Empirical CDFs and QQ evidence

Try

Four recorded proportions are 0.10, 0.25, 0.25 and 0.80. Compare their distribution with a fully specified uniform law on [0,1]. A histogram would require choosing bins. Instead ask, at every threshold x, what fraction of observations is at or below x. This gives the empirical cumulative distribution function, abbreviated ECDF.

For n observations x₁,…,xₙ with n≥1, define Fₙ(x)=(number of i with xᵢ≤x)/n. Equivalently it is the average of the indicators 1❴xᵢ≤x❵, each equal to 1 when its condition holds and 0 otherwise. The empirical law puts mass 1/n at each observation, counting repeated observations separately. This is a descriptive construction even without a random-sampling model. An inferential interpretation will additionally need a sampling model.

At x=0.25 the ECDF is 3/4, but immediately to its left it is 1/4. Draw this jump before proceeding. Return to CDFs and atoms if the distinction between F(x) and its left limit F(x−) is unfamiliar.

Predict

The reference CDF F₀(x) is 0 below 0, x on [0,1], and 1 above 1. Predict whether the largest vertical discrepancy occurs before or after the repeated value 0.25. Also predict whether a plot containing only the right endpoints of jumps could miss the largest discrepancy in another sample.

A CDF is right-continuous: Fₙ includes an observation at its exact threshold. A repeated value has one combined jump, not several different thresholds. Here that jump has height 2/4. Ties are legitimate descriptive data; under an exactly observed iid continuous law they have probability zero. Rounding, discrete measurement or dependence can change that model. Do not erase a tie merely to make a continuous-null test applicable.

Experiment

Compare the exact step graph with the reference line. Coordinates use the same probability scale on both axes. At each jump the open circle marks the left limit and the filled circle the actual ECDF value.

Sort the observations as x₍₁₎≤⋯≤x₍ₙ₎. Complete the following calculation by hand, especially the two rows sharing x=0.25. Entries that are negative are retained during the calculation; the column maxima are the relevant discrepancies.

The live experiment below starts with a different, untied sample: .1,.3,.6,.9. First use its specified Uniform[0,1] mode and predict D=.2. Then enter the worked tied sample .1,.25,.25,.8 and inspect both sides of its jumps. Keep the continuous-model warning separate from the exact descriptive discrepancy. The simulation controls are explained in the next lesson; no simulation is required to verify the table.

Live experiment: discrepancy and calibration

Predict which side of the empirical jump gives the largest gap. Then compare a specified model with a model fitted to the same observations. Inputs and simulations are temporary; record your evidence in the lesson response.

Observations, separated by commas Reference procedure Specified Uniform[0,1]Specified exponential rateExponential rate fitted to observations

D⁺ = 0.2; D⁻ = 0.15; D = 0.2.

Inspect ordered ranks, gap calculations and QQ coordinates

Rank i	xᵢ	F(xᵢ)	i/n−F	F−(i−1)/n	QQ position	Model quantile
1	0.1	0.1	0.15	0.1	0.125	0.125
2	0.3	0.3	0.2	0.05	0.375	0.375
3	0.6	0.6	0.15	0.1	0.625	0.625
4	0.9	0.9	0.1	0.15	0.875	0.875

Positions are (i−½)/n. Within a tied block, only the first lower rank and last upper rank are actual ECDF jump limits; intermediate ranks still produce the correct maximum formula.

SeedReplicates (1–2,000)Simulate null referenceCancel

Reset experiment

The reference assumes independent observations from the stated continuous null. The rank-test proof assumes genuinely independent null draws; this reproducible xorshift32 simulation and floating-point CDF arithmetic approximate that ideal. It is not an exact KS distribution evaluator or evidence that the null is true. Choosing a model after inspecting results requires a new validation plan. Simulations stop on cancellation, hiding, leaving this panel or navigation.

The positive-direction maximum is D⁺=1/2; the negative-direction maximum is D⁻=1/10. The largest absolute discrepancy is D=1/2. This is an exact descriptive result, not yet a p-value.

Understand

Why both sides of every jump suffice

For a continuous reference CDF F₀ define D⁺=sup over real x of (Fₙ(x)−F₀(x)), D⁻=sup over real x of (F₀(x)−Fₙ(x)), and D=max(D⁺,D⁻). The supremum means the least upper bound; a left-limit gap need not be attained at the jump itself.

Between consecutive distinct observations Fₙ is constant and F₀ is nondecreasing. Consequently Fₙ−F₀ is largest at the left endpoint, just after its jump, while F₀−Fₙ is largest in the limit just before the next jump. The exterior intervals are handled by F₀ tending to 0 and 1. Continuity gives the same reference value on both sides of each observation. If a tied block occupies positions a through b, its left ECDF value is (a−1)/n and its right value is b/n. Thus its two candidates are b/n−F₀(x) and F₀(x)−(a−1)/n.

Equivalently, keeping every ordered row gives

D⁺=max over i of (i/n−F₀(x₍ᵢ₎)),

D⁻=max over i of (F₀(x₍ᵢ₎)−(i−1)/n).

Within each tied block the last row maximizes the first expression and the first row maximizes the second. Intermediate artificial ranks do not alter either maximum. Both maxima are nonnegative: the last row supplies 1−F₀(x₍ₙ₎)≥0 and the first supplies F₀(x₍₁₎)≥0. This proves the formula, including ties and observations outside a bounded reference support. If F₀ has jumps, its own left limits also matter: these formulas must not be imported unchanged.

What a QQ plot asks instead

A quantile–quantile plot compares ordered observations with reference quantiles. Choose and disclose plotting probabilities pᵢ=(i−1/2)/n and put the point (Q₀(pᵢ),x₍ᵢ₎) in the plot, where Q₀(p)=inf❴x:F₀(x)≥p❵. Horizontal coordinates are theoretical quantiles; vertical coordinates are observations, in matching units. For our uniform example the points are (1/8,1/10), (3/8,1/4), (5/8,1/4), (7/8,4/5).

A specified-model identity line is a comparison, not a line fitted to force agreement. A location/scale line estimated after seeing data changes the question. Plotting positions are a declared convention; another convention changes finite-sample points. No p-value or rejection threshold follows merely from seeing a nearly straight plot. Use it to locate possible departures and propose further checks.

Pointwise evidence is not automatically uniform

Under iid sampling with true CDF F, each Fₙ(x) is an average of Bernoulli variables with mean F(x) at a fixed x. Its variance is F(x)(1−F(x))/n≤1/(4n), so Chebyshev proves convergence in probability at that x. Checking finitely many thresholds allows a union bound. Neither argument, without another step, controls the supremum over all real thresholds at once. Uniform empirical convergence is a separate theorem; no general uniform theorem or exact KS reference distribution is being assumed from a finite plot.

Apply

Use observations 1,4,4,9 and the specified Uniform[0,10] law. Compute both jump gaps and all QQ coordinates using the same plotting-position convention. Then try observations 0.7,0.8,0.9 against Uniform[0,1]: does looking only after jumps now underestimate D?

Review the changed calculations

The first transformed values are .1,.4,.4,.9. The D⁺ candidates are .15,.10,.35,.10; D⁻ candidates are .10,.15,−.10,.15. Hence D⁺=.35, D⁻=.15 and D=.35. QQ coordinates are (1.25,1), (3.75,4), (6.25,4), (8.75,9). For the second sample, D⁺=1/10 but D⁻=7/10, approached immediately before the first observation. A right-jump-only calculation would miss the dominant gap.

Continue

Retain the sorted sample, tie blocks, reference CDF, both gap columns and plotting convention. Separately record why the observations could or could not follow the proposed sampling model. Continue to fixed-null KS and calibration for the statistical reference calculation. Describing a discrepancy and calibrating its sampling distribution are different proof obligations.