Statistical inference for \(L^2\)-distances to uniformity


The paper deals with the asymptotic behaviour of estimators, statistical tests and confidence intervals for \(L^2\)-distances to uniformity based on the empirical distribution function, the integrated empirical distribution function and the integrated empirical survival function. Approximations of power functions, confidence intervals for the \(L^2\)-distances and statistical neighbourhood-of-uniformity validation tests are obtained as main applications. The finite sample behaviour of the procedures is illustrated by a simulation study.

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The authors thank the referees for constructive comments and suggestions.

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Correspondence to L. Baringhaus.



We introduce the entire function

$$\begin{aligned} d(z)=\frac{1}{2}\left( \cosh (z)\sin (z)+\sinh (z)\cos (z)\right) ,~z\in \mathbb {C}, \end{aligned}$$

and note that the real positive zeros of d are just the positive solutions \(\kappa _j,\,j\ge 1,\) of the equation (2.1). We aim to derive the representation of d as an infinite product involving the zeros of d. Obviously, 0 is a simple zero of d. Observing

$$\begin{aligned}\sin (-z)&=-\sin (z),~\sinh (-z)=-\sinh (z),\\ \cos (-z)&=\cos (z),~\cosh (-z)=\cosh (z) \end{aligned}$$


$$\begin{aligned} \sin (iz)=i\sinh (z),~\sinh (iz)=i\sin (z),~\cos (iz)=\cosh (z),~\cosh (iz)=\cos (z) \end{aligned}$$

for each \(z\in \mathbb {C},\) we note that if \(0\ne z_0\in \mathbb {C}\) is a zero of d,  also \(-z_0,~iz_0,\) and \(-iz_0\) are zeros of d. Thus, with \(\kappa _k,k\in \mathbb {N},\) as the real positive zeros of d,  we recognise that \(-\kappa _k,k\in \mathbb {N},\) \(i\kappa _k,k\in \mathbb {N},\) and \(-i\kappa _k,k\in \mathbb {N},\) are zeros of d. Adding the zero 0, we assert that this is the whole set of zeros of d. To see that the assertion is true, it suffices to prove the following lemma.

Lemma 4

There is no zero \(z=x+iy\) of d with real positive x and real positive y.


Assume, there is a zero \(z=x+iy\) of d with real positive x and real positive y. Then,

$$\begin{aligned} \cosh (x+y)\sin (x-y)+\sinh (x-y)\cos (x+y)&=0,\\ \cosh (x-y)\sin (x+y)+\sinh (x+y)\cos (x-y)&=0. \end{aligned}$$

Due to \(\sin u +\sinh u>0\) for real positive u,  it is \(x\ne y.\) We can (and do) assume without loss of generality that \(x>y.\) Putting \(a=x+y,~b=x-y\) we have

$$\begin{aligned} \begin{aligned} \cosh a\sin b+\sinh b\cos a&=0,\\ \cosh b\sin a+\sinh a\cos b&=0. \end{aligned} \end{aligned}$$

From this, we deduce that

$$\begin{aligned} \begin{aligned} \cosh ^2 a \sin ^2 b&= \sinh ^2 b \cos ^2 a,\\ \cosh ^2 b \sin ^2 a&= \sinh ^2 a \cos ^2 b. \end{aligned} \end{aligned}$$

Adding the corresponding terms on the right and the left hand side in (6.2) and using (6.2) we obtain

$$\begin{aligned} \cosh ^2 a=\sinh ^2 b+ \sin ^2 a +\cos ^2 b \end{aligned}$$


$$\begin{aligned} \cosh ^2 b=\sinh ^2 a+ \sin ^2 b +\cos ^2 a. \end{aligned}$$

It follows that

$$\begin{aligned} \cosh 2a + \cos 2a =\cosh 2b + \cos 2b. \end{aligned}$$

This is impossible, because the function \(\cosh u + \cos u,\,u>0,\) is strictly increasing. \(\square \)

Theorem 4

(Gao et al. (2003)) The Laplace Transform of the limit distribution is given by

$$\begin{aligned} L(t)&=\left( \frac{2z(t)}{\sinh (z(t))+\sin (z(t))}\right) ^{1/2},~t\ge 0, \end{aligned}$$

where \(z(t)=2^{1/2}(2t)^{1/4}\) for \(t\ge 0.\)


Note that

$$\begin{aligned} f(z)=\frac{d(z)}{z},~z\in \mathbb {C}, \end{aligned}$$

is an entire function of order 1 and that \(f(0)=1.\) Due to

$$\begin{aligned} \left( j-\frac{1}{2}\right) \pi<\kappa _j<j\pi ,~j\ge 1, \end{aligned}$$

the genus of the canonical product of the primary factors of f is 1. So, by definition, f is of genus 1. By Hadamard’s factorisation theorem, see, e.g., Titchmarsh (1939), Sec. 8.24, it follows that f has the representation

$$\begin{aligned} f(z)=\exp (Q(z))\prod _{k=1}^\infty \left( 1-\frac{z^4}{\kappa _k^4}\right) ,~ z\in \mathbb {C}, \end{aligned}$$

where \(Q(z)=\alpha z, z\in \mathbb {C},\) with \(\alpha \in \mathbb {C}.\) Since \(f(z)=f(-z)\) for \(z\in \mathbb {C}\) it is \(\alpha =0.\) Thus,

$$\begin{aligned} f(z)=\prod _{k=1}^\infty \left( 1-\frac{z^4}{\kappa _k^4}\right) ,~z\in \mathbb {C}, \end{aligned}$$

and the Laplace transform of

$$\begin{aligned} T=\sum _{k=1}^\infty \frac{1}{\kappa _k^4}W_k^2 \end{aligned}$$

is seen to be

$$\begin{aligned} L(t)&=\prod _{k=1}^\infty \left( 1+\frac{2t}{\kappa _k^4}\right) ^{-1/2}\\&=\left( f\left( (-2t)^{1/4}\right) \right) ^{-1/2}\\&= \left( \frac{2z(t)}{\sinh (z(t))+\sin (z(t))}\right) ^{1/2},~t\ge 0. \end{aligned}$$

\(\square \)

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Baringhaus, L., Gaigall, D. & Thiele, J.P. Statistical inference for \(L^2\)-distances to uniformity. Comput Stat 33, 1863–1896 (2018).

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  • Integrated empirical distribution (survival) function
  • Goodness-of-fit tests for uniformity
  • Numerical inversion of Laplace transforms
  • Coverage probability
  • Equivalence test
  • Neighbourhood-of-uniformity validation test