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For a periodic even chain, eliminating alternate spins gives $Z_N(K)=A^{N/2}{Z}_{N/2}(K^{\prime })$. Thus dimensionless free energy per original spin obeys $f(K)=-\frac{1}{2}\log A+\frac{1}{2}f(K^{\prime })$. The exact solution $f(K)=-\log (2\cosh K)$ satisfies this relation. The multiplicative constant cancels from normalized retained-spin probabilities but records eliminated-spin entropy; discarding it loses thermodynamics even if the interaction recursion is correct.

The Ising model is a spin model with an Ising spin ${\sigma }_v\in \{-1,1\}$ at every graph vertex $v$. For a finite graph with vertex set $V$ and edge set $E$, a uniform coupling $J$ and field parameter $h$ give the HamiltonianEach edge is counted once. Here $T>0$ is temperature, $k_B$ is the Boltzmann constant, and $\beta$ is inverse temperature; $J$ and $h$ have units of energy. For magnetic applications, $h$ includes the magnetic moment multiplying the applied magnetic field. The canonical ensemble assigns the Boltzmann distributionThe partition function $Z$ makes the probabilities sum to one. Positive $J$ rewards equal neighboring Ising spins, modeling ferromagnetism; negative $J$ rewards opposite neighbors, giving an antiferromagnetic Ising model. Positive $h$ rewards $+1$, while negative $h$ rewards $-1$.^[ref]

A single local rule can organize a whole graph: rewarding agreement on each edge makes all-equal Ising spin patterns favorable even though no edge directly couples distant vertices. The four-spin Ising ring makes this visible with only sixteen microstates. It also separates three different questions: whether neighbors agree, whether the mean spin magnetization is nonzero, and how large its thermal fluctuations are.

A motivating experiment is the University of Colorado Boulder's Curie-temperature demonstration. An iron rod suspended from a twisting wire on a stand is initially held near a permanent magnet by ferromagnetism. A gas burner supplied from a propane cylinder heats the rod; the readout is its motion away from the magnet, followed by return as it cools. Repeated cooling supplies a reversible comparison on the same sample. The source explicitly calls for heating the rod while keeping the magnet cool, since heating the magnet can reduce its strength. The inference is that warming the sample weakens its magnetic attraction across the Curie temperature. This is a qualitative force readout, with no calibrated temperature or magnetization trace supplied: it motivates competition between alignment and thermal fluctuations, but does not identify $J$, measure individual Ising spins, or realize a four-spin Ising ring.^[ref]

Ernst Ising's 1925 paper examined a linear chain of two-orientation magnetic elements with neighboring interactions. In its calculation for a large chain, the zero-field mean magnetization vanishes and the model has no spontaneous magnetization. The inspected English translation states this immediately after equation (8) in §2. This historical result concerns the chain; it does not establish the behavior of a two-dimensional Ising model.^[ref]

The Ising ring bond correlation is the neighboring two-point correlation function $c_1=⟨{\sigma }_i{\sigma }_{i+1}⟩$ in the four-spin Ising ring. All four edges are equivalent. At zero field each mean Ising spin is zero by spin inversion symmetry, so $c_1$ also equals the connected correlation function. Agreement gives bond product $+1$ and disagreement $-1$; thus $P({\sigma }_i={\sigma }_{i+1})=(1+c_1)/2$.

Differentiating the finite partition function with respect to $K=\beta J$ inserts the sum of four bond products. ThereforeHere $U$ is internal energy. At $K=(\ln 2)/4$, the class-weight calculation gives $c_1=(4-1)/17=3/17$ and $U=-12J/17$. The agreement probability for a selected bond is $10/17$, despite the exactly zero mean spin magnetization.

A second derivative gives the energy variance, and hence the heat capacity $C_H=dU/dT$ at fixed $J$ and $h=0$:To see the variance identity directly, a derivative of an expected value under weights $e^{KS}$ inserts $S-⟨S⟩$, where $S={\sum }_i{\sigma }_i{\sigma }_{i+1}=-H/J$. The factor $K^2$ follows from $dK/dT=-K/T$. At the same exact example, $\mathrm{Var}(H)/J^2=1216/289$. The heat capacity vanishes both as $K\to 0$ and as $K\to \infty$: at high temperature its dimensionless factor $K^2$ vanishes, while at low temperature excited microstates become exponentially rare.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

For the nonzero spectral decomposition $\rho ={\sum }_{a=1}^r{\lambda }_a|e_a⟩⟨e_a|$, every pure-state quantum state ensemble of $\rho$ is specified by the displayed column linear isometry. To prove this, write $w_j=\sqrt{p_j}{\varphi }_j$. Positivity forces each $w_j$ into the support of $\rho$, and the coefficients $U_{ja}=⟨e_a|w_j⟩/\sqrt{{\lambda }_a}$ have orthonormal columns because ${\sum }_j|w_j⟩⟨w_j|=\rho$. Conversely this column identity reconstructs $\rho$; the squared vector norms are the probabilities and sum to one. Complete the columns to unitary bases after adding zero ensemble vectors to obtain the usual unitary-mixing formulation.

For two individuals carrying labelled copies $(a,b)$ and $(c,d)$, condensed identity coefficients give the probabilities of nine identity by descent patterns, identifying states that differ only by exchanging the two homologues within either individual. They retain within-individual autozygosity, which three ordinary outbred sharing coefficients omit. One convention isEach block consists of copies identical by descent, and different blocks have distinct ancestral origins. For siblings whose parents are otherwise outbred first cousins, these coefficients are $(1,0,2,1,2,1,15,30,12)/64$. This follows by conditioning on whether the parents share one ancestral copy and enumerating the four independent Mendelian segregation choices to their two children. The kinship coefficient is ${\Delta }_1+({\Delta }_3+{\Delta }_5+{\Delta }_7)/2+{\Delta }_8/4$, here $9/32$. Weir's discussion of identity states uses this convention.

Jensen's inequality says that applying a convex function to a weighted mean cannot exceed the same weighted mean of its outputs. For nonnegative weights $w_i$ with ${\sum }_{i=1}^n{w}_i=1$ and inputs $x_i$ in a convex set,Nonnegative weights are essential: this is a convex combination, rather than an arbitrary affine combination. For two inputs it is precisely the defining inequality for a convex function; repeatedly grouping inputs gives the finite weighted version.

Take $f(x)=x^2$, weights $1/4$ and $3/4$, and inputs $1$ and $3$. Their weighted mean is $2.5$, so the two sides areThe Jensen gap is $0.75$. The inequality measures the effect of replacing distinct inputs by one representative input before applying a nonlinear function.

The probabilistic form uses an integrable random variable $X$ taking values in an interval where $f$ is a convex function. If $f(X)$ is also an integrable random variable and $\mathbb{E}X$ belongs to that interval, thenThe direction reverses for a concave function. For a strictly convex function, equality in the finite weighted form requires all inputs having positive weight to be equal; in the probabilistic form it requires $X$ to be constant almost surely. An affine function gives equality for any admissible distribution. ^[ref].

For the quadratic function $f(x)=x^2$ and a random variable with a finite second moment, the difference between the two sides is exactly its variance:Consider two equally likely values $2-d$ and $2+d$. Their expected value stays at $2$ as $d$ increases, but their variance is $d^2$. Thus the expected value of the square grows from $4$ to $4+d^2$. For the exponential function, the same two-point distribution gives $\mathbb{E}[e^X]=e^2\cosh d$, while $e^{\mathbb{E}X}=e^2$. Both are examples of Jensen's inequality, but their Jensen gaps depend differently on the spread of the inputs.

Another useful choice is the concave function $\log x$ on positive inputs. Its version of Jensen's inequality gives ${\sum }_i{w}_i\log x_i\le \log ({\sum }_i{w}_i{x}_i)$. Applying the increasing exponential function shows that the weighted geometric mean cannot exceed the weighted arithmetic mean. For the equally weighted inputs $1$ and $9$, these are $3$ and $5$. This recovers the arithmetic-geometric mean inequality and illustrates how choosing one function can generate an entire family of bounds.

A short proof idea explains the probabilistic form. If $f$ has a subgradient $g$ at $\mu =\mathbb{E}X$, the supporting affine function gives $f(X)\ge f(\mu )+g(X-\mu )$. Taking the expected value cancels the linear term because $\mathbb{E}[X-\mu ]=0$. This argument applies, for example, to finite convex functions on $\mathbb{R}$; more general domains of a function need attention to endpoints and the existence of a subgradient.

Johan Jensen's 1906 paper “Sur les fonctions convexes et les inégalités entre les valeurs moyennes” is an early source for the connection between convex functions and inequalities for means. ^[ref].

Joint entropy is the information entropy of an entire random vector, treating its coordinates together as one outcome. For random variables with finite alphabets and joint probability mass function $p$,It is measured in bits. The distinction from separate information entropies is whether shared uncertainty is counted once or repeatedly. Two copies of the same fair bit have joint entropy one bit, since only $00$ and $11$ can occur. Two independent fair bits have joint entropy two bits, since all four pairs are possible and equiprobable.

For a less symmetric example, assign probabilities $1/2,1/4,1/4$ to $(X,Y)=(0,0),(1,0),(1,1)$ respectively. ThenHere $X$ is fair, while $\mathbb{P}(Y=1)=1/4$. Thus $H(X)=1$ and $H(Y)=h_2(1/4)\approx 0.811278$ bits, where $h_2$ is binary entropy. Adding the separate information entropies would give approximately $1.811278$ bits, overcounting by the mutual information $I(X;Y)\approx 0.311278$ bits.

The chain rule for information entropy makes the shared information explicit:In this example $Y$ is certain given $X=0$ and fair given $X=1$, so its average conditional entropy is $H(Y\mid X)=\frac{1}{2}\cdot 0+\frac{1}{2}\cdot 1=1/2$ bit. Sending the first coordinate changes how much uncertainty remains in the second. A prefix code with codewords 0, 10, 11 for the three possible pairs has average length $3/2$ bits, exactly the joint entropy in this stipulated law.

For finite alphabets,The lower bound follows because knowing the full random vector determines every coordinate; the upper bound follows from conditioning reduces entropy in the chain rule for information entropy. Equality in the upper bound means joint independence. The difference from that upper bound is total correlation, whose noisy parity total correlation example shows why separate pairwise comparisons need not find all the dependence.

For a proper prior distribution $\pi (\theta )$ independent of coalescent times $T$, let $q(T)={\prod }_{j=2}^n{\lambda }_j{e}^{-{\lambda }_j{T}_j}$ with ${\lambda }_j=\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ and $L=\sum j{T}_j$. Observing $S=k$ under an infinite sites mutation model gives posterior density $\pi (\theta )q(T)e^{-\theta L/2}(\theta L/2{)}^k/(k!m_k)$, where $m_k$ is the prior predictive probability of $S=k$. Multiplication by the conditional Poisson mass and normalization is an application of Bayes' theorem.

// Target: statistical-inference.bigb

For discrete random variables $X,Y$, their joint probability mass function assigns a probability to each possible ordered pair. Its entries are nonnegative and sum to one. Marginalization gives the probability mass functions $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ and $p_Y(y)={\sum }_x{p}_{X,Y}(x,y)$. In a table with rows indexed by $x$ and columns by $y$, these are the row and column totals. The joint probability distribution records how the coordinates occur together, while the marginal distributions describe each separately.

For a finite-intensity set of mean $m$, independent red marking with probability $r$ gives counts $R,G$ withDisjoint source sets give independent pairs of these counts. Refining overlapping test sets into disjoint cells then proves independence of the two whole Poisson point processes, not only their counts on one set.

For $P_{ii}<1$, observe a discrete-time Markov chain only when its state changes. The probability that its next different state is $j$ is ${\sum }_{r\ge 0}{P}_{ii}^r{P}_{ij}=P_{ij}/(1-P_{ii})$. Its holding times are finite geometric waiting times. This construction deletes self-transitions and differs from the existing continuous-time jump-chain concept.

At distinct observed event times $t_1<t_2<\cdots$, let $n_j$ be the risk set size immediately before $t_j$, and let $d_j$ count events at $t_j$. The Kaplan–Meier estimator of the survival function isThe empty product is one. Each factor estimates a conditional survival probability among those still observed and event-free; their product joins the successive steps. This is nonparametric statistics: no specified shape for the survival distribution is imposed. Use independent censoring and the appropriate risk sets to interpret the result as an estimate for the intended population.

The Kaplan–Meier estimator is a right-continuous function with downward jumps at event times. Right censoring changes later denominators without causing a jump. A tick on the curve marks right censoring, rather than a failure. For a tied group of $d_j$ events, use the one factor $(n_j-d_j)/n_j$; subjects right-censored at that recorded time remain in $n_j$ for this factor. The fraction $d_j/n_j$ estimates a discrete hazard, not a continuous hazard function in inverse-time units. With no censoring, the factors telescope to the fraction of the original cohort whose observed survival times exceed $t$, the complement of its empirical distribution function.

Edward L. Kaplan and Paul Meier published ^[ref]. Their paper addresses lifetime estimation when observation ends before failure and explicitly calls attention to the assumption of independence between lifetime and potential observation loss. The date 1958 is the original publication year; the publisher's later online-posting date is not the discovery date.

In his ^[ref], Kaplan traces Meier's interest to cancer-duration data in 1952 and his own to the lifetimes of vacuum tubes in undersea telephone-cable equipment a year later. The journal encouraged them to combine their manuscripts. The shared mathematical question was how to estimate a survival function when some lifetimes are still running or disappear from observation.

Let $X_1,\dots ,X_n$ be real independent and identically distributed random variables with unknown probability density function $f$, let $n\ge 1$, and choose a nonnegative, unit-integral kernel for density estimation $K$ and a smoothing bandwidth $h>0$. The kernel density estimator isPlace a bump at every observation and add their contributions. Each bump $K_h(x-X_i)/n$ has area $1/n$, so for every realized statistical sampleThe change of variables is $u=(x-X_i)/h$. Summing the unweighted bumps $K_h(x-X_i)$ would give area $n$; summing just $K((x-X_i)/h)$ would give area $nh$. Those raw sums do not have the normalization of a probability density function.

Unlike a histogram, the construction does not require choosing bin origins, and a Gaussian density kernel gives a smooth curve. Its appeal is that identical local contributions can combine into a probability density function with several peaks without first choosing their number. This makes kernel density estimation useful for exploring an unknown probability distribution and estimating interval probabilities, while the smoothing bandwidth controls how much detail is visible.

Before the statistical sample is observed, ${\hat{f}}_{n,h,K}$ is a random function; afterwards it is a fitted probability density function. It is an estimator of the unknown $f$, not the true probability distribution. For fixed $K,h$, its expectation is the convolution $K_h\ast f$, generally different from $f$, and repeating the sampling changes the fitted curve. The exact mean integrated squared error of a kernel density estimator separates this smoothing bias of an estimator from sampling variance. Allowing signed kernels of order ell retains the formula but can lose nonnegativity.

The kinship coefficient ${\varphi }_{XY}$ is the probability that one randomly selected allele copy from each of $X,Y$ is identical by descent. For a noninbred individual, ${\varphi }_{XX}=1/2$, and more generally ${\varphi }_{XX}=(1+F_X)/2$. For distinct individuals with parents $M_X,P_X$, Mendelian segregation gives ${\varphi }_{XY}=[{\varphi }_{M_{X}Y}+{\varphi }_{P_{X}Y}]/2$. The offspring's inbreeding coefficient equals the parents' kinship coefficient.

// Target: biology.bigb

For the outgoing negative-energy branch $q=-Q$, matching the field and its spatial derivative gives these signed charge-current ratios. Their sum is one although reflection exceeds one. The ideal sharp step has a singular stationary matching denominator at $E=V_0/2$ in the supercritical range. It is not a finite single-particle probability prediction; wavepackets, source preparation and field quantization matter for its physical interpretation.

In a sufficiently strong electrostatic step, the positive-energy incident channel overlaps a negative-kinetic-energy channel on the far side. For a charged scalar field the outgoing antiparticle branch has negative charge current, allowing reflected particle current larger than the incident current while signed charge is conserved. The external source supplies energy for particle-antiparticle creation. This requires a multiparticle interpretation; negative transmitted charge current is not a negative probability.

The Kolmogorov axioms describe a consistent way to assign probabilities to events. Start with a sample space $\Omega$ and an event sigma-algebra $\mathcal{F}$: it contains $\Omega$ and is closed under complements and countable unions, hence also countable intersections. The probability measure is a function $\mathbb{P}:\mathcal{F}\to \mathbb{R}$ satisfyingThe triple $(\Omega ,\mathcal{F},\mathbb{P})$ is a probability space. The third axiom is countable additivity; the infinite series means the limit of a sequence of nonnegative partial sums. It applies to pairwise disjoint events, not to overlapping events and not to arbitrary uncountable unions.^[ref]

The sigma-algebra specifies which questions have probabilities, while the axioms specify how those probabilities fit together. For a finite sample space one can take the entire power set; a continuous model commonly uses a Borel sigma-algebra. Nonnegativity and normalization do not require equally likely outcomes or independence: those are additional modelling choices.

For an exact six-outcome model, take $\Omega =\{1,2,3,4,5,6\}$, $\mathcal{F}=\mathcal{P}(\Omega )$, and assign $\mathbb{P}(\{k\})=k/21$. The total is one because $1+2+3+4+5+6=21$. Define the probability of an event by summing the weights of its outcomes. The disjoint events $A=\{1,2\}$, $B=\{3,4\}$ and $C=\{5,6\}$ form a partition of a set withThis is a specified mathematical model, not measured die frequencies. The figure makes probability literal area: each region has the same height and a width proportional to its assigned mass. Regrouping disjoint events preserves their total area.

Several familiar rules are consequences, not further axioms. Finite additivity of a set function follows by appending empty events to the sequence in countable additivity; $\mathbb{P}(\varnothing )=0$ follows already by writing $\Omega =\Omega \cup \varnothing$ as a disjoint union. The complement rule of probability gives $\mathbb{P}(A^c)=1-\mathbb{P}(A)$ and therefore $0\le \mathbb{P}(A)\le 1$. Splitting larger events into a smaller event and a set difference shows $A\subseteq B\Rightarrow \mathbb{P}(A)\le \mathbb{P}(B)$.

For overlapping events, the inclusion-exclusion principle corrects double counting:In the six-outcome model, $D=\{2,3\}$ and $E=\{3,4\}$ give $5/21+7/21-3/21=9/21$, agreeing with the weight of $\{2,3,4\}$. This decomposition is useful because a complicated event can be assembled from simpler questions with known probabilities.

On a finite sample space, a family of pairwise disjoint events has only finitely many nonempty members, so finite additivity of a set function already handles every countable disjoint union. On an infinite sample space, the countable axiom additionally controls what is left beyond every finite stage. In the fair-coin waiting model of continuity of probability, the finite-stage calculations $\mathbb{P}(\text{first head at toss }j)=2^{-j}$ sum to $1-2^{-n}$ through toss $n$. Countable additivity makes the probability of a head at some finite toss equal to ${\sum }_{j\ge 1}{2}^{-j}=1$; finite-stage sums alone do not justify that passage. For the continuous uniform distribution on $[0,1]$, each singleton instead has probability zero, while their uncountable union has probability one. Countable, rather than unrestricted, additivity accommodates both models.

Historically, Kolmogorov's 1933 book Grundbegriffe der Wahrscheinlichkeitsrechnung established the influential axiomatic framework connecting probability theory with measure theory. The three conditions above are the modern presentation on a sigma-algebra. Shafer and Vovk's historical study explains that the original presentation began with finite additivity on an algebra of sets and added continuity for decreasing events with empty intersection, which supplies the countable extension.^[ref]

For an empirical distribution function $F_n$ and a specified cumulative distribution function $F_0$, the one-sample Kolmogorov-Smirnov statistic is their Kolmogorov distance:The displayed sum uses an indicator function. With this convention, $F_n$ is right-continuous. For distinct observations, its value and its left one-sided limit at the $i$th order statistic are $F_n(X_{(i)})=i/n$ and $F_n(X_{(i)}^{-})=(i-1)/n$.

For continuous $F_0$, both jump sides matter. Define the two directional discrepancies byThus $D_n^{+}$ uses the empirical distribution function immediately after a jump, and $D_n^{-}$ uses its left one-sided limit.^[ref] A supremum need not be attained.

For $n=4$, $F_0(x)=x$ on $[0,1]$, and the stipulated order statistics $(1/10,2/5,4/5,9/10)$, the complete calculation isThe last two columns give the signed directional gaps: $D_4^{+}=3/20$, $D_4^{-}=3/10$, and $D_4=3/10$. In contrast, checking only $|F_4(X_{(i)})-F_0(X_{(i)})|$ gives $3/20$ and misses the largest gap approached as $x\uparrow 4/5$. At $x=4/5$ itself, the gap has already fallen to $1/20$.

The finite formula follows from the behavior between jumps: the empirical distribution function $F_n$ is constant and the cumulative distribution function $F_0$ is monotone, so the largest directional gaps occur at the two interval ends, possibly as one-sided limits. The two tails introduce no larger gap. Checking every jump side therefore gives the exact supremum.

A common normalized Kolmogorov-Smirnov statistic is $\sqrt{n}{D}_n$, which equals $3/5$ in this example. Its large-sample reference law comes from the Kolmogorov-Smirnov theorem. The unscaled $D_n$ remains the largest difference of cumulative probabilities, not a p-value.

There is a useful exact exception to the usual warning about fitting statistical parameters. Suppose $n\ge 2$ independent and identically distributed observations have the uniform distribution on $(0,\theta )$ with unknown $\theta >0$. Let $M={\max }_i{X}_i$ and let $J$ be the label of its almost surely unique maximizing observation. Conditional on $M=m$ and $J=j$, the surviving labelled observations $(X_i{)}_{i\ne j}$ are independent with the uniform distribution on $(0,m)$; after division by $m$ they are independent and identically distributed uniforms on $(0,1)$. These are conditional distributions defined by probability density functions, since the event $M=m$ has probability zero.

To check this, the joint probability density function of $M$ and $J=j$ is $m^{n-1}/{\theta }^n$ for $0<m<\theta$. Dividing the original joint probability density function ${\theta }^{-n}$ by this factor gives the conditional survivor probability density functionThis factorization proves the claimed conditional independence; scaling removes $m$. Keep the surviving labels in a fixed order, or apply an independent random permutation. The sorted survivors instead have the joint law of uniform order statistics and exhibit statistical dependence; sorting does not affect their empirical distribution function. Delete the maximum itself, which would otherwise become a forced value $1$, and compute the Kolmogorov-Smirnov statistic using sample size $n-1$ and jump heights $i/(n-1)$. This gives an exact conditional Kolmogorov-Smirnov test under the uniform-endpoint statistical model. It does not justify fitting an arbitrary endpoint or other statistical parameters and using the ordinary $n$-sample calibration.

The one-sample Kolmogorov-Smirnov test checks goodness of fit by comparing an empirical distribution function $F_n$ with a specified cumulative distribution function $F_0$. Its Kolmogorov-Smirnov statistic $D_n$ is the largest vertical gap anywhere between the two curves. Every threshold asks the same concrete question: what proportion of the observations are at or below it, and what probability does the proposed probability distribution assign there? A large discrepancy supplies evidence against the null hypothesis.

For example, four stipulated observations $(1/10,2/5,4/5,9/10)$ compared with the uniform distribution on $[0,1]$ give $D_4=3/10$. Just before $4/5$, only two observations have been counted, so the empirical distribution function approaches $1/2$, while $F_0$ approaches $4/5$. The Kolmogorov-Smirnov statistic section calculates every jump-side gap. These are exact illustrative numbers, not experimental measurements or a claim of statistical significance.

The ordinary one-sample calibration assumes independent and identically distributed real observations and a simple hypothesis $H_0:F=F_0$, with $F_0$ a continuous function fixed before inspecting these observations. All statistical parameters of $F_0$ must be specified.^[ref] Under that null hypothesis, the probability integral transform gives $U_i=F_0(X_i)$ with the uniform distribution on $(0,1)$, retaining independence. Consequently the sampling distribution of $D_n$ depends on the sample size $n$, but not on the particular continuous $F_0$. This is the useful sense in which the calibration is distribution-free.

For the exponential distribution of rate one, $F_0(x)=1-e^{-x}$ for $x\ge 0$. Take $x_i=-\log (1-u_i)$ for the four stipulated $u_i$ above. The probability integral transform changes the horizontal coordinates but preserves both the order statistics and their cumulative heights, so the same Kolmogorov-Smirnov statistic $3/10$ appears on both scales. More generally, for continuous $F_0$, the jump-side formulas depend only on $F_0(X_{(i)})$. Strict increase everywhere is unnecessary: intervals on which $F_0$ is constant have probability zero under the null hypothesis. This explains the invariance without assuming a strictly positive probability density function.

The observed Kolmogorov-Smirnov statistic is a discrepancy, whereas its p-value is a tail probability under the null hypothesis:Here $D_n^{\mathrm{new}}$ is computed from a fresh sample of the same sample size. Choose a significance level $\alpha$ before the test and reject when the calibrated p-value is at most $\alpha$. The value $D_4=3/10$ alone supplies neither a p-value nor a rejection decision. Failure to reject does not establish that the null hypothesis is true. A finite-sample calibration and the asymptotic Kolmogorov-Smirnov theorem are distinct; a cutoff for $D_n$ must be rescaled before use with $\sqrt{n}{D}_n$.^[ref]

Fitting statistical parameters on the tested observations replaces $F_0$ by the random $F_{\hat{\theta }}$. The resulting transformed observations need not be independent and identically distributed uniforms, so the ordinary reference law fails. The Lilliefors test addresses a particular fitted normal distribution. More general parametric bootstrap calibration must simulate from the fitted statistical model and repeat the fitting in every replicate; holding the fitted statistical parameters fixed during those comparisons calibrates a different procedure. Such a parametric bootstrap requires its own validity justification rather than inheriting exactness from the simple-null test.

For a discrete probability distribution, $F_0(X)$ is generally not uniform. Equal observations also arise from rounding, so tied data require a calibration that accounts for the discrete probability distribution or the rounding rule. The discrepancy remains meaningful, but continuous-null p-values are not automatically valid. Statistical dependence likewise changes the reference law. The two-sample Kolmogorov-Smirnov test compares two empirical distribution functions and has a different calibration.^[ref]