**The Error Function and Normal Distribution: Norm(x) and Erf(x)**

**Introduction**

The error function is defined as:

erf(x) = 2/√π * ∫( e^(-t^2) dt, 0, x)

norm(x) = 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, x)

We were assuming that the mean μ = 0 and variance σ = 1.

**Derivation: norm(x) in terms of erf(x)**

norm(x)

= 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, x)

= 1/√(2*π) * ∫( e^(-z^2/2) dx, -∞, 0) + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

= 1/2 + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

—————————–

Substitution:

t^2 = z^2/2

t = z/√2

√2 * t = z

√2 dt = dz

z = 0, t = 0

z = x, t = x/√2

—————————–

= 1/2 + 1/√(2*π) * ∫( e^(-z^2/2) dx, 0, x)

= 1/2 + 1/√(2*π) * ∫(√2 * e^(-t^2) dt, 0, x/√2)

= 1/2 + 1/√π * √π/2 * 2/√π * ∫(√2 * e^(-t^2) dt, 0, x/√2)

= 1/2 + (1/√π * √π/2) * (2/√π * ∫(√2 * e^(-t^2) dt, 0, x/√2))

= 1/2 + 1/2 * 2/√π * ∫(√2 * e^(-t^2) dt, 0, x/√2)

= 1/2 + 1/2 *erf(x/√2)

norm(x) = 1/2 + 1/2 *erf(x/√2)

norm(x) – 1/2 = 1/2 *erf(x/√2)

2 norm(x) – 1 = erf(x/√2)

Let t = x/√2, then:

2 norm(√2 * t) – 1 = erf(t)

**Summary**

norm(x) = 1/2 + 1/2 * erf(x/√2)

erf(t) = 2 * norm(√2 * t) – 1

with μ = 0 and σ = 1.

**Examples**

norm(1) = 1/2 + 1/2 * erf(1/√2) ≈ 0.841344746069

erf(1) = 2 * norm(√2) – 1 ≈ 0.84270079295

Eddie

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