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Normal differences have limits at their central quantiles

Statement

Assume the Axiom of Choice. Use $D$ , $μ_{D}$ , and $σ_{D}$ from Method agreement and limits of agreement, and suppose the law of $D$ is $N (μ_{D}, σ_{D}^{2})$ with $σ_{D} \geq 0$ . Put $Φ (x) : = N (0, 1) ((- \infty, x]), z_{.975} : = \inf {x \in R : Φ (x) \geq .975} .$ For $D$ , put $F_{D} (x) : = P (D \leq x)$ and, for $0 < p < 1$ , define the lower $p$ -quantile by $q_{p} (D) : = \inf {x \in R : F_{D} (x) \geq p} .$ Then $z_{.975}$ is the unique real number with $Φ (z_{.975}) = .975$ , and $z_{.975} > 0$ . The exact central normal quantiles satisfy $q_{.025} (D) = μ_{D} - z_{.975} σ_{D}, q_{.975} (D) = μ_{D} + z_{.975} σ_{D} .$ If $σ_{D} > 0$ , the closed interval $[q_{.025} (D), q_{.975} (D)]$ has probability $0.95$ . If $σ_{D} = 0$ , both quantiles equal $μ_{D}$ and the singleton has probability $1$ . Thus $μ_{D} \pm 1.96 σ_{D}$ is the customary rounded display; the exact endpoints use $z_{.975}$ .

Facts & Assumptions

Given: AC; a square-integrable paired difference $D = A - B$ under a declared probability model; the law identity $L (D) = N (μ_{D}, σ_{D}^{2})$ with $σ_{D} \geq 0$ ; $F_{D} (x) = P (D \leq x)$ ; and the definitions of $Φ$ , $z_{.975}$ , and $q_{p} (D)$ in the Statement.

[A1]

The Axiom of Choice says that every family of nonempty sets has a choice function. The Axiom of Choice

[F1]

Assume AC. The standard normal law has density $ϕ (t) = e^{- t^{2} / 2} / \sqrt{2 π}$ , and $N (m, σ^{2})$ is the law of $x \mapsto m + σ x$ under $N (0, 1)$ for $σ \geq 0$ ; when $σ = 0$ , $N (m, 0) = δ_{m}$ . Standard normal and normal laws

[F2]

Under AC, $ϕ$ is positive and Borel measurable on $R$ and has Lebesgue integral one. The standard normal density has total mass one

[F3]

The Axiom of Countable Choice says every sequence of nonempty sets has a choice sequence. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F4]

A one-dimensional box with equal endpoints has Lebesgue measure zero; the box theorem assumes countable choice. A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

[F5]

The nonnegative integral over a measurable null set is zero. A nonnegative integral over a null set vanishes

[F6]

The real exponential maps $R$ into $(0, \infty)$ . The power-series, product-limit, IVP, functional-equation, and Picard definitions agree

[F7]

The real exponential is smooth with derivative equal to itself. The exponential function is smooth and $(\exp)^{'} = \exp$

[F8]

The mean value theorem gives $f (b) - f (a) = f^{'} (c) (b - a)$ for some $c \in (a, b)$ when $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$

[F9]

Probability measures are continuous from below and, for decreasing events, continuous from above. Basic identities for a probability measure

[F10]

For a differentiable change of variable on a compact interval and a continuous integrand, the oriented integral obeys the substitution formula. Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$

[F11]

The paired difference is $D = A - B$ , with bias $μ_{D} = E [D]$ and spread $σ_{D} = \sqrt{Var (D)}$ . Method agreement and limits of agreement

[F12]

A continuous real function on a closed bounded interval attains every value between its endpoint values; the constructive theorem uses no additional choice principle. Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$

[F13]

The nonnegative integral is monotone and homogeneous for positive scalars (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F14]

A continuous real function on a compact nondegenerate interval is bounded and Riemann integrable. A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion

[F15]

Under countable choice, a bounded Riemann integrable function on a compact interval has the same Lebesgue and Riemann integrals. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral

Proof

technique · direct

1.1A1F1F2F3F4F5givenconstruct

AC in [A1] applies to the normal-law construction [F1]. It also supplies countable choice: for any sequence of nonempty sets, apply AC to the family of its values and compose the resulting choice function with the sequence, which is the property in [F3]. Thus the countable-choice hypothesis in [F4] is available. Taking the one-dimensional box with equal endpoints $x = x$ in [F4] gives $λ ({x}) = 0$ ; the density in [F1] is nonnegative measurable by [F2], so [F5] gives $N (0, 1) ({x}) = 0$ . Thus the standard normal law has no atoms.

1.2F1F2F4F6F7F8F9F13algebra

For $x < y$ , probability subtraction and the density definition give $Φ (y) - Φ (x) = N (0, 1) ((x, y]) = \int_{(x, y]} ϕ d λ$ . Let $M = \max {∣ x ∣, ∣ y ∣}$ . Positivity [F6], derivative [F7], and the mean value theorem [F8] show that the exponential is increasing. Hence on $[x, y]$ , $α : = e^{- M^{2} / 2} / \sqrt{2 π} \leq ϕ (t) \leq β : = 1 / \sqrt{2 π}$ , with $α > 0$ . By [F4, F13], $0 < α (y - x) \leq Φ (y) - Φ (x) \leq β (y - x)$ . Therefore $Φ$ is strictly increasing and continuous.

2.1F1F7F9F10F14F15step 1.1step 1.2algebra

By [F9], $Φ (n) \to 1$ and $Φ (- n) \to 0$ as $n \to \infty$ . The density is continuous by [F7] and even. On each compact interval below, [F14, F15] identify its Riemann and Lebesgue integrals, using countable choice from step 1.1. For $n > ∣ x ∣$ , substitution $u = - t$ in [F10] gives $\int_{- n}^{- x} ϕ (t) d t = \int_{x}^{n} ϕ (u) d u$ . Passing to the tail limits gives $Φ (- x) = P (Z \geq x)$ for $Z \sim N (0, 1)$ . By the no-atom result in step 1.1, $P (Z \geq x) = 1 - Φ (x)$ , so $Φ (- x) = 1 - Φ (x)$ and $Φ (0) = 1 / 2$ .

3.1F12step 1.2step 2.1choosealgebra

By step 2.1, choose an integer $b > 0$ with $Φ (b) > .975$ , while $Φ (0) = 1 / 2 < .975$ . Step 1.2 makes $Φ$ continuous on the finite closed interval $[0, b]$ , so [F12] gives $z \in (0, b)$ with $Φ (z) = .975$ . Strict increase from step 1.2 makes $z$ unique, and step 2.1 gives $Φ (- z) = .025$ . Strict increase makes these the unique lower quantiles at those levels, so $z = z_{.975}$ , the set ${x : Φ (x) \geq .975}$ has infimum $z$ , and the set ${x : Φ (x) \geq .025}$ has infimum $- z$ .

4.1F1F11step 1.1step 3.1algebra

If $σ_{D} > 0$ , [F1] gives $F_{D} (x) = Φ ((x - μ_{D}) / σ_{D})$ , since $μ_{D} + σ_{D} t \leq x$ exactly when $t \leq (x - μ_{D}) / σ_{D}$ . The increasing affine change of variable carries the quantiles from step 3.1 to $μ_{D} - z_{.975} σ_{D}$ and $μ_{D} + z_{.975} σ_{D}$ . By [F1] and step 1.1, this affine image of the standard normal law has no atoms. The endpoint CDF values are $.025$ and $.975$ , so the closed interval between them has probability $.975 - .025 = .95$ , as stated.

5.1F1F11algebra∎

If $σ_{D} = 0$ , [F1] gives $N (μ_{D}, 0) = δ_{μ_{D}}$ . Its distribution function is $0$ below $μ_{D}$ and $1$ at or above $μ_{D}$ , so both lower quantiles are $μ_{D}$ and the singleton has probability $1$ . This is exactly the formula in the Statement with $σ_{D} = 0$ .

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Sources

J. M. Bland and D. G. Altman, Statistical methods for assessing agreement between two methods of clinical measurement, Lancet 1986; 327(8476):307–310, complete author-hosted reproduction (standard reference, not scraped)
M. Bland, An Introduction to Medical Statistics, 4th ed., Oxford University Press 2015, §20.5 (standard reference, not scraped)