Propagating uncertainties

Builds on: Sections S3 and D3
Reading time: 45 min
Exercises: 6
Olympiad problems: 1

The Large Magellanic Cloud, the hazy patch of light near the south celestial pole that we met in Section O1.6, contains thousands of Cepheids. Once the line of Sections D2 and D3 tells us the absolute magnitude of a Cepheid from its period, comparing it with how bright the star looks gives the distance of the whole galaxy. The line itself is uncertain, though, and so are the magnitudes, and the question every data-analysis problem asks at the end is how uncertain the distance comes out. In this section we carry uncertainties through the functions that olympiad problems use most, logarithms, magnitudes, parallaxes and angles, then through formulas with several measured values, and finally from a hand-drawn line to a distance.

D4.1One measured value

Before you readMake a guess first

The flux of a star is measured with an uncertainty of 10%. Roughly how uncertain is its magnitude?

When a result $f$ depends on one measured value $x$ , the rule for small changes of (S.12) gives its uncertainty. Over the short stretch from $x - σ_{x}$ to $x + σ_{x}$ the curve of $f$ is nearly straight, so $f$ is uncertain by the slope of the curve times $σ_{x}$ , which is $σ_{f} = | d f / d x | σ_{x}$ . The functions of olympiad problems give a few rules worth knowing by heart. The derivative of $\log x$ is $1 / (x \ln 10)$ , so the uncertainty of a logarithm is $0.434$ times the fractional uncertainty of $x$ . A magnitude is $- 2.5$ times the logarithm of a flux, so it is uncertain by $2.5 \times 0.434 = 1.086$ times the fractional uncertainty of the flux. The absolute magnitude worked out from a parallax $ϖ$ contains $5 \log ϖ$ , so it is uncertain by $2.17$ times the fractional uncertainty of the parallax, the rule we used for the Cepheids in Section D2.

ResultUncertainties through common functions(D.3)

\begin{aligned} σ_{\log x} & = 0.434 \frac{σ_{x}}{x} \\ σ_{m} & = 1.086 \frac{σ_{F}}{F} \\ σ_{M} & = 2.17 \frac{σ_{ϖ}}{ϖ} \\ σ_{\sin θ} & = | \cos θ | σ_{θ} \end{aligned}

$x, F, ϖ, θ$	a measured value, a flux, a parallax and an angle
$σ$	the uncertainty of the quantity in its subscript
$m, M$	a magnitude worked out from the flux, and an absolute magnitude worked out from the parallax

The rules hold when each uncertainty is small compared with its value. In the last line

σ_{θ}

must be in radians. The numbers are

1 / \ln 10 = 0.434

2.5 / \ln 10 = 1.086

and

5 / \ln 10 = 2.17

For RT Aur, with a parallax of 2.40 ± 0.19 mas, the fractional uncertainty is $0.19 / 2.40 = 0.079$ , so its absolute magnitude is uncertain by $2.17 \times 0.079 = 0.17$ , the value in the table of Figure D.1.

Figure D.3An uncertainty carried through a curveinteractive

ValueUncertainty

Result
By the rule of small changes
From the ends of the band

The figure shows the distance worked out from a parallax, or the magnitude worked out from a flux, against the measured value. The heavy band on the horizontal axis is the measured value plus or minus its uncertainty, and the dashed lines carry its two ends to the curve and across to the vertical axis, where the heavy band shows the true spread of the result. Beside it, the dashed band is the spread that the rule of small changes predicts from the slope of the thin tangent line. The words below set , which gives a distance of 417 pc and a spread of about 33 pc either way, and .

For the parallax of 2.40 ± 0.19 mas the ends of the band give distances of 452 and 386 pc, while the rule predicts 417 ± 33 pc, and the two agree well. With the larger uncertainty the curve bends noticeably within the band, and the result spreads much further on one side than on the other.

D4.2Several measured values

Before you readMake a guess first

A result is the sum of two measured values, uncertain by 3 and by 4 units, and their errors are independent. How uncertain is the sum?

When a result $f$ depends on several measured values, each one contributes as it would on its own, as if the others were known exactly. The contribution of $x$ is the change in $f$ when only $x$ moves by $σ_{x}$ , which is the derivative of $f$ with respect to $x$ with the other values held fixed, called the partial derivative $\partial f / \partial x$ , times $σ_{x}$ . Independent errors partly cancel, so the contributions combine in quadrature, as in (S.16), which is the special case for sums, products and powers. Adding the contributions instead gives the maximum error, which is larger and more cautious. The IOAA syllabus accepts either, and an answer should say which one it uses.

ResultCombining the contributions(D.4)

\begin{aligned} σ_{f} & = \sqrt{{(\frac{\partial f}{\partial x} σ_{x})}^{2} + {(\frac{\partial f}{\partial y} σ_{y})}^{2}} \\ Δ f_{max} & = | \frac{\partial f}{\partial x} | σ_{x} + | \frac{\partial f}{\partial y} | σ_{y} \end{aligned}

$x, y$	measured values with uncertainties $σ_{x}$ and $σ_{y}$
$\partial f / \partial x$	the derivative of $f$ with respect to $x$ when $y$ is held fixed
$σ_{f}$	the uncertainty of $f$ when the errors of $x$ and $y$ are independent
$Δ f_{max}$	the maximum error of $f$

More measured values add more terms in the same way. The first line needs independent errors, and both lines need uncertainties that are small compared with the values.

The distance modulus $μ = m - M$ , the apparent magnitude of a star minus its absolute magnitude, is a simple case. Its partial derivatives are $+ 1$ and $- 1$ , so with $m$ uncertain by 0.05 and $M$ by 0.19 the distance modulus is uncertain by $\sqrt{{0.05}^{2} + {0.19}^{2}} = 0.20$ magnitudes, or at most by $0.05 + 0.19 = 0.24$ . The larger contribution dominates, and improving the smaller one would hardly help.

D4.3From the line to the answer

Before you readMake a guess first

Lines turned about the centre C of the points give the slope of the Cepheid relation as −2.48 ± 0.35. Where along the line is the absolute magnitude it predicts most certain?

To use the Cepheid line we read the absolute magnitude of a star from its period and compare it with the star's apparent magnitude. The slope and intercept of the line are both uncertain, but their errors are not independent. The steepest and shallowest lines cross at C, so at C they agree, and a distance $Δ x$ along the axis away from C they differ by $Δ x$ times the difference of their slopes. The uncertainty of the value read from the line at $x$ is therefore $| x - \bar{x} | σ_{m}$ , which is the same as reading the value from the steepest and shallowest lines and taking half the difference. Combining the uncertainties of the slope and the intercept in quadrature instead would count the same error twice.

Example D.8The distance to the Large Magellanic Cloud from one Cepheid

The Cepheid HV899 in the Large Magellanic Cloud has a period of 30.90 days and a mean apparent magnitude of 13.07, already corrected for dust. Use the line $M_{V} = - 2.48 \log (P / day) - 1.44$ , whose slope is uncertain by 0.35, with C at $\bar{x} = 0.925$ , to find the distance of the star and its uncertainty. The distance modulus gives the distance in parsecs as $d = 10^{(μ + 5) / 5}$ , as Chapter 4 explains.

Solution

The logarithm of the period is $\log 30.90 = 1.490$ , so the line gives $M_{V} = - 2.48 \times 1.490 - 1.44 = - 3.70 - 1.44 = - 5.14$ . The star lies $1.490 - 0.925 = 0.565$ from C along the axis, so $M_{V}$ is uncertain by $0.565 \times 0.35 = 0.20$ .

The distance modulus is $μ = 13.07 - (- 5.14) = 18.21$ , uncertain by the same 0.20, since the apparent magnitude is precise. The distance is

d = 10^{(18.21 + 5) / 5} = 10^{4.642} pc = 44 000 pc .

The derivative of $10^{(μ + 5) / 5}$ is $(\ln 10 / 5)$ times itself, so the fractional uncertainty of the distance is $0.461 \times 0.20 = 0.092$ , and $σ_{d} = 0.092 \times 44 000 = 4 000$ pc.

AnswerThe star, and with it the Large Magellanic Cloud, is 44 ± 4 thousand parsecs away. Most of the uncertainty comes from the slope of the line, because the star's period is far from the periods of the five Cepheids that set it.

D4.4When the uncertainty is large

Before you readMake a guess first

A star has a parallax of 0.10″ ± 0.04″, which gives a distance of 10 pc. What range of distances does the uncertainty allow?

All the rules of this section assume that the uncertainty is small, so that the curve is nearly straight across the band. When the uncertainty is a large fraction of the value, say more than about a tenth, the curve bends within the band, as Figure D.3 shows for a parallax of 1.00 ± 0.40 mas. The honest answer is then to work out the result at both ends of the band and quote the two sides separately. For a parallax of 0.10″ ± 0.04″ the distance is 10 pc, and the ends give 16.7 and 7.1 pc, so we write it as $10_{- 2.9}^{+ 6.7}$ pc. The rule of small changes would have given ±4 pc, which is too little on one side and too much on the other.

Example D.9The sine of a measured anglefinish it

An angle is measured as 30° ± 2°. How uncertain is its sine?

The start

By the last line of (D.3), $σ_{\sin θ} = | \cos θ | σ_{θ}$ with $σ_{θ}$ in radians. Two degrees is $2 \times π / 180 = 0.0349$ rad, and $\cos 30 ° = 0.866$ .

Your stepNow find

σ_{\sin θ}

Common errorsSection D4

An uncertainty in degrees put straight into $| \cos θ | σ_{θ}$ gives a result 57 times too large. The angle's uncertainty must be in radians.
Adding the uncertainties of independent values gives the maximum error, which is larger than the usual uncertainty. Either is accepted, but the answer should say which one it is.
The slope and intercept of one line are not independent. The uncertainty of a value read from the line comes from the steepest and shallowest lines at that point, and combining the uncertainties of slope and intercept in quadrature counts the same error twice.
A magnitude has no fractional uncertainty to speak of, since a magnitude of zero is a perfectly good value. Its uncertainty is given in magnitudes, and it corresponds to a fractional uncertainty of the flux.
When an uncertainty is a large fraction of its value, the rule of small changes gives a symmetric answer to a lopsided question. Working out the result at both ends of the band gives the true range.

ExercisesSection D46 questions

D4.1
The flux of a star is uncertain by 5%. How uncertain is its magnitude?
mag
D4.2
A period is measured as 250 ± 10 days. How uncertain is $\log (P / day)$ ?
D4.3
A result is the difference of two values uncertain by 0.3 and 0.4, with independent errors. How uncertain is it?
D4.4
A Cepheid in another galaxy has $\log (P / day) = 0.30$ . Using the Cepheid line of this section, with C at $\bar{x} = 0.925$ and a slope uncertain by 0.35, how uncertain is the absolute magnitude it predicts?
mag
D4.5
A star has a parallax of 0.050″ ± 0.025″. What is the largest distance the uncertainty allows?
pc
D4.6
A result is worked out from two values, one uncertain by 1% and the other by 6%, and it is proportional to their product. Which would most improve the result?

Olympiad problemsPast IOAA problems, restated1 problem

2017The distance to the Large Magellanic Cloud24 points

Adapted from IOAA 2017 (Phuket), data-analysis problem 1, part 3. We have restated it in our own words and kept the original data. The relations to use are the ones found in Sections D2 and D3, which take the place of the relations found in the earlier parts of the original problem. The solution was last checked on 20 October 2026.

The table gives the periods $P$ , the mean apparent magnitudes $⟨ V ⟩$ , already corrected for dust, and the mean Wesenheit magnitudes $⟨ W ⟩$ of five Cepheids in the Large Magellanic Cloud.

Star	$P$ (day)	$⟨ V ⟩$ (mag)	$⟨ W ⟩$ (mag)
HV12199	2.63	16.08	14.56
HV12203	2.95	15.93	14.40
HV12816	9.10	14.30	12.80
HV899	30.90	13.07	10.97
HV2257	39.36	12.86	10.54

Use the relations $M_{V} = - 2.48 \log (P / day) - 1.44$ and $M_{W} = - 3.43 \log (P / day) - 2.34$ to find the distance modulus and the distance of each star in both bands. Then find the mean distance to the Large Magellanic Cloud and its standard deviation for each band, say whether the two mean distances differ by more than their scatter and whether the two standard deviations differ, and decide which band gives the better distance. The distance in parsecs is $d = 10^{(μ + 5) / 5}$ . In the original paper this part was worth 24 marks.

I solved it without the solution

AnswerThe Large Magellanic Cloud is 46 600 ± 4 300 pc away from the V magnitudes and 47 400 ± 900 pc from the Wesenheit magnitudes. The two distances agree within their scatter, but the Wesenheit distances scatter about five times less, because they are free of the uneven dimming by dust, so the Wesenheit band gives the better distance. The official solution, with slightly different relations, finds 46 200 ± 4 100 pc and 48 100 ± 900 pc and the same conclusions.