Astronomy · Comparing galaxies

Comparing samples without ignoring selection

Follow a finite galaxy ledger from its parent populations through conditional selection, inverse inclusion weighting and a separate comparison at common luminosity-bin weights.

Continue in the approved chapter order. Necessary vocabulary, quantities and assumptions are explained in the lesson.

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After reading, you can

Distinguish parent-population fractions from ratios of expected selected counts.
Reconstruct a declared expectation ledger using known positive inclusion probabilities.
Explain why common-bin standardization answers a different question from restoring each parent’s composition.

Which population does the fraction describe?

A statement such as “this environment has fewer disk galaxies” needs a parent population, a classification rule and a selection description. A catalogue fraction is calculated from objects that entered the catalogue, which need not have the same composition as the intended parent population.

The classification itself is an inference from data with finite depth and resolution. Projected structure, colour, activity and stellar mass remain different properties. Comparing a disk fraction requires a consistent structural criterion rather than quietly replacing structure with a colour label.

Observing band, angular resolution, depth and surface-brightness sensitivity affect the available measurements. Samples at different distances may also require explicitly defined rest-frame band comparisons. Environment assignment introduces its own membership choices and uncertainties.

The following original exercise avoids those additional uncertainties by stipulating perfect disk-versus-other classification, two fixed luminosity bins and known inclusion probabilities. Its purpose is to distinguish three different numerical questions, not to claim that a simple weighting formula repairs every real catalogue.

A parent difference can come from population composition

Let hypothetical environments P and Q each contain exactly 100 parent galaxies. P has 80 in a lower-luminosity bin and 20 in a higher-luminosity bin; Q has 20 lower and 80 higher. These are finite stipulated parents, not observed estimates or an evolutionary sequence.

Within each luminosity bin, the disk fractions are identical in P and Q: 0.8 in the lower bin and 0.2 in the higher bin. Thus P contains 64 + 4 = 68 disks, while Q contains 16 + 16 = 32. Their parent disk fractions really are 0.68 and 0.32 under this ledger.

That aggregate parent difference arises because the two environments contain different proportions of the two bins. It does not contradict the equality of their stipulated within-bin fractions. Asking about each whole parent and asking about otherwise matched bin composition are different questions.

The bins are defined only by the supplied luminosity measure. Equal luminosity bins do not automatically mean equal stellar mass, age or gas content. No environmental mechanism or causal process has been assigned to these populations.

Selection changes the expected catalogue composition

Assign inclusion probability s = 0.25 to every lower-bin galaxy and s = 1 to every higher-bin galaxy. These probabilities are known, positive and identical for disks and other objects within a bin and in both environments. Expected selected counts are the parent counts multiplied by the applicable probability.

P then has expected selected counts 16 lower-bin disks, 4 lower-bin others, 4 higher-bin disks and 16 higher-bin others. The ratio of its expected selected disk count to expected selected total is 20/40 = 0.50. Q has corresponding counts 4, 1, 16 and 64, giving 20/85 = 4/17 ≈ 0.235294118.

These entries are an expectation ledger. They are not a particular random catalogue realization. A ratio of expected counts is not generally the expectation of a ratio computed from a random selected catalogue, and neither expression should be called an observed fraction without actual data.

The same inclusion rule changes each parent’s representation of the two bins. It does not alter an individual galaxy’s stipulated type. Selection and a genuine difference in parent composition can coexist, as they do here.

Original finite parents and their expected selected ledgers. Disk classification is perfect. Known inclusion probabilities are 0.25 in the lower bin and 1 in the higher bin, independent of structural type within each bin. Entries after selection are expectations.
Environment / bin	Parent disks	Parent others	Inclusion probability	Expected selected disks	Expected selected others
P / lower	64	16	0.25	16	4
P / higher	4	16	1	4	16
Q / lower	16	4	0.25	4	1
Q / higher	16	64	1	16	64

Two weighting operations answer two questions

Divide each expected selected count by its known inclusion probability. For P’s lower bin, 16/0.25 = 64 disks and 4/0.25 = 16 others; the higher-bin counts are unchanged. Doing the same in Q restores its declared parent counts. The reconstructed parent fractions are therefore 0.68 and 0.32, not a forced equality.

This exact reconstruction concerns the supplied expectation ledger with correct, positive probabilities. For a random realized catalogue, inverse-inclusion weighting gives a statistical estimator whose uncertainty and ratio properties require analysis. Dividing by a guessed probability or by zero is not a valid way to reveal an unknown missing population.

Now ask a different question: what disk fraction would each environment have if the two luminosity bins received common weights of 0.5 and 0.5? Standardizing the stipulated within-bin fractions gives 0.5(0.8) + 0.5(0.2) = 0.50 for both P and Q.

Common-bin standardization intentionally changes the aggregate composition being summarized. It does not claim that either original parent had that composition, erase the parent difference or prove that environment has no causal role. It describes the stated conditional comparison at the supplied bin weights.

Three explicitly different summaries of the original ledger. Selected summaries are ratios of expectations; parent restoration uses known positive inclusion probabilities; standardization uses common one-half weights for each luminosity bin.
Summary question	P	Q
Declared whole-parent disk fraction	68/100 = 0.68	32/100 = 0.32
Ratio of expected selected disk count to total	20/40 = 0.50	20/85 = 4/17
Restore each parent from its expectation ledger	0.68	0.32
Common lower/higher weights (0.5, 0.5)	0.50	0.50

Report the comparison that the data support

A practical comparison should identify its target population, observing band, property definitions, environment assignment and inclusion process. It should also describe measurement errors, missing classifications and the regions of parameter space where the selection model has been tested.

Skibba and collaborators separate morphology, colour and environmental relationships in an explicitly selected observational setting. Blanton’s low-surface-brightness work illustrates why detectability matters, while population modelling and group-finding studies expose further inference and membership assumptions. These sources support careful comparison, not a universal correction coefficient.

Matching or weighting observed properties can help ask a more specific question. It cannot establish that all unobserved differences are absent, that the adopted categories are physically complete or that a correlation is causal. A result should retain the limits of the conditioning variables and of the selection probabilities.

This chapter has separated shapes, light mixtures, detectability, environment, interaction forces and sample comparison. The next chapter examines gas, dust and star formation, where different observing tracers will again weight different physical components and timescales.

Worked examples

Recover the declared P parent

What does inverse inclusion weighting recover from P’s expected selected counts?

Divide lower-bin counts 16 disks and 4 others by 0.25 to obtain 64 and 16.
Keep higher-bin counts 4 disks and 16 others because their inclusion probability is 1.
Sum disks 68 and all objects 100, giving the declared parent fraction 0.68.

It reconstructs P’s supplied whole-parent expectation ledger and disk fraction 0.68, rather than imposing a common composition.

Change the comparison question explicitly

Why do P and Q both give 0.50 after common-bin standardization?

Their supplied disk fractions within each bin are the same: 0.8 lower and 0.2 higher.
Assign common weights 0.5 and 0.5, rather than each parent’s original bin proportions.
Compute 0.5 × 0.8 + 0.5 × 0.2 = 0.50 in each environment.

The standardized summaries agree because the stipulated within-bin fractions agree at the chosen common weights. Their original whole-parent fractions remain different.

Try the reasoning

Is the difference between the stipulated parent fractions 0.68 and 0.32 merely a nonexistent observational illusion?
Show the answer and reasoning
No.
It is a real difference in these declared finite parents, caused by their different bin compositions.
Is a ratio of expected selected counts automatically an observed fraction or the expectation of a random catalogue’s ratio?
Show the answer and reasoning
No.
Those are distinct statistical quantities and require separate evidence or analysis.
Can division by an inclusion probability of zero restore an unseen bin?
Show the answer and reasoning
No.
No observations of that bin are supplied by the selection; additional assumptions or data are required.
Does equality after luminosity-bin standardization prove equal stellar masses or absence of causal environmental effects?
Show the answer and reasoning
No.
The operation controls only the supplied bin composition and does not measure all other properties or establish causation.

What to carry forward

A catalogue, its parent and a composition-matched comparison can have different fractions.
Known positive inclusion probabilities reconstruct the supplied expectation ledger without erasing its parent differences.
Conditional standardization and causal explanation are distinct tasks with explicit assumptions.

References

Teaching resources and research supporting this lesson’s explanations and examples.