Estimation of the regression slope by means of Gini’s cograduation index

Abstract

The simple linear model $Y_i = \alpha + \beta \, x_i + \epsilon _i$ $(i=1,2, \ldots ,N \ge 2)$ is considered, where the $x_i$’s are given constants and $\epsilon _1, \epsilon _2 , \ldots , \epsilon _N$ are independent identically distributed (iid) with continuous distribution function F. An estimator $\tilde{\beta }$ of the slope parameter is proposed, based on a stochastic process which makes use of Gini’s cograduation index. The properties of $\tilde{\beta }$ and of the related confidence interval are studied. Some comparisons are given, in terms of asymptotic relative efficiency, with other estimators of $\beta$ including that obtained with the method of least squares.

Appendix

To prove Theorem 2 of Sect. 7, some preliminary results will be considered. Let f be a probability density function with support in $\mathfrak {R}$ and define the two probability measures

$$\begin{aligned} P_N(A) = \int _A \, \prod _{1 \le i \le N} f(y_i) \quad \text{ and } \quad Q_N(A) = \int _A \, \prod _{1 \le i \le N} f \left( y_i + \frac{b}{T} (x_i-\bar{x}) \right) \end{aligned}$$

where $x_1< x_2< \cdots < x_N$ as usual, $b \ne 0$ is finite, A is any event and

$$\begin{aligned} T^2 = \sum _{1 \le i \le N} (x_i - \bar{x})^2 \qquad M = \max _{1 \le i \le N} (x_i - \bar{x})^2 . \end{aligned}$$

Lemma 1

[Hájek and Šidák (1967, p. 208); Hájek (1962, p. 1134)] If the vector

$$\begin{aligned} \left( \sqrt{N} \, G(\underline{Y};0) \, , \, \log \frac{\displaystyle {\prod _{1 \le i \le N}} \, f \left( Y_i + \frac{b}{T} (x_i - \bar{x}) \right) }{\displaystyle {\prod _{1 \le i \le N}} f(Y_i)} \right) \end{aligned}$$

converges, with the measure $P_N$, to the Gaussian distribution with parameters

$$\begin{aligned} \left( \mu \, , \, - \frac{1}{2} \sigma _2^2 \, , \, \sigma _1^2 \, , \, \sigma _2^2 \, , \, \sigma _{12} \right) \end{aligned}$$

then the variable

$$\begin{aligned} \sqrt{N} \, G(\underline{Y};0) \end{aligned}$$

converges, with the measure $Q_N$, to the Gaussian distribution with mean $\mu + \sigma _{12}$ and variance $\sigma _1^2.$

Lemma 2

[Hájek and Šidák (1967, p. 213); Hájek (1962, p. 1136)]. If $\displaystyle I(f) = \int \left( \frac{f'}{f} \right) ^2 f < \infty$ and if $\displaystyle {\frac{T^2}{M} \rightarrow \infty },$ then

$$\begin{aligned}&{P_N\lim }_{N \rightarrow \infty } \Bigg ( \log \frac{\displaystyle {\prod _{1 \le i \le N}} f \left( Y_i + \frac{b}{T} (x_i-\bar{x}) \right) }{\displaystyle {\prod _{1 \le i \le N}} \, f(Y_i)} - \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \frac{f'(Y_i)}{f(Y_i)} \\&\quad + \frac{b}{2} \, I(f) \Bigg ) =0 \end{aligned}$$

where $P_N \lim$ denotes the limit in $P_N$-probability.

Remark to Lemma 2

According to the measure $P_N,$ the variable

$$\begin{aligned} \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \, \frac{f'(Y_i)}{f(Y_i)} \end{aligned}$$

has variance

$$\begin{aligned} \text{ Var } \left( \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \, \frac{f'(Y_i)}{f(Y_i)} \right) = b^2 \, I(f) < \infty \end{aligned}$$

and expectation (Hájek 1962, p. 1125)

$$\begin{aligned} \text{ E } \left( \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \, \frac{f'(Y_i)}{f(Y_i)} \right) = 0 . \end{aligned}$$

Moreover, the variable

$$\begin{aligned} \sum _{1 \le i \le N} Z_{2i} = \sum _{1 \le i \le N} \frac{\displaystyle {\frac{b}{T} \, (x_i - \bar{x}) \, \frac{f'(Y_i)}{f(Y_i)}}}{\sqrt{b^2 I (f)}} \end{aligned}$$

satisfies the Lindeberg–Feller condition. Indeed, after defining

$$\begin{aligned} Z_{2i} (\delta ) = Z_{2i} \, \, s \left( |Z_{2i}| - \delta \right) \end{aligned}$$

where $\delta >0$ and s(x) equals 1 if $x\ge 0$ and 0 elsewhere, such a condition can be written as

$$\begin{aligned} \sum _{1 \le i \le N} \text{ E } \left( Z_{2i}^2 (\delta ) \right) = \sum _{1 \le i \le N} \int _{|z| \ge \delta } z^2 \, d P_N \left\{ \frac{b}{T} (x_i - \bar{x}) \frac{f'(Y_i)}{f(Y_i)} \le z \, \sqrt{b^2 \, I(f)} \right\} \rightarrow 0 . \end{aligned}$$

However, by putting $t = z \, \sqrt{b^2 \, I(f)},$ one gets

$$\begin{aligned}&\sum _{1 \le i \le N} \frac{1}{b^2 \, I(f)} \, \int _{|t| \ge \delta \sqrt{b^2 \, I(f)}} \, t^2 \, { d P_N} \left\{ \frac{b}{T}(x_{i}-\bar{x}) \frac{f'(Y_i)}{f(Y_i)} \le t \right\} \\&\quad = \frac{1}{T^2} \, \sum _{1 \le i \le N} \frac{(x_i- \bar{x})^2}{I(f)} \int _{|y| \ge \delta \sqrt{I(f)} \left| \frac{T}{x_i - \bar{x}} \right| } \, y^2 \, d P_N \left\{ \frac{f'(Y_i)}{f(Y_i)} \le y \right\} \rightarrow 0 \end{aligned}$$

because, by hypothesis,

$$\begin{aligned} \frac{T^2}{M} \rightarrow + \infty \quad \Rightarrow \quad \left| \frac{T}{x_i- \bar{x}} \right| \rightarrow + \infty . \end{aligned}$$

Lemma 3

If $F'=f$ and if

$$\begin{aligned} \hat{G}(\underline{Y};0) = \frac{2N}{D} \sum _{1 \le i \le N} \left( \left| 1 - \frac{i}{N} - F(Y_i) \right| - \left| \frac{i}{N} - F(Y_i) \right| \right) \end{aligned}$$

then

$$\begin{aligned} {P_N \lim }_{N \rightarrow \infty } \, \sqrt{N} \, (G(\underline{Y};0) - \hat{G}(\underline{Y};0)) = 0. \end{aligned}$$

Proof

By using the identity $|x| = 2x \, s(x) - x$ $(x \in \mathfrak {R}),$ the definition in (2) and the expression of $\hat{G}(\underline{Y}; 0),$ one gets

$$\begin{aligned} \sqrt{N} (G(\underline{Y};0) - \hat{G} (\underline{Y};0)) = A_N + B_N + C_N + D_N, \end{aligned}$$

where

$$\begin{aligned} A_N&=\frac{4 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \left( 1- \frac{i}{N} - F(Y_i) \right) \left[ s(N+1-i-R(Y_i)) \right. \\&\quad \left. - s(N-i-N\, F(Y_i)) \right] \\ B_N&=- \frac{4 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \left( \frac{i}{N} - F(Y_i) \right) \left[ s(i-R(Y_i)) - s(i-N\, F(Y_i)) \right] \\ C_N&=\frac{4 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \left( \frac{R(Y_i)}{N} - F(Y_i) \right) \left[ s(i - R(Y_i)) - s(N+1-i-R(Y_i)) \right] \\ D_N&=\frac{2 \, N^{1/2}}{D} \, \sum _{1 \le i \le N} s(N+1-i-R(Y_i)) . \end{aligned}$$

It follows that

$$\begin{aligned} \text{ E } \{ |D_N| \}&\le \frac{2 \, N^{3/2}}{D} \, \rightarrow 0 \\ \text{ E } \{ |B_N| \}&\le \frac{4 \, N^{3/2}}{D} \sum _{1 \le i \le N} \text{ E } \left\{ \left| \frac{i}{N} - F(Y_i) \right| \, \, \left| s(i-R(Y_i)) - s(i-NF(Y_i)) \right| \right\} . \end{aligned}$$

By using the joint distribution of $(Y_i, R(Y_i)),$ that is

$$\begin{aligned}&\text{ Pr } \{ R(Y_i) = r \, ; \, y< Y_i < y+\hbox {d}y \} = \frac{1}{N} \, g_{Y_{(r)}} (y) \, \hbox {d}y \\&\quad = \frac{1}{N} \frac{N!}{(N-r)! \, (r-1)!} \, [F(y)]^{r-1} \, [1-F(y)]^{N-r} \, f(y) \, \hbox {d}y \quad r=1,2, \ldots , N; y \in \mathfrak {R}, \end{aligned}$$

one gets

$$\begin{aligned}&\text{ E } \left\{ \left| \frac{i}{N} - F(Y_i) \right| \, \, \left| s(i-R(Y_i)) - s(i-NF(Y_i)) \right| \right\} \\&= \frac{1}{N} \sum _{1 \le r \le N} \int _{- \infty }^{+ \infty } \left| \frac{i}{N} - F(y) \right| \, \, \left| s(i-r) - s(i-NF(y)) \right| \, g_{Y_{(r)}} (y) \, \hbox {d}y \\&= \sum _{1 \le r \le N} \int _0^1 \left| \frac{i}{N} - v \right| \, \, \left| s(i-r) - s(i-Nv) \right| {N-1 \atopwithdelims ()r-1} \, v^{r-1} \, (1-v)^{N-r} \, dv \\&=\int _0^1 \left| \frac{i}{N} - v \right| \left[ \left( 1- s(i-Nv) \right) \, \mathrm{Pr}\{ U_{(i)} > v \} + s(1-Nv) \, \mathrm{Pr}\{ U_{(i)} \le r \} \right] \, dv , \end{aligned}$$

where $U_{(i)}$ is the i-th order statistic of a $(N-1)$-sized random sample drawn from a uniform population in (0, 1). By partitioning the integration interval, after some trivial passages, one gets

so that

$$\begin{aligned} \text{ E } \{ |B_N| \} \le \frac{2 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \frac{i(N-i)}{N^2(N+1)} \quad \rightarrow 0 . \end{aligned}$$

By following similar steps, one can prove that

$$\begin{aligned} \text{ E } \{ |A_N| \} \rightarrow 0 . \end{aligned}$$

Now let

$$\begin{aligned} S_{i,N} = \left( \frac{R(Y_i)}{N} - F(Y_i) \right) \left[ s ( i - R(Y_i) ) - s(N+1-i-R(Y_i)) \right] \qquad i =1, \ldots , N . \end{aligned}$$

and simply consider that

$$\begin{aligned} \text{ E } (C_N) = \frac{4 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \text{ E } \left( S_{i,N} \right) =0 . \end{aligned}$$

Moreover,

$$\begin{aligned} \text{ Var } (C_N) = \frac{16 \, N^3}{D^2} \sum _{1 \le i \le N} \text{ E } \left( S_{i,N}^2 \right) + \frac{16 \, N^3}{D^2} \sum _{i \ne j} \, \text{ E } \left( S_{i,N} \, S_{j,N} \right) \end{aligned}$$

(19)

and

$$\begin{aligned} \sum _{1 \le i \le N} \text{ E } (S_{i,N}^2)= & {} \sum _{1 \le i \le N} \frac{1}{N} \sum _{1 \le r \le N} \int _{-\infty }^{+ \infty } \left( \frac{r}{N} - F(y) \right) ^2 \cdot \\&\cdot \, \left[ s(i-r) -s(N+1-i-r) \right] ^2 \, g_{Y_{(r)}} (y) \, \hbox {d}y \\\le & {} \sum _{1 \le r \le N} \int _{-\infty }^{+\infty } \left( \frac{r}{N} - F(y) \right) ^2 \, g_{Y_{(r)}}(y)\,\hbox {d}y \\\le & {} \sum _{1 \le r \le N} \text{ E } \left\{ \left( F(Y_{(r)}) -\frac{r}{N} \right) ^2 \right\} \, \rightarrow \, A < +\infty \end{aligned}$$

being that

$$\begin{aligned} \text{ E } \left\{ \left( F(Y_{(r)}) - \frac{r}{N} \right) ^2 \right\} = \frac{r(N-r+1)}{(N+2)\, (N+1)^2} + \frac{r^2}{N^2 \, (N+1)^2} \qquad \forall r = 1,2, \ldots , N . \end{aligned}$$

The first summand in the right-hand side of (19) thus tends to zero. Moreover,

$$\begin{aligned}&\left| \sum _{i \ne j} \text{ E } (S_{i,N} \, S_{j,N}) \right| \\&\quad = \left| \frac{1}{N(N-1)} \, \sum _{i \ne j} \, \sum _{r \ne k} \int _{-\infty }^{+\infty } \int _{-\infty }^{+\infty } \left( \frac{r}{N} - F(x) \right) \, \left( \frac{k}{N} - F(y) \right) \right. \\&\quad \left( s(i-r) - s(N+1-i-r) \right) \, \left( s(j-k) - s (N+1-j-k) \right) \\&g_{Y_{(r)},Y_{(k)}} (x,y) \, \hbox {d}x \, \hbox {d}y \Big | \\&\quad = \left| \frac{1}{N \, (N-1)} \sum _{r \ne k} \left[ \int _{-\infty }^{+\infty } \int _{-\infty }^{+\infty } \left( \frac{r}{N} - F(x) \right) \, \left( \frac{k}{N} - F(y) \right) \, g_{Y_{(r)}, Y_{(k)}} (x,y) \right. \right. \\&\quad \hbox {d}x \, \hbox {d}y \Big ] \sum _{i \le i \le N} \left( s(i-r) -s(N+1-i-r) \right) \, \left( s(i-k) - s(N+1-i-k) \right) \big | . \end{aligned}$$

Now, as

$$\begin{aligned}&\int _{-\infty }^{+\infty } \int _{-\infty }^{+\infty } \left( \frac{r}{N} - F(x) \right) \, \left( \frac{k}{N} - F(y) \right) \, g_{Y_{(r)}, Y_{(k)}} (x,y) \, \hbox {d}x \, \hbox {d}y \\&\quad = \text{ Cov } \left\{ F(Y_{(r)}) \, , \, F(Y_{(k)}) \right\} + \frac{rk}{N^2 \, (N+1)^2} \\&\quad = \left\{ \begin{array}{l@{\quad }l} \dfrac{r(N+1-k)}{(N+2)\, (N-1)^2} + \dfrac{rk}{N^2 \, (N+1)^2}>0 &{}\quad r<k \\ \dfrac{k(N+1-r)}{(N+2)\, (N+1)^2} + \dfrac{rk}{N^2 \, (N+1)^2}>0 &{}\quad r>k \end{array} \right. \end{aligned}$$

one obtains

$$\begin{aligned}&\left| \sum _{i \ne j} \text{ E } (S_{i,N} \, S_{j,N}) \right| \\&\quad \le \frac{1}{N-1} \sum _{r \ne k} \left\{ \text{ Cov } (F(Y_{(r)}) \, , \, F(Y_{(k)})) + \frac{rk}{N^2 \, (N+1)^2} \right\} \, \rightarrow \, B < + \infty \end{aligned}$$

so that the second summand in (19) tends to zero as well. The proof follows then by a suitable application of Thchebycheff’s inequality to the four variables. $\square$

Remark to Lemma 3

Lemma 3 makes it possible to obtain the asymptotic distribution of Gini’s cograduation index under indifference in an alternative way with respect to a former paper (Cifarelli and Regazzini 1977). Indeed, Lemma 3 assures that $\sqrt{N} \, G(\underline{Y};0)$ is asymptotically distributed as $\sqrt{N} \, \hat{G} (\underline{Y}; 0),$ for which the classical limit theorems can be applied, because it can be regarded as a sum of independent variables. As a matter of fact, the variable

$$\begin{aligned} \sqrt{N} \, \hat{G}(\underline{Y}; 0) = \frac{2 \, N^{3/2}}{D} \, \sum _{1 \le i \le N} \left( \left| 1-\frac{i}{N} - F(Y_i) \right| - \left| \frac{i}{N} - F(Y_i) \right| \right) \end{aligned}$$

has mean and variance

$$\begin{aligned} \text{ E } (\sqrt{N} \, \hat{G}(\underline{Y};0)) = 0 \qquad \text{ Var } (\sqrt{N} \, \hat{G}(\underline{Y};0)) \simeq \frac{2}{3} \end{aligned}$$

and, by letting

$$\begin{aligned} \sum _{1 \le i \le N} \dfrac{\frac{2 \, N^{3/2}}{D} \, \left( | 1 - \frac{i}{N} - F(Y_i) | - | \frac{i}{N} - F(Y_i) | \right) }{\sqrt{\frac{2}{3}}} = \sum _{1 \le i \le N} Z_{1i} \end{aligned}$$

and, for every $\delta >0,$

$$\begin{aligned} Z_{1i} (\delta ) = Z_{1i} \, s(|Z_{1i} - \delta |), \end{aligned}$$

the Lindeberg condition is satisfied, i.e.,

$$\begin{aligned} \sum _{1 \le i \le N} \text{ E } (Z_{1i}^2 (\delta )) \, \rightarrow \, 0 \qquad \forall \delta >0. \end{aligned}$$

Lemma 4

If $F' = f,$ $I(f) < +\infty ,$ $\int |f'| < + \infty$ and $\frac{T^2}{M} \rightarrow + \infty ,$ then the vector

$$\begin{aligned} \left( \sqrt{N} \, G (\underline{Y};0) \, , \, \log \frac{\displaystyle {\prod _{1 \le i \le N}} \, f \left( Y_i + \frac{b}{T} (x_i - \bar{x}) \right) }{\displaystyle {\prod _{1 \le i \le N}} f(Y_i)} \right) \end{aligned}$$

converges in distribution, with the measure $P_N,$ to the bivariate Gaussian with parameters

$$\begin{aligned} \left( 0, \quad -\frac{b^2}{2} I(f), \quad \frac{2}{3} , \quad b^2 I(f), \quad \sigma _{12} \right) \end{aligned}$$

where

Proof

By Lemmas 2 and 3, it suffices to show that the vector

$$\begin{aligned} \left( \sqrt{N} \, \hat{G} (\underline{Y};0) \, , \, \frac{b}{T} \sum _{1 \le i \le N} (x_i -\bar{x}) \dfrac{f'(Y_i)}{f(Y_i)} \right) \end{aligned}$$

converges in distribution to the bivariate Gaussian with parameters

$$\begin{aligned} \left( 0, \quad 0, \quad \frac{2}{3} , \quad b^2 I(f), \quad \sigma _{12} \right) . \end{aligned}$$

By means of the remarks following Lemmas 2 and 3, the limiting distribution surely takes the first four parameters listed above. Moreover, consider that

$$\begin{aligned}&\text{ Cov } \left\{ \sqrt{N} \hat{G} (\underline{Y};0)\, , \, \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \dfrac{f'(Y_i)}{f(Y_i)} \right\} \\&\quad = \frac{2 \, N^{3/2}}{D \, T} \, b \, \int _{-\infty }^{+\infty } f'(y) \, \left[ \sum _{1 \le i \le N} (x_i - \bar{x}) \left( \left| 1 - \frac{i}{N} - F(y) \right| - \left| \frac{i}{N} - F(y) \right| \right) \right] \, \hbox {d}y \\&\quad = 4b \, \int _{- \infty }^{+ \infty } f'(y) \left[ \frac{N^{3/2}}{D \, T} \left( \sum _{i=1}^{[N\, (1-F(y))]} (x_i - \bar{x}) \left( 1-F(y) - \frac{i}{N} \right) \right. \right. \\&\qquad - \left. \left. \sum _{i=1}^{[N\, F(y)]} (x_i - \bar{x}) \left( F(y) - \frac{i}{N} \right) \right) \right] \, \hbox {d}y . \end{aligned}$$

By passing to the limit (with N) under the integral sign, one gets

$$\begin{aligned} \sigma _{12}&= \lim _{N \rightarrow + \infty } \text{ Cov } \left\{ \sqrt{N} \hat{G} (\underline{Y};0)\, , \, \frac{b}{T} \sum _{1 \le i \le N} (x_i - \bar{x}) \dfrac{f'(Y_i)}{f(Y_i)} \right\} \\&= 4b \, \int _{- \infty }^{+ \infty } \left[ \psi (1-F(y)) - \psi (F(y)) \right] \, f'(y) \, \hbox {d}y \\&= 4b \, \int _0^1 \left[ \psi (1-v) - \psi (v) \right] \, \dfrac{f'(F^{-1}(v))}{f(F^{-1}(v))} \, dv. \end{aligned}$$

To prove that the limiting distribution is Gaussian, one can then show that, for every real $\lambda _1$ and $\lambda _2,$ the following variable is asymptotically normally distributed:

$$\begin{aligned} \lambda _1 \, \sqrt{N} \, \hat{G} (\underline{Y};0)+ \lambda _2 \, \frac{b}{T} \sum _{1 \le i \le N} (x_i -\bar{x}) \dfrac{f'(Y_i)}{f(Y_i)} . \end{aligned}$$

However, as both the variables

$$\begin{aligned} \lambda _1 \, \sqrt{N} \, \hat{G} (\underline{Y};0) \quad \text{ and } \quad \lambda _2 \, \frac{b}{T} \sum _{1 \le i \le N} (x_i -\bar{x}) \dfrac{f'(Y_i)}{f(Y_i)} \end{aligned}$$

satisfy the Lindeberg condition, one obtains the aimed result as in Hájek and Šidák (1967, p. 218). $\square$

The proof of Theorem 2 in Sect. 7 now immediately follows from Lemma 1 and Lemma 4. Indeed, for every real z, 

$$\begin{aligned} Q_N \left\{ \sqrt{N} \, G(\underline{Y};0) \le z \right\}&= \int _{\left\{ \sqrt{N} \, G(\underline{y} ; 0) \le z \right\} } \prod _{1 \le i \le N} \, f \left( y_i + \frac{b}{T} (x_i - \bar{x}) \right) \\&= \int _{\left\{ \sqrt{N} \, G \left( \underline{y} ; \frac{b}{T} \right) \le z \right\} } \prod _{1 \le i \le N} \, f(y_i) \\&= P_N \left\{ \sqrt{N} \, G \left( \underline{Y} ; \frac{b}{T} \right) \le z \right\} \end{aligned}$$

and, by Lemmas 1 and 4,

$$\begin{aligned} \lim _{N \rightarrow + \infty } Q_N \left\{ \sqrt{N} \, G(\underline{Y};0) \le z \right\} = \phi \left( \frac{z-\sigma _{12}}{\sqrt{2/3}} \right) . \end{aligned}$$