Kernel Density Estimation Based on the Distinct Units in Sampling with Replacement

Abstract

This paper considers the problem of estimating density functions using the kernel method based on the set of distinct units in sampling with replacement. Using a combined design-model-based inference framework, which accounts for the underlying superpopulation model as well as the randomization distribution induced by the sampling design, we derive asymptotic expressions for the bias and integrated mean squared error (MISE) of a Parzen-Rosenblatt-type kernel density estimator (KDE) based on the distinct units from sampling with replacement. We also prove the asymptotic normality of the distinct units KDE under both design-based and combined inference frameworks. Additionally, we give the asymptotic MISE formulas of several alternative estimators including the estimator based on the full with-replacement sample and estimators based on without-replacement sampling of similar cost. Using the MISE expressions, we discuss how the various estimators compare asymptotically. Moreover, we use Mote Carlo simulations to investigate the finite sample properties of these estimators. Our simulation results show that the distinct units KDE and the without-replacement KDEs perform similarly but are all always superior to the full with-replacement sample KDE. Furthermore, we briefly discuss a Nadaraya-Watson-type kernel regression estimator based on the distinct units from sampling with replacement, derive its MSE under the combined inference framework, and demonstrate its finite sample properties using a small simulation study. Finally, we extend the distinct units density and regression estimators to the case of two-stage sampling with replacement.

Appendix: Additional Proofs

Proof of Corollary 2.2.

The estimator $\hat {f}_{n}(y;h)$ can be seen as a sample mean of a fixed size simple random sample drawn with replacement from the finite population. Therefore, $\hat {f}_{n}(y;h)$ is design-unbiased for the finite population smooth f_U(y;h) and its design variance is given by (cf. Cochran, 1977, pg. 30)

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\{\hat{f}_n(y)|\mathbf{Y}_{\text{U}}\} &=&\frac{1}{n}\left( 1-\frac{1}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2. \end{array} $$

Now, the bias of $\hat {f}_{n}(y;h)$ under combined inference is identical to the bias of $\hat {f}_{\nu }(y;h)$ (see Theorem 2.2), whereas the combined variance can be written as

$$ \begin{array}{@{}rcl@{}}\mathbb{V}\{\hat{f}_n(y)\}& = &\mathrm{E}_{\mathcal{\xi}}[\mathrm{V}_{\mathcal{D}}\{\hat{f}_n(y)|\mathbf{Y}_{\text{U}}\}] + \mathrm{V}_{\mathcal{\xi}}[\mathrm{E}_{\mathcal{D}}\{\hat{f}_{\nu}(y)|\mathbf{Y}_{\text{U}}\}].\end{array} $$

Using (14) and (15), it is not difficult to see that

$$ \begin{array}{@{}rcl@{}}\mathbb{V}\{\hat{f}_n(y)\}&=&\frac{1}{nh}\left[\left( 1-\frac{1}{N}\right)+\frac{n}{N}\right]d_K f(y)+o\left( \frac{1}{Nh}\right). \end{array} $$

The result follows upon collecting the squared bias and variance of $\hat {f}_{n}(y;h)$ and integrating over y.□

Proof of Corollary 2.3.

The estimator $\hat {f}_{\nu , \text { wor}}(y;h)$ is the sample mean of a fixed size simple random sample drawn without replacement from the finite population. Therefore, $\hat {f}_{\nu , \text { wor}}(y;h)$ is design-unbiased for the finite population smooth f_U(y;h) and its design variance is given by (cf. Cochran, 1977, pg. 23)

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\{\hat{f}_{\nu, \text{ wor}}(y)|\mathbf{Y}_{\text{U}}\} &=&\frac{1}{\nu}\left( 1-\frac{\nu}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2. \end{array} $$

The rest of the proof follows the lines of the proof of Corollary 2.2.□

Proof of Corollary 2.4.

As in the proof of Corollary 2.3, the estimator $\hat {f}_{\nu _{0}, \text { wor}}(y;h)$ is design-unbiased for f_U(y;h) and its design variance is given by

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\{\hat{f}_{\nu_0, \text{ wor}}(y)|\mathbf{Y}_{\text{U}}\}&=&\left( \frac{1}{\nu_0}-\frac{1}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2. \end{array} $$

(A.1)

Now, using $\nu _{0}=\mathrm {E}_{\nu }(\nu )=N\left (1-\left \{1-1/N\right \}^{n}\right )$ to substitute in (A.1) and simplifying, we get

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\{\hat{f}_{\nu_0, \text{ wor}}(y)|\mathbf{Y}_{\text{U}}\}&=&\frac{1}{n}\mathcal{C}^{**}_{N,n}\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2,\end{array} $$

where $\mathcal {C}^{**}_{N,n}=(n/N)(1-1/N)^{n}/\{1-(1-1/N)^{n}\}$. The rest of the proof follows the lines of the proof of Corollary 2.2.□

Proof of Corollary 2.5.

Given λ, the estimator $\hat {f}_{\nu _{r}, \text { wor}}(y;h)$ is design-unbiased for f_U(y;h) and its design variance is given by

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\{\hat{f}_{\nu_r, \text{ wor}}(y)|\mathbf{Y}_{\text{U}}\} &=&\left( \frac{1}{\nu_r}-\frac{1}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2. \end{array} $$

(A.2)

Further, note that since λ = 1 with probability p = (ν₀ −⌊ν₀⌋), and λ = 0 with probability (1 − p), we can write

$$ \begin{array}{@{}rcl@{}}\mathrm{E}_\lambda \left( \frac{1}{\nu_r}\right)&=&\mathrm{E}_\lambda \left( \frac{1}{(1-\lambda)\lfloor \nu_0 \rfloor+\lambda\left\lceil \nu_0 \right\rceil}\right)\\ &=&\frac{p}{\left\lceil \nu_0 \right\rceil}+\frac{(1-p)}{\lfloor \nu_0 \rfloor}=\frac{1-\nu_0+2\lfloor \nu_0 \rfloor}{\lfloor \nu_0 \rfloor\left\lceil \nu_0 \right\rceil}. \end{array} $$

(A.3)

To reach the second equality in (A.3), use the fact that $\left \lceil \nu _{0} \right \rceil -\lfloor \nu _{0} \rfloor =1$ and some basic algebra. Using (A.2) and (A.3), we get

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{P}}\{\hat{f}_{\nu_r, \text{ wor}}(y)|\mathbf{Y}_{\text{U}}\}&=&\frac{1}{n}\left( \mathcal{C}^{***}_{N,n}-\frac{n}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[K_h\left( y-Y_i\right)-f_{\text{U}}(y)\right]^2. \end{array} $$

The rest of the proof follows the lines of the proof of Corollary 2.2.□

Proof of Lemma 4.1.

Observe that,

$$ \begin{array}{@{}rcl@{}}\mathrm{E}\left\{\frac{1}{\nu}\sum\limits_{i=1}^{\nu}\hat{f}_{-i}(Y^{*}_i)\right\}&=& \mathrm{E}_{\nu}\left\{\mathrm{E}_{\mathcal{\xi}}\left[\frac{1}{\nu}\sum\limits_{i=1}^{{\nu}}\hat{f}_{-i}(Y^{*}_i)\Big|\nu\right]\right\}.\end{array} $$

But,

$$ \begin{array}{@{}rcl@{}}\mathrm{E}_{\mathcal{\xi}}\left[\frac{1}{\nu}\sum\limits_{i=1}^{\nu}\hat{f}_{-i}(Y^{*}_i)\Big|\nu\right]&=& \mathrm{E}_{\mathcal{\xi}}\left[\frac{1}{\nu(\nu-1)}\sum\limits_{i=1}^{\nu}\sum\limits_{j\not=i}^{\nu}K_h(Y^{*}_i-Y^{*}_j)\Big|\nu \right]\\ &\overset{iid}{=}& \mathrm{E}_{\mathcal{\xi}}\left[K_h(Y^{*}_1-Y^{*}_2)\right]\\ &=& \mathrm{E}_{Y^{*}_1}\left\{\mathrm{E}_{Y^{*}_2}\left[K_h(Y^{*}_1-Y^{*}_2)\right]\right\}\\ &=& \mathrm{E}_{Y^{*}_1}\left\{\int K_h(Y^{*}_1-y)f(y) dy\right\}.\end{array} $$

Therefore,

$$ \begin{array}{@{}rcl@{}}\mathrm{E}\left\{\frac{1}{\nu}\sum\limits_{i=1}^{\nu}\hat{f}_{-i}(Y^{*}_i)\right\} &=& \mathrm{E}_{\mathcal{\xi}}\left\{\int K_h(Y^{*}_1-y)f(y) dy\right\}\\&=&\mathrm{E}_{\mathcal{\xi}}\left\{\int \hat{f}_{\nu}(y)f(y) dy\Big|\nu\right\}=\mathrm{E}\left\{\int \hat{f}_{\nu}(y)f(y) dy\right\},\end{array} $$

and the proof is complete. □

Proof 13 (Proof of (4.5)).

$$ \begin{array}{@{}rcl@{}}d_{\hat{f}^{\prime\prime}_\nu}&=&{\int}_{\mathbb{R}}\left\{(\nu h^{3})^{-1}\sum\limits_{j=1}^{\nu}K^{\prime\prime}\left( \frac{y-y^{*}_j}{h}\right)\right\}^2dy\\ &=&(\nu h^{3})^{-2}{\int}_{\mathbb{R}}\left\{\sum\limits_{i=1}^{\nu}\left[K^{\prime\prime}\left( \frac{y-y^{*}_i}{h}\right)\right]^2+2\underset{i <j}{\sum\sum}K^{\prime\prime}\left( \frac{y-y^{*}_i}{h}\right)K^{\prime\prime}\left( \frac{y-y^{*}_j}{h}\right)\right\}dy\\ &=&(\nu h^{3})^{-2}\sum\limits_{i=1}^{\nu}{\int}_{\mathbb{R}}\left[K^{\prime\prime}\left( \frac{y-y^{*}_i}{h}\right)\right]^2 dy + 2(\nu h^{3})^{-2}\underset{i <j}{\sum\sum}{\int}_{\mathbb{R}}K^{\prime\prime}\left( \frac{y-y^{*}_i}{h}\right)K^{\prime\prime} \left( \frac{y-y^{*}_j}{h}\right) dy\\ &=&{\nu}^{-1}h^{-6}{\int}_{\mathbb{R}}\left[K^{\prime\prime}(z)\right]^2 hdz+2{\nu}^{-2}h^{-6}\underset{i <j}{\sum\sum}{\int}_{\mathbb{R}}K^{\prime\prime}(z)K^{\prime\prime}\left( z+\frac{y^{*}_i-y^{*}_j}{h}\right) hdz\\ &=&({\nu}h^{5})^{-1}d_{K^{\prime\prime}}+2{\nu}^{-2}h^{-5}\underset{i <j}{\sum\sum}\phi(c_{ij}), \end{array} $$

where $\phi (c_{ij})={\int \limits }_{\mathbb {R}}K^{\prime \prime }(z)K^{\prime \prime }(z+h^{-1}\{y^{*}_{i}-y^{*}_{j}\})dz$. □

Proofs of (6.7), (6.8) and (6.9).

For (6.7), notice that

$$ \begin{array}{@{}rcl@{}}\mathbb{E}(Q)&=&\mathrm{E}_{\mathcal{\xi}}\left[\mathrm{E}_\nu \left\{\mathrm{E}_{\mathcal{D}}\left[\frac{1}{\nu}\sum\limits_{i=1}^{\nu}Y^{*}_iK_b(x-X^{*}_i)|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}},\nu\right]|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right]\\ &=&\mathrm{E}_{\mathcal{\xi}}\left[\mathrm{E}_\nu \left\{\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right]\\ &=&\mathrm{E}_{\mathcal{\xi}}\left[\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)\right]\\ &=&\mathrm{E}_{\mathcal{\xi}}\left[Y_1K_b(x-X_1)\right] \end{array} $$

(A.4)

$$ \begin{array}{@{}rcl@{}} &=&\iint_{\mathbb{R}}y_1K_b(x-x_1)t(y_1|x_1)g(x_1)dy_1dx_1\\ &=&{\int}_{\mathbb{R}}m(x_1)K_b(x-x_1)g(x_1)dx_1\\ &=&{\int}_{\mathbb{R}}m(x+bu)K(u)g(x+bu)du\\ &=&{\int}_{\mathbb{R}}K(u)[m(x)+bu \ m^{\prime}(x)+\frac{1}{2}b^2u^2m^{\prime\prime}(x)+o(b^2)][g(x)\\&&+bu g^{\prime}(x)+\frac{1}{2}b^2u^2g^{\prime\prime}(x)+o(b^2)]du\\ &=&m(x)g(x)+b^2m^{\prime}(x)g^{\prime}(x)c_K+\frac{1}{2}b^2m(x)g^{\prime\prime}(x)c_K\\&&+\frac{1}{2}b^2m^{\prime\prime}(x)g(x)c_K+o(b^2)\\ &=& m(x)g(x)+\frac{1}{2}b^2[2m^{\prime}(x)g^{\prime}(x)+m(x)g^{\prime\prime}(x)\\&&+m^{\prime\prime}(x)g(x)]c_K+o(b^2), \end{array} $$

(A.5)

Next, we prove (6.8) as follows. Note that

$$ \begin{array}{@{}rcl@{}}\mathbb{V}(Q) &=& \mathrm{E}_{\mathcal{\xi}}\left[\mathrm{V}_{\mathcal{P}}\left\{Q|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right] + \mathrm{V}_{\mathcal{\xi}}\left[\mathrm{E}_{\mathcal{P}}\left\{Q|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right] =: L_1 + L_2.\end{array} $$

(A.6)

Using (A.4), we have

$$ \begin{array}{@{}rcl@{}}L_2&=&\frac{1}{N}\mathrm{V}_{\mathcal{\xi}}\left[Y_1K_b(x-X_1)\right]=\frac{1}{N}\mathrm{E}_{\mathcal{\xi}}\left[Y_1K_b(x-X_1)\right]^2\\&&-\frac{1}{N}E^2_{\mathcal{\xi}}\left[Y_1K_b(x-X_1)\right] =: L_{21}-L_{22}.\end{array} $$

Observe that,

$$ \begin{array}{@{}rcl@{}}L_{21}&=&\frac{1}{N}\iint_{\mathbb{R}}y^2_1K^2_b(x-x_1)t(y_1|x_1)g(x_1)dy_1dx_1\\ &=&\frac{1}{N}{\int}_{\mathbb{R}}K^2_b(x-x_1)g(x_1)\left[{\int}_{\mathbb{R}}y^2_1t(y_1|x_1)dy_1\right]dx_1\\ &=&\frac{1}{N}{\int}_{\mathbb{R}}\left[\sigma^2(x_1)+m^2(x_1)\right]K^2_b(x-x_1)g(x_1)dx_1\\ &=&\frac{1}{Nb}{\int}_{\mathbb{R}}\left[\sigma^2(x+bu)+m^2(x+bu)\right]K^2(u)g(x+bu)du\\ &=&\frac{1}{Nb}\left[\sigma^2(x)g(x)+m^2(x)g(x)\right]d_K+o\left( \frac{1}{Nb}\right).\end{array} $$

From (A.5), we have

$$ \begin{array}{@{}rcl@{}}L_{22}&=&\frac{1}{N}\left[m(x)g(x)+O(b^2)\right]^2.\end{array} $$

Therefore,

$$ \begin{array}{@{}rcl@{}}L_{2}&=&\frac{1}{Nb}\left[\sigma^2(x)g(x)+m^2(x)g(x)\right]d_K-\frac{1}{N}\left[m(x)g(x)+O(b^2)\right]^2+o\left( \frac{1}{Nb}\right)\\ &=&\frac{1}{Nb}\left[\sigma^2(x)+m^2(x)\right]g(x)d_K+o\left( \frac{1}{Nb}\right). \end{array} $$

(A.7)

Moreover, given ν, Q can be seen as the sample mean of a fixed size simple random sample drawn without replacement from the finite population. Therefore, Q is design-unbiased for the finite population mean of the variable Y K_b(x − X), and its design variance is given by

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{D}}\left[Q|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}},\nu\right]&=&\left( \frac{1}{\nu}-\frac{1}{N}\right)\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[Y_iK_b(x-X_i)-\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)\right]^2,\end{array} $$

and, hence,

$$ \begin{array}{@{}rcl@{}}\mathrm{V}_{\mathcal{P}}\left[Q|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right]&=&\left[\mathrm{E}_\nu \left( \frac{1}{\nu}\right)-\frac{1}{N}\right]\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[Y_iK_b(x-X_i)-\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)\right]^2\\&&+\mathrm{V}_\nu \left[\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)\right]\\ &=&\left[\mathrm{E}_\nu \left( \frac{1}{\nu}\right) - \frac{1}{N}\right]\frac{1}{N-1}\sum\limits_{i=1}^{N}\left[Y_iK_b(x-X_i)-\frac{1}{N}\sum\limits_{i=1}^{N}Y_iK_b(x-X_i)\right]^2. \end{array} $$

Consequently,

$$ \begin{array}{@{}rcl@{}}L_1&=&\left[\mathrm{E}_\nu \left( \frac{1}{\nu}\right)-\frac{1}{N}\right]\mathrm{V}_{\mathcal{\xi}}\left\{Y_1K_b(x-X_1)\right\}\\ &=&\frac{1}{nb}\mathcal{C}_{N,n}\left[\sigma^2(x)+m^2(x)\right]g(x)d_K+o\left( \frac{1}{Nb}\right), \end{array} $$

(A.8)

Using (A.7) and (A.8) in (A.6), we get (6.8).

For (6.9), observe that

$$ \begin{array}{@{}rcl@{}} \mathbb{E}(QW)&=&\mathbb{E}\left[\left\{\frac{1}{\nu}\sum\limits_{i=1}^{\nu}y^{*}_iK_b(x-x^{*}_i)\right\}\left\{\frac{1}{\nu}\sum\limits_{i=1}^{\nu}K_b(x-x^{*}_i)\right\}\right]\\ &=&\mathbb{E}\left[\frac{1}{\nu^2}\sum\limits_{i=1}^{\nu}y^{*}_iK^2_b(x-x^{*}_i)\right]+\mathbb{E}\left[\frac{1}{\nu^2}\underset{i\not=j}{\sum\sum}y^{*}_iK_b(x-x^{*}_i)K_b(x-x^{*}_j)\right]\\ &=:&B_1+B_2. \end{array} $$

(A.9)

First,

$$ \begin{array}{@{}rcl@{}}B_1&=&\mathrm{E}_\nu \left( \frac{1}{\nu}\right)\mathrm{E}_{\mathcal{\xi}}\left[Y_1K^2_b(x-X_1)\right]\\ &=&\mathrm{E}_\nu \left( \frac{1}{\nu}\right){\int}_{\mathbb{R}}m(x_1)K^2_b(x-x_1)g(x_1)dx_1\\ &=&\mathrm{E}_\nu \left( \frac{1}{\nu}\right)\left[\frac{1}{b}m(x)g(x)d_K+o\left( \frac{1}{Nb}\right)\right]\\ &=&\frac{1}{nb}\left[\mathcal{C}_{N,n}+\frac{n}{N}\right]m(x)g(x)d_K+o\left( \frac{1}{Nb}\right). \end{array} $$

(A.10)

Second,

$$ \begin{array}{@{}rcl@{}} B_2&=&\mathrm{E}_{\mathcal{\xi}}\left[\mathrm{E}_\nu \left\{\mathrm{E}_{\mathcal{D}}\left[\frac{1}{\nu^2}\underset{i,j\in U, i\not=j}{\sum\sum}I_iI_jY_iK_b(x-X_i)K_b(x-X_j)|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}},\nu\right]|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right]\\ &=&\mathrm{E}_{\mathcal{\xi}}\left[\mathrm{E}_\nu \left\{\frac{1}{\nu^2}\underset{i,j\in U, i\not=j}{\sum\sum}\pi_{ij}Y_iK_b(x-X_i)K_b(x-X_j)|\mathbf{X}_{\text{U}},\mathbf{Y}_{\text{U}}\right\}\right]\\ &=&\mathrm{E}_{\mathcal{\xi}}\left[\mathrm{E}_\nu \left\{\frac{\nu(\nu-1)}{\nu^2N(N-1)}\right\}\underset{i,j\in U, i\not=j}{\sum\sum}Y_iK_b(x-X_i)K_b(x-X_j)\right]\\ &=&\left\{1-\mathrm{E}_\nu \left( \frac{1}{\nu}\right)\right\}\frac{1}{N(N-1)}\underset{i,j\in U, i\not=j}{\sum\sum}\mathrm{E}_{\mathcal{\xi}}\left[Y_iK_b(x-X_i)\right]\mathrm{E}_{\mathcal{\xi}}\left[K_b(x-X_j)\right]\\ &=&\left\{1-\mathrm{E}_\nu \left( \frac{1}{\nu}\right)\right\}\left[m(x)g(x)+\frac{1}{2}b^2\{m(x)g^{\prime\prime}(x)+2m^{\prime}(x)g^{\prime}(x)+m^{\prime\prime}(x)g(x)\}c_K+o(b^2)\right]\\ &&\qquad\qquad\qquad\qquad\times \left[g(x)+\frac{1}{2}b^2g^{\prime\prime}(x)c_K+o(b^2)\right]\\ &=& \left\{1 - \frac{1}{n}\left( \mathcal{C}_{N,n} + \frac{n}{N}\right)\right\}\left[m(x)g^2(x) + \frac{1}{2}b^2\left\{2m(x)g^{\prime\prime}(x) + 2m^{\prime}(x)g^{\prime}(x) + m^{\prime\prime}(x)g(x)\right\}g(x)c_K\right.\\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad+o(b^2)\vphantom{\frac{1}{2}}\right]\\ &=&m(x)g^2(x)+\frac{1}{2}b^2\left[2m(x)g^{\prime\prime}(x)+2m^{\prime}(x)g^{\prime}(x)+m^{\prime\prime}(x)g(x)\right]g(x)c_K+o(b^2)\\&&-O\left( \frac{1}{n}\right). \end{array} $$

(A.11)

Finally, using (A.10) and (A.11) in (A.9), we get (6.9).□