Cosmology · Consistency, tensions and claims of new physics

Cross-probe consistency and correlated uncertainty

Compare two exact finite calibration-error laws with identical marginal variances but different uncertainty in their difference.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

Track a shared error through two declared response signs.
Calculate the variance of a comparison and an average.
Interpret finite one-standard-deviation coverage without assuming Gaussian errors.

A comparison needs the joint uncertainty model

Before comparing estimates, establish which physical quantity, epoch and assumptions each estimate refers to. A reported error on each estimate alone does not describe their joint uncertainty.

A common calibration can move two quantities together or affect them with opposite responses. Its contribution to their difference depends on those response signs.

Covariance records this joint variation under a specified probability law. It is not an optional correction determined by how similar the method names sound.

Historical supernova methods supply a concrete common-calibration example. The following numbers define original finite error experiments and represent no observed discrepancy or cosmological significance.

Keep fixed bias and ensemble error distinct

Choose arbitrary unit U and hold an unknown quantity θ fixed. Supply A = θ+c+e_A and B = θ+λc+e_B, where known response coefficient λ is separately +1 or −1.

For each case, independently draw c = ±2U/5, e_A = ±U/10 and e_B = ±U/5, with equal probability for each sign. The eight ordered sign triples are equiprobable.

The same symmetric random calibration c is shared by A and B. Its response in B changes sign between the two deliberately separate experiments.

This is a repeated-draw ensemble error model. A single unknown fixed common bias would not acquire the same repeated-sampling variance without a supplied probability model.

The same marginal errors can yield different comparison errors

The marginal variances are Var(A) = 17U²/100 and Var(B) = U²/5 in both cases. Their covariance is 4λU²/25.

For D = A−B, use Var(D) = Var(A)+Var(B)−2Cov(A,B). When λ = +1, the common c cancels and Var(D) = U²/20.

When λ = −1, D contains 2c as well as the individual errors, giving Var(D) = 69U²/100. Thus the difference uncertainty changes while both marginal variances stay fixed.

The corresponding standard deviations are about 0.223607U and 0.830662U. They are conditional summaries of the stated finite laws, not significance levels for any observed residual.

Two separately supplied response laws; variance and covariance entries are in U²
λ	Var(A)/U²	Var(B)/U²	Cov(A,B)/U²	Var(D)/U²	SD(D)/U	Var(M)/U²
+1	17/100	1/5	4/25	1/20	0.223607	69/400
−1	17/100	1/5	−4/25	69/100	0.830662	1/80

A standard deviation does not prescribe a universal probability

For λ = +1, D/U has four distinct values −0.3,−0.1,0.1,0.3, each with probability 1/4. Only the two values ±0.1 lie within one standard deviation of the mean zero.

For λ = −1, its eight values are ±0.5,±0.7,±0.9,±1.1, each with probability 1/8. The four values ±0.5 and ±0.7 lie within the corresponding one-standard-deviation interval.

Therefore P(|D|≤SD(D)) = 1/2 in both cases. No Gaussian law or universal 68% rule follows from the variances.

The finite probability masses are part of the declared experiment. Smoothing them into an unspecified continuous distribution or attaching a posterior to θ would introduce additional assumptions.

This supplied finite repeated-error experiment fixes common physical θ in an arbitrary positive unit U. Independently draw c/U = ±0.4, e_A/U = ±0.1 and e_B/U = ±0.2, with each sign equally likely; all eight ordered triples have probability 1/8. A = θ+c+e_A and B = θ+λc+e_B are two separately supplied response laws, λ = +1 or −1. D = A−B. Both panels use linear horizontal D/U axes from −1.2 to 1.2 and the same zero-based vertical probability-mass axis from 0 to 0.3. Panel A, labeled λ = +1, uses solid blue stems and hollow circles. It has four masses, each 1/4, at D/U = −0.3, −0.1, 0.1, 0.3. Panel B, labeled λ = −1, uses dashed orange stems and hollow squares. It has eight masses, each 1/8, at D/U = −1.1, −0.9, −0.7, −0.5, 0.5, 0.7, 0.9, 1.1. These are point masses, not a probability density, histogram of measured counts or Gaussian curve. Both cases have Var(A)/U² = 17/100 and Var(B)/U² = 1/5. Cov(A,B)/U² = 4λ/25, giving Var(D)/U² = 1/20 or 69/100 and SD(D)/U approximately 0.223607 or 0.830662, respectively. Exactly one half of the probability lies within one standard deviation of zero in either law: the two inner masses ±0.1 in A or four inner masses ±0.5 and ±0.7 in B. The mean M = (A+B)/2 instead has Var(M)/U² = 69/400 or 1/80. A single fixed unknown bias is not automatically a draw from this zero-mean law. No observed discrepancy, significance, posterior, Gaussian 68% rule or independent new evidence is supplied. Separate panel labels, stem patterns, marker shapes and printed complete supports permit reading without color. Original Ark teaching model.

The same shared error affects other combinations differently

Let M = (A+B)/2. Its variance is [Var(A)+Var(B)+2Cov(A,B)]/4, giving 69U²/400 when λ = +1 and U²/80 when λ = −1.

The same-sign calibration cancels in the difference but remains in the average. The opposite-sign calibration instead cancels in the average and remains in the difference.

A small difference can therefore coexist with shared uncertainty in an inferred common value. Agreement and combination are different uses of the same joint model.

A real cross-probe assessment retains the measurement response, shared information, calibration and sampling assumptions. The mathematical cancellation here neither supplies independent evidence nor identifies a new physical explanation.

Worked reasoning

Calculate the same-sign difference variance

What is Var(A−B) for λ = +1?

Keep marginal variances 17U²/100 and 20U²/100.
Subtract twice the common covariance 16U²/100: (17+20−32)U²/100.
Obtain U²/20; the complete finite support gives one-standard-deviation content 1/2.

Var(D) = U²/20, conditional on the declared common response and error law.

Try the reasoning

What is Var(D) for λ = −1?
Show the answer and reasoning
69U²/100.
The negative cross-covariance contributes positively when it is subtracted twice.
Do the two marginal standard deviations fully specify this comparison?
Show the answer and reasoning
No.
The response sign changes the cross-covariance and therefore the difference variance.
Is the one-standard-deviation interval automatically a 68% interval here?
Show the answer and reasoning
No; its exact finite-law content is 50%.
Enumerating the supplied masses gives half inside the interval in either case.

What to carry forward

A probe comparison depends on shared responses and covariance.
Identical marginal errors can accompany different difference uncertainties.
Standard deviations alone do not specify Gaussianity, coverage or significance.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.