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For a fully specified continuous null distribution $F_0$, compare the empirical distribution function $F_n$ with $F_0$. The probability integral transform makes $F_0(Y_i)$ iid uniform under the null, giving distribution-free calibration. Three common discrepancies areThese define the Kolmogorov-Smirnov test, Cramér–von Mises criterion and Anderson–Darling test, respectively. Reject for large values. The supremum detects the largest discrepancy, the unweighted integral combines discrepancies throughout the distribution, and the reciprocal weight emphasizes the tails.

The Donsker theorem for empirical distribution functions gives $\sqrt{n}(F_n-F_0)\Rightarrow B^0\circ F_0$ for a Brownian bridge $B^0$, with covariance $\min (u,v)-uv$. Consequently the standard continuous-null limits are ${\sup }_{0\le u\le 1}|B^0(u)|$, ${\int }_0^1{B}^0(u{)}^{2}du$, and ${\int }_0^1{B}^0(u{)}^2/[u(1-u)]du$ for $\sqrt{n}{D}_n,C_n,A_n$, respectively; the weighted last limit also requires control near the endpoints. Exact finite-sample or asymptotic null quantiles provide critical values.

If parameters in $F_{\theta }$ are estimated from the same observations, these fully specified-null critical values generally fail. A first-order expansion isso the estimated parameter changes the limiting process. One must use the fitted-model null law, adjusted asymptotic theory, or a parametric bootstrap that re-estimates the parameters in each replicate.

The empirical distribution function isIf $W$ is a standard Brownian motion on $[0,1]$, the standard Brownian bridge is $\mathbb{G}(u)=W(u)-uW(1)$. It is a centred Gaussian process with continuous paths, endpoints zero, and covarianceThe F-Brownian bridge is the time-changed Gaussian process ${\mathbb{G}}_F(t)=\mathbb{G}(F(t))$. Its covariance is $F(\min (s,t))-F(s)F(t)$, exactly the covariance of the empirical distribution process ${\alpha }_n=\sqrt{n}(F_n-F)$.

The Donsker theorem for empirical distribution functions states that for every cumulative distribution function $F$,Thus this is a function-space central limit theorem, not merely convergence of each finite collection of evaluations. The limit is a tight centred Gaussian process. No continuity of $F$ is required for this statement; a jump of $F$ can give a jump of ${\mathbb{G}}_F$.

For precision, ${\ell }^{\infty }(\mathbb{R})$ is generally nonseparable, and the finite-sample process need not be a Borel-measurable random element for its norm topology. The displayed convergence in distribution has the usual empirical-process interpretation: for every bounded continuous real functional $H$, the outer expectation and inner expectation of $H({\alpha }_n)$ converge to $\mathbb{E}H({\mathbb{G}}_F)$. The outer expectation is the infimum of expectations of measurable majorants, and the inner expectation is its negative applied to $-H({\alpha }_n)$. For measurable real statistics this gives ordinary convergence in distribution. The limit is tight and separably supported: $g\mapsto g\circ F$ maps the separable space of continuous bridge paths continuously into ${\ell }^{\infty }(\mathbb{R})$.

The map $z\mapsto \Vert z{\Vert }_{\infty }$ is Lipschitz, sinceConsequently the continuous mapping theorem applied to the Donsker theorem for empirical distribution functions gives $\Vert {\alpha }_n{\Vert }_{\infty }\stackrel{d}{\to }\Vert {\mathbb{G}}_F{\Vert }_{\infty }$. The statistic on the left is measurable: both $F_n$ and $F$ are right-continuous, so its supremum equals the supremum over rational $t$.

If $F$ is continuous, its limits at $-\infty$ and $+\infty$ are zero and one. The intermediate value theorem therefore shows that $F(\mathbb{R})$ contains $(0,1)$, even when $F$ is not strictly increasing. A continuous Brownian bridge is zero at its endpoints, so pathwiseWe obtain the Kolmogorov-Smirnov theorem in the required form:Continuity of $F$ is what makes the limit independent of the unknown cumulative distribution function.

Choose $c_{\alpha }$ with $\mathbb{P}(Z\le c_{\alpha })=1-\alpha$. The law of $Z$ has no atoms. For example, the Brownian reflection principle and conditioning a Brownian motion on its endpoint give $\mathbb{P}(\max \mathbb{G}>a)=e^{-2a^2}$ for $a>0$: the reflected endpoint density at zero is the normal density at $2a$, divided by its value at zero. Hence both one-sided maxima have continuous laws, and an atom of $Z$ would require an atom of one of them. Also $\mathbb{P}(Z=0)=0$, since $\mathbb{G}(1/2)$ is nondegenerate. Thus such quantiles give the desired limiting coverage.

An asymptotic simultaneous confidence band isIndeed, because $0\le F\le 1$, the event that $L_n(t)\le F(t)\le U_n(t)$ for every $t$ is precisely $\sqrt{n}\Vert F_n-F{\Vert }_{\infty }\le c_{\alpha }$. Its probability tends to $1-\alpha$. This is simultaneous coverage over all real $t$, rather than separate pointwise confidence intervals.

A real functional $\Phi$ is Hadamard differentiable at $s_0$ if there exists a continuous linear functional $L={\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\ne 0$, $t_j\to 0$ and $h_j\to h$ in the normed vector space,If $\Phi$ is defined only on a subset, require $s_0+t_j{h}_j$ to lie in its domain. The variation of the directions $h_j$ is essential: convergence along each fixed direction alone is a weaker differentiability condition. Tangential Hadamard differentiability restricts the limit directions to a specified subspace; here the derivative is defined on all of $S$.

We prove the functional delta method using compact sets, so do not need an unjustified application of the Skorokhod representation theorem on an arbitrary nonseparable normed vector space. As usual for function-space convergence in distribution, the limiting random element is tight, and the real statistics are measurable, or convergence is understood through outer expectations. These are automatic under the usual separable complete random-element formulation; the empirical-process formulation used below has a tight separably supported limit.

For $t\ne 0$ putHadamard differentiability says $R_{t_j}(h_j)\to 0$ for every convergent sequence of directions, since continuity gives $L(h_j)\to L(h)$. It implies a Hadamard remainder near a compact set: for every compact $K\subset S$ and $ϵ>0$ there are $a>0$ and $\delta >0$ such thatTo prove it, suppose it failed. Choose $t_j\to 0$ and $h_j$ at distance less than $1/j$ from $K$ with $|R_{t_j}(h_j)|\ge ϵ$. Choose $k_j\in K$ with $\Vert h_j-k_j\Vert <2/j$. A subsequence of $k_j$ converges by compactness, so the corresponding $h_j$ converge too. The defining derivative limit contradicts the lower bound on $R_{t_j}(h_j)$.

Now let $Z_n=r_n(X_n-s_0)\stackrel{d}{\to }X$ and $t_n=1/r_n$. Given $\eta >0$, tightness of a probability measure provides a compact $K$ with $\mathbb{P}(X\in K)>1-\eta$. For the open $\delta$-neighbourhood $K^{\delta }$, the Portmanteau theorem givesOn $K^{\delta }$ the preceding uniform remainder is less than $ϵ$ for all sufficiently large $n$. Consequently ${\mathrm{limsup}}_n\mathbb{P}(|R_{t_n}(Z_n)|\ge ϵ)\le \eta$. Since $\eta$ is arbitrary, this remainder tends to zero in probability. We have proved the actual expansionContinuity of $L$ and the continuous mapping theorem give $L(Z_n)\stackrel{d}{\to }L(X)$. The vanishing remainder and Slutsky theorem therefore yieldThis derivation uses the derivative definition, rather than assuming a uniform Fréchet remainder over every bounded set. If outer expectations are needed, the same compact-neighbourhood argument uses outer probabilities and gives the corresponding empirical-process functional delta method.

For the statistical application, use the sample's empirical distribution function and the plug-in estimatorHere the printed $L^{\infty }$ denotes bounded pointwise functions with the supremum norm, namely ${\ell }^{\infty }(\mathbb{R})$, rather than equivalence classes modulo Lebesgue null sets. Assume Hadamard differentiability at the true $F$ and the usual measurability of the scalar estimator. The Donsker theorem for empirical distribution functions and the proved functional delta method giveNo continuity of $F$ is needed for this application. Since the derivative is continuous and linear and the bridge is a centred Gaussian process, $Z_F$ is a centred Gaussian real random variable, possibly identically zero.

Finally, a finite real weak limit ensures the required boundedness in probability. Choose $M$ so large that $\mathbb{P}(|Z_F|\ge M)<ϵ/2$. Applying the closed-set part of the Portmanteau theorem to $\{z:|z|\ge M\}$ givesHence there is $n_0$ such that, for every $n\ge n_0$,In particular the upper-tail condition in the question holds. The plug-in estimator has estimation error $O_{\mathbb{P}}(n^{-1/2})$, even when its limiting variance is zero.