On the efficiency of Gini’s mean difference

Abstract

The asymptotic relative efficiency of the mean deviation with respect to the standard deviation is 88 % at the normal distribution. In his seminal 1960 paper A survey of sampling from contaminated distributions, J. W. Tukey points out that, if the normal distribution is contaminated by a small $\epsilon$-fraction of a normal distribution with three times the standard deviation, the mean deviation is more efficient than the standard deviation—already for $\epsilon < 1\,\%$. In the present article, we examine the efficiency of Gini’s mean difference (the mean of all pairwise distances). Our results may be summarized by saying Gini’s mean difference combines the advantages of the mean deviation and the standard deviation. In particular, an analytic expression for the finite-sample variance of Gini’s mean difference at the normal mixture model is derived by means of the residue theorem, which is then used to determine the contamination fraction in Tukey’s 1:3 normal mixture distribution that renders Gini’s mean difference and the standard deviation equally efficient. We further compute the influence function of Gini’s mean difference, and carry out extensive finite-sample simulations.

Appendices

Appendix 1: Proofs

Towards the proof of Theorem 1, we spare a few words about the derivation of the corresponding result for the normal distribution. When evaluating the integral $J$, cf. (9), for the standard normal distribution, one encounters the integral

$$\begin{aligned} I_1 = \int _{-\infty }^{\infty } x^2 \phi (x) \varPhi (x)^2 dx, \end{aligned}$$

where $\phi$ and $\varPhi$ denote the density and the cdf of the standard normal distribution, respectively. Nair (1936) gives the value $I_1 = 1/3 + 1/(2\pi \sqrt{3})$, resulting in $J = \sqrt{3}/(2 \pi ) - 1/6$, but does not provide a proof. The author refers to the derivation of a similar integral (integral 8 in Table I, Nair 1936, p. 433), where we find the result as well as the derivation doubtful, and to an article by Hojo (1931), which gives numerical values for several integrals, but does not contain an explanation for the value of $I_1$ either. We therefore include a proof here. Writing $\varPhi (x)$ as the integral of its density and changing the order of the integrals in thus obtained three-dimensional integral yields

$$\begin{aligned} I_1 = (2\pi )^{-3/2} \int _{y=-\infty }^0 \int _{z=\infty }^0 \int _{x=-\infty }^\infty x^2 e^{x^2/2} e^{(y+x)^2/2} e^{(z+x)^2/2} \, d x\, d z\, d y. \end{aligned}$$

Solving the inner integral, we obtain

$$\begin{aligned} I_1 = (18\pi \sqrt{3})^{-1} \int _{y=0}^\infty \int _{z=0}^\infty [ (y+z)^2 + 3 ] \exp \left\{ - \frac{1}{3} \left[ y^2 + z^2 - y z \right] \right\} \, d z\, d y. \end{aligned}$$

Introducing polar coordinates $\alpha , r$ such that $y = r \cos \alpha$, $z = r \sin \alpha$, and solving the integral with respect to $r$, we arrive at

$$\begin{aligned} I_1 = \frac{1}{4\pi \sqrt{3}} \int _{\alpha =0}^\pi \frac{4+\sin \alpha }{(2-\sin \alpha )^2} \, d \alpha . \end{aligned}$$

This remaining integral may be solved by means of the residue theorem (e.g. Ahlfors 1966, p. 149). Substituting $\gamma = e^{i \alpha }$ and using $\sin \alpha = (e^{i \alpha } - e^{- i \alpha })/(2i)$, we transform $I_1$ into the following line integral in the complex plane,

$$\begin{aligned} I_1 = \frac{1}{4\pi \sqrt{3}} \int _{\varGamma _0} \frac{\gamma ^2 + 8 i \gamma -1 }{(\gamma ^2- 4 i \gamma -1 )^2} \, d \gamma , \end{aligned}$$

(10)

where $\varGamma _0$ is the upper unit half circle in the complex plane, cp. Fig. 4. Let us call $h$ the integrand in (10), its poles (both of order two) are $\gamma _{1/2} = (2\pm \sqrt{3})i$, so that $\gamma _2$ lies within the closed upper half unit circle $\varGamma$. The residue of $h$ in $\gamma _2$ is $-\sqrt{3} i /2$. Integrating $h$ along $\varGamma _1$, i.e. the real line from $-$1 to 1, cf. Fig. 4, and applying the residue theorem to the closed line integral along $\varGamma$ completes the derivation.

Fig. 4

Residue theorem: the line integral over $h$ along the closed curve $\varGamma =\varGamma _0\, \cup \, \varGamma _1$ is determined by the residue of $h$ in $\gamma _2$

Proof

(Proof of Theorem 1)

Evaluating the integral $J$ for the normal mixture distribution, we arrive after lengthy calculations at

$$\begin{aligned} J= & {} \Big [ \epsilon ^3 \lambda ^2 + (1-\epsilon )^3 \Big ] \Big [ 2 A(1) + C(1) + E(1) \Big ] \ - \ (\epsilon \lambda ^2 + 1 - \epsilon )B \\&\quad + \ \epsilon ^2(1-\epsilon ) \Big [ 2 (2+ \lambda ^2) A(1/\lambda ) + C(\lambda ) + 2\lambda ^2 D(1/\lambda )+ \lambda (2+ \lambda ^2)E(1/\lambda ) \Big ] \\&\quad + \ \epsilon (1-\epsilon )^2 \Big [ 2 (2\lambda ^2+ 1 ) A(\lambda ) + \lambda ^2C(1/\lambda ) + 2 D(\lambda )+ (\lambda ^{-1} + 2 \lambda )E(\lambda ) \Big ], \end{aligned}$$

where

$$\begin{aligned} A(\lambda )= & {} \int _{-\infty }^\infty x \phi ^2(x) \varPhi (x/\lambda ) dx \ = \ \frac{1}{4 \pi \sqrt{ 1 + 2 \lambda ^2 }}, \quad \\ B= & {} \int _{-\infty }^\infty x^2 \phi (x) \varPhi (x) dx \ = \ \frac{1}{2}, \\ C(\lambda )= & {} \int _{-\infty }^\infty x^2 \phi (x) \varPhi ^2(x/\lambda ) dx \ = \ \frac{1}{4} + \frac{\lambda }{\pi (1 + \lambda ^2)\sqrt{ 2 + \lambda ^2 }}\\&+ \frac{1}{2 \pi } \arctan \left( \frac{1}{\lambda \sqrt{2+\lambda ^2}}\right) , \\ D(\lambda )= & {} \int _{-\infty }^\infty x^2 \phi (x) \varPhi (x) \varPhi (x/\lambda ) dx \ = \ \frac{1}{4} + \frac{3\lambda ^2 + 1}{4 \pi (1 + \lambda ^2)\sqrt{ 2 \lambda ^2 + 1 }}\\&+ \frac{1}{2 \pi } \arctan \left( \frac{1}{\sqrt{2\lambda ^2+ 1}}\right) , \\ E(\lambda )= & {} \int _{-\infty }^\infty \phi ^2(x) \phi (x/\lambda ) dx \ = \ \frac{1}{2 \pi \sqrt{ 1 + 2 \lambda ^2 }}, \end{aligned}$$

for all $\lambda > 0$. As before, $\phi$ and $\varPhi$ denote the density and the cdf of standard normal distribution. The tricky integrals are $C(\lambda )$ and $D(\lambda )$, which, for $\lambda = 1$, both reduce to the integral $I_1$ above. Proceeding as before for the integral $I_1$, solving the respective two inner integrals yields

$$\begin{aligned}&C(\lambda ) = \frac{\lambda ^3}{2 \pi \sqrt{2+\lambda ^2}} \int _0^{\pi /2} \frac{ 3 + \lambda ^2 + \sin (2\alpha )}{ \{1 + \lambda ^2 - \sin (2 \alpha )\}^2} d\alpha , \\&D(\lambda ) = \frac{1}{2 \pi \sqrt{1+2\lambda ^2}} \int _0^{\pi /2} \frac{ 2 + \lambda ^2 (2+ \sin (2\alpha )) + (3\lambda ^4 - \lambda ^2 - 2) \sin ^2(\alpha )}{ \{ 2 - \sin (2\alpha ) + (\lambda ^2-1) \sin ^2(\alpha ) \}^2 } d \alpha . \end{aligned}$$

These integrals are again solved by the residue theorem, which completes the proof. $\square$

For the proof of Theorem 2, the following identities are helpful:

$$\begin{aligned}&\textstyle \int x \left( 1 + \frac{x^2}{\beta } \right) ^\alpha dx \ = \ \frac{\beta }{2(\alpha +1)} \left( 1 + \frac{x^2}{\beta }\right) ^{\alpha +1}, \quad \alpha \ne -1, \ \beta \ne 0. \end{aligned}$$

(11)

$$\begin{aligned}&\textstyle \int _{-\infty }^\infty \left( 1 + \frac{x^2}{\nu }\right) ^{-\nu } dx \ = \ \frac{1}{c_{2\nu -1}} \sqrt{\frac{\nu }{2\nu -1}}, \quad \nu > 0, \end{aligned}$$

(12)

$$\begin{aligned}&\textstyle \int _{-\infty }^\infty \left( 1 + \frac{x^2}{\nu }\right) ^{-\frac{3\nu -1}{2}} dx \ = \ \frac{1}{c_{3\nu -2}} \sqrt{\frac{\nu }{3\nu -2}}, \quad \nu > 0, \end{aligned}$$

(13)

where $c_{\nu }$ is the scaling factor of the $t_\nu$ density, cf. Table 1. The identities (12) and (13) can be obtained by transforming the respective left-hand sides into a $t_\nu$-densities by substituting $y = ((2\nu -1)/\nu )^{1/2} \, x$ and $y = ((3\nu -2)/\nu )^{1/2}\, x$, respectively.

Proof

(Proof of Theorem 2) For computing $g$, we evaluate (7), successively making use of (11) and (12), and obtain

$$\begin{aligned} g \ = \ 4 \frac{ \nu \, c_\nu ^2}{\nu -1} \int _{-\infty }^{\infty } \Big ( 1 + \frac{x^2}{\nu } \Big )^{-\nu } \, dx \ = \ \frac{4 \, \nu ^{3/2} \, c_{\nu }^2}{(\nu -1)\, \sqrt{2\nu -1}\, c_{2\nu -1}}, \end{aligned}$$

which can be written as in Theorem 2 by using $B(x,y) = \varGamma (x)\varGamma (y)/\varGamma (x+y)$. For evaluating $J$, we write $J$ as $J = \int _\mathbb {R}A(x) f_{\nu }(x)\, dx$ with $f_\nu$ being the $t_\nu$ density and

$$\begin{aligned} A(x)= & {} \int _{-\infty }^x \int _x^{\infty } x z f_\nu (z) f_\nu (y) \,dz\, dy \ - \ \int _{-\infty }^x \int _x^{\infty } y z f_\nu (z) f_\nu (y) \,dz\, dy \\&- \int _{-\infty }^x \int _x^{\infty } x^2 f_\nu (z) f_\nu (y) \,dz\, dy \ + \ \int _{-\infty }^x \int _x^{\infty } x y f_\nu (z) f_\nu (y) \,dz\, dy \\= & {} A_1(x) - A_2(x) - A_3(x) + A_4(x). \end{aligned}$$

Using (11), we obtain

$$\begin{aligned} A_1(x) + A_4(x) \ = \ \frac{c_\nu \, \nu \, x}{\nu -1}\left( 1 + \frac{x^2}{\nu } \right) ^{-\frac{\nu -1}{2}} \, \int _{-x}^x f_\nu (y) \, dy, \end{aligned}$$

and

$$\begin{aligned} - A_2(x) \ = \ \left( \frac{c_\nu \,\nu }{\nu -1} \right) ^2 \, \left( 1+ \frac{x^2}{\nu }\right) ^{-\nu +1}. \end{aligned}$$

Hence, $J = B_1 + B_2 - B_3$ with

$$\begin{aligned} \textstyle&B_1 \ = \ \int _{-\infty }^{\infty } \frac{c_\nu \, \nu \, x}{\nu -1} \left( 1 + \frac{x^2}{\nu } \right) ^{-\frac{\nu -1}{2}} f_\nu (x) \, \int _{-x}^x f_\nu (y) \, dy \, dx, \\&\textstyle B_2 \ = \ \int _{-\infty }^{\infty } \left( \frac{c_\nu \,\nu }{\nu -1} \right) ^2 \, \left( 1+ \frac{x^2}{\nu }\right) ^{-\nu +1} \! f_\nu (x) \, dx, \\&B_3 \ = \ \int _{-\infty }^{\infty } x^2 F_\nu (x) \left( 1 - F_\nu (x) \right) f_\nu (x) \, dx \ = \ \frac{\nu }{2(\nu -2)} - \int _{-\infty }^{\infty } x^2 f_\nu (x) F_\nu ^2(x) \, dx, \end{aligned}$$

where $F_\nu$ is the cdf of the $t_\nu$ distribution. By employing (11) and (13), we find

$$\begin{aligned} B_1 \ = \ B_2 \ = \ \frac{2}{c_{3\nu -2}}\, \left( \frac{c_\nu \,\nu }{\nu -1} \right) ^2 \, \sqrt{\frac{\nu }{3\nu -2}} \end{aligned}$$

and arrive, again by employing $B(x,y) = \varGamma (x)\varGamma (y)/\varGamma (x+y)$, at the expression for $J$ given in Theorem 2. $\square$

The remaining integral

$$\begin{aligned} K_\nu \ = \int _{-\infty }^{\infty } x^2 f_\nu (x) F_\nu ^2(x) \, dx \end{aligned}$$

cannot be solved by the same means as the analogous integral $I_1$ for the normal distribution, and we state this as an open problem. However, this one-dimensional integral can easily be approximated numerically, and the expression is quickly entered into a mathematical software like R (R Development Core Team 2010).

Proof

(Proof of Proposition 1) We have

$$\begin{aligned}&g(F_{\epsilon ,x}) = 2 \int _{-\infty }^{\infty } \int _{y}^{\infty } (z - y)\, d F_{\epsilon ,x} (z)\, d F_{\epsilon ,x}(y), \\&\quad = (1-\epsilon )^2 g(F) + 2 \epsilon (1-\epsilon ) \int _{-\infty }^{\infty } (x-z) \left\{ 1\!\!1_{ (-\infty ,x] } (y) - 1\!\!1_{ [x,\infty ) } (y) \right\} \, d F(y) \end{aligned}$$

and hence

$$\begin{aligned}&I\!F(x,g(\cdot ); F) = \lim _{\epsilon \searrow 0} \frac{1}{\epsilon } \{ g(F_{\epsilon ,x}) - g(F) \} \\&\quad = \ - 2 g(F) + 2 \left\{ x [ F(x) + F(x-) - 1 ] + E[X 1\!\!1_{ \{ X\ge x \} } ] - E[X 1\!\!1_{ \{ X \le x \} } ] \right\} , \end{aligned}$$

which completes the proof. $\square$

With the influence function known, it is also possible use the relationship

$$\begin{aligned} ASV(s_n;F) = \int _\mathbb {R}I\!F(x,s,F)^2 F(dx) \end{aligned}$$

instead of referring to the terms given in Sect. 2 to compute the asymptotic variance of the estimators. This leads to the same integrals.

Appendix 2: Miscellaneous

Lemma 1

For $X_1,\ldots , X_n$ being independent and $U(a,b)$ distributed for $a,b \in \mathbb {R}$, $a < b$, we have for the sample mean deviation (about the median)

$$\begin{aligned} E(d_n) = {\left\{ \begin{array}{ll} (b-a)/4 &{} \quad \hbox {for odd } n \, (n \ge 3), \\ \displaystyle \frac{b-a}{4}\frac{n^2}{n^2-1} &{} \quad \hbox { for even } n. \end{array}\right. } \end{aligned}$$

Proof

For notational convenience we restrict our attention to the case $a=0$, $b=1$. Let $X_{(i)}$ denote the $i$th order statistic, $1 \le i \le n$. The random variable $X_{(i)}$ has a Beta$(\alpha ,\beta )$ distribution with parameters $\alpha = i$ and $\beta = n+1-i$, and hence $E(X_{(i)}) = i/(n+1)$. If $n$ is odd, we write $d_n$ as $d_n = (n-1)^{-1} \sum _{i=1}^{\lfloor n/2 \rfloor } (X_{(n+1-i)} - X_{(i)})$ and obtain

$$\begin{aligned} E(d_n) = \frac{1}{n-1} \sum _{i=1}^{\lfloor n/2 \rfloor } \left( \frac{n+1-i}{n+1} - \frac{i}{n+1} \right) = \frac{1}{4}. \end{aligned}$$

If $n$ is even, we have $d_n = (n-1)^{-1} \sum _{i=1}^{n/2} (X_{(n+1-i)} - X_{(i)})$, and hence

$$\begin{aligned} E(d_n) = \frac{1}{n-1} \sum _{i=1}^{n/2} \left( \frac{n+1-i}{n+1} - \frac{i}{n+1} \right) = \frac{n^2}{4(n^2-1)}, \end{aligned}$$

which completes the proof. $\square$