sums of independent normals are normal (via transforms)

Stack two independent bell-shaped quantities — two independent measurement errors, two independent stock returns — and ask what the total looks like. Astonishingly, it is another perfect bell curve, with no new shape, no skew, no fatter tails. The normal distribution is closed under addition of independent copies, and transforms make this otherwise-tricky fact a one-line calculation.

Here is the proof in full. Let X ~ Normal(mu_1, sigma_1^2) and Y ~ Normal(mu_2, sigma_2^2) be independent. The mgf of a normal is M(t) = e^(mu t + sigma^2 t^2/2). By the product rule for independent sums, M_{X+Y}(t) = M_X(t) M_Y(t) = e^(mu_1 t + sigma_1^2 t^2/2) times e^(mu_2 t + sigma_2^2 t^2/2). Multiplying exponentials adds the exponents, giving e^((mu_1 + mu_2)t + (sigma_1^2 + sigma_2^2)t^2/2). But that is exactly the mgf of a Normal(mu_1 + mu_2, sigma_1^2 + sigma_2^2). By the uniqueness theorem, X + Y must BE that normal. The means add and the variances add — which is the whole answer, derived without ever touching a convolution integral. (The identical argument runs with characteristic functions if you prefer the always-existing tool.)

This closure property is why the normal is the natural limiting shape in the central limit theorem, and why so much of statistics lives comfortably in the normal world: sums, averages, and linear combinations of independent normals stay normal, so you always know the family you are in. Two honest reminders. The summands must be independent — for dependent normals the variance of the sum picks up a covariance term, though it is still normal if they are jointly normal. And it is special to the normal: adding two independent uniforms does not give a uniform (you get a triangular shape), so closure under addition is a privilege, not the rule.