Small area estimation under transformed nested-error regression models

Abstract

The empirical best linear unbiased prediction (EBLUP) based on the nested error regression model (Battese et al. in J Am Stat Assoc 83:28–36, 1988, NER) has been widely used for small area mean estimation. Its so-called optimality largely depends on the normality of the corresponding area level and unit level error terms. To allow departure from normality, we propose a transformed NER model with an invertible transformation, and employ the maximum likelihood method to estimate the underlying parameters of the transformed NER model. Motivated by Duan’s (J Am Stat Assoc 78:605–610, 1983) smearing estimator, we propose two small area mean estimators depending on whether all the population covariates or only the population covariate means are available in addition to sample covariates. We conduct two design-based simulation studies to investigate their finite-sample performance. The simulation results indicate that compared with existing methods such as the empirical best linear unbiased prediction, the proposed estimators are nearly the same reliable when the NER model is valid and become more reliable in general when the NER model is violated. In particular, our method does benefit from incorporating auxiliary covariate information.

Appendix

To study the consistency of the MLEs, we make the following assumptions.

• (C1) There exists constants $0<c_1 <c_2$ such that the transformed response $h(y; \lambda _0)$ with $\lambda _0\in [c_1, c_2]$ and the transformation in (3) satisfy the NER model in (1).

• (C2) The small area number m keeps fixed and that as n goes to infinity, $n_k/n=\rho _k +o(1)$ for $\rho _k\in (0,1)$.

• (C3) Assume that $\Sigma _x=\sum _{k=1}^m \rho _k \Sigma _{xk}$ is nonsingular, where $\Sigma _{xk}$ is the variance matrix of $\mathbf{x}_{kj}$ in small area k.

Apparently condition (C1) requires $y_{kj}$ is an increasing function of $\mathbf{x}_{kj}^{\mathrm {\scriptscriptstyle \top }}\beta + v_k +\varepsilon _{kj}$ since $h(y; \lambda _0)$ is increasing when $\lambda _0>0$. If $y_{kj}$ is an decreasing function of $\mathbf{x}_{kj}^{\mathrm {\scriptscriptstyle \top }}\beta + v_k +\varepsilon _{kj}$, it must be an increasing function of $\mathbf{x}_{kj}^{\mathrm {\scriptscriptstyle \top }}(-\beta ) +(- v_k)+( - \varepsilon _{kj})$. Then condition (C1) is still satisfied except that the parameter $\beta$ has an opposite sign. Condition (C2) is imposed to provide a justification of the proposed estimation method for $\lambda$. Also we can check whether there exists system bias in the proposed method although the sample sizes $n_k$ are generally very small in the literature of small area estimation.

Proof of Theorem 1

We begin by proving the consistency of ${\hat{\lambda }}$. Let $(\lambda _0, \gamma _0)$ be the true value of $(\lambda , \gamma )$. It is sufficient to show that as n is large, (1)

$$\begin{aligned} S= \frac{1}{n} \frac{\partial \ell (\lambda _0, \gamma _0)}{\partial \lambda } = o_p(1) \end{aligned}$$

(19)

and (2) $\frac{1}{n} \frac{\partial ^2 \ell (\lambda _0, \gamma _0)}{\partial \lambda \partial \lambda ^{\mathrm {\scriptscriptstyle \top }}}$ is positive definite. We shall prove only (1) since (2) can be proved along the same line of proving (1) but with more tedious derivation.

From (7), we have

$$\begin{aligned} S = \frac{1}{n} \frac{\partial \ell (\lambda _0, \gamma _0)}{\partial \lambda } = \frac{1}{\lambda _0}+ \frac{1}{n} \sum _{k=1}^m \sum _{j=1}^{n_k}\log (| y_{kj}|) -\frac{1}{2} \frac{\partial {\tilde{\sigma }}_e^2(\lambda _0, \gamma _0)/\partial \lambda }{{\tilde{\sigma }}_e^2(\lambda _0, \gamma _0) }. \end{aligned}$$

To simplify S, we need to investigate ${\tilde{\sigma }}_e^2(\lambda _0, \gamma _0)$ and $\partial {\tilde{\sigma }}_e^2(\lambda _0, \gamma _0)/\partial \lambda$. It follows from $\mathbf{A}_k(\gamma )= \mathbf{I}_{n_k} - \frac{\gamma }{1+\gamma n_k}\mathbf{1}_{n_k} \mathbf{1}_{n_k}^{{\mathrm {\scriptscriptstyle \top }}}$ that for any fixed $\gamma >0$,

$$\begin{aligned} \frac{1}{n_r}\mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{x}_{r}= & {} \frac{1}{n_r}\sum _{j=1}^{n_r} \mathbf{x}_{rj}\mathbf{x}_{rj}^{\mathrm {\scriptscriptstyle \top }}- \frac{n_r \gamma }{1+\gamma n_r} {\bar{\mathbf{x}}}_{r}{\bar{\mathbf{x}}}_{r}^{\mathrm {\scriptscriptstyle \top }}\\= & {} \frac{1}{n_r} \sum _{i=1}^{n_r} ( \mathbf{x}_{rj}-{\bar{x}}_r)( \mathbf{x}_{rj}-{\bar{x}}_r)^{\mathrm {\scriptscriptstyle \top }}+ \frac{ {\bar{\mathbf{x}}}_{r}{\bar{\mathbf{x}}}_{r}^{\mathrm {\scriptscriptstyle \top }}}{1+n_r\gamma } \\= & {} \Sigma _{xr} + O_p(n^{-1/2}), \end{aligned}$$

$$\begin{aligned} \frac{1}{n_r}\mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{y}_{r}^{(\lambda _0)}= & {} \frac{1}{n_r}\sum _{j=1}^{n_r} \mathbf{x}_{rj}y_{rj}^{(\lambda _0)} - \frac{n_r \gamma }{1+\gamma n_r} {\bar{\mathbf{x}}}_{r}{\bar{y}}_{r}^{(\lambda _0)} \\= & {} \frac{1}{n_r}\sum _{j=1}^{n_r} (\mathbf{x}_{rj} - {\bar{x}}_r)(y_{rj}^{(\lambda _0)} - {\bar{y}}_{r}^{(\lambda _0)}) + \frac{ {\bar{\mathbf{x}}}_{r}{\bar{y}}_{r}^{(\lambda _0)} }{1+\gamma n_r} \\= & {} \Sigma _{xr}\beta _0 + O_p(n^{-1/2}), \end{aligned}$$

$$\begin{aligned} \frac{1}{n_r}\{\mathbf{y}_{r}^{(\lambda )}\}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{y}_{r}^{(\lambda )}= & {} \frac{1}{n_r}\sum _{j=1}^{n_r} \{y_{rj}^{(\lambda )}\}^2 - \frac{n_r \gamma }{1+\gamma n_r} \{{\bar{y}}_{r}^{(\lambda )}\}^2 \\= & {} \frac{1}{n_r}\sum _{j=1}^{n_r} \{ (\mathbf{x}_{rj} - {\bar{\mathbf{x}}}_r)^{\mathrm {\scriptscriptstyle \top }}\beta _0 + ( \varepsilon _{rj} -{\bar{\varepsilon }}_r) \} ^2 + O_p(n^{-1 }) \\= & {} \beta _0^{{\mathrm {\scriptscriptstyle \top }}}\Sigma _{xr}\beta _0 + \sigma _e^2 + O_p(n^{-1/2}). \end{aligned}$$

The above three equalities imply that

$$\begin{aligned} \frac{1}{n } \sum _{r=1}^m \mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{x}_{r}= & {} \sum _{r=1}^m \rho _r \Sigma _{xr} + O_p(n^{-1/2}), \end{aligned}$$

(20)

$$\begin{aligned} \frac{1}{n } \sum _{r=1}^m \mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{y}_{r}^{(\lambda _0)}= & {} \sum _{r=1}^m \rho _r \Sigma _{xr}\beta _0 + O_p(n^{-1/2}), \end{aligned}$$

(21)

$$\begin{aligned} \frac{1}{n } \sum _{r=1}^m \{\mathbf{y}_{r}^{(\lambda )}\}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{y}_{r}^{(\lambda )}= & {} \beta _0^{{\mathrm {\scriptscriptstyle \top }}} \sum _{r=1}^m \rho _r \Sigma _{xr}\beta _0 + \sigma _e^2 + O_p(n^{-1/2}). \end{aligned}$$

By these three equalities, we immediately have

$$\begin{aligned} {\tilde{\sigma }}_e^2(\lambda _0, \gamma _0)= & {} \sigma _e^2 + O_p(n^{-1/2}). \end{aligned}$$

(22)

To calculate $\partial {\tilde{\sigma }}_e^2(\lambda _0, \gamma _0)/\partial \lambda$, we denote $z_{kj} \equiv \partial y_{rj}^{(\lambda _0)} / \partial \lambda = y_{rj}^{(\lambda _0)} \log (|y_{kj}|).$ Then

$$\begin{aligned} \frac{\partial }{\partial \lambda } \frac{1}{n }\sum _{s=1}^m \mathbf{x}_{s}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_s(\gamma )\mathbf{y}_s^{(\lambda _0)}= & {} \sum _{r=1}^m \frac{n_r}{n} \Big [\frac{1}{n_r}\sum _{j=1}^{n_r} (\mathbf{x}_{rj} - {\bar{x}}_r)(y_{rj}^{(\lambda )} - {\bar{y}}_{r}^{(\lambda )}) + \frac{ {\bar{\mathbf{x}}}_{r}{\bar{y}}_{r}^{(\lambda )} }{1+\gamma n_r} \Big ] \\= & {} \sum _{r=1}^m \rho _r {\mathbb {C}\mathrm{ov}}(\mathbf{x}_{rj}, z_{rj}) + O_p(n^{-1/2}) \\= & {} \Sigma _{xz}+ O_p(n^{-1/2}), \\ \frac{\partial }{\partial \lambda } \frac{1}{n } \sum _{r=1}^m \{\mathbf{y}_{r}^{(\lambda )}\}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{y}_{r}^{(\lambda )}= & {} \sum _{r=1}^m \rho _r \{ 2\beta _0^{{\mathrm {\scriptscriptstyle \top }}} {\mathbb {C}\mathrm{ov}}(\mathbf{x}_{rj}, z_{rj}) + 2 {\mathbb {C}\mathrm{ov}}(z_{rj}, \varepsilon _{rj} ) \}\\= & {} 2\beta _0^{{\mathrm {\scriptscriptstyle \top }}}\Sigma _{xz} + 2 \Sigma _{z\varepsilon } + O_p(n^{-1/2}). \end{aligned}$$

It can be found that

$$\begin{aligned} \tilde{\sigma }_e^2(\lambda , \gamma )= & {} \frac{1}{n} \sum _{k=1}^m \{\mathbf{y}_k^{(\lambda )} \}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_k(\gamma ) \mathbf{y}_k^{(\lambda )} \\&- \{\frac{1}{n}\sum _{s=1}^m\mathbf{x}_{s}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_s(\gamma )\mathbf{y}_s^{(\lambda )}\}^{\mathrm {\scriptscriptstyle \top }}\left\{ \frac{1}{n}\sum _{r=1}^m\mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma )\mathbf{x}_{r}\right\} ^{-1} \frac{1}{n}\sum _{s=1}^m\mathbf{x}_{s}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_s(\gamma )\mathbf{y}_s^{(\lambda )}. \end{aligned}$$

Accordingly

$$\begin{aligned} \frac{\partial }{\partial \lambda } \tilde{\sigma }_e^2(\lambda _0, \gamma _0)= & {} 2 \frac{1}{n} \sum _{k=1}^m \{ \frac{\partial }{\partial \lambda } \mathbf{y}_k^{(\lambda _0)} \}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_k(\gamma _0) \mathbf{y}_k^{(\lambda _0)} - 2\{\frac{1}{n}\sum _{s=1}^m\mathbf{x}_{s}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_s(\gamma _0)\frac{\partial }{\partial \lambda }\mathbf{y}_s^{(\lambda _0)}\}^{\mathrm {\scriptscriptstyle \top }}\nonumber \\&\times \left\{ \frac{1}{n}\sum _{r=1}^m\mathbf{x}_{r}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_r(\gamma _0)\mathbf{x}_{r}\right\} ^{-1} \frac{1}{n}\sum _{s=1}^m\mathbf{x}_{s}^{{\mathrm {\scriptscriptstyle \top }}}\mathbf{A}_s(\gamma _0)\mathbf{y}_s^{(\lambda _0)} + o_p(1) \nonumber \\= & {} 2\beta _0^{{\mathrm {\scriptscriptstyle \top }}}\Sigma _{xz} + 2 \Sigma _{z\varepsilon } - 2\beta _0^{\mathrm {\scriptscriptstyle \top }}\Sigma _x \Sigma _x^{-1} \Sigma _{xz} + o_p(1) \nonumber \\= & {} 2 \Sigma _{z\varepsilon } + o_p(1). \end{aligned}$$

(23)

Putting (23) into (19), we obtain

$$\begin{aligned} S= & {} \frac{1}{\lambda _0}+ \sum _{k=1}^m \rho _k {\mathbb {E}}\{\log (| y_{kj}|)\} -\frac{1}{2} \frac{\Sigma _{z\varepsilon } }{ \sigma _e^2 }+ o_p(1). \end{aligned}$$

To prove $S=o_p(1)$, it is sufficient to prove for $r=1,2,\ldots , m$ that

$$\begin{aligned} \frac{1}{\lambda _0}+{\mathbb {E}}\{\log (| y_{rj}|)\} - \frac{1}{ \sigma _e^2}{\mathbb {C}\mathrm{ov}}(z_{rj}, \varepsilon _{rj} ) = 0, \end{aligned}$$

which is true as shown in Lemma 1. This proves the consistency of ${\hat{\lambda }}$.

Note that Eqs. (20) and (21) holds for any $\gamma >0$. Since ${\hat{\lambda }}$ is consistent, we immediately obtain that ${\hat{\beta }}$ is an consistent estimator of $\beta _0$. By re-studying the proof of (22), we find that it is still true when $\gamma _0$ is replaced by any positive $\gamma$ and $\lambda _0$ is replaced by its consistent estimator ${\hat{\lambda }}$. This completes the proof of Theorem 1. $\square$

Lemma 1

Under the assumptions for the transformed NER model, it holds that

$$\begin{aligned} \frac{1}{\lambda _0}+{\mathbb {E}}_{x, \varepsilon }\left\{ \log (| y_{kj}|)\right\} - \frac{1}{\sigma _e^2}{\mathbb {C}\mathrm{ov}}_{x, \varepsilon }(z_{kj}, \varepsilon _{kj} ) = 0, \end{aligned}$$

(24)

where ${\mathbb {E}}_{x, \varepsilon }$ and ${\mathbb {C}\mathrm{ov}}_{x, \varepsilon }$ denote expectation and covariance conditionally on $(\mathbf{x}, \varepsilon )$.

Proof

Denote the left-hand side of Eq. (24) by $\Delta$. By assumption, the response $y_{kj}^{(\lambda _0)}$ conditionally on $v_k$ and $\mathbf{x}_{kj}$ follows $N(\mathbf{x}_{kj}^{\mathrm {\scriptscriptstyle \top }}\beta +v_k, \sigma _e^2)$. Since ${\mathbb {E}}(\varepsilon _{kj})=0$ and $z_{kj} = y_{rj}^{(\lambda _0)} \log (|y_{kj}|) = \lambda _0^{-1} y_{rj}^{(\lambda _0)} \log (|y_{kj}^{(\lambda _0)} |),$ it follows that

$$\begin{aligned} \Delta= & {} \frac{1}{\lambda _0}+ \frac{1}{\lambda _0} \int _{-\infty }^{\infty } \log (| t|) \phi \left( \frac{t-a}{\sigma _e}\right) \frac{dt}{\sigma _e} - \frac{1}{\lambda _0} \int _{-\infty }^{\infty } t(t-a) \log (| t|) \phi \left( \frac{t-a}{\sigma _e}\right) \frac{dt}{\sigma _e^3}, \end{aligned}$$

where we denote $a=\mathbf{x}_{kj}^{\mathrm {\scriptscriptstyle \top }}\beta +v_k$ for short. By transforming $u=t/\sigma _e$ and $b=a/\sigma _e$, we further have

$$\begin{aligned} \Delta= & {} \frac{1}{\lambda _0}+ \frac{1}{\lambda _0} \int _{-\infty }^{\infty } \log (|u|\sigma _e) \phi (u-b) du - \frac{1}{\lambda _0} \int _{-\infty }^{\infty } u(u-b) \log (|u|\sigma _e) \phi (u-b) du \\= & {} \frac{1}{\lambda _0}+ \frac{1}{\lambda _0} \int _{-\infty }^{\infty } \{1- u(u-b) \} \log (|u| ) \phi (u-b) du. \end{aligned}$$

Using $d\{\phi (u-b)u\}= \{1- u(u-b) \} \phi (u) du$ and integration by parts, we have

$$\begin{aligned} \Delta= & {} \frac{1}{\lambda _0}+ \frac{1}{\lambda _0} \int _{-\infty }^{\infty } \log (|u| ) d\{\phi (u-b)u\} \\= & {} \frac{1}{\lambda _0}+ \frac{1}{\lambda _0} \log (|u| ) \phi (u-b)u \Big |_{-\infty }^{\infty } - \frac{1}{\lambda _0} \int _{-\infty }^{\infty } \phi (u-b) du \\= & {} 0. \end{aligned}$$

$\square$