Robust density power divergence estimates for panel data models

Abstract

The panel data regression models have become one of the most widely applied statistical approaches in different fields of research, including social, behavioral, environmental sciences, and econometrics. However, traditional least-squares-based techniques frequently used for panel data models are vulnerable to the adverse effects of data contamination or outlying observations that may result in biased and inefficient estimates and misleading statistical inference. In this study, we propose a minimum density power divergence estimation procedure for panel data regression models with random effects to achieve robustness against outliers. The robustness, as well as the asymptotic properties of the proposed estimator, are rigorously established. The finite-sample properties of the proposed method are investigated through an extensive simulation study and an application to climate data in Oman. Our results demonstrate that the proposed estimator exhibits improved performance over some traditional and robust methods in the presence of data contamination.

Appendices

Appendix

A The Estimating Equations

Using Equation (17) of Supplementary Material, the DPD measure in Eq. (8) can be simplified as

$$\begin{aligned} \begin{aligned} {\widehat{d}}_\gamma (f_\theta , g)&= (2\pi )^{-\frac{T \gamma }{2}} |{{\Omega }}|^{-\frac{\gamma }{2}} (1+\gamma )^{-\frac{1}{2}} - \frac{1+\gamma }{N \gamma } \sum _{i=1}^{N} f_\theta ^{\gamma }(y_i | x_i) + c(\gamma )\\&= (2\pi )^{-\frac{T \gamma }{2}} |{{\Omega }}|^{-\frac{\gamma }{2}} (1+\gamma )^{-\frac{1}{2}} \left[ 1 - \frac{(1+\gamma )^{3/2} }{N\gamma } \sum _{i=1}^{N}\exp \left[ -\frac{\gamma }{2} B_i\right] \right] + c(\gamma ), \end{aligned} \end{aligned}$$

(18)

where $B_i = (y_i - x_i \beta )^T {{\Omega }}^{-1} (y_i - x_i \beta )$. Using the Sherman–Morrison formula (Sherman and Morrison 1950), we get

$$\begin{aligned} {{\Omega }}^{-1} = \frac{1}{\sigma _\epsilon ^2} {{I}}_T - \frac{\sigma _\alpha ^2 e_T e_T^T}{\sigma _\epsilon ^2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )},\ \ \ |{{\Omega }}| = \sigma _\epsilon ^{2(T-1)} ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 ). \end{aligned}$$

(19)

It further simplifies $B_i$ as follows

$$\begin{aligned} B_i = \frac{1}{\sigma _\epsilon ^2}\sum _{t=1}^{T} (y_{it} - x_{it} \beta )^2 - \frac{\sigma _\alpha ^2 }{\sigma _\epsilon ^2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} \left\{ \sum _{t=1}^{T} (y_{it} - x_{it} \beta ) \right\} ^2 . \end{aligned}$$

(20)

The estimating equations of $\theta$ is obtained from equation $\frac{\partial }{\partial \theta } {\widehat{d}}_\gamma (f_\theta , g) =0$, and the equations corresponding to $\beta$, $\sigma _\alpha ^2$ and $\sigma _\epsilon ^2$ are simplified as

$$\begin{aligned} \begin{aligned}&\sum _{i=1}^{N} \sum _{t=1}^{T} x_{it} (y_{it} - x_{it} \beta ) \exp \left[ -\frac{\gamma }{2} B_i\right] = \frac{T \sigma _\alpha ^2 }{( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} \sum _{i=1}^{N} \sum _{t=1}^{T} {\bar{x}}_i (y_{it} - x_{it} \beta ) \exp \left[ -\frac{\gamma }{2} B_i\right] ,\\&\gamma T |{{\Omega }}|^{-1} \sigma _\epsilon ^{2(T-1)} \left\{ (1+\gamma )^{-\frac{1}{2}} - \frac{1+\gamma }{N \gamma } \sum _{i=1}^{N}\exp \left[ -\frac{\gamma }{2} B_i\right] \right\} \\&\quad = - \frac{(1+\gamma )}{N ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \sum _{i=1}^{N}\exp \left[ -\frac{\gamma }{2} B_i\right] \left\{ \sum _{t=1}^{T} (y_{it} - x_{it} \beta ) \right\} ^2 ,\\&\gamma T |{{\Omega }}|^{-1} \sigma _\epsilon ^{2(T-2)} \left\{ \sigma _\epsilon ^2 + (T-1) \sigma _\alpha ^2 \right\} \left\{ (1+\gamma )^{-\frac{1}{2}} - \frac{1+\gamma }{N \gamma } \sum _{i=1}^{N}\exp \left[ -\frac{\gamma }{2} B_i\right] \right\} \\&\ \ = \frac{(1+\gamma ) }{N} \sum _{i=1}^{N}\exp \left[ -\frac{\gamma }{2} B_i\right] \left[ - \frac{1}{\sigma _\epsilon ^4}\sum _{t=1}^{T} (y_{it} - x_{it} \beta )^2 + \frac{\sigma _\alpha ^2 (2\sigma _\epsilon ^2 + T \sigma _\alpha ^2 )}{\sigma _\epsilon ^4 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \left\{ \sum _{t=1}^{T} (y_{it} - x_{it} \beta ) \right\} ^2 \right] , \end{aligned} \end{aligned}$$

(21)

where ${\bar{x}}_i = \frac{1}{T}\sum _{t=1}^T x_{it}$. The MDPDE of $\theta$ is obtained by solving the above system of equations. One may use an iterative algorithm for this purpose or directly minimize the DPD measure in Eq. (8) with respect to $\theta \in \Theta _0$.

B Regularity Conditions

For the asymptotic distribution of the MDPDE, we need the following assumptions:

• (A1) The true density g(y|x) is supported over the entire real line $\mathbb {R}$.

• (A2) There is an open subset $\omega \in \Theta _0$ containing the best fitting parameter $\theta$ such tat ${{J}}$ is positive definite for all $\theta \in \omega$.

• (A3) There exist functions $M_{jkl}(x, y)$ such tat $|\partial ^3 \exp [(y - x \beta )^T {{\Omega }}^{-1} (y - x \beta )] /\partial \theta _j \partial \theta _k \partial \theta _l | \le M_{jkl}(x, y)$ for all $\theta \in \omega$, where $\int _x \int _y |M_{jkl}(x, y)| g(y|x) h(x) dy dx < \infty$ for all j, k and l.

Note that these regularity conditions hold good for the contaminated model, defined in Lemma 1, when $\eta (\gamma )$ is sufficiently small.

C Proof of Theorem 1

Proof

The proof of the first part closely follows the consistency of the maximum likelihood estimator with the line of modifications as given in Theorem 3.1 of Ghosh and Basu (2013). For brevity, we only present the detailed proof of the second part.

Let $\widehat{\theta }$ be the MDPDE of $\theta$. Then

$$\begin{aligned} \frac{\partial }{\partial \theta } {\widehat{d}}_\gamma (f_\theta , g) = \frac{\partial }{\partial \theta } \left[ \frac{1}{N} \sum _{i=1}^{N} \int _y f^{1+\gamma }_\theta (y|x_i) dy - \frac{1+\gamma }{N \gamma } \sum _{i=1}^{N} f_\theta ^{\gamma }(y_i | x_i)\right] =0. \end{aligned}$$

(22)

Thus, it can be written as the estimating equation of an M-estimator as follows

$$\begin{aligned} \sum _{i=1}^N \Psi _{\widehat{\theta }}(y_i|x_i) = 0, \end{aligned}$$

(23)

where

$$\begin{aligned} \Psi _{\theta }(y_i|x_i) = u_\theta (y_i|x_i) f_\theta ^\gamma (y_i|x_i) - \int _y u_\theta (y|x_i) f_\theta ^{1+\gamma }(y_i|x_i) dy. \end{aligned}$$

(24)

Let $\theta _g$ be the true value of $\theta$, then $E\left( \sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i)\right) = 0$ gives

$$\begin{aligned} \sum _{i=1}^N \Bigg [\int _y u_{\theta _g}(y|x_i) f_{\theta _g}^\gamma (y|x_i) g(y|x_i) dy - \int _y u_{\theta _g}(y|x_i) f_{\theta _g}^{1+\gamma }(y_i|x_i) dy \Bigg ] = 0 . \end{aligned}$$

(25)

Taking a Taylor series expansion of Eq. (23), we get

$$\begin{aligned} \begin{aligned} \frac{1}{N}\sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i)&+ \frac{1}{N}\sum _{i=1}^N \frac{\partial }{\partial \theta }\Psi _\theta (y_i|x_i)\Big |_{\theta = \theta _g} (\widehat{\theta } -\theta _g) + R_N = 0,\\ \text{ or } \sqrt{N}(\widehat{\theta } -\theta _g)&= - \left[ \frac{1}{N}\sum _{i=1}^N \frac{\partial }{\partial \theta }\Psi _\theta (y_i|x_i)\Big |_{\theta = \theta _g} \right] ^{-1} \left[ \frac{1}{\sqrt{N}}\sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i) + \sqrt{N}R_N\right] ,\\ \end{aligned} \end{aligned}$$

(26)

where $R_N$ is the remainder term. Using the weak law of large numbers (WLLN), we have

$$\begin{aligned} \begin{aligned}&\frac{1}{N}\sum _{i=1}^N \frac{\partial }{\partial \theta }\Psi _\theta (y_i|x_i) \\&\overset{p}{\rightarrow }\ \lim _{N \rightarrow \infty } E\left[ \frac{1}{N}\sum _{i=1}^N \frac{\partial }{\partial \theta }\Psi _\theta (y_i|x_i) \right] \\&\overset{p}{\rightarrow }\ \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{i=1}^N E\left[ \frac{\partial }{\partial \theta } \left( u_\theta f_\theta ^\gamma - \int u_\theta f_\theta ^{1+\gamma } \right) \right] \\&\overset{p}{\rightarrow }\ \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{i=1}^N E\left[ -I_\theta f_\theta ^\gamma + \gamma u_\theta u_\theta ^T f_\theta ^\gamma - \int \left\{ - I_\theta f_\theta ^{1+\gamma } + (1+\gamma ) u_\theta u_\theta ^T f_\theta ^{1+\gamma } \right\} \right] \\&\overset{p}{\rightarrow }\ \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{i=1}^N \left[ - \int I_\theta f_\theta ^\gamma g + \gamma \int u_\theta u_\theta ^T f_\theta ^\gamma g + \int I_\theta f_\theta ^{1+\gamma } - (1+\gamma ) \int u_\theta u_\theta ^T f_\theta ^{1+\gamma } \right] \\&\overset{p}{\rightarrow }\ - \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{i=1}^N \left[ \int u_\theta u_\theta ^T f_\theta ^{1+\gamma } + \int \left( I_\theta -\gamma u_\theta u_\theta ^T \right) (g - f_\theta ) f_\theta ^\gamma \right] . \end{aligned} \end{aligned}$$

(27)

So

$$\begin{aligned} \frac{1}{N}\sum _{i=1}^N \frac{\partial }{\partial \theta }\Psi _\theta (y_i|x_i)\Big |_{\theta = \theta _g} \overset{p}{\rightarrow }\ - \lim _{N \rightarrow \infty } \frac{1}{N}\sum _{i=1}^N {{J}}^{(i)} = - {{J}}. \end{aligned}$$

(28)

From Eq. (25), we get

$$\begin{aligned} \begin{aligned}&E\left[ \frac{1}{\sqrt{N}}\sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i) \right] \\&\quad = \frac{1}{\sqrt{N}} \sum _{i=1}^N \Bigg [\int _y u_{\theta _g}(y|x_i) f_{\theta _g}^\gamma (y|x_i) g(y|x_i) dy - \int _y u_{\theta _g}(y|x_i) f_{\theta _g}^{1+\gamma }(y_i|x_i) dy \Bigg ]\\&\quad = 0. \end{aligned} \end{aligned}$$

(29)

Now,

$$\begin{aligned} \begin{aligned} V\left[ \frac{1}{\sqrt{N}}\sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i) \right]&= \frac{1}{N} \sum _{i=1}^N V\left[ \Psi _{\theta _g}(y_i|x_i) \right] \\&= \frac{1}{N} \sum _{i=1}^N \Bigg [\int _y u_{\theta _g}(y|x_i) u_{\theta _g}^T(y|x_i) f_{\theta _g}^{2\gamma }(y|x_i) g(y|x_i) dy - \xi ^{(i)} \xi ^{(i)T}\Bigg ]\\&= \frac{1}{N} \sum _{i=1}^N {{K}}^{(i)}. \end{aligned} \end{aligned}$$

(30)

Following Section 5 of Ferguson (1996) or Section 2.7 of Lehmann (1999) and using Eqs. (29) and (30), the central limit theorem (CLT) for the independent but not identical random variables gives

$$\begin{aligned} \frac{1}{\sqrt{N}}\sum _{i=1}^N \Psi _{\theta _g}(y_i|x_i) \overset{a}{\sim }\ N\left( 0, {{K}}\right) . \end{aligned}$$

(31)

Under regularity condition (A3), it can be easily shown that the reminder term $\sqrt{N}R_N = o_p(1).$ Therefore, combining Eqs. (28) and (31), we get from Eq. (26)

$$\begin{aligned} \sqrt{N}(\widehat{\theta } -\theta _g) \overset{a}{\sim }\ N\left( 0, {{J}}^{-1}{{K}}{{J}}^{-1}\right) . \end{aligned}$$

(32)

This completes the proof. $\square$

1.1 D ${{J}}$ and ${{K}}$ Matrices at the model

Let us write the score function as

$$\begin{aligned} u_\theta (y_i | x_i) = (u^T_\beta (y_i | x_i), u^T_{\sigma _\alpha ^2}(y_i | x_i), u^T_{\sigma _\epsilon ^2}(y_i | x_i))^T. \end{aligned}$$

(33)

Suppose ${\bar{x}}_i = \frac{1}{T}\sum _{t=1}^T x_{it}$. Then, it can be shown that

$$\begin{aligned} \begin{aligned} u_\beta (y_i | x_i)&= \frac{\partial }{\partial \beta } \log f_\theta (y_i | x_i) = \frac{1}{\sigma _\epsilon ^2}\sum _{t=1}^{T} x_{it} (y_{it} - x_{it} \beta ) - \frac{ T {\bar{x}}_i \sigma _\alpha ^2 }{\sigma _\epsilon ^2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} \sum _{t=1}^{T} (y_{it} - x_{it} \beta ),\\ u_{\sigma _\alpha ^2}(y_i | x_i)&= \frac{\partial }{\partial \sigma _\alpha ^2} \log f_\theta (y_i | x_i) = -\frac{T}{2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} + \frac{ 1}{2( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \left\{ \sum _{t=1}^{T} (y_{it} - x_{it} \beta ) \right\} ^2,\\ u_{\sigma _\epsilon ^2}(y_i | x_i)&= \frac{\partial }{\partial \sigma _\epsilon ^2} \log f_\theta (y_i | x_i) = -\frac{1}{2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} \left[ (T-1) \sigma _\epsilon ^{2} ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 ) + 1\right] \\&\quad + \frac{1}{2\sigma _\epsilon ^4}\sum _{t=1}^{T} (y_{it} - x_{it} \beta )^2 - \frac{\sigma _\alpha ^2 (2\sigma _\epsilon ^2 + T \sigma _\alpha ^2 )}{2\sigma _\epsilon ^4 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \left\{ \sum _{t=1}^{T} (y_{it} - x_{it} \beta ) \right\} ^2. \end{aligned} \end{aligned}$$

(34)

Note that if the true distribution g(y|x) is a member of the model family $f_\theta (y|x)$ for some $\theta \in \Theta _0$, then

$$\begin{aligned} \begin{aligned}&{{J}}^{(i)} = \int _y u_\theta (y | x_i) u^T_\theta (y | x_i) f_\theta ^{1+\gamma }(y | x_i) dy, \\&{{K}}^{(i)} = \int _y u_\theta (y | x_i) u^T_\theta (y | x_i) f_\theta ^{2\gamma +1}(y | x_i) dy - \xi ^{(i)} \xi ^{(i)T} , \\&\xi ^{(i)} = \int _y u_\theta (y | x_i) f_\theta ^{\gamma +1}(y | x_i) dy. \end{aligned} \end{aligned}$$

(35)

In this case, the symmetric matrix ${{J}}^{(i)}$ can be partitioned as

$$\begin{aligned} {{J}}^{(i)} = \begin{bmatrix} {{J}}_\beta ^{(i)} &{} {{J}}_{\beta , \ \sigma _\alpha ^2}^{(i)} &{} {{J}}_{\beta , \ \sigma _\epsilon ^2}^{(i)}\\ . &{} {{J}}_{\sigma _\alpha ^2}^{(i)} &{} {{J}}_{\sigma _\alpha ^2, \ \sigma _\epsilon ^2}^{(i)}\\ . &{} . &{} {{J}}_{\sigma _\epsilon ^2}^{(i)} \end{bmatrix}, \end{aligned}$$

(36)

where

$$\begin{aligned} \begin{aligned} {{J}}_\beta ^{(i)}&= M \sigma _\epsilon ^{-4} \Bigg [ \sigma _\epsilon ^2 \sum _{t=1}^{T} x_{it} x_{it}^T + T^2 \sigma _\alpha ^2 \left( \frac{ T \sigma _\alpha ^2 }{( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )} -1 \right) {\bar{x}}_i {\bar{x}}_i^T \Bigg ] ,\\ {{J}}_{\sigma _\alpha ^2}^{(i)}&= \frac{M T^2 ( \gamma ^2 + 2 )}{4 (1+\gamma ) ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} ,\\ {{J}}_{\sigma _\epsilon ^2}^{(i)}&= \frac{ M T^2 (\gamma - 1) \left[ \sigma _\epsilon ^2 + (T-1) \sigma _\alpha ^2 \right] ^2}{4\sigma _\epsilon ^4 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \\&\quad + \frac{MT}{4\sigma _\epsilon ^8} \left[ (T+2) \sigma _\epsilon ^4 + 2(T+2) \sigma _\epsilon ^2\sigma _\alpha ^2 + 3T \sigma _\alpha ^4 \right] + \frac{3MT^2\sigma _\alpha ^4 (2\sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2}{4\sigma _\epsilon ^8 (1+\gamma ) ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \\&\quad - \frac{ T M (1+\gamma )\sigma _\alpha ^2 (2\sigma _\epsilon ^2 + T \sigma _\alpha ^2 )}{2\sigma _\epsilon ^8 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \Big [ (T+2) \sigma _\epsilon ^4 + (T^2 + 2T +3) \sigma _\epsilon ^2 \sigma _\alpha ^2 + 3(T^2 -T + 1 ) \sigma _\alpha ^4 \Big ] ,\\ {{J}}_{\beta , \ \sigma _\alpha ^2}^{(i)}&= 0,\ \ {{J}}_{\beta , \ \sigma _\epsilon ^2}^{(i)} = 0,\\ {{J}}_{\sigma _\alpha ^2, \ \sigma _\epsilon ^2}^{(i)}&= \frac{T M (1+\gamma ) }{4 \sigma _\epsilon ^4 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} \Big [ 2(T+1) \sigma _\epsilon ^4 + (2T^2 + T +3) \sigma _\epsilon ^2 \sigma _\alpha ^2 + 3(T^2 -T +1) \sigma _\alpha ^4 \Big ] \\&\quad - \frac{3MT^2\sigma _\alpha ^2 (2\sigma _\epsilon ^2 + T \sigma _\alpha ^2 )}{4\sigma _\epsilon ^4 (1+\gamma ) ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} - \frac{M T^2 \left[ (T-1) \sigma _\alpha ^2 + \sigma _\epsilon ^2 \right] }{2\sigma _\epsilon ^2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^2} , \end{aligned} \end{aligned}$$

(37)

and

$$\begin{aligned} M = (2\pi )^{-\frac{T \gamma }{2}} (1+\gamma )^{-\frac{T+2}{2}} \sigma _\epsilon ^{-\gamma (T-1)} ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )^{-\frac{\gamma }{2}} . \end{aligned}$$

(38)

Similarly, $\xi ^{(i)}$ can be partitioned as $\xi ^{(i)} = \left( \xi _\beta ^{(i)T}, \xi _{\sigma _\alpha ^2}^{(i)} , \xi _{\sigma _\epsilon ^2}^{(i)} \right) ^T$, and it is shown that

$$\begin{aligned} \xi _\beta ^{(i)} = 0,\ \ \ \xi _{\sigma _\alpha ^2}^{(i)} = -\frac{MT \gamma }{2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 )}, \text{ and } \xi _{\sigma _\epsilon ^2}^{(i)} = -\frac{M T \gamma \left[ \sigma _\epsilon ^2 + (T-1) \sigma _\alpha ^2 \right] }{2 \sigma _\epsilon ^2 ( \sigma _\epsilon ^2 + T \sigma _\alpha ^2 ) }. \end{aligned}$$

(39)

Note that if we write the matrix ${{J}}^{(i)}$ as a function of $\gamma$, i.e., ${{J}}^{(i)} \equiv {{J}}^{(i)}(\gamma )$, then we have

$$\begin{aligned} {{K}}^{(i)} = {{J}}^{(i)}(2\gamma ) - \xi ^{(i)} \xi ^{(i)T}. \end{aligned}$$

(40)

Moreover, $\xi _\beta ^{(i)}$ is constant for all values of $i=1, 2, \cdots , N$. Therefore, ${{K}}$ can be written as

$$\begin{aligned} {{K}}= \frac{1}{N} \sum _{i=1}^N {{J}}^{(i)}(2\gamma ) - \xi ^{(i)} \xi ^{(i)T}. \end{aligned}$$

(41)