Estimation of parameters of logistic regression for two-stage randomized response technique

Abstract

When a survey study is related to sensitive issues such as political orientation, sexual orientation, and income, respondents may not be willing to reply truthfully, which leads to bias results. To protect the respondents’ privacy and improve their willingness to provide true answers, Warner (J Am Stat Assoc 60:63–69, 1965) proposed the randomized response (RR) technique in which respondents select a question by means of a random device in order to ensure that they maintain privacy. Huang (Stat Neerl 58:75–82, 2004) extended the RR design of Warner (1965) to propose a two-stage RR design. Not only can this method be used to estimate the population proportion of persons with a sensitive characteristic, but also estimate the honest answer rate in the first stage. This work develops a covariate extension of the two-stage RR design of Huang (2004) by applying logistic regression to investigate the effects of covariates on a sensitive characteristic and an honest response. Simulation experiments are conducted to study the finite-sample performance of the maximum likelihood estimators of the logistic regression parameters. The proposed methodology is applied to analyze the survey data of sexuality of freshmen at Feng Chia University in Taiwan in 2016.

Appendix

1.1 Proof of Lemma 1

$\varvec{\Psi }_i(\Theta )$ in (4) can be expressed as

$$\begin{aligned} \varvec{\Psi }_i(\Theta )&=\frac{Y_{i1}}{g_{i1}(\Theta )}\left( \frac{\partial {g_{i1}}(\Theta )}{\partial \Theta }\right) + \frac{Y_{i2}}{g_{i2}(\Theta )}\left( \frac{\partial {g_{i2}}(\Theta )}{\partial \Theta }\right) \\&\quad -\frac{1-Y_{i1}-Y_{i2}}{1-g_{i1}(\Theta )-g_{i2}(\Theta )}\left( \frac{\partial {g_{i1}}(\Theta )}{\partial \Theta } + \frac{\partial {g_{i2}}(\Theta )}{\partial \Theta }\right) \\&=\frac{\left[ Y_{i1}-g_{i1}(\Theta )\right] g_{i2}(\Theta )[1-g_{i2}(\Theta )]}{g_{i1}(\Theta )g_{i2}(\Theta )[1-g_{i1}(\Theta )-g_{i2}(\Theta )]} \left( \frac{\partial {g_{i1}}(\Theta )}{\partial \Theta }\right) \nonumber \\&\quad + \frac{\left[ Y_{i2}-g_{i2}(\Theta )\right] g_{i1}(\Theta )[1-g_{i1}(\Theta )]}{g_{i1}(\Theta )g_{i2}(\Theta )[1-g_{i1}(\Theta )-g_{i2}(\Theta )]} \left( \frac{\partial g_{i2}(\Theta )}{\partial \Theta }\right) \nonumber \\&\quad + \frac{\left[ Y_{i1}-g_{i1}(\Theta )\right] g_{i1}(\Theta )g_{i2}(\Theta )}{g_{i1}(\Theta )g_{i2}(\Theta )[1-g_{i1}(\Theta )-g_{i2}(\Theta )]} \left( \frac{\partial {g_{i2}}(\Theta )}{\partial \Theta }\right) \nonumber \\&\quad + \frac{\left[ Y_{i2}-g_{i2}(\Theta )\right] g_{i1}(\Theta )g_{i2}(\Theta )}{g_{i1}(\Theta )g_{i2}(\Theta )[1-g_{i1}(\Theta )-g_{i2}(\Theta )]} \left( \frac{\partial {g_{i1}}(\Theta )}{\partial \Theta }\right) \nonumber \\&=\left( \frac{\partial {g_{i1}}(\Theta )}{\partial \Theta }, \frac{\partial {g_{i2}}(\Theta )}{\partial \Theta }\right) \frac{1}{det(\varvec{V}_{i}(\Theta ))}\\&\quad \begin{bmatrix} g_{i2}(\Theta )[1-g_{i2}(\Theta )] &{} g_{i1}(\Theta )g_{i2}(\Theta ) \\ g_{i1}(\Theta )g_{i2}(\Theta ) &{} g_{i1}(\Theta )[1-g_{i1}(\Theta )] \end{bmatrix} [\varvec{Y}_i-\varvec{g}_i(\Theta )]^T\\&=\left( \frac{\partial \varvec{g_i}(\Theta )}{\partial \Theta }\right) {\varvec{V}_i^{-1}(\Theta )}[\varvec{Y}_i-\varvec{g}_i(\Theta )]^T,\ i=1,2,\dots , n, \end{aligned}$$

where $det(\varvec{V}_i(\Theta ))= g_{i1}(\Theta )g_{i2}(\Theta )(1-g_{i1}(\Theta )-g_{i2}(\Theta ))$ and $\varvec{V}_i(\Theta )$ is given in (8). Hence the score function $\varvec{U}_n(\Theta )$ can be written as

$$\begin{aligned} \varvec{U}_n(\Theta ) =\sum _{i=1}^n\varvec{\Psi }_i(\Theta ) =\sum _{i=1}^{n}\left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_{i}^{-1}(\Theta )[\varvec{Y}_i-\varvec{g}_i(\Theta )]^T. \end{aligned}$$

(9)

1.2 Proof of Theorem 1

1.2.1 (a) Proof of consistency of ${\widehat{\Theta }}$

Because from condition (C1) and the inverse function theorem of Foutz (1977), $\varvec{U}_n(\Theta )=\varvec{0}$ has a unique solution, the ML estimator ${\widehat{\Theta }}$ is a consistent estimator of $\Theta$.

1.2.2 (b) Proof of asymptotic normality of $\sqrt{n}({\widehat{\Theta }}-\Theta )$

Let $\varvec{{\mathcal {U}}}_n={\frac{1}{\sqrt{n}}}\varvec{U}_n(\Theta )=\frac{1}{\sqrt{n}}\sum _{i=1}^n\varvec{\Psi }_i(\Theta )$. By a Taylor’s expansion of $\varvec{{\mathcal {U}}}_n({\widehat{\Theta }})$ at $\Theta$, we can have

$$\begin{aligned} \varvec{0}=\varvec{{\mathcal {U}}}_n({\widehat{\Theta }})&=\varvec{{\mathcal {U}}}_n(\Theta ) +\frac{\partial \varvec{{\mathcal {U}}}_n(\Theta )}{\partial \Theta ^T} ({\widehat{\Theta }}-\Theta )+o_p\left( \sqrt{n}[({\widehat{\Theta }}-\Theta )]^{\otimes 2}\right) \nonumber \\&=\varvec{{\mathcal {U}}}_n(\Theta )+\frac{\partial \varvec{{\mathcal {U}}}_n(\Theta )}{\sqrt{n}\partial \Theta ^T}\sqrt{n}({\widehat{\Theta }}-\Theta )+o_p(\varvec{1}), \end{aligned}$$

(10)

where

$$\begin{aligned} \frac{\partial \varvec{{\mathcal {U}}}_n(\Theta )}{\sqrt{n}\partial \Theta ^T}&=\frac{1}{n}\sum _{i=1}^{n}\frac{\partial \varvec{\Psi }_i(\Theta )}{\partial \Theta ^T}\\&=\frac{1}{n}\sum _{i=1}^{n}\frac{\partial \left[ \left( \dfrac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta )\right] }{\partial \Theta ^T} [\varvec{Y}_i-\varvec{g}_i(\Theta )]^T\\&\quad -\frac{1}{n}\sum _{i=1}^{n}\left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta ) \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) ^T. \end{aligned}$$

Because $(Y_{i1},Y_{i2},1-Y_{i1}-Y_{i2})|\varvec{X}_i\sim {Mult}\left( 1,g_{i1}(\Theta ),g_{i2}(\Theta ),1-g_{i1}(\Theta )-g_{i2}(\Theta )\right)$, $i=1,2,\dots$, we have

$$\begin{aligned} E\left\{ [\varvec{Y}_i-\varvec{g}_i(\Theta )]|\varvec{X}_i\right\}&=\varvec{0}, \end{aligned}$$

(11)

$$\begin{aligned} E[\varvec{Y}_i-\varvec{g}_i(\Theta )]&=E\left\{ E\left\{ [\varvec{Y}_i-\varvec{g}_i(\Theta )]|\varvec{X}_i\right\} \right\} =\varvec{0}, \nonumber \\ Var\left\{ [\varvec{Y}_i-\varvec{g}_i(\Theta )]|\varvec{X}_i\right\}&=E\left\{ [\varvec{Y}_i-\varvec{g}_i(\Theta )][\varvec{Y}_i-\varvec{g}_i(\Theta )]^T|\varvec{X}_i\right\} \nonumber \\&=\begin{bmatrix} g_{i1}(\Theta )[1-g_{i1}(\Theta )] &{}\quad -g_{i1}(\Theta )g_{i2}(\Theta ) \\ -g_{i1}(\Theta )g_{i2}(\Theta ) &{}\quad g_{i2}(\Theta )[1-g_{i2}(\Theta )] \end{bmatrix} =\varvec{V}_i(\Theta ). \end{aligned}$$

(12)

Because $(\varvec{Y}_i,\varvec{X}_i)$, $i=1,2,\ldots ,n$, are independent and identically distributed and $E[\varvec{Y}_i-\varvec{g}_i(\Theta )]=\varvec{0}$, it can be shown according to condition (C2) and the weak law of large numbers that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\left\{ \frac{\partial }{\partial \Theta ^T}\left[ \left( \dfrac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta )\right] \right\} [\varvec{Y}_i- \varvec{g}_i(\Theta )]^T\overset{p}{\longrightarrow }\varvec{0}, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^n\left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta ) \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) ^T\overset{p}{\longrightarrow }E\left[ \left( \frac{\partial \varvec{g}_1(\Theta )}{\partial \Theta }\right) \varvec{V}_1^{-1}(\Theta )\left( \frac{\partial \varvec{g}_1(\Theta )}{\partial \Theta }\right) ^T\right] . \end{aligned}$$

Hence

$$\begin{aligned} -\frac{\partial \varvec{{\mathcal {U}}}_n(\Theta )}{\sqrt{n}\partial \Theta ^T} \overset{p}{\longrightarrow }E\left[ \left( \frac{\partial \varvec{g}_1(\Theta )}{\partial \Theta }\right) \varvec{V}_1^{-1}(\Theta )\left( \frac{\partial \varvec{g}_1(\Theta )}{\partial \Theta }\right) ^T\right] =\varvec{\Delta }^{-1}. \end{aligned}$$

From (10), $\sqrt{n}({\widehat{\Theta }}-\Theta )$ can be expressed as

$$\begin{aligned} \sqrt{n}({\hat{\Theta }}-\Theta )&=\left( -\frac{\partial \varvec{{\mathcal {U}}}_n(\Theta )}{\sqrt{n}\partial \Theta ^T}\right) ^{-1}\varvec{{\mathcal {U}}}_n(\Theta )+o_p(\varvec{1}) \nonumber \\&=\left[ \varvec{\Delta }+o_p(\varvec{1})\right] \varvec{{\mathcal {U}}}_n(\Theta )+o_p(\varvec{1}) \nonumber \\&=\varvec{\Delta }\varvec{{\mathcal {U}}}_n(\Theta )+o_p(\varvec{1}). \end{aligned}$$

(13)

Because from (11) and (12) we can have

$$\begin{aligned} E\left[ \varvec{\Psi }_i(\Theta )\right] =E\left[ \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_{i}^{-1}(\Theta )[\varvec{Y}_i-\varvec{g}_i(\Theta )]^T\right] =\varvec{0} \end{aligned}$$

and

$$\begin{aligned} Var\left[ \varvec{\Psi }_i(\Theta )\right]&=E\left[ \varvec{\Psi }_i(\Theta )\varvec{\Psi }_i^T(\Theta )\right] \\&=E\left\{ E\left[ \varvec{\Psi }_i(\Theta )\varvec{\Psi }_i^T(\Theta )|\varvec{X}_i\right] \right\} \\&=E\left[ \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta ) E\left\{ [\varvec{Y}_i-\varvec{g}_i(\Theta )]^T[\varvec{Y}_i-\varvec{g}_i(\Theta )]|\varvec{X}_i\right\} \right. \\&\quad \left. \varvec{V}_i^{-1}(\Theta )\left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) ^T\right] \\&=E\left[ \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) \varvec{V}_i^{-1}(\Theta ) \left( \frac{\partial \varvec{g}_i(\Theta )}{\partial \Theta }\right) ^T\right] =\varvec{\Delta }^{-1},\ i=1,2,\dots ,n, \end{aligned}$$

it can be shown via the central limit theorem that

$$\begin{aligned} \varvec{{\mathcal {U}}}_n(\Theta ) =\frac{1}{\sqrt{n}}\sum _{i=1}^{n}\varvec{\Psi }_i(\Theta ) \overset{d}{\longrightarrow }{\mathcal {N}}\left( \varvec{0},E[\varvec{\Psi }_1(\Theta )\varvec{\Psi }_1^{T}(\Theta )]\right) . \end{aligned}$$

Therefore

$$\begin{aligned} \varvec{\Delta }\varvec{{\mathcal {U}}}_n(\Theta ) = \varvec{\Delta }\left[ \frac{1}{\sqrt{n}}\sum _{i=1}^{n}\varvec{\Psi }_i(\Theta )\right] \overset{d}{\longrightarrow }{\mathcal {N}}(\varvec{0},\varvec{\Delta }), \end{aligned}$$

and by Slutsky’s theorem $\sqrt{n}({\widehat{\Theta }}-\Theta )=\varvec{\Delta }\varvec{{\mathcal {U}}}_n(\Theta ) +o_p(\varvec{1})\overset{d}{\longrightarrow }N(\varvec{0},\varvec{\Delta })$ to finish the proof.