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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]

The saturated model estimates each Poisson mean separately as $Y_i$, permitting the boundary value zero when $Y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson deviancewith $0\log (0/{\hat{\mu }}_i)=0$. A zero observation contributes $2{\hat{\mu }}_i$. If the model includes an intercept, the score equation gives ${\sum }_i{Y}_i={\sum }_i{\hat{\mu }}_i$, so the summed linear terms cancel; the unsimplified expression applies generally.

For goodness of fit, the null hypothesis is that the specified independent Poisson mean model is correct; the alternative is an unrestricted collection of means. When a quadratic likelihood approximation is valid, for example with fixed $n$ and large expected counts, $D$ has approximately a chi-squared distribution with $n-p$ statistical degrees of freedom. Reject for a large upper-tail deviance. Merely increasing the number of sparse observations does not make this saturated-model reference reliable: the alternative dimension also increases. With small fitted means, replication-based checks or a parametric bootstrap are more defensible. This qualification concerns the absolute goodness-of-fit deviance; a fixed-dimensional nested likelihood-ratio comparison can have a better chi-squared approximation.

For the Poisson variance function $V(\mu )=\mu$, the code computes the Pearson chi-squared statisticand the Pearson dispersion estimatorThe first quantity measures goodness of fit; the second estimates the dispersion parameter, which equals one in a correctly specified Poisson regression. A standard rough calculation substitutes the residual deviance for the Pearson statistic and givesIf the reported upper-tail probability $4.908651\times 10^{-11}$ is inverted numerically, the actual Pearson statistic used by the code is about $82.93$, giving $\hat{\varphi }\simeq 5.18$. Either calculation reveals severe overdispersion.