Manuscript Psy03-D@1 · Psy03-D-v1.tex · theorem · 3e18951d

Theorem: Certified ratio interval

Sequence-level statistical certificates › Ratios of population means · lines 1108–1137 · proof included or linked

Psy03-D@1#thm:ratio, Psy03-D@1#eq:ratiointerval

Suppose simultaneous intervals satisfy

μ_{p} \in [p_{L}, p_{U}], μ_{n} \in [n_{L}, n_{U}],

with nonnegative endpoints. If $n_{L} > 0$ , then

Π_{old} \in [max {0, \frac{p _{L}}{n _{U}}}, min {1, \frac{p _{U}}{n _{L}}}] .

If $n_{L} = 0$ , this construction does not certify a positive population denominator and the ratio is unresolved.

The same statement applies to $Π_{repeat ∣ error}$ using its normalized numerator and error count.

Proof. Division is increasing in a nonnegative numerator and decreasing in a positive denominator. Use $0 \leq μ_{p} \leq μ_{n}$ for truncation at one.

Displayed formulas in this passage

unlabeled · line 1111
$μ_{p} \in [p_{L}, p_{U}], μ_{n} \in [n_{L}, n_{U}],$
Psy03-D@1#eq:ratiointerval · line 1117
$Π_{old} \in [max {0, \frac{p _{L}}{n _{U}}}, min {1, \frac{p _{U}}{n _{L}}}] .$

Theorem: Certified ratio interval

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