A new powerful version of the BUS test of normality

Abstract

In this paper we introduce a modified version of the BUS test, which we call NBUS (New Borovkov–Utev Statistic). This latter defines a family of goodness of fit tests that can be used to detect normality against alternative hypothesis of which all moments up to the fifth exist. The test statistic depends on empirical moments and real parameters that have to be chosen appropriately. The good abilities of the NBUS with respect to BUS and other powerful normality tests are illustrated by means of a Monte Carlo experiment for finite samples. Besides, we show how an adaptation of NBUS for testing departing from normality due only to kurtosis, leads to comparable performances with classical tests based on the fourth moment.

Appendices

Appendix: Proof of Proposition 1

For $m_{3}=0$ the matrix:

$$\begin{aligned} \mathbf {M}_{Q}=\left( \begin{array}{ccc} 1 &{}\quad 0 &{}\quad -\frac{\sqrt{6}}{6}\left( m_{4}-3\right) \\ 0 &{}\quad 2 &{}\quad -\frac{2\sqrt{3}}{3}\frac{m_{5}}{m_{4}-1} \\ -\frac{\sqrt{6}}{6}\left( m_{4}-3\right) &{}\quad -\frac{2\sqrt{3}}{3}\frac{m_{5}}{ m_{4}-1} &{}\quad \frac{2m_{5}^{2}}{3\left( m_{4}-1\right) ^{2}}+\frac{1}{6} m_{4}\left( m_{4}+3\right) \end{array} \right) \end{aligned}$$

has characteristic polynomial:

$$\begin{aligned} \mathcal {P}\left( x\right)&= x^{3}-\left( \frac{1}{6}m_{4}\left( m_{4}+3\right) +3+\frac{2}{3}\frac{m_{5}^{2}}{\left( m_{4}-1\right) ^{2}} \right) x^{2}+ \\&\quad +\,\left( \frac{1}{3}m_{4}^{2}+\frac{5}{2}m_{4}+\frac{1}{2}+\frac{2}{3}\frac{ m_{5}^{2}}{\left( m_{4}-1\right) ^{2}}\right) x-3\left( m_{4}-1\right) . \end{aligned}$$

Since

$$\begin{aligned} \mathcal {P}\left( 1\right) =\frac{1}{6}\left( m_{4}-3\right) ^{2} \end{aligned}$$

$\xi =1$ is an eigenvalue of $\mathbf {M}_{Q}$ if and only if $m_{4}=3$. Furthermore, if $m_{4}\ne 3$ then $\mathcal {P}\left( 1\right) >0$, hence there exists a positive eigenvalue of $\mathbf {M}_{Q}$ smaller than $1$. If $m_{4}=3$, since

$$\begin{aligned} \mathcal {P}^{\prime }\left( x\right)&= 3x^{2}-2\left( \frac{1}{6} m_{5}^{2}+6\right) x+\frac{1}{6}m_{5}^{2}+11, \\ \mathcal {P}^{\prime \prime }\left( x\right)&= 6x-2\left( \frac{1}{6} m_{5}^{2}+6\right) \end{aligned}$$

we find

$$\begin{aligned} \mathcal {P}^{\prime }\left( 1\right) =\frac{1}{6}\left( 12-m_{5}^{2}\right) ,\quad \quad \mathcal {P}^{\prime \prime }\left( 1\right) =-\frac{1}{3} m_{5}^{2}-6<0. \end{aligned}$$

These results imply that:

• 1. if $m_{5}^{2}>12$, then there exists an eigenvalue smaller than $1$;

• 2. if $m_{5}^{2}<12$, the smaller eigenvalues of $\mathbf {M}_{Q}$ is $1$;

• 3. if $m_{5}^{2}=12$, the smaller eigenvalue of $\mathbf {M}_{Q}$ is $1$ that is a double root of the characteristic equation of $\mathbf {M}_{Q}$.

Proof of Proposition 2

Note first of all that $A^{2}>\xi _{2}$ for any $m_{4}$, $B$ and $D$. Indeed, this inequality is true if and only if

$$\begin{aligned} m_{4}\left( m_{4}+3\right) D^{2}-A^{2}<\sqrt{\Delta }. \end{aligned}$$

(13)

If the left hand side of (13) is negative, then the inequality is always satisfied while, if it is non negative, (13) is equivalent to:

$$\begin{aligned} -4A^{2}D^{2}\left( m_{4}-3\right) ^{2}<0, \end{aligned}$$

which is always satisfied. This result implies that if $\xi ^{G}=A^{2}$ then $\xi ^{G}>\xi _{\min }$. The conclusion is obvious when $\xi ^{G}=4B^{2}$. Finally, we suppose that $\xi ^{G}=18D^{2}$ for which the inequality $\xi ^{G}>\xi _{\min }$ is equivalent to $18D^{2}>\xi _{2}$. Then in this case we have $18D^{2}<A^{2}$ and $\xi _{2}<4B^{2}$. Assume now that these two last inequalities are satisfied. Then, $18D^{2}>\xi _{2}$ if and only if

$$\begin{aligned} \xi _{2}-18D^{2}=A^{2}+\left( m_{4}\left( m_{4}+3\right) -36\right) D^{2}- \sqrt{\Delta }<0. \end{aligned}$$

(14)

Note that

$$\begin{aligned} A^{2}+\left( m_{4}\left( m_{4}+3\right) -36\right) D^{2}=A^{2}-18D^{2}+\left( m_{4}-3\right) \left( m_{4}+6\right) D^{2}. \end{aligned}$$

Assume that $m_{4}<3$. Then, either $A^{2}-18D^{2}\le \left( 3-m_{4}\right) \left( m_{4}+6\right) D^{2}$ and (14) is satisfied or $A^{2}-18D^{2}>\left( 3-m_{4}\right) \left( m_{4}+6\right) D^{2}$, which leads by straightforward calculations to

$$\begin{aligned} \left( m_{4}-3\right) \left( 18D^{2}-A^{2}+2\left( m_{4}-3\right) D^{2}\right) >0 \end{aligned}$$

(15)

which again implies that (14) is satisfied.

Assume now that $m_{4}\ge 3$. Then (14) is satisfied if and only if

$$\begin{aligned} m_{4}>3+\frac{A^{2}-18D^{2}}{2D^{2}}. \end{aligned}$$

The derivation of an approximated distribution for the test statistic $\mathfrak {N}_{n,3}$

In order to compute the null distribution of the test statistic $\mathfrak {N} _{n,3}$ for finite sample sizes, one could refer to the classical formula for the density of a transformation of a continuous r.v., when the transformation is defined for $x\ge 1$ by

$$\begin{aligned} g\left( x\right) {=}A^{2}-\frac{1}{2}\left( A^{2}+x\left( x+3\right) D^{2}{-} \sqrt{\left( A^{2}{+}x\left( x+3\right) D^{2}\right) ^{2}-36\left( x{-}1\right) A^{2}D^{2}}\right) . \end{aligned}$$

If $A=1$, $B=1/\sqrt{2}$ and $D=1/\sqrt{6}$, then $g\left( x\right)$ is non negative, decreasing for $x<3$ and increasing for $x>3$. Hence the density of $\mathfrak {N}_{n,3}/n=g\left( \widehat{m}_{4}\right)$ can be derived using the following expression:

$$\begin{aligned} \sum _{j=1}^{2}f\left( g_{j}^{-1}\left( \widehat{m}_{4}\right) \right) \frac{1 }{g^{\prime }\left( g_{j}^{-1}\left( \widehat{m}_{4}\right) \right) } \end{aligned}$$

where $f$ is the density of $\widehat{m}_{4}$ and $g_{j}$ is the restriction of $g$ to the $j$-th interval of monotonicity.

Since we have not an explicit expression for $f$, we can use one of the approximations available in the literature (see e.g. Thode 2002, Ch. 3). For instance, when we apply the approximation based on the Pearson Type IV curve, we have

$$\begin{aligned} f\left( x\right) =c\left( 1+\frac{x^{2}}{a^{2}} \right) ^{-m}e^{-v\arctan \frac{x}{a}}\quad \quad x\in \mathbb {R} \end{aligned}$$

where the coefficients are given by the following expressions. Defining

$$\begin{aligned} s=6\frac{1+\beta _{1}-\beta _{2}}{3\beta _{1}-2\beta _{2}+6} \end{aligned}$$

we have

$$\begin{aligned} a&= \frac{\sigma }{4}\sqrt{16\left( s-1\right) -\beta _{1}\left( s-2\right) ^{2}} \\ m&= \frac{s+2}{2} \\ v&= \frac{s\left( s-2\right) \sqrt{\beta _{1}}}{\sqrt{16\left( s-1\right) -\beta _{1}\left( s-2\right) ^{2}}} \\ c&= \left( ae^{-v\frac{\pi }{2}}\int _{0}^{\pi }e^{vt}\left( \sin \left( t\right) \right) ^{s}dt\right) \end{aligned}$$

where the moment coefficients $\sigma$, $\beta _{1}$ and $\beta _{2}$ for the statistic $\widehat{m}_{4}$ are given at page 51 of Thode (2002).

Proof of Proposition 3

The proof is based on a direct application of the Second Order Delta Method when we consider the function:

$$\begin{aligned} g\left( \widehat{m}_{4}\right)&= \frac{1}{2}\Bigg ( A^{2}+\widehat{m} _{4}\left( \widehat{m}_{4}+3\right) D^{2}\\&-\sqrt{\left( A^{2}+\widehat{m} _{4}\left( \widehat{m}_{4}+3\right) D^{2}\right) ^{2}-36\left( \widehat{m} _{4}-1\right) A^{2}D^{2}}\Bigg ). \end{aligned}$$

Using the second order Taylor expansion at $\widehat{m}_{4}=3$, we have

$$\begin{aligned} g\left( \widehat{m}_{4}\right) =A^{2}+\frac{24A^{2}D^{2}}{A^{2}-18D^{2}} \left( \widehat{m}_{4}-3\right) ^{2}+o_{P}\left( \left( \widehat{m} _{4}-3\right) ^{2}\right) \end{aligned}$$

and thus

$$\begin{aligned} n\left( g\left( \widehat{m}_{4}\right) -A^{2}\right) =n\frac{24A^{2}D^{2}}{ A^{2}-18D^{2}}\left( \widehat{m}_{4}-3\right) ^{2}+o_{P}\left( n\left( \widehat{m}_{4}-3\right) ^{2}\right) . \end{aligned}$$

Since under the hypothesis of normality $\sqrt{n}\left( \widehat{m} _{4}-3\right)$ converges in distribution to a standard normal r.v., it follows that

$$\begin{aligned} n\left( \widehat{m}_{4}-3\right) ^{2}\overset{\mathcal {L}}{\longrightarrow } \chi ^{2}\left( 1\right) \quad \hbox {and}\quad n\left( \widehat{m}_{4}-3\right) ^{2}=O_{P}\left( 1\right) . \end{aligned}$$

Therefore, by Slutsky Theorem, it follows

$$\begin{aligned} n\frac{18D^{2}-A^{2}}{24A^{2}D^{2}}\left( A^{2}-\widehat{\mathbf {\xi }} \right) \overset{\mathcal {L}}{\longrightarrow }\chi ^{2}\left( 1\right) . \end{aligned}$$

With the choice $A=1$ and $D=1/\sqrt{6}$ the result follows.

The distribution of NBUS test statistic for $k=2$

We study the distribution of $\mathfrak {N}_{n,2}$ under $\mathcal {H}_{0}$. As the power of the test does not change with $B$ and $C$, it is not restrictive to set $A=1$ and $B=1/2$ (that is $C=2$), a choice that gives a simplified form to $\mathfrak {N}_{n,2}$ useful for further computations. Since $\mathfrak {N}_{n,2}$ is a transformation of $\widehat{m}_{3}$, it is possible to deduce an approximation of the critical value of the test for finite sample sizes, as illustrated in the following remark.

Remark 3

An approximation of the critical value of the test at level $\alpha$ when $n\ge 8$ is given by

$$\begin{aligned} \frac{t_{1-\alpha /2}^{2}s^{2}\left( \nu -2\right) }{8\nu }\left( \sqrt{1+ \frac{16\nu }{t_{1-\alpha /2}^{2}s^{2}\left( \nu -2\right) }}-1\right) \end{aligned}$$

(16)

where $t_{1-\alpha /2}$ is the $1-\alpha /2$ order quantile of the Student’s t r.v. with $v$ degrees of freedom, and

$$\begin{aligned} \nu =\frac{4\kappa -6}{\kappa -3},\quad \quad \kappa =3+\frac{36\left( n-7\right) \left( n^{2}+2n-5\right) }{\left( n-2\right) \left( n+5\right) \left( n+7\right) \left( n+9\right) },\quad \quad s^{2}=\dfrac{6\left( n-2\right) }{\left( n+1\right) \left( n+3\right) }. \end{aligned}$$

The steps to obtain the result above are the following.

We have to consider the r.v.$~Y=g\left( X\right)$ where

$$\begin{aligned} g\left( x\right) =\frac{1}{8}\left[ \sqrt{ x^{2}\left( x^{2}+16\right) }-x^{2}\right] \quad x\in \mathbb {R }. \end{aligned}$$

(17)

Such function is bounded between $0$ and $1$, even and strictly increasing on the interval $\left( 0,+\infty \right)$. Therefore, the quantile of order $1-\alpha$ of $Y$ can be deduced directly from the ones of order $1-\alpha /2$ of $X$. Since the law of the r.v. $X=\widehat{m}_{3}$ is well approximated (for $n\ge 8$) by the one of a scaled Student’s t distribution with $\nu$ degrees of freedom (see Thode 2002, p. 50):

$$\begin{aligned} \widehat{m}_{3}=s\left( \dfrac{\nu }{\nu -2}\right) ^{-1/2}T_{\nu } \end{aligned}$$

where

$$\begin{aligned} \nu =\frac{4\kappa -6}{\kappa -3},\quad \kappa =3+\frac{36\left( n-7\right) \left( n^{2}+2n-5\right) }{\left( n-2\right) \left( n+5\right) \left( n+7\right) \left( n+9\right) },\quad s^{2}=\dfrac{6\left( n-2\right) }{\left( n+1\right) \left( n+3\right) }, \end{aligned}$$

it follows that the quantile of order $1-\alpha$ of $Y$ is obtained by inserting the expression of the quantile of order $1-\alpha /2$ of $\widehat{ m}_{3}$, that is $t_{1-\alpha /2}s\left( \dfrac{\nu }{\nu -2}\right) ^{-1/2}$ (where $t_{1-\alpha /2}$ is the $1-\alpha /2$ order quantile of $T_{\nu }$ ), in (17).

For the sake of completeness, we derive the explicit expression of the density of $Y$ and its cumulative distribution function (cdf). Since the inverse of the restriction of $g$ to the interval of monotonicity $\left( 0,+\infty \right)$ is equal to

$$\begin{aligned} \gamma _{1}\left( y\right) =\frac{2y}{\sqrt{1-y}}\quad \quad 0<y<1, \end{aligned}$$

by straightforward calculations we obtain the expression of the density of the r.v. $Y$:

$$\begin{aligned} f\left( y\right)&= \frac{2}{s}\left( \frac{\nu }{\nu -2}\right) ^{1/2}\frac{ \Gamma \left( \frac{\nu +1}{2}\right) }{\Gamma \left( \frac{\nu }{2}\right) \sqrt{\nu \pi }}\left( 1+\frac{4}{s^{2}\left( \nu -2\right) }\frac{y^{2}}{1-y }\right) ^{-\left( \nu +1\right) /2} \\&\quad \times \, \left| \frac{2-y}{\left( 1-y\right) \sqrt{1-y}}\right| \quad \quad 0<y<1. \end{aligned}$$

The cdf of $Y$ on the interval $\left( 0,1\right)$ is given by

$$\begin{aligned} F\left( y\right) =2G\left( \frac{2y}{\sqrt{1-y}}\frac{1}{s}\left( \frac{\nu }{\nu -2}\right) ^{1/2}\right) -1 \end{aligned}$$

where $G$ is the cdf of the Student’s t with $\nu$ degree of freedom.

For large samples, the asymptotic law of $\mathfrak {N}_{n,2}$ is provided by the following:

Proposition 5

Under the hypothesis of normality one has

$$\begin{aligned} \mathfrak {N}_{n,2}\overset{\mathcal {L}}{\longrightarrow }\sqrt{\frac{3}{2}}H \end{aligned}$$

where $H$ is the Daniel’s half-normal distribution (see Daniel 1959).

Proof of Proposition 5

We have to consider the transformation (17). Since $g\left( x\right) =\left| x\right| /2+o\left( \left| x\right| \right)$ for $x\rightarrow 0$, it follows that

$$\begin{aligned} \sqrt{n}\left( 1-\widehat{\xi }\right) =\sqrt{n}\frac{\left| \widehat{m} _{3}\right| }{2}+o\left( \sqrt{n}\left| \widehat{m}_{3}\right| \right) . \end{aligned}$$

Recalling that for a Gaussian sample $\sqrt{n}\left( \widehat{m} _{3}-m_{3}\right)$ converges in distribution to a centered Gaussian r.v. with variance equal to $6$, we obtain

$$\begin{aligned} \sqrt{n}\left| \widehat{m}_{3}\right| \overset{\mathcal {L}}{ \longrightarrow }\sqrt{6}\left| \mathcal {N}\left( 0,1\right) \right| \quad \hbox {and then}\quad \sqrt{n}\left| \widehat{m} _{3}\right| =O_{P}\left( 1\right) . \end{aligned}$$

Finally, recalling that $H=\left| \mathcal {N}\left( 0,1\right) \right|$ is a half-normal distribution, by Slutsky Theorem we get

$$\begin{aligned} 2\sqrt{\frac{n}{6}}\left( 1-\widehat{\mathbf {\xi }}\right) \overset{\mathcal { L}}{\longrightarrow }H. \end{aligned}$$

$\square$

Proof of Proposition 4

Consider

$$\begin{aligned} \mathfrak {B}_{n,2}=\sqrt{\frac{n}{2}}\left( \widehat{U}_{n,2}-1\right) = \sqrt{\frac{n}{2}}\frac{1-\widehat{\xi }}{\widehat{\xi }} \end{aligned}$$

where $\widehat{\xi }$ is defined in (12). The first partial derivatives of

$$\begin{aligned} g\left( x,y\right) =1-\xi \left( x,y\right) =1-\frac{1}{2\left( y-x^{2}-1\right) }\left( y+3-\sqrt{16x^{2}+\left( y-5\right) ^{2}}\right) \end{aligned}$$

for $x=0$ and $y=3$ are zero and

$$\begin{aligned} \frac{\partial ^{2}}{\partial x^{2}}g\left( 0,3\right) =1,\quad \frac{ \partial ^{2}}{\partial y^{2}}g\left( 0,3\right) =0,\quad \frac{\partial }{ \partial x\partial y}g\left( 0,3\right) =0. \end{aligned}$$

Since in the normal case $m_{3}=0$, $m_{4}=3$, and $\left( \sqrt{n}\left( \widehat{m}_{3}-m_{3}\right) ,\sqrt{n}\left( \widehat{m}_{4}-m_{4}\right) \right)$ is asymptotically normally distributed with zero mean and covariance matrix

$$\begin{aligned} \Sigma =\left( \sigma _{ij}\right) _{\begin{array}{c} 1\le i\le 2 \\ 1\le j\le 2 \end{array}}=\left( \begin{array}{cc} 6 &{} 0 \\ 0 &{} 24 \end{array} \right) \end{aligned}$$

a direct application of the Second Order Delta Method gives

$$\begin{aligned} n\left( g\left( \widehat{m}_{3},\widehat{m}_{4}\right) -g\left( m_{3},m_{4}\right) \right) \overset{\mathcal {L}}{\longrightarrow }3\chi ^{2}\left( 1\right) . \end{aligned}$$

Hence (recall that under assumption of normality, $\xi =1$) one has:

$$\begin{aligned} n\left( 1-\widehat{\xi }\right) \overset{\mathcal {L}}{\longrightarrow }3\chi ^{2}\left( 1\right) . \end{aligned}$$

Since, for a Gaussian population, $\widehat{m}_{3}\overset{P}{ \longrightarrow }0$, $\widehat{m}_{4}\overset{P}{\longrightarrow }3$ and $\xi \left( x,y\right)$ is continuous at $\left( 0,3\right)$, it follows $\widehat{\xi }\overset{P}{\longrightarrow }1$. Therefore, thanks to Slutsky Theorem the statements follows.