Abstract
The single phrase “this instrument is accurate” names two different quantities. One is the offset from the true value; the other is the scatter. This paper asks whether the two can be reduced by the same operation—the answer is no. Repeating reduces only one of them; calibrating reduces only the other. And which one is the bias and which one is the scatter turned out not to be a property of the instrument, but a consequence of what one chooses to repeat. No new mathematical theorem and no new law is claimed.
First, the tools were built here and checked against known values. Evaluating $\mathrm{MSE}=b^2+\sigma^2$ with $10^6$ random draws gives $\mathbf{0.042462}$, agreeing with the exact $0.042500$ to four decimal places (Section 2).
Second, this is the core of the paper. Repetition reduces only one of the two. Averaging $10^6$ readings of an instrument with bias $0.05$ and scatter $0.20$ leaves the error at $\mathbf{0.050000}$—it never falls below $0.05$ (Section 3).
Third, two instruments with the same error part company under averaging. At $n=1$ both read $0.206155$; at $n=10^4$ they read $\mathbf{0.050040}$ and $\mathbf{0.200001}$—a factor of $3.9968$ (Section 4).
Fourth, calibration removes the bias alone. The same calibration is worth a factor of $1.0308$ at $n=1$ and a factor of $\mathbf{25.02}$ at $n=10^4$—its value differs by a factor of $24$ (Section 5).
Fifth, calibration has a price. If the standard is read only $m$ times, calibration imports an error of its own. For $m\le 16$ it was more accurate not to calibrate at all (Section 5).
Sixth, what counts as bias is decided by what one repeats. For the same $100$ measurements, one laboratory measuring $100$ times gives $0.350571$, while $100$ laboratories measuring once each give $\mathbf{0.040311}$—a factor of $\mathbf{8.70}$ (Section 7).
Scope (what is not claimed)
No new mathematical theorem and no new law is claimed—that the mean squared error splits into a squared bias and a variance, that the variance of a sample mean is $\sigma^2/n$, that the error variance of uniform quantization is $q^2/12$, and that variance separates into levels, are all standard. No theory of measurement is built—the tools used are one decomposition, one sum of two variances, and random numbers. Estimation theory is not entered—unbiasedness, efficiency and the Cramer-Rao fence are treated in Papers 133 and 171. What this paper looks at is an instrument, not an estimator, and the bias here belongs to the instrument, not to the sample. Inverse problems are not entered—Papers 106 and 159 treat them. This paper looks only in the forward direction, at settings where the true value is known. No standard is interpreted clause by clause—the standards are cited only as evidence that the vocabulary itself splits in two, and no question of conformity is discussed. No single Japanese rendering is settled on—the mapping between the Japanese words for accuracy, trueness and precision differs between sources. This paper does not try to repair that disagreement, and writes $b$ and $\sigma$ instead. Normality is not assumed—the decomposition in Section 2 is an identity and does not depend on the shape of the distribution; the random numbers only confirm it. The numbers involve sampling and discretization—the Monte Carlo uses $10^6$ draws, the dither test uses $200$ trials, the resolution is $q=0.10$, and the random seed is fixed. Relation to earlier papers: Paper 133 showed that the Cramer-Rao fence is unbiasedness—this paper follows, in numbers, what lies outside that fence: the case where a bias remains. Paper 171 set unbiasedness and efficiency side by side as separate properties—this paper writes the same two for an instrument rather than an estimator. Paper 106 treated the Wiener filter as the minimizer of the mean squared error—this paper looks at the same $\mathrm{MSE}$ from the side of decomposition rather than minimization. Paper 112 counted resolution as one of six distinct roots—Section 6 here observes that this same resolution behaves exactly like a bias in one respect: it does not fall with repetition. Paper 190 measured “rare” on a logarithmic scale—this paper likewise reports the unreachable limit not as a ratio but as the $n$ at which the crossover happens. Paper 232 counted four different claims written with the same equals sign—this paper counts two or more different quantities written with the same word “accuracy”. Paper 186 showed that a proxy stops being a proxy once it becomes the target—Section 7 here observes that changing the unit of repetition exchanges bias and scatter. What is added is building the decomposition of the $\mathrm{MSE}$ here and checking it against random numbers, putting a concrete number, $m=16.67$, on the break-even point of calibration, observing that the value of one and the same calibration ranges from a factor of $1.03$ to a factor of $25.02$ depending on $n$, computing that the same $100$ measurements differ by a factor of $8.70$ according to how they are divided, and stating that “bias or scatter” is settled not by the instrument but by the unit of repetition.
1. Notation and Placement
Setting: a quantity with true value $\theta$ is measured by a single instrument. Write a reading as
$$X = \theta + b + \varepsilon,\qquad E[\varepsilon]=0,\quad \mathrm{sd}(\varepsilon)=\sigma .$$Here $b$ is the part that departs in the same direction every time, and $\varepsilon$ is the part that changes every time.
Question: can the two be reduced by the same operation?
The answer is no. Repeating reduces $\sigma$ alone; calibrating reduces $b$ alone. There are two knobs, and each is connected to something different.
On vocabulary: in English, accuracy is divided into trueness and precision. The Japanese renderings are assigned differently by different sources—one word may be the umbrella term in one text and the name of the scatter alone in another. This paper settles on no rendering and writes $b$ and $\sigma$. The disagreement among the renderings is itself evidence that one word is naming two things.
2. Building the Tools Here and Checking Them against Known Values
Check 2.1: take $b=0.05$ and $\sigma=0.20$ and evaluate $E[(X-\theta)^2]$ with random numbers. The value to compare against is $b^2+\sigma^2 = 0.042500$.
| Draws | Value from random numbers | Difference from exact |
|---|---|---|
| $10^5$ | $0.042206$ | $-0.000294$ |
| $10^6$ | $\mathbf{0.042462}$ | $\mathbf{-0.000038}$ |
Observation 2.2 At $10^6$ draws the difference is $0.000038$. The tools used below are calibrated by this.
Observation 2.3: the decomposition is an identity.
$$E[(X-\theta)^2] \;=\; \bigl(E[X]-\theta\bigr)^2 + \mathrm{Var}(X) \;=\; b^2+\sigma^2 .$$No property of the distribution has been used—normality is not assumed. This is the same figure as Paper 171, which wrote that the fence around least squares is linearity rather than normality: here too, what does the work is not the distribution but only the definitions of expectation and variance.
3. The Core: Repetition Reduces Only One of the Two
Setting: measure $n$ times with the same instrument and average. The bias stays at $b$ under averaging, while the scatter shrinks to $\sigma/\sqrt{n}$.
Check 3.1: computed with $b=0.05$ and $\sigma=0.20$. The last column is $\sqrt{b^2+\sigma^2/n}$.
| $n$ | Bias | Scatter | Error |
|---|---|---|---|
| $1$ | $0.0500$ | $0.200000$ | $0.206155$ |
| $4$ | $0.0500$ | $0.100000$ | $0.111803$ |
| $16$ | $0.0500$ | $\mathbf{0.050000}$ | $0.070711$ |
| $100$ | $0.0500$ | $0.020000$ | $0.053852$ |
| $10^4$ | $0.0500$ | $0.002000$ | $0.050040$ |
| $10^6$ | $0.0500$ | $0.000200$ | $\mathbf{0.050000}$ |
Observation 3.2 The left column does not move. Multiplying $n$ by $10^6$ leaves it at $0.05$—repetition does not touch this column.
Check 3.3: solve backwards for the number of repetitions needed to reach a target error, $n = \sigma^2/(\text{target}^2-b^2)$.
| Target error | Repetitions needed |
|---|---|
| $0.070711$ | $16$ |
| $0.055$ | $77$ |
| $0.051$ | $397$ |
| $0.0501$ | $3997$ |
| $0.05001$ | $39997$ |
| $0.05$ | unreachable, however many are taken |
Observation 3.4 The closer the target is to $0.05$, the more the required count grows by orders of magnitude. Each further decimal place costs roughly a factor of $10$ in repetitions, and at $0.05$ itself the denominator becomes $0$—this is not “slow” but “unreachable”.
Observation 3.5 The crossover is at $n^{*}=\sigma^2/b^2$. Here $0.04/0.0025=\mathbf{16}$—for $n<16$ the scatter dominates, and for $n>16$ the bias does. Which of the two should be repaired first is exchanged by how many measurements one intends to take.
Observation 3.6 Do not report a ratio. “How many times worse is the bias than the scatter” moves with $n$ and is therefore not a single number—what should be reported is the $n$ at which they cross (the same practice as Paper 190).
4. Two Instruments with the Same Error Part Company under Averaging
Setting: take two instruments. A has $b=0.05$, $\sigma=0.20$; B has $b=0.20$, $\sigma=0.05$. At $n=1$ the errors of the two are identical.
Check 4.1:
| $n$ | Error of A | Error of B | Ratio |
|---|---|---|---|
| $1$ | $0.206155$ | $0.206155$ | $\mathbf{1.0000}$ |
| $10$ | $0.080623$ | $0.200624$ | $2.4884$ |
| $100$ | $0.053852$ | $0.200062$ | $3.7151$ |
| $10^4$ | $\mathbf{0.050040}$ | $\mathbf{0.200001}$ | $\mathbf{3.9968}$ |
| $\infty$ | $0.050000$ | $0.200000$ | $4.0000$ |
Observation 4.2 Even when the single-reading error quoted on the data sheet is the same, a difference in use makes them differ by a factor of $4$. A data sheet that writes “accuracy” as one number cannot express this difference.
Check 4.3: for four instruments, count how much averaging $10^4$ readings is worth.
| Instrument | $b$ | $\sigma$ | Error at $n=1$ | Value of averaging |
|---|---|---|---|---|
| Small bias, small scatter | $0.01$ | $0.02$ | $0.022361$ | $2.2356$ |
| Small bias, large scatter | $0.01$ | $0.50$ | $0.500100$ | $\mathbf{44.7303}$ |
| Large bias, small scatter | $0.50$ | $0.02$ | $0.500400$ | $\mathbf{1.0008}$ |
| Large bias, large scatter | $0.50$ | $0.50$ | $0.707107$ | $1.4141$ |
Observation 4.4 The second and third instruments have single-reading errors of $0.500100$ and $0.500400$, which are all but the same. Yet the same average over $10^4$ readings is worth a factor of $44.73$ to one and only $1.0008$ to the other—the return on the same operation differs by a factor of $\mathbf{44}$.
Observation 4.5 The separator—what has to be changed for these to become different things. Not the single-reading error, but how that error is composed. Hand over one number without the composition, and what to do next cannot be decided.
5. Which of the Two Calibration Removes, and What Calibration Costs
Setting: measure a standard whose true value is known, and subtract the departure. This is calibration.
Check 5.1: the ideal case, in which the bias is known exactly. Then $b=0.05\to 0$, while $\sigma$ stays at $0.20$ and does not move.
| $n$ | Before calibration | After calibration | Value |
|---|---|---|---|
| $1$ | $0.206155$ | $0.200000$ | $\mathbf{1.0308}$ |
| $10^4$ | $0.050040$ | $0.002000$ | $\mathbf{25.02}$ |
Observation 5.2 The value of the same calibration ranges from a factor of $1.03$ to a factor of $25.02$ according to $n$. If only one reading is taken, calibration is worth almost nothing; if $10^4$ readings are taken, calibration is almost the whole of the gain—the value of calibration is set not by the instrument but by how many measurements are still to come.
Check 5.3: a real calibration estimates the departure from $m$ readings of the standard. The estimate imports an error of its own: $\sigma^2/m + u^2$, where $u=0.01$ is the uncertainty of the standard itself. Compare this with $b^2=0.0025$, the case of not calibrating.
| $m$ | Error imported | Against $b^2$ | Floor after calibration |
|---|---|---|---|
| $4$ | $0.01010000$ | worse | $0.100499$ |
| $9$ | $0.00454444$ | worse | $0.067412$ |
| $16$ | $0.00260000$ | worse | $0.050990$ |
| $17$ | $0.00245294$ | better | $\mathbf{0.049527}$ |
| $25$ | $0.00170000$ | better | $0.041231$ |
| $100$ | $0.00050000$ | better | $0.022361$ |
| $10^4$ | $0.00010400$ | better | $\mathbf{0.010198}$ |
Observation 5.4 The break-even point is $m=\sigma^2/(b^2-u^2)=\mathbf{16.67}$. If the standard is read $16$ times or fewer, it was more accurate not to calibrate—calibration is not an operation that improves things unconditionally.
Observation 5.5 However large $m$ is made, the floor stops at $u=0.01$. The uncertainty of the standard itself cannot be moved by repetitions on this side—calibration does not remove the bias; it replaces one's own bias with the bias of the standard.
Observation 5.6 The two knobs do not meet. Repetition is connected only to $\sigma$, and calibration only to $b$. However far one is turned, the floor set by the other does not move.
6. Bias Is Not the Only Thing That Does Not Fall with Repetition
Setting: a display with graduation $q=0.10$ reads a true value of $20.037$. The reading is rounded to the nearest graduation.
Check 6.1: read the same value repeatedly on an ideal instrument with no scatter.
| Readings | Averaged reading | Error |
|---|---|---|
| $1$ | $20.000000$ | $-0.037000$ |
| $100$ | $20.000000$ | $-0.037000$ |
| $10^6$ | $20.000000$ | $\mathbf{-0.037000}$ |
Observation 6.2 It never falls. With zero scatter, averaging only returns the same value again—here the error due to resolution behaves entirely as a bias.
Check 6.3: add scatter of standard deviation $\sigma=0.20$ before the same display reads (that $\sigma$ is twice the graduation $q$). The error is measured over $200$ trials.
| Readings | Error | $\sigma/\sqrt{n}$ |
|---|---|---|
| $1$ | $0.199772$ | $0.200000$ |
| $100$ | $0.020192$ | $0.020000$ |
| $10^4$ | $\mathbf{0.002142}$ | $0.002000$ |
Observation 6.4 Now it falls as $\sqrt{n}$. The averaged answer comes closer to the true value when scatter is added—the same graduation on the same display became a bias in one case and a scatter in the other.
Observation 6.5 What separated them was not the instrument. It was whether the scatter is larger than the graduation—at $\sigma=0$ it behaves as a bias, and at $\sigma\gg q$ as a scatter. Paper 112 counted resolution as one of six distinct roots—what this section adds is the single point that the same resolution moves to the bias side or to the scatter side.
7. What Counts as Bias Is Decided by What You Repeat
Setting: several laboratories measure the same sample. The scatter now has two storeys: the scatter on repetition within one laboratory, $\sigma_r$, and the standing departure of each laboratory, $\sigma_L$. Write the combination as $\sigma_R$.
Check 7.1: take $\sigma_r=0.20$ and $\sigma_L=0.35$.
| Quantity | Value |
|---|---|
| Scatter within one laboratory $\sigma_r$ | $0.200000$ |
| Departure of each laboratory $\sigma_L$ | $0.350000$ |
| Combination $\sigma_R=\sqrt{\sigma_L^2+\sigma_r^2}$ | $\mathbf{0.403113}$ |
| Ratio $\sigma_R/\sigma_r$ | $2.015564$ |
Check 7.2: increase the number of repetitions within one laboratory.
| $n$ | Error from the true value |
|---|---|
| $1$ | $0.403113$ |
| $4$ | $0.364005$ |
| $16$ | $0.353553$ |
| $100$ | $0.350571$ |
| $10^4$ | $\mathbf{0.350006}$ |
Observation 7.3 It stops at $0.35$. So long as one stands inside a single laboratory, that laboratory's departure is a bias—it does not fall with repetition. The table has the same shape as the one in Section 3.
Check 7.4: take the same $100$ measurements and change only how they are divided between laboratories and repetitions. Every row totals $100$.
| Division | Total | Error |
|---|---|---|
| $1$ laboratory $\times$ $100$ readings | $100$ | $\mathbf{0.350571}$ |
| $2$ laboratories $\times$ $50$ readings | $100$ | $0.248294$ |
| $5$ laboratories $\times$ $20$ readings | $100$ | $0.157797$ |
| $10$ laboratories $\times$ $10$ readings | $100$ | $0.112472$ |
| $25$ laboratories $\times$ $4$ readings | $100$ | $0.072801$ |
| $100$ laboratories $\times$ $1$ reading | $100$ | $\mathbf{0.040311}$ |
Observation 7.5 The same $100$ measurements differ by a factor of $\mathbf{8.70}$. The number of measurements is the same and the effort spent is the same—what differs is only what was repeated.
Observation 7.6 This is the core of the paper. The laboratory departure $\sigma_L$ is a bias when one stands inside a single laboratory (it does not fall with repetition) and a scatter when one stands across laboratories (it falls with the number of laboratories). One and the same quantity becomes either. “Bias or scatter” is not a property of the instrument but a question of where the unit of repetition has been placed.
Observation 7.7 A point to state honestly—if every laboratory calibrates against the same standard, $\sigma_L$ does shrink. But that means the departures between laboratories were a collection of biases. What remains after the shrinking is the true scatter at that level—calibration is also the experiment that separates the two.
8. Roll Call and Closing
| Quantity | Falls with repetition? | Falls with calibration? |
|---|---|---|
| Bias $b$ | no | yes |
| Scatter within one laboratory $\sigma_r$ | yes | no |
| Departure of each laboratory $\sigma_L$ | depends on the unit of repetition | shrinks against a common standard |
| Resolution $q$ | depends on the size of the scatter | no |
Observation 8.1 Of the four rows, only the top two admit an answer written as “yes” or “no”. The lower two become either, according to the conditions.
- The tools were calibrated (Check 2.1): a difference of $0.000038$ from the exact value at $10^6$ draws.
- The decomposition is an identity (Observation 2.3): no property of the distribution is used.
- Repetition reduces only one (Check 3.1): averaging $10^6$ readings still stops at $0.050000$.
- The crossover is at $n^{*}=16$ (Observation 3.5): which to repair first is exchanged by $n$.
- Two instruments with the same error part company (Check 4.1): a factor of $3.9968$ at $n=10^4$.
- The return on averaging differs by a factor of $44$ (Check 4.3): $44.7303$ against $1.0008$.
- The value of calibration is set by $n$ (Check 5.1): $1.0308$ against $25.02$.
- The break-even point of calibration is $m=16.67$ (Observation 5.4): at $16$ readings or fewer it is more accurate not to calibrate.
- Resolution moves to either side (Checks 6.1 and 6.3): it stays at $-0.037000$, or it falls as $\sqrt{n}$.
- The same $100$ measurements differ by a factor of $8.70$ (Check 7.4): $0.350571$ against $0.040311$.
Closing: “this instrument is accurate” was not saying one thing. It named the departure from the true value and the scatter—the first does not fall with repetition, and the second does not fall with calibration. The two knobs do not meet, and however far one is turned the floor set by the other does not move. And which of them is the bias and which the scatter turned out not to belong to the instrument. Add scatter and the resolution turns into scatter; move the unit of repetition to the laboratory and the laboratory's departure turns into scatter. One thing separates them—writing down what it is that is being repeated. Write it down, and which knob to turn next is settled. Do not write it down, and one cannot see that $100$ measurements have been spent at a loss of a factor of $8.70$.
References
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Reference
Yamagishi, Y. (2026). “Accuracy” Is Two Things, and Calibration Removes Only One [Paper 255]. YOI: https://www.yamagishiyuuki.com/yoi/255
DOI: 10.5281/zenodo.22168883
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 255. This paper is a preprint and has not been peer reviewed.