, yet they are dependent, because |X| = |Y|. So this also illustrates that uncorrelated normal marginals do not imply independence.

# Numerical illustration (not a proof)

out/analysis.py draws n = 100000 pairs from this construction with numpy.random.default_rng(20261029). It writes them to out/counterexample.csv and summarises them in out/results.json:

• the marginals x and y agree with N(0,1) (small KS distances);
• about half of the x+y values are exactly 0;
• the KS distance between x+y and its fitted normal is about 0.25. This matches the theoretical value of 1/2 * (jump of 1/2 at 0).