Why is the sum of two independent normal random variables exactly normal?

## What the moment generating function is The moment generating function of a random variable `X` is `M_X(t) = E[exp(t*X)]`, a function of a real parameter `t`. It is useful for two reasons. First, it converts addition into multiplication: if `X` and `Y` are independent, then `exp(t*(X+Y)) = exp(t*X) * exp(t*Y)` and independence lets the expectation factor, giving `M_{X+Y}(t) = M_X(t) * M_Y(t)`. Second, it is a fingerprint. The uniqueness theorem says that if two variables have MGFs that agree on an open interval containing `t = 0`, they have the same distribution. Together, those two facts turn a hard convolution integral into an algebra exercise. ## The normal case A normal variable with mean `m` and variance `v` has `M(t) = exp(m*t + v*t^2/2)`. Take independent `X` with mean `m1` and variance `v1`, and `Y` with mean `m2` and variance `v2`. Then `M_{X+Y}(t) = exp(m1*t + v1*t^2/2) * exp(m2*t + v2*t^2/2) = exp((m1 + m2)*t + (v1 + v2)*t^2/2)`. Read the result: it has precisely the shape `exp(m*t + v*t^2/2)` with `m = m1 + m2` and `v = v1 + v2`. By uniqueness, the sum is normal with mean `m1 + m2` and variance `v1 + v2`. Note the word *exactly*: this is an algebraic identity that holds for two summands or two hundred, with no appeal to any large-sample argument. Two normals added give a normal, full stop. ## Variances add, standard deviations do not If `X` has standard deviation 3 and `Y` has standard deviation 4, the sum has variance `9 + 16 = 25` and therefore standard deviation 5 — not 7. The same holds for the difference. Writing `X - Y` as `X + (-Y)` and noting that `-Y` is normal with mean `-m2` and the same variance `v2`, we get `X - Y` normal with mean `m1 - m2` and variance `v1 + v2`. Subtracting means but adding variances is the single most common slip on this material: subtraction removes signal, but it accumulates noise from both sources. ## The same trick for Poissons A Poisson variable with rate `r` has `M(t) = exp(r*(exp(t) - 1))`. Multiply two independent ones: `exp(r1*(exp(t) - 1)) * exp(r2*(exp(t) - 1)) = exp((r1 + r2)*(exp(t) - 1))`, which is the MGF of a Poisson with rate `r1 + r2`. So counts from two independent streams pool into a single Poisson stream with the rates added — the reason you can aggregate independent event counts across shards or hours without leaving the family. The same machinery shows binomials with a common success probability add their trial counts, and that gammas with a common scale add their shape parameters. ## Where the argument needs care **Independence is essential.** The factorisation step is the only place independence enters, and it is load-bearing. Here is a clean counterexample. Let `X` be standard normal and let `S` be an independent random sign, `+1` or `-1` with probability one half each, and set `Y = S * X`. By symmetry `Y` is standard normal too. But `X + Y` equals `2X` when `S = +1` and equals `0` when `S = -1`, so the sum puts probability one half on the single value zero. A distribution with an atom is certainly not normal. Two marginally normal variables therefore need not sum to a normal; what suffices is that they are *jointly* normal, of which independence is the simplest special case. **Existence is not automatic.** An MGF may fail to exist for any nonzero `t` when the tails are heavy — the Cauchy distribution is the standard example. The remedy is the characteristic function `E[exp(i*t*X)]`, which always exists, multiplies the same way under independence, and carries the same uniqueness theorem. Every MGF argument in this area has a characteristic-function version that is technically safer. **Uniqueness is doing real work.** Showing that two functions agree is only useful because an MGF determines a distribution. Without the uniqueness theorem, matching MGFs would prove nothing about the shape of the sum. ## How to present it in an interview A strong answer is four steps: define the MGF; state that it multiplies for independent summands; substitute the normal MGF and collect the exponent; invoke uniqueness. If the interviewer pushes for an alternative, mention that the direct route is a convolution integral of two normal densities, which works but requires completing the square in the exponent — the MGF argument is the same computation in disguise, with the algebra made easy.