Statistical Linear Calibration in Data with Measurement Errors

Abstract

The direct and inverse regression-based estimators are used for linear statistical calibration. The direct regression finds the relationship between the dependent and independent variables, and inverse regression uses it for calibration. When both the dependent and independent variables in linear calibration are subjected to measurement errors, the resultant estimators become biased and inconsistent. Assuming the availability of replicated observations on the independent variable, a calibration estimator is presented which has zero large sample asymptotic bias. The large sample efficiency properties of the calibration estimators are derived and analyzed assuming the random errors are not necessarily normally distributed.

Appendix

Let us first introduce the following notation

$$\begin{aligned} A= & {} I_{np} - \frac{1}{np} e_{np} e_{np}'\nonumber \\ B= & {} \frac{1}{p} (I_n \otimes \ e_p') - \frac{1}{np} e_n e_{np}'\nonumber \\ C= & {} I_n - \frac{1}{n} e_n e_n'\\ D= & {} \frac{1}{p} (I_n \otimes \ e_p e_p') - \frac{1}{np} e_{np} e_{np}'\nonumber \end{aligned}$$

where $\otimes$ denotes the Kronecker product operator of matrices and e denotes a column vector with all elements unity and its suffix indicating the number of elements in it.

The following properties of these matrices may be noted:

$$\begin{aligned} AD=D,\quad \ BD = D, \quad \ pB'B=D, \quad \ pBB' = C,\nonumber \\ tr A = (np-1), \quad \ tr C = tr D = (n-1),\\ A e_{np} = 0, \quad \ B e_{np} = 0, \quad \ e_n'B = 0, \quad \ C e_n = 0.\nonumber \end{aligned}$$

(5.1)

Further, if Z denotes a $m\times 1$ random vector such that its elements are independently and identically distributed with mean 0, variance $\sigma ^2_z$, third moment $\sigma ^3_z \gamma _{1z}$ and fourth moment $\sigma ^4_z (\gamma _{2z} + 3)$, we have

$$\begin{aligned} E(Z'GZ)= & {} \sigma ^2_z tr G\nonumber \\ E(Z'GZ.Z)= & {} \sigma ^3_z \gamma _{1z} (I *G)e_m\\ E(Z'GZ.Z'MZ)= & {} \sigma ^4_z [\gamma _{2z} tr H (I*G) + (tr G)(tr H) + 2 tr GH]\nonumber \end{aligned}$$

(5.2)

where G and H are $m\times m$ symmetric matrices with nonstochastic elements and $*$ denotes the Hadamard product operator of matrices.

If we define

$$\begin{aligned} f_{xx}= & {} \frac{2}{n^{1/2}} X' B v + \frac{n^{1/2}}{p} \left( \frac{v'A v}{n} -p\sigma _v^2 \right) \nonumber \\ f_{xy}= & {} \frac{1}{n^{1/2}} \left[ \frac{1}{\beta } (X'Cu + u'B v) + X' B v \right] \nonumber \\ f_{yy}= & {} \frac{2}{\beta n^{1/2}} X' CY + \frac{n^{1/2}}{\beta ^2} \left( \frac{u'Cu}{n} - \sigma _u^2 \right) \\ f= & {} n^{1/2} \left[ \frac{v' (A-D)v}{n(p-1)} -\sigma _v^2 \right] \nonumber \\ f_x= & {} \frac{1}{p n^{1/2}} \sum \limits _i\sum \limits _j v_{ij}\nonumber \\ f_y= & {} \frac{1}{\beta n^{1/2}} \sum \limits _i u_i,\nonumber \end{aligned}$$

it is observed that all these quantities are of order $O_p(1)$ where $u=(u_1,u_2,\ldots ,u_n)$ and $v=(v_{11},v_{12},\ldots ,v_{np})$ are column vectors of order $n\times 1$ and $np \times 1$ respectively.

Now, from (2.6), we can express

$$\begin{aligned} (\hat{X}_c - X)= & {} \sigma _v d + \frac{f_x}{n^{1/2}} + \frac{s^2 + \sigma _v^2 + \frac{f_{xx}}{n^{1/2}}}{s^2 + \frac{f_{xy}}{n^{1/2}}} \left( \frac{U}{\beta } - \sigma _v d - \frac{f_y}{n^{1/2}} \right) \nonumber \\= & {} \sigma _v d + \frac{f_x}{n^{1/2}} + \left( \frac{U}{\beta } - \sigma _v d - \frac{f_y}{n^{1/2}} \right) \left( \frac{1}{\lambda _x} + \frac{f_{xx}}{s^2 n^{1/2}}\right) \left( 1 + \frac{f_{xy}}{s^2 n^{1/2}} \right) ^{-1}\nonumber \end{aligned}$$

where $s^2$ and $\lambda _x$ are defined in (3.1).

Expanding and retaining terms to order $O_p(n^{-1/2})$, we get

$$\begin{aligned} (\hat{X}_c - X) = \frac{1}{\lambda _x} \left[ \frac{U}{\beta } - (1-\lambda _x) \sigma _v d \right] + \frac{1}{n^{1/2}} \hat{\xi }_c + O_p\left( \frac{1}{n}\right) \end{aligned}$$

where

$$\begin{aligned} \hat{\xi }_c = f_x - \frac{1}{\lambda _x} f_y + \frac{1}{s^2} \left( \frac{U}{\beta } - \sigma _v d \right) \left( f_{xx} - \frac{1}{\lambda _x} f_{xy} \right) . \end{aligned}$$

Similarly, from (2.7), we obtain

$$\begin{aligned} (\hat{X}_c^* - X)= & {} \sigma _v d + \frac{f_x}{n^{1/2}} + \left( \frac{U}{\beta } - \sigma _v d - \frac{f_y}{n^{1/2}} \right) \left[ 1 + \frac{f_{xx} - f}{s^2 n^{1/2}} \left( 1+ \frac{f_{xy}}{s^2 n^{1/2}}\right) ^{-1} \right] \\= & {} \frac{U}{\beta } + \frac{1}{n^{1/2}} \hat{\xi }_c^* + O_p\left( \frac{1}{n}\right) \nonumber \end{aligned}$$

where

$$\begin{aligned} \hat{\xi }_c^* = f_x - f_y + \frac{1}{s^2} \left( \frac{U}{\beta } - \sigma _v d \right) \left( f_{xx} - f_{xy} -f \right) . \end{aligned}$$

Likewise, it is easy to see from (2.8) that

$$\begin{aligned} (\hat{X}_I - X)= & {} \sigma _v d + \frac{f_x}{n^{1/2}} + \left( \frac{U}{\beta } - \sigma _v d - \frac{f_y}{n^{1/2}} \right) \\&\quad \left( \frac{\beta ^2 f_{xy}}{(\beta ^2s^2 + \sigma _u^2) n^{1/2}} + \frac{\beta ^2s^2}{\beta ^2s^2 + \sigma _u^2} \right) \left( 1 + \frac{\beta ^2 f_{yy}}{(\beta ^2s^2 + \sigma _u^2) n^{1/2}} \right) ^{-1}\nonumber \\= & {} \sigma _v d+ \frac{f_x}{n^{1/2}} + \left( \frac{U}{\beta } - \sigma _v d - \frac{f_y}{n^{1/2}}\right) \left( \lambda _y + \frac{\lambda _y f_{xy}}{s^2 n^{1/2}} \right) \left( 1 - \frac{\lambda _y f_{yy}}{s^2 n^{1/2}} + \ldots \right) \nonumber \\= & {} \frac{\lambda _y U}{\beta } + (1-\lambda _y)\sigma _v d + \frac{1}{n^{1/2}} \hat{\xi }_I + O_p(n^{-1})\nonumber \end{aligned}$$

where

$$\begin{aligned} \hat{\xi }_I = f_x - \lambda _y f_y + \frac{\lambda _y}{s^2} \left( \frac{U}{\beta } - \sigma _v d \right) \left( f_{xy} - \lambda _y f_{yy} \right) . \end{aligned}$$

Observing that

$$\begin{aligned} E(f_{xx}) = -\frac{\sigma _v^2}{n^{1/2}p}, \quad E(f_{yy}) = - \frac{\sigma _u^2}{n^{1/2} \beta ^2},\nonumber \\ E(f_{xy}) = E(f) = E(f_x) = E(f_y) = 0, \nonumber \end{aligned}$$

the LSABs of $\hat{X}_c$, $\hat{X}_I$ and $\hat{X}_c^*$ to $O(n^{-1/2})$ are given by

$$\begin{aligned} E(\hat{X}_c - X)= & {} \frac{1}{\lambda _x} \left[ \frac{1}{\beta } E(U) - (1- \lambda _x) \sigma _v d \right] + \frac{1}{n^{1/2}} E(\hat{\xi }_c)\\= & {} - \left( \frac{1-\lambda _x}{\lambda _x} \right) \sigma _v d \nonumber \\ E(\hat{X}_c^* - X)= & {} \frac{1}{\beta } E(U) + \frac{1}{n^{1/2}} E(\hat{\xi }_c^*)\\= & {} 0\nonumber \\ E(\hat{X}_I - X)= & {} \frac{\lambda _y}{\beta } E(U) - (1- \lambda _y) \sigma _v d + \frac{1}{n^{1/2}} E(\hat{\xi }_I)\\= & {} (1-\lambda _y) \sigma _v d \nonumber \end{aligned}$$

which are the results stated in Theorem 1.

By virtue of the distributional properties of u and v, the following results are obtained:

$$\begin{aligned} E(f_{xx}^2)= & {} \frac{s^4 (1-\lambda _x)}{p \lambda _x} \left[ 4 + \left( \frac{1 - \lambda _x}{\lambda _x}\right) (2-\gamma _{2v}) \right] + O \left( \frac{1}{n} \right) \\ E(f_{xy}^2)= & {} \frac{s^4 (1-\lambda _x)}{p \lambda _x \lambda _y} \left[ 1 + p \lambda _x \left( \frac{1 - \lambda _y}{1-\lambda _x}\right) \right] + O \left( \frac{1}{n} \right) \\ E(f_{yy}^2)= & {} \frac{s^4 (1-\lambda _y)}{\lambda _y} \left[ 4 + \left( \frac{1 - \lambda _y}{\lambda _y}\right) (2+\gamma _{2v}) \right] + O \left( \frac{1}{n} \right) \\ E(f^2)= & {} \frac{s^4 (1-\lambda _x)^2}{p \lambda _x^2} \left[ 2 \left( \frac{p}{p-1}\right) + \gamma _{2v} \right] + O \left( \frac{1}{n} \right) \\ E(f_{xx} f_{xy})= & {} \frac{2s^4 (1-\lambda _x)}{p\lambda _x} + O \left( \frac{1}{n} \right) \\ E(f_{xx} f)= & {} \frac{s^4 (1-\lambda _x)^2}{p \lambda _y^2} (2+\gamma _{2v}) + O \left( \frac{1}{n} \right) \\ E(f_{xy} f_{yy})= & {} \frac{2s^4 (1-\lambda _y)}{\lambda _y}\\ E(f_{x} f_{xx})= & {} \frac{s^2 \sigma _v \gamma _{1v} (1-\lambda _x)}{p \lambda _x} + O \left( \frac{1}{n} \right) \\ E(f_x f)= & {} \frac{s^2 \sigma _v \gamma _{1v} (1-\lambda _x)}{p \lambda _x}\\ E(f_y f_{yy})= & {} \frac{s^2 \sigma _u \gamma _{1u} (1-\lambda _y)}{ \beta \lambda _y} + O \left( \frac{1}{n} \right) \\ E(f_x^2)= & {} \frac{\sigma _v^2}{p}\\ E(f_y^2)= & {} \frac{\sigma _v^2 \lambda _x (1-\lambda _y)}{\lambda _y (1- \lambda _x)} \end{aligned}$$

and

$$ E(f_{xy} f) =E(f_x f_{xy}) = E(f_x f_{yy}) = E(f_y f_{xx}) = E(f_y f_{xy}) = E(f_y f) = E(f_x f_y) =0 $$

where repeated use has been made of (5.1) and (5.2)

Utilizing these results and we obtain the LSAVs to order $O(n^{-1})$ as follows:

$$\begin{aligned} E[\hat{X}_c - E(\hat{X}_c)]^2= & {} \frac{1}{n} E(\hat{\xi }_c^2)\\ E[\hat{X}_c^* - E(\hat{X}_c^*)]^2= & {} \frac{1}{n} E(\hat{\xi }_c^{*2})\\ E[\hat{X}_I - E(\hat{X}_I)]^2= & {} \frac{1}{n} E(\hat{\xi }_I^2) \end{aligned}$$

which gives the results mentioned in Theorem 2.