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Calibration and measurement uncertainty — Examples

1 · Prerequisites

2 · Summary

Work a two-point calibration example with correlated uncertain reference standards and a separate repeatability contribution from a later reading. A second stipulated calibration batch shows how a changed curve can affect the reported estimate; an explicit equal-weight pooled least-squares fit shows a separate discrepancy. The calculation is illustrative and makes no universal process or coverage claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Calibration curve with uncertain reference standards and batch drift

Example

Consider a two-point calibration curve. At the estimated reference values ${\hat{r}}_{1} = 10$ and ${\hat{r}}_{2} = 20$ , suppose the corresponding instrument responses are treated as exact values $1$ and $21$ . Here the inverse model has domain $U = R^{3}$ . For a later response $Y$ , the inverse curve estimates the measurand by $q = f (R_{1}, R_{2}, Y) = R_{1} + \frac{Y - 1}{20} (R_{2} - R_{1}) .$ Use a finite input uncertainty model on $(Ω, F, P)$ , where $Ω = {- 1, 1}^{3}$ , $F = 2^{Ω}$ , and each of the eight points has probability $1 / 8$ . For $Z_{i} (ϵ) = ϵ_{i}$ , set $R_{1} = 10 + 0.1 Z_{1}, R_{2} = 20 + 0.1 Z_{1} + \sqrt{0.03} Z_{2}, Y = 13 + 0.2 Z_{3} .$ The shared $Z_{1}$ component represents covariance in the reference-standard values; the separate $Y$ term represents repeatability uncertainty in the later reading. These are stipulated input uncertainty summaries for this example, not an empirical calibration study.

For a second calibration batch, suppose the response curve instead passes through $(10, 1)$ and $(20, 23)$ . The change in slope is a stipulated batch drift: it shows what applying a different batch's curve can do, but it is not a universal model of measurement-process drift. For the pooled comparison, define the pooled response curve as the equal-weight least-squares line with an intercept fitted to the four nominal calibration points $(10, 1), (20, 21), (10, 1), (20, 23)$ , one point for each batch-standard pair.

Facts & Assumptions

Given: The finite input model and the two stipulated calibration response curves in the Example.

[F1]

Measurands, calibration, and measurement uncertainty: the input covariance matrix records covariance between reference-standard values and a later reading; the combined standard uncertainty is attached to the first-order linearized output.

[F2]

First-order propagation of measurement uncertainty: the variance of that output is $\nabla f (\hat{x})^{T} Σ \nabla f (\hat{x})$ , with all off-diagonal covariance terms retained.

[F3]

Moments, variance, and covariance on a probability space: for square-integrable real random variables, covariance is the expectation of the centered product.

[A1]

$(Ω, F, P)$ is the finite probability space specified in the Example, and the input variables $R_{1}, R_{2}, Y$ are the functions specified there.

[A2]

The first calibration batch's stipulated response values are $1$ at ${\hat{r}}_{1} = 10$ and $21$ at ${\hat{r}}_{2} = 20$ ; the inverse relation in the Example uses response gap $20$ .

[A3]

The second batch's stipulated response values are $1$ at $10$ and $23$ at $20$ , corresponding to slope $2.2$ through those two points.

[A4]

The pooled calibration rule is equal-weight ordinary least squares in response-on-quantity form with a free intercept over the four nominal calibration points listed in the Example.

Verification

1.1givenA1A2algebra

Averaging over the eight sign triples gives $E [Z_{i}] = 0$ , $E [Z_{i}^{2}] = 1$ , and $E [Z_{i} Z_{j}] = 0$ for $i \neq j$ : each sign appears equally often, its square is always one, and reversing one coordinate cancels each cross-product pair. Also $R_{2} - R_{1} = 10 + \sqrt{0.03} Z_{2} > 9.8$ , while $Y \in [12.8, 13.2] \subset (1, 21)$ , so the ordered reference values and inverse curve are valid throughout the finite input support.

1.2givenA2algebra

At the input means, the first batch's inverse curve gives $\hat{q} = 10 + (13 - 1) (20 - 10) / 20 = 16$ . Its gradient in the order $(R_{1}, R_{2}, Y)$ is $(1 - \frac{Y - 1}{20}, \frac{Y - 1}{20}, \frac{R_{2} - R_{1}}{20}) = (0.4, 0.6, 0.5)$ .

1.3A2A3A4algebra

In the second batch, a measurand value $q = 16$ produces response $1 + 2.2 (16 - 10) = 14.2$ . The first batch's inverse curve maps this response to $10 + (14.2 - 1) / 2 = 16.6$ , while the second batch's curve maps it to $10 + (14.2 - 1) / 2.2 = 16$ , so the first-curve mismatch is $0.6$ . For the pooled fit, the four nominal quantities and responses have means $\overset{ˉ}{q} = 15$ and $\overset{ˉ}{r} = 11.5$ . Their centered sums are $S_{q q} = \sum (q - \overset{ˉ}{q})^{2} = 4 (5^{2}) = 100$ and $S_{q r} = \sum (q - \overset{ˉ}{q}) (r - \overset{ˉ}{r}) = 2 (- 5) (- 10.5) + 5 (9.5) + 5 (11.5) = 210$ .

2.1F3A1step 1.1algebra

The input means are $(10, 20, 13)$ . By [F3] and step 1.1, $Var (R_{1}) = 0.01$ , $Var (R_{2}) = 0.01 + 0.03 = 0.04$ , $Var (Y) = 0.04$ , and $Cov (R_{1}, R_{2}) = 0.01$ ; the covariances with $Y$ are zero. Therefore $Σ = (\begin{array}{ccc} 0.01 & 0.01 & 0 \\ 0.01 & 0.04 & 0 \\ 0 & 0 & 0.04 \end{array})$ .

3.1F1F2step 2.1step 1.2algebra

Apply [F2] to step 2.1 and the gradient in step 1.2: $u_{c}^{2} = {0.4}^{2} (0.01) + {0.6}^{2} (0.04) + {0.5}^{2} (0.04) + 2 (0.4) (0.6) (0.01) = 0.0308$ , so $u_{c} = \sqrt{0.0308} \approx 0.1755$ in measurand units. The covariance contribution is positive and would be lost by treating the two reference values as independent.

4.1A4step 1.3algebra∎

For an equal-weight line $r = a + b q$ , write each residual as $(r - \overset{ˉ}{r}) - b (q - \overset{ˉ}{q}) + (\overset{ˉ}{r} - a - b \overset{ˉ}{q})$ . The centered terms sum to zero, so the sum of squared residuals is $S_{r r} - 2 b S_{q r} + b^{2} S_{q q} + 4 (\overset{ˉ}{r} - a - b \overset{ˉ}{q})^{2}$ , where $S_{r r} = \sum (r - \overset{ˉ}{r})^{2}$ . For each $b$ this is minimized in $a$ at $a = \overset{ˉ}{r} - b \overset{ˉ}{q}$ . Since $S_{q q} = 100 > 0$ , the remaining quadratic in $b$ is uniquely minimized because it equals $S_{r r} - S_{q r}^{2} / S_{q q} + S_{q q} (b - S_{q r} / S_{q q})^{2}$ . Thus $b = S_{q r} / S_{q q} = 2.1$ , giving $a = \overset{ˉ}{r} - b \overset{ˉ}{q} = - 20$ and the pooled curve $r = 1 + 2.1 (q - 10)$ . Its inverse maps response $14.2$ to $10 + (14.2 - 1) / 2.1 = 114 / 7 \approx 16.286$ , which is $2 / 7$ above the second-batch estimate $16$ . These discrepancies are specific to the stipulated curves.

Remarks

The value $u_{c}$ is a first-order standard uncertainty, not a 95% interval or a coverage claim. The batch comparison is illustrative; process stability and the adequacy of either curve require empirical assessment.