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Standard-normal CDF, tail, and quantile identities for power calculations

Statement

Assume AC. Let Z∼N(0,1) and Φ(x)=P(Z≤x). Then Φ is continuous and strictly increasing, with limits 0 and 1 at −∞ and +∞; P(Z=x)=0 for every x∈R; and Φ(−x)=1−Φ(x). For each q∈(0,1), there is a unique real zq with Φ(zq)=q, and z1−q=−zq, in particular z1/2=0.

Facts & Assumptions

Given: AC, the standard normal law N(0,1), and its cumulative distribution function Φ.

[A1]

The standard-normal law is constructed under AC, and AC implies countable choice. The Axiom of Choice Standard normal and normal laws The Axiom of Choice implies countable choice

[F1]

The standard-normal law is the probability measure with density g(t)=e−t2/2/2π; this density is positive, Borel, and has integral one. Standard normal and normal laws The standard normal density has total mass one

[F3]

A continuous real-valued function on a nonempty compact interval attains finite maximum and minimum values. Extreme value theorem: a continuous real function on a nonempty compact subset of R attains a greatest and a least value

[F4]

Probability is continuous on increasing event sequences and on decreasing event sequences, the latter because all probabilities are finite. Continuity from below for measures Continuity from above when one set has finite measure

[F7]

Substitution u=−t in a convergent improper integral is valid for the monotone differentiable bijection t↦−t; its derivative reverses orientation. Change of variable in an improper integral

[F8]

If A⊆B, then P(B∖A)=P(B)−P(A), and P(Ac)=1−P(A). Basic identities for a probability measure

[F9]

The CDF is Φ(x)=P(Z≤x). Cumulative distribution function of a real random variable

Proof

1.1A1F1F2F3F8F9algebra

For a<b, [F8] and [F9] give Φ(b)−Φ(a)=P(a<Z≤b)=∫(a,b]g(t) dt. By [F2] and [F3], g has a minimum c on [a,b]; [F1] makes c>0, so Φ(b)−Φ(a)≥c(b−a)>0, proving strict increase. For a fixed x, [F2] and [F3] bound g by some M<∞ on [x−1,x+1]; whenever ∣y−x∣<1, ∣Φ(y)−Φ(x)∣≤M∣y−x∣, proving continuity at x. Since x was arbitrary, Φ is continuous.

2.1F1F4F5step 1.1choose

The events {Z≤k} increase to the whole space and {Z≤−k} decrease to the empty set as positive integers k→∞. By [F1] and [F4], Φ(k)→1 and Φ(−k)→0. Given q∈(0,1), choose k with Φ(−k)<q<Φ(k); [F5] on [−k,k] gives zq with Φ(zq)=q, and strict increase from step 1.1 makes it unique.

3.1A1F1F6F7F8step 2.1algebra∎

By [A1] and [F6], P(Z=x)=∫{x}g dλ=0 for every x. The density g is even; applying [F7] gives Φ(−x)=∫x∞g(t) dt. Since P(Z=x)=0, [F8] identifies this upper tail with 1−Φ(x). Thus Φ(−x)=1−Φ(x). Taking x=zq and using the uniqueness in step 2.1 gives Φ(−zq)=1−q and z1−q=−zq; at q=1/2, this yields z1/2=0.

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