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Bounded-error model comparison: a derivative lower bound confines compatible parameters

Statement

Let $I \subseteq R$ be an interval with at least two elements (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $θ_{0} \in I$ , and let $f : I \to R$ be of class $C^{1}$ on $I$ in the sense that $f$ is differentiable at every point of $I$ and $f^{'}$ is continuous on $I$ (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Assume $f (θ_{0}) = 0$ and that $f^{'} (θ) \geq m > 0$ for every $θ \in I$ . Let a recorded scalar $\hat{y} \in R$ satisfy $∣ \hat{y} ∣ \leq s$ with $s \geq 0$ (Absolute value in an ordered field), and let a separately specified calibration-error model bound be $∣ y - \hat{y} ∣ \leq δ$ with $δ \geq 0$ for a value $y$ assigned to the same quantity. If a model requires $y = f (θ)$ for some $θ \in I$ , then every $θ \in I$ compatible with those two simultaneous bounds satisfies

$∣ θ - θ_{0} ∣ \leq \frac{s + δ}{m};$

equivalently, if a proposed $θ$ violates this bound, then the specified model-plus-error-bound pair is incompatible with the premises. The conclusion is a deterministic conditional statement about the stated bounds; it is not a sampling law, a statistical inference, or a claim about the historical reports of the causal-response page.

Facts & Assumptions

Given: An interval $I \subseteq R$ with at least two elements, a point $θ_{0} \in I$ , a function $f : I \to R$ differentiable at every point of $I$ with $f^{'}$ continuous on $I$ , the normalisation $f (θ_{0}) = 0$ , the lower bound $f^{'} \geq m > 0$ on $I$ , a recorded scalar $\hat{y} \in R$ with $∣ \hat{y} ∣ \leq s$ , $s \geq 0$ , a calibration bound $∣ y - \hat{y} ∣ \leq δ$ , $δ \geq 0$ , and the model requirement $y = f (θ)$ for a point $θ \in I$ .

[L1]

Every interval is order-convex: if $x, y \in I$ and $x \leq z \leq y$ then $z \in I$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Hence for any two points $a < b$ of $I$ the whole closed interval $[a, b]$ is contained in $I$ .

[L2]

For an ordered field $F$ and $x, y \in F$ one has $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ (The triangle inequality), with the absolute value of Absolute value in an ordered field: $∣ u ∣ = u$ for $u \geq 0$ , $∣ u ∣ = - u$ for $u < 0$ , so $∣ u ∣ \geq 0$ with equality only for $u = 0$ .

[L3]

A function differentiable at a limit point $c$ of its domain is continuous at $c$ , and consequently a function differentiable at every point of a set is continuous at every point of that set (A function differentiable at $c$ is continuous at $c$ ). Since $I$ has at least two elements, every point of $I$ is a limit point of $I$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): given distinct $x, y \in I$ , an arbitrary small one-sided step from $x$ towards $y$ stays in $I$ .

[L4]

Differentiation is a local operation and continuity is preserved by restriction (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point): for $a < b$ in $I$ , a punctured neighbourhood of an interior point $c \in (a, b)$ lies in $[a, b]$ , so the difference quotient of the restriction $f ∣_{[a, b]}$ at $c$ agrees with that of $f$ and the restriction is differentiable at $c$ with the same derivative $f^{'} (c)$ ; and by [L3] the restriction is continuous at every point of $[a, b]$ .

[L5]

Mean value theorem (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 $f (b) - f (a) = f^{'} (c) (b - a)$ ): if $a < b$ and $g : [a, b] \to R$ is continuous on $[a, b]$ and differentiable at every point of $(a, b)$ , then there is $ξ \in (a, b)$ with $g (b) - g (a) = g^{'} (ξ) (b - a)$ .

Proof

1.1L2algebra

Write $y = \hat{y} + (y - \hat{y})$ and apply the triangle inequality of [L2] together with the two assumed bounds: $∣ y ∣ \leq ∣ \hat{y} ∣ + ∣ y - \hat{y} ∣ \leq s + δ$ .

1.2L2algebra

If $θ = θ_{0}$ then $∣ θ - θ_{0} ∣ = 0 \leq (s + δ) / m$ , because $s \geq 0$ , $δ \geq 0$ give $s + δ \geq 0$ and $m > 0$ makes $(s + δ) / m \geq 0$ .

1.3L1L3L4L5

Suppose $θ \neq θ_{0}$ and put $a = \min (θ, θ_{0})$ , $b = \max (θ, θ_{0})$ , so $a < b$ and ${a, b} = {θ, θ_{0}}$ . By [L1] the closed interval $[a, b]$ is contained in $I$ ; by [L3] the function $f$ is continuous on $I$ , and by the restriction clause of [L4] the restriction $g = f ∣_{[a, b]}$ is continuous on $[a, b]$ ; by the locality clause of [L4] $g$ is differentiable at every point of $(a, b)$ with $g^{'} = f^{'}$ there. The mean value theorem [L5] therefore yields $ξ \in (a, b)$ with $g (b) - g (a) = g^{'} (ξ) (b - a)$ . Since $g (b) - g (a)$ equals $f (θ) - f (θ_{0})$ up to the sign that reverses $b - a$ as well, and $g^{'} (ξ) = f^{'} (ξ)$ , this is exactly $f (θ) - f (θ_{0}) = f^{'} (ξ) (θ - θ_{0})$ .

2.1step 1.3L2algebra

At the intermediate point of step 1.3 one has $f^{'} (ξ) \geq m > 0$ , so $f (θ) - f (θ_{0})$ has the sign of $θ - θ_{0}$ , and therefore $∣ f (θ) - f (θ_{0}) ∣ = f^{'} (ξ) ∣ θ - θ_{0} ∣ \geq m ∣ θ - θ_{0} ∣$ .

3.1step 2.1step 1.1algebra

Since $y = f (θ)$ and $f (θ_{0}) = 0$ , step 2.1 gives $m ∣ θ - θ_{0} ∣ \leq ∣ y ∣$ , and step 1.1 gives $∣ y ∣ \leq s + δ$ . Hence $m ∣ θ - θ_{0} ∣ \leq s + δ$ , and dividing by $m > 0$ yields $∣ θ - θ_{0} ∣ \leq (s + δ) / m$ .

4.1step 1.2step 3.1∎

Every $θ \in I$ compatible with the model requirement and the two bounds falls under one of the two cases: $θ = θ_{0}$ satisfies the bound by step 1.2, and $θ \neq θ_{0}$ satisfies it by step 3.1. Equivalently, if a proposed $θ$ has $∣ θ - θ_{0} ∣ > (s + δ) / m$ , then no assignment satisfying both stated bounds can also satisfy $y = f (θ)$ , so the specified model-plus-error-bound pair is incompatible with the premises; no probability, sampling, or statistical inference has been used.

Depends on

Dependency tree · 2 levels

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 5, Theorem 5.10 (the mean value theorem)