The normal distribution, derived

Take the binomial probabilities P(k) = C(n, k)p^kq^n−k and divide each bar by the one before it. The factorials in the coefficients nearly all cancel:

While this ratio is above 1 the bars are rising, below 1 they are falling, and the peak is where it crosses 1, near k = np.

Write k = np + x, where x is the distance from the centre. Then n − k = nq − x and k + 1 = np + x + 1, so

The approximation drops the small +q and uses p + q = 1; it is good when x is small compared with npq, which holds across the bulk of the distribution once n is large.

The logarithm of a ratio is a difference of logarithms, and ln(1 − h) ≈ −h for small h. So each step to the right changes ln P by about −x ÷ npq: the further from the centre, the faster the fall. Add the steps from the centre out to x:

The log of the probabilities is a downward parabola, and a curve whose logarithm is a parabola is exactly e^{−x²/(2npq)}. For n = 100 and p = 0.5, the bar at 55 over the bar at 50 is 0.6091, and the formula gives e^−25/50 = 0.6065.

The centre is np, the spread is σ = √(npq), and the shape is the same bell whatever p is, as long as n is large. Nothing here depends on coins: any count built from many small independent pieces behaves this way, which is the central limit theorem of the inference module. The curve only needs a constant in front so that its area is 1, and the next derivation finds it.

A density must have total area 1. The bell e^−z²/2 alone has some other area, I, so we divide by I. But e^−z²/2 has no antiderivative made of ordinary functions, so I cannot be found by integrating directly.

The trick is to find I² instead. Multiply two copies of the integral, one in x and one in y:

This is the volume under a three-dimensional bell, a surface whose height depends only on the distance r from the origin, since r² = x² + y².

Because the height depends only on r, slice the surface into thin rings around the origin. A ring of radius r and width dr is a loop of length 2πr, with the height e^−r²/2 all the way round, so its volume is 2πr · e^−r²/2 dr. Add rings in the lab and watch the total settle.

The ring volumes can be added exactly because r · e^−r²/2 is the slope of −e^−r²/2:

So I = √(2π) ≈ 2.5066, and the density that has area 1 is e^−z²/2 ÷ √(2π). At its peak it equals 1 ÷ √(2π) = 0.3989.

Take φ(z) = e^−z²/2 ÷ √(2π). It is symmetric about 0, so z · φ(z), which changes sign across 0, has an area that cancels: E[Z] = 0.

The slope of φ is φ′(z) = −z φ(z). So E[Z²] = ∫ z · (zφ) dz = ∫ z · (−φ′) dz. Integrate by parts, taking z as the part to differentiate:

The bracket vanishes because φ falls to zero faster than z grows, and the remaining integral is the total area, 1. With E[Z] = 0, Var(Z) = 1.

Let X = μ + σZ. Its mean is μ + σ · 0 = μ and its variance σ² · 1 = σ². Change variable in the density: z = (x − μ) ÷ σ gives dz = dx ÷ σ, so the density of X is

The 1 ÷ σ keeps the area at 1 after the stretch.

The event a < X < b is the same event as (a − μ) ÷ σ < Z < (b − μ) ÷ σ, so the two probabilities are equal. One table of Φ, the standard normal distribution function, therefore answers every normal question. Move the ends in the lab and the shaded areas stay equal.

Between −1 and 1 on the z scale the area is 0.6827, between −2 and 2 it is 0.9545, and between −3 and 3 it is 0.9973: the 68-95-99.7 rule. The curve bends from concave to convex at z = ±1, where φ″ = (z² − 1)φ is zero, which is why σ is the distance from the centre to the shoulders of the bell.