Using Controlled Feeding Study for Biomarker Development in Regression Calibration for Disease Association Estimation

Abstract

Correction for systematic measurement error in self-reported data is an important challenge in association studies of dietary intake and chronic disease risk. The regression calibration method has been used for this purpose when an objectively measured biomarker is available. However, a big limitation of the regression calibration method is that biomarkers have only been developed for a few dietary components. We propose new methods to use controlled feeding studies to develop valid biomarkers for many more dietary components and to estimate the diet disease associations. Asymptotic distribution theory for the proposed estimators is derived. Extensive simulation is performed to study the finite sample performance of the proposed estimators. We applied our method to examine the associations between the sodium/potassium intake ratio and cardiovascular disease incidence using the Women’s Health Initiative cohort data. We discovered positive associations between sodium/potassium ratio and the risks of coronary heart disease, nonfatal myocardial infarction, coronary death, ischemic stroke, and total cardiovascular disease.

Appendices

Appendix A: Technical Details

1.1 A.1: Bias Factor Derivation

In this subsection, we derived the asymptotic bias of ${\varvec{\theta }}$ for method 1 under rare disease assumption by checking how $E({\hat{X}}|Q,{\varvec{V}})$ is biased away from $E(X|Q,{\varvec{V}})$. For the first step regression model of ${\tilde{X}}$ on $({\varvec{W}}, {\varvec{V}})$, we have

$$\begin{aligned} {\hat{X}}= & {} E({\tilde{X}}|{\varvec{W}},{\varvec{V}})=E(X|{\varvec{W}},{\varvec{V}})=E(X|{\varvec{V}})+\left\{ {\varvec{W}}-E({\varvec{W}}|{\varvec{V}})\right\} ^\top \Sigma _{{\varvec{W}}{\varvec{W}}|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}^\top . \end{aligned}$$

where $\Sigma _{{\varvec{W}}{\varvec{W}}|V}=Var({\varvec{W}}|{\varvec{V}})$ and $\Sigma _{X{\varvec{W}}|{\varvec{V}}}=Cov(X,{\varvec{W}}|{\varvec{V}})$ are conditional variance covariance matrices. Plugging this into the second step regression model, we have

$$\begin{aligned} E({\hat{X}}|Q, {\varvec{V}})= & {} E\left\{ E\left({\hat{X}}|X, Q, {\varvec{V}}\right) | Q,{\varvec{V}}\right\} =E\left\{ E\left( {\hat{X}}|X,{\varvec{V}}\right) |Q,{\varvec{V}}\right\} \\= & {} E\left[ E\left[ E(X|{\varvec{V}})+\left\{ {\varvec{W}}-E({\varvec{W}}|{\varvec{V}})\right\} ^\top \Sigma _{{\varvec{W}}{\varvec{W}}|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}^\top |X,{\varvec{V}}\right] |Q,{\varvec{V}}\right] \\= & {} E\left[ E(X|{\varvec{V}})+\left\{ E({\varvec{W}}|X,{\varvec{V}})-E({\varvec{W}}|{\varvec{V}})\right\} ^\top \Sigma _{{\varvec{W}}{\varvec{W}}|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}^\top |Q,{\varvec{V}}\right] \\= & {} E\left[ E(X|{\varvec{V}})+\left\{ X-E(X|{\varvec{V}})\right\} \Sigma _{XX|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}\Sigma _{{\varvec{W}}{\varvec{W}}|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}^\top |Q,{\varvec{V}}\right] \\= & {} E\left[ E(X|{\varvec{V}})+\left\{ X-E(X|{\varvec{V}})\right\} \rho _{{\varvec{V}}}|Q,{\varvec{V}}\right] \\= & {} \rho _{{\varvec{V}}}E(X|Q,{\varvec{V}})+(1-\rho _{{\varvec{V}}})E(X|{\varvec{V}}). \end{aligned}$$

When the bias factor $\rho _{{\varvec{V}}}$ is a constant over ${\varvec{V}}$, we simply denote it as $\rho$ and we have $\rho \theta _z^{*}=\theta _z$, or $\theta _z^{*}=\rho ^{-1}\theta _z$ with appropriate adjustment for V. Explicitly, we have

$$\begin{aligned} \rho= & {} 1-\Sigma _{XX|{\varvec{V}}}^{-1}\left( \Sigma _{XX|{\varvec{V}}}-\Sigma _{X{\varvec{W}}|{\varvec{V}}}\Sigma _{{\varvec{W}}{\varvec{W}}|{\varvec{V}}}^{-1}\Sigma _{X{\varvec{W}}|{\varvec{V}}}^\top \right) \\= & {} 1-\frac{Var(X|{\varvec{W}},{\varvec{V}})}{Var(X|{\varvec{V}})}=R^2_{X,{\varvec{W}}|{\varvec{V}}}. \end{aligned}$$

If we further have $E(X|{\varvec{V}})={\varvec{V}}\delta$ is a linear function of V, then $(1-\rho )\delta \theta _z^{*}+\theta _v^{*}=\theta _v$, or $\theta _v^{*}=\theta _v-\frac{(1-\rho )\delta }{\rho }\theta _z$.

1.2 A.2: Proof of Theorem 1

Before proof of main theorem 1, we first proof the lemma below which give the asymptotics for the third step regression given the ${\hat{\gamma }}$ from the second step.

Lemma 1

Assume $n_3/n_2\rightarrow C_{2}<\infty$ and $\theta ^{*}$ is the unique solution to $EU(\theta ,\gamma ^{*})=0$ for an estimating function $U(\theta ,\gamma )$. If ${\hat{\theta }}$ solve the estimating equation $0=n_3^{-1}\sum _{i=1}^{n_3} U_i(\theta ,{\hat{\gamma }})$ where $\sqrt{n_2}({\hat{\gamma }}-\gamma ^{*})\rightarrow N(0,\Sigma _{\gamma })$ and ${\hat{\gamma }}$ is independent of $U_i(\theta ,\gamma )$, then we have $\sqrt{n_3}({\hat{\theta }}-\theta ^{*})\rightarrow N(0, \Sigma _{\theta })$ where

$$\begin{aligned} \Sigma _{\theta }=I_{\theta }^{-1}(J_{\theta }+C_2I_{\gamma }\Sigma _{\gamma }I_{\gamma }^\top )I_{\theta }^{-\top }, \end{aligned}$$

where $I_{\theta }=E\left( -\frac{\partial U(\theta ,\gamma )}{\partial \theta }|_{\theta ^{*},\gamma ^{*}}\right)$, $I_{\gamma }=E\left( -\frac{\partial U(\theta ,\gamma )}{\partial \gamma }|_{\theta ^{*},\gamma ^{*}}\right)$ and $J_{\theta }=Var(U(\theta ^{*},\gamma ^{*}))$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\theta }={\hat{I}}_{\theta }^{-1}\left( {\hat{J}}_{\theta }+\frac{n_3}{n_2}{\hat{I}}_{\gamma }{\hat{\Sigma }}_{\gamma }{\hat{I}}_{\gamma }^\top \right) {\hat{I}}_{\theta }^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\theta }=n_3^{-1}\sum _{i=1}^{n_3}\left( -\frac{\partial U_i(\theta ,\gamma )}{\partial \theta }|_{{\hat{\theta }},{\hat{\gamma }}}\right)$, ${\hat{I}}_{\gamma }=n_3^{-1}\sum _{i=1}^{n_3}\left( -\frac{\partial U_i(\theta ,\gamma )}{\partial \gamma }|_{{\hat{\theta }},{\hat{\gamma }}}\right)$, ${\hat{J}}_{\theta }=n_3^{-1}\sum _{i=1}^{n_3}U_i({\hat{\theta }},{\hat{\gamma }})U^\top _i({\hat{\theta }},{\hat{\gamma }})$ and ${\hat{\Sigma }}_{\gamma }$ is a consistent estimator of $\Sigma _{\gamma }$.

Proof

We can derive the asymptotic for ${\hat{\theta }}$ as below:

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_3} U_i({\hat{\theta }},{\hat{\gamma }})\\= & {} \sum _{i=1}^{n_3} \left\{ U_i (\theta ^{*},\gamma ^{*})+\frac{\partial U_i(\theta ,\gamma )}{\partial \theta }|_{\theta ^{*},\gamma ^{*}}({\hat{\theta }}-\theta ^{*}) +\frac{\partial U_i(\theta ,\gamma )}{\partial \gamma }|_{\theta ^{*},\gamma ^{*}}({\hat{\gamma }}-\gamma ^{*})+o(||{{{\hat{\theta }}}-\theta ^{*}}||,||{{{\hat{\gamma }}}-\gamma ^{*}}||)\right\} . \end{aligned}$$

With Uniform Law of Large Number (ULLN), we have

$$\begin{aligned} n_3^{-1}\sum _{i=1}^{n_3}\frac{\partial U_i(\theta , \gamma )}{\partial \theta }|_{\theta ^{*}, \gamma ^{*}}=E\left( \frac{\partial U(\theta ,\gamma )}{\partial \theta }|_{\theta ^{*},\gamma ^{*}}\right) +o_p(1)=-I_{\theta }+o_p(1).\\ \end{aligned}$$

Similarly, we have

$$\begin{aligned} n_3^{-1}\sum _{i=1}^{n_3}\frac{\partial U_i(\theta , \gamma )}{\partial \gamma }|_{\theta ^{*}, \gamma ^{*}}=-I_{\gamma }+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\theta ^{*},\gamma ^{*})=0$, we have

$$\begin{aligned} 0= & {} n_3^{-1}\sum _{i=1}^{n_3} \left\{ U_i(\theta ^{*},\gamma ^{*})-EU(\theta ^{*},\gamma ^{*})\right\} -I_{\theta }({\hat{\theta }}-\theta ^{*}) -I_{\gamma }({\hat{\gamma }}-\gamma ^{*})\\&+o(||{{{\hat{\theta }}}-\theta ^{*}}||,||{{{\hat{\gamma }}}-\gamma ^{*}}||)+o_p(1)*||{{{\hat{\gamma }}}-\gamma ^{*}}||+o_p(1)*||{{{\hat{\theta }}}-\theta ^{*}}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_3}({\hat{\theta }}-\theta ^{*})=I_{\theta }^{-1}\left\{ n_3^{-1/2}\sum _{i=1}^{n_3} U_i (\theta ^{*},\gamma ^{*})-I_{\gamma }\sqrt{n_3}({\hat{\gamma }}-\gamma ^{*})\right\} +o_p(1). \end{aligned}$$

By Central Limit Theorem (CLT), we have

$$\begin{aligned} n_3^{-1/2}\sum _{i=1}^{n_3} \left\{ U_i(\theta ^{*},\gamma ^{*})-EU(\theta ^{*},\gamma ^{*})\right\} \Rightarrow _d N(0,J_{\theta }) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\theta }=I_{\theta }^{-1}\left\{ J_{\theta }+C_2I_{\gamma }\Sigma _{\gamma }I_{\gamma }^\top \right\} I_{\theta }^{-\top }. \end{aligned}$$

By ULLN and continuous mapping theorem, we have ${\hat{I}}_{\theta }=I_{\theta }+o_p(1)$, ${\hat{I}}_{\gamma }=I_{\gamma }+o_p(1)$ and ${\hat{J}}_{\theta }=J_{\theta }+o_p(1)$. By assumption ${\hat{\Sigma }}_{\gamma }=\Sigma _{\gamma }+o_p(1)$ and $n_3/n_2\rightarrow C_{2}<\infty$, using continuous mapping theorem, we have ${\hat{\Sigma }}_{\theta }=\Sigma _{\theta }+o_p(1)$. $\square$

Now we provide the detail forms for the quantities involved in a Cox regression model.

Lemma 2

Assume $n_3/n_2\rightarrow C_{2}<\infty$ and $\theta ^{*}$ is the unique solution to

$$\begin{aligned} 0=U(\theta ,\gamma ^{*})=E\int _0^{\tau }\left[ \left( \begin{array}{c} Z^{*}\\ {\varvec{V}}\end{array}\right) -\frac{E\left[ \left( \begin{array}{c} Z^{*}\\ {\varvec{V}}\end{array}\right) Y(t)\exp \left\{ (Z^{*},{\varvec{V}}^\top )\theta \right\} \right] }{E\left[ Y(t)\exp \left\{ (Z^{*},{\varvec{V}}^\top )\theta \right\} \right] }\right] dN(t). \end{aligned}$$

where $Z^{*}=\mathbb {X}\gamma ^{*}$. If ${\hat{\theta }}$ solve the estimating equation

$$\begin{aligned} 0=n_3^{-1}\sum _{i=1}^{n_3}\int _0^{\tau }\left[ \left( \begin{array}{c} {\hat{Z}}_{i}\\ {\varvec{V}}_i\end{array}\right) -\sum _{j=1}^{n_3}\frac{Y_j(t)\exp \left\{ ({\hat{Z}}_{j},{\varvec{V}}_j^\top )\theta \right\} }{\sum _{k=1}^{n_3} Y_k(t)\exp \left\{ ({\hat{Z}}_{k},{\varvec{V}}_k^\top )\theta \right\} }\left( \begin{array}{c} {\hat{Z}}_{j}\\ {\varvec{V}}_j\end{array}\right) \right] dN_i(t), \end{aligned}$$

where ${\hat{Z}}_i=\mathbb {X}_i{\hat{\gamma }}$ and $\sqrt{n_2}({\hat{\gamma }}-\gamma ^{*})\rightarrow N(0,\Sigma _{\gamma })$, then we have $\sqrt{n_3}({\hat{\theta }}-\theta ^{*})\rightarrow N(0, \Sigma _{\theta })$ where

$$\begin{aligned} \Sigma _{\theta }=I_{\theta }^{-1}(J_{\theta }+C_2I_{\gamma }\Sigma _{\gamma }I_{\gamma }^\top )I_{\theta }^{-\top }, \\ I_{\theta }= & {} E\left[ Y_i(t)\exp \left\{ (Z_i^{*},{\varvec{V}}_i^\top )\theta \right\} \left( \begin{array}{c} Z_i^{*}\\ V_i\end{array}\right) ^{\otimes 2}\right] \\ J_{\theta }= & {} E\int _0^{\tau }\left[ \left( \begin{array}{c} Z_{i}^{*}\\ {\varvec{V}}_i\end{array}\right) -\frac{s^{(1)}(\theta ,t)}{s^{(0)}(\theta ,t)}\right] ^{\otimes 2}dN_i(t)\\ I_{\gamma }= & {} -E\left[ \int _0^{\tau }\left\{ \left( \begin{array}{c} \mathbb {X}_i \\ 0\end{array}\right) -\frac{A(\theta ,t)}{s^{(0)}(\theta ,t)}+\frac{s^{(1)}(\theta ,t)G(\theta ,t)}{s^{(0)}(\theta ,t)^2}\right\} dN_i(t)\right] .\\ \end{aligned}$$

where $a^{\otimes 2}=aa^T$ and

$$\begin{aligned} A(\theta ,t)= & {} E\left[ Y_i(t)\left( \begin{array}{c} 1+\theta _ZZ_i^{*}\\ \theta _Z{\varvec{V}}_i\end{array}\right) \exp \left\{ (Z_i^{*},{\varvec{V}}_i^\top )\theta \right\} \mathbb {X}_i\right] ,\\ G(\theta ,t)= & {} E\left[ Y_i(t)\exp \left\{ (Z_i^{*},{\varvec{V}}_i^\top )\theta \right\} \theta _Z\mathbb {X}_i\right] ,\\ s^{(0)}(\theta ,t)= & {} E\left[ Y_i(t)\exp \left\{ (Z_i^{*},{\varvec{V}}_i^\top )\theta \right\} \right] ,\\ s^{(1)}(\theta ,t)= & {} E\left[ Y_i(t)\exp \left\{ (Z_i^{*},{\varvec{V}}_i^\top )\theta \right\} \left( \begin{array}{c} Z_i^{*}\\ V_i\end{array}\right) \right] . \end{aligned}$$

This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\theta }={\hat{I}}_{\theta }^{-1}\left( {\hat{J}}_{\theta }+\frac{n_3}{n_2}{\hat{I}}_{\gamma }{\hat{\Sigma }}_{\gamma }{\hat{I}}_{\gamma }^\top \right) {\hat{I}}_{\theta }^{-\top }, \end{aligned}$$

where

$$\begin{aligned} {\hat{I}}_{\theta }= & {} n_3^{-1}\sum _{i=1}^{n_3}\left[ Y_i(t)\exp \left\{ ({\hat{Z}}_i,{\varvec{V}}_i^\top ){\hat{\theta }}\right\} \left( \begin{array}{c} {\hat{Z}}_i\\ V_i\end{array}\right) ^{\otimes 2}\right] \\ {\hat{J}}_{\theta }= & {} n_3^{-1}\sum _{i=1}^{n_3}\Delta _i \left[ \left( \begin{array}{c} {\hat{Z}}_{i}\\ {\varvec{V}}_i\end{array}\right) -\sum _j\frac{Y_j(T_i)\exp \left\{ ({\hat{Z}}_{j},{\varvec{V}}_j^\top ){\hat{\theta }}\right\} }{\sum _k Y_k(T_i\exp \left\{ ({\hat{Z}}_{k},{\varvec{V}}_k^\top ){\hat{\theta }}\right\} }\left( \begin{array}{c} {\hat{Z}}_{j}\\ {\varvec{V}}_j\end{array}\right) \right] ^{\otimes 2}\\ {\hat{I}}_{\gamma }= & {} n_3^{-1}\sum _{i=1}^{n_3} \Delta _i \left\{ \left( \begin{array}{c} \mathbb {X}_i \\ 0\end{array}\right) -\frac{{\hat{A}}({\hat{\theta }},T_i)}{s^{(0)}({\hat{\theta }},T_i)}+\frac{{\hat{s}}^{(1)}({\hat{\theta }},T_i){\hat{G}}({\hat{\theta }},T_i)}{{\hat{s}}^{(0)}({\hat{\theta }},T_i)^2}\right\} \\ {\hat{A}}(\theta ,t)= & {} n_3^{-1}\sum _{i=1}^{n_3}\left[ Y_i(t)\left( \begin{array}{c} 1+\theta _Z{\hat{Z}}_i\\ \theta _Z{\varvec{V}}_i\end{array}\right) \exp \left\{ ({\hat{Z}}_i,{\varvec{V}}_i^\top )\theta \right\} \mathbb {X}_i\right] ,\\ {\hat{G}}(\theta ,t)= & {} n_3^{-1}\sum _{i=1}^{n_3}\left[ Y_i(t)\exp \left\{ ({\hat{Z}}_i,{\varvec{V}}_i^\top )\theta \right\} \theta _Z\mathbb {X}_i\right] , \end{aligned}$$

${\hat{\Sigma }}_{\gamma }$ is a consistent estimator of $\Sigma _{\gamma }$ and $n_3/n_2 \rightarrow {C_2}$.

Proof

Denote $U_i(\theta ,\gamma )=\int _0^{\tau }\left[ \left( \begin{array}{c} Z_{i}\\ {\varvec{V}}_i\end{array}\right) -\frac{E\left[ Y_j(t)\exp \left\{ (Z_{j},{\varvec{V}}_j^\top )\theta \right\} \left( \begin{array}{c} Z_{j}\\ {\varvec{V}}_j\end{array}\right) \right] }{E \left[ Y_k(t)\exp \left\{ (Z_{k},{\varvec{V}}_k^\top )\theta \right\} \right] }\right] dN_i(t)$ where $Z_i=\mathbb {X}_i\gamma$. Then by definition of $I_{\theta }$, $I_{\gamma }$, $J_{\theta }$ in Lemma 1, we can compute the form of these terms as stated above. The convergence of ${\hat{\theta }}$ and ${\hat{\gamma }}$ as long as ULLN and continuous mapping ensure that we have

$$\begin{aligned} {\hat{A}}({\hat{\theta }},t)= & {} A(\theta ^{*},t)+o_p(1)\\ {\hat{G}}({\hat{\theta }},t)= & {} G(\theta ^{*},t)+o_p(1)\\ {\hat{s}}^{(0)}({\hat{\theta }},t)= & {} s^{(0)}(\theta ^{*},t)+o_p(1)\\ {\hat{s}}^{(1)}({\hat{\theta }},t)= & {} s^{(1)}(\theta ^{*},t)+o_p(1) \end{aligned}$$

which lead to the convergence ${\hat{I}}_{\theta }=I_{\theta }+o_p(1)$, ${\hat{I}}_{\gamma }=I_{\gamma }+o_p(1)$ and ${\hat{J}}_{\theta }=J_{\theta }+o_p(1)$. Applying Lemma 1, we have the asymptotic for ${\tilde{\theta }}$ that solve the equation $0=n_3^{-1}\sum _{i=1}^{n_3}U_i(\theta ,{\hat{\gamma }})$. Now we just need to show ${\tilde{\theta }}$ and ${\hat{\theta }}$ is asymptotically equivalent, which is guaranteed by applying ULLN to get:

$$\begin{aligned}&n_3^{-1}\sum _{i=1}^{n_3}Y_j(t)\exp \left\{ ({\hat{Z}}_{j},{\varvec{V}}_j^\top )\theta \right\} =s^{(0)}(\theta ,t)+o_p(1)\\&n_3^{-1}\sum _{i=1}^{n_3}Y_j(t)\exp \left\{ ({\hat{Z}}_{j},{\varvec{V}}_j^\top )\theta \right\} \left( \begin{array}{c} {\hat{Z}}_{j}\\ {\varvec{V}}_j\end{array}\right) =s^{(1)}(\theta ,t)+o_p(1) \end{aligned}$$

Here we would like to comment that for method 2–4, $Z^{*}$ has the same expression $E(Z|Q,{\varvec{V}})$ and thus the $I_{\theta }$ are the same though their estimated version ${\hat{I}}_{\theta }$ are different. $\square$

Now we handle the estimation of $\gamma$ in the following lemma.

Lemma 3

Assume $n_2/n_1\rightarrow C_{1}<\infty$ and $\gamma ^{*}$ is the unique solution to $EU(\gamma ,\beta ^{*})=0$ for an estimating function $U(\gamma ,\beta )=[(Q_i,{\varvec{V}}_i^\top )^T{\hat{X}}_i-(Q_i,{\varvec{V}}_i^\top )^T(Q_i,{\varvec{V}}_i^\top )\gamma ]$. If ${\hat{\gamma }}$ solve the estimating equation $0=n_3^{-1}\sum _{i=1}^{n_3} U_i(\gamma ,{\hat{\beta }})$ where $\sqrt{n_1}({\hat{\beta }}-\beta ^{*})\rightarrow N(0,\Sigma _{\beta })$, then we have $\sqrt{n_2}({\hat{\gamma }}-\gamma ^{*})\rightarrow N(0, \Sigma _{\gamma })$ where

$$\begin{aligned} \Sigma _{\gamma }=I_{\gamma }^{-1}(I_{\gamma }+C_1I_{\beta }\Sigma _{\beta }I_{\beta }^\top )I_{\gamma }^{-\top }, \end{aligned}$$

where $I_{\gamma }=E\left( -\frac{\partial U(\gamma ,\beta )}{\partial \gamma }|_{\gamma ^{*},\beta ^{*}}\right) =E[(Q_i,{\varvec{V}}_i^\top )^\top (Q_i,{\varvec{V}}_i^\top )]$ and $I_{\beta }=E\left( -\frac{\partial U(\gamma ,\beta )}{\partial \beta }|_{\gamma ^{*},\beta ^{*}}\right)$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\gamma }={\hat{I}}_{\gamma }^{-1}\left( {\hat{I}}_{\gamma }+\frac{n_2}{n_1}{\hat{I}}_{\beta }{\hat{\Sigma }}_{\beta }{\hat{I}}_{\beta }^\top \right) {\hat{I}}_{\gamma }^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\gamma }=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma ,\beta )}{\partial \gamma }|_{{\hat{\gamma }},{\hat{\beta }}}\right)$, ${\hat{I}}_{\beta }=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma ,\beta )}{\partial \beta }|_{{\hat{\gamma }},{\hat{\beta }}}\right)$ and ${\hat{\Sigma }}_{\beta }$ is a consistent estimator of $\Sigma _{\beta }$.

Proof

We can derive the asymptotic for ${\hat{\gamma }}$ as below:

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_2} U_i({\hat{\gamma }},{\hat{\beta }})\\= & {} \sum _{i=1}^{n_2} \left\{ U_i (\gamma ^{*},\beta ^{*})+\frac{\partial U}{\partial \gamma }|_{\gamma ^{*},\beta ^{*}}({\hat{\gamma }}-\gamma ^{*}) +\frac{\partial U}{\partial \beta }|_{\gamma ^{*},\beta ^{*}}({\hat{\beta }}-\beta ^{*})+o(||{{{\hat{\gamma }}}-\gamma ^{*}}||,||{{{\hat{\beta }}}-\beta ^{*}}||)\right\} . \end{aligned}$$

With ULLN,

$$\begin{aligned} n_2^{-1}\sum _{i=1}^{n_2}\frac{\partial U_i(\gamma , \beta )}{\partial \gamma }|_{{\hat{\gamma }}, {\hat{\beta }}}=E\left( \frac{\partial U}{\partial \gamma }|_{{\hat{\gamma }},{\hat{\beta }}}\right) +o_p(1) \end{aligned}$$

and with continuous mapping theorem, we have

$$\begin{aligned} E\left( \frac{\partial U}{\partial \gamma }|_{{\hat{\gamma }},{\hat{\beta }}}\right) =E\left( \frac{\partial U}{\partial \gamma }|_{\gamma ^{*},\beta ^{*}}\right) +o_p(1) \end{aligned}$$

So we have

$$\begin{aligned} n_2^{-1}\sum _{i=1}^{n_2}\frac{\partial U_i(\gamma , \beta )}{\partial \gamma }|_{{\hat{\gamma }}, {\hat{\beta }}}=-I_{\gamma }+o_p(1). \end{aligned}$$

Similarly, we have

$$\begin{aligned} n_2^{-1}\sum _{i=1}^{n_2}\frac{\partial U_i(\gamma , \beta )}{\partial \beta }|_{{\hat{\gamma }}, {\hat{\beta }}}=-I_{\beta }+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\gamma ^{*},\beta ^{*})=0$, we have

$$\begin{aligned} 0= & {} n_2^{-1}\sum _{i=1}^{n_2} \left\{ U_i(\gamma ^{*},\beta ^{*})-EU(\gamma ^{*},\beta ^{*})\right\} -I_{\gamma }({\hat{\gamma }}-\gamma ^{*}) -I_{\beta }({\hat{\beta }}-\beta ^{*})\\&+o(||{{{\hat{\gamma }}}-\gamma ^{*}}||,||{{{\hat{\beta }}}-\beta ^{*}}||)+o_p(1)*||{{{\hat{\beta }}}-\beta ^{*}}||+o_p(1)*||{{{\hat{\gamma }}}-\gamma ^{*}}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_2}({\hat{\gamma }}-\gamma ^{*})=I_{\gamma }^{-1}\left\{ n_2^{-1/2}\sum _{i=1}^{n_2} U_i (\gamma ^{*},\beta ^{*})-I_{\beta }\sqrt{n_2}({\hat{\beta }}-\beta ^{*})\right\} +o_p(1). \end{aligned}$$

By CLT, we have

$$\begin{aligned} n_2^{-1/2}\sum _{i=1}^{n_2} \left\{ U_i(\gamma ^{*},\beta ^{*})-EU(\gamma ^{*},\beta ^{*})\right\} \Rightarrow _d N(0,Var(U_i(\gamma ^{*},\beta ^{*}))) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\gamma }=I_{\gamma }^{-1}\left\{ Var(U(\gamma ,\beta ))+C_2I_{\beta }\Sigma _{\beta }I_{\beta }^\top \right\} I_{\gamma }^{-\top }. \end{aligned}$$

As we have shown ${\hat{I}}_{\gamma }=I_{\gamma }+o_p(1)$, ${\hat{I}}_{\beta }=I_{\beta }+o_p(1)$ and by assumption ${\hat{\Sigma }}_{\beta }=\Sigma _{\beta }+o_p(1)$ and $n_2/n_1\rightarrow C_{1}<\infty$, using continuous mapping theorem, we have ${\hat{\Sigma }}_{\gamma }=\Sigma _{\gamma }+o_p(1)$. $\square$

Applying the general form to each specific regression method, we can obtain the asymptotic results we need.

Lemma 4

Assume $n_2/n_1\rightarrow C_{1}<\infty$ and $\gamma ^{*}_1$ is the unique solution to $EU(\gamma _1,\beta ^{*}_1)=0$ for an estimating function $U(\gamma _1,\beta _1)$. If ${\hat{\gamma }}_1$ solve the estimating equation $0=n_2^{-1}\sum _{i=1}^{n_2} U_i(\gamma _1,{\hat{\beta }}_1)$ where $\sqrt{n_1}({\hat{\beta }}_1-\beta ^{*}_1)\rightarrow N(0,\Sigma _{\beta _1})$, then we have $\sqrt{n_2}({\hat{\gamma }}_1-\gamma ^{*}_1)\rightarrow N(0, \Sigma _{\gamma _1})$ where

$$\begin{aligned} \Sigma _{\gamma _1}=I_{\gamma _1}^{-1}(I_{\gamma _1}+C_1I_{\beta _1}\Sigma _{\beta _1}I_{\beta _1}^\top )I_{\gamma _1}^{-\top }, \end{aligned}$$

where $I_{\gamma _1}=E\left( -\frac{\partial U(\gamma _1,\beta _1)}{\partial \gamma _1}|_{\gamma ^{*}_1,\beta ^{*}_1}\right)$ and $I_{\beta _1}=E\left( -\frac{\partial U(\gamma _1,\beta _1)}{\partial \beta _1}|_{\gamma ^{*}_1,\beta ^{*}_1}\right)$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _1}={\hat{I}}_{\gamma _1}^{-1}\left( {\hat{I}}_{\gamma _1}+\frac{n_2}{n_1}{\hat{I}}_{\beta _1}{\hat{\Sigma }}_{\beta _1}{\hat{I}}_{\beta _1}^\top \right) {\hat{I}}_{\gamma _1}^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\gamma _1}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _1,\beta _1)}{\partial \gamma _1}|_{\hat{\gamma _1},\hat{\beta _1}}\right)$, ${\hat{I}}_{\beta _1}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _1,\beta _1)}{\partial \beta _1}|_{\hat{\gamma _1},\hat{\beta _1}}\right)$ and ${\hat{\Sigma }}_{\beta _1}$ is a consistent estimator of $\Sigma _{\beta _1}$.

Proof

We need to derive asymptotic for ${\hat{\beta }}_1$ and then apply 3. The asymptotic of ${\hat{\beta }}_1$ can be derived as below:

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_1} U_i({\hat{\beta }}_1)\\= & {} \sum _{i=1}^{n_1} \left\{ U_i (\beta ^{*}_1) +\frac{\partial U}{\partial \beta _1}|_{\beta _1^{*}}({\hat{\beta }}_1-\beta ^{*}_1)+o(|| {{\hat{\beta }}_1-\beta ^{*}_1}||)\right\} . \end{aligned}$$

where $U_i(\beta _1^{*})=(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top X_i^{*}-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top (1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )\beta _1^{*}.$ With ULLN,

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _1)}{\partial \beta _1}|_{{\hat{\beta }}_1}=E\left( \frac{\partial U}{\partial \beta _1}|_{{\hat{\beta }}_1}\right) +o_p(1) \end{aligned}$$

and with continuous mapping theorem, we have

$$\begin{aligned} E\left( \frac{\partial U}{\partial \beta _1}|_{{\hat{\beta }}_1}\right) =E\left( \frac{\partial U}{\partial \beta _1}|_{\beta ^{*}_1}\right) +o_p(1) \end{aligned}$$

So we have

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _1)}{\partial \beta _1}|_{{\hat{\beta }}_1}=-I_{\beta _1}+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\beta ^{*}_1)=0$, we have

$$\begin{aligned} 0= & {} n_1^{-1}\sum _{i=1}^{n_1} \left\{ U_i(\beta ^{*}_1)-EU(\beta ^{*}_1)\right\} -I_{\beta _1}(\hat{\beta _1}-\beta ^{*}_1)\\&+o_p(1)||{{\hat{\beta }}_1-\beta ^{*}_1}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_1}({\hat{\beta }}_1-\beta ^{*}_1)=I_{\beta _1}^{-1}\left\{ n_1^{-1/2}\sum _{i=1}^{n_1} U_i (\beta ^{*}_1) \right\} +o_p(1). \end{aligned}$$

By CLT, we have

$$\begin{aligned} n_1^{-1/2}\sum _{i=1}^{n_2} \left\{ U_i(\beta ^{*}_1)-EU(\beta ^{*}_1)\right\} \Rightarrow _d N(0,Var(U_i(\beta ^{*}_1))) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\beta _1}=I_{\beta _1}^{-1}\left\{ Var(U(\beta _1)) \right\} I_{\beta _1}^{-\top }. \end{aligned}$$

By assumption, ${\hat{\Sigma }}_{\beta _1}=\Sigma _{\beta _1}+o_p(1)$ which is a consistent estimator of $\Sigma _{\beta _1}$. Then by applying Lemma 3, we have

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _1}={\hat{I}}_{\gamma _1}^{-1}\left( {\hat{I}}_{\gamma _1}+\frac{n_2}{n_1}{\hat{I}}_{\beta _1}{\hat{\Sigma }}_{\beta _1}{\hat{I}}_{\beta _1}^\top \right) {\hat{I}}_{\gamma _1}^{-\top } \end{aligned}$$

$\square$

Lemma 5

Assume $n_2/n_1\rightarrow C_{1}<\infty$ and $\gamma ^{*}_2$ is the unique solution to $EU(\gamma _2,\beta ^{*}_2)=0$ for an estimating function $U(\gamma _2,\beta _2)$. If ${\hat{\gamma }}_2$ solve the estimating equation $0=n_2^{-1}\sum _{i=1}^{n_2} U_i(\gamma _2,{\hat{\beta }}_2)$ where $\sqrt{n_1}({\hat{\beta }}_2-\beta ^{*}_2)\rightarrow N(0,\Sigma _{\beta _2})$, then we have $\sqrt{n_2}({\hat{\gamma }}_2-\gamma ^{*}_2)\rightarrow N(0, \Sigma _{\gamma _2})$ where

$$\begin{aligned} \Sigma _{\gamma _2}=I_{\gamma _2}^{-1}(I_{\gamma _2}+C_1I_{\beta _2}\Sigma _{\beta _2}I_{\beta _2}^\top )I_{\gamma _2}^{-\top }, \end{aligned}$$

where $I_{\gamma _2}=E\left( -\frac{\partial U(\gamma _2,\beta _2)}{\partial \gamma _2}|_{\gamma ^{*}_2,\beta ^{*}_2}\right)$ and $I_{\beta _2}=E\left( -\frac{\partial U(\gamma _2,\beta _2)}{\partial \beta _2}|_{\gamma ^{*}_2,\beta ^{*}_2}\right)$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _2}={\hat{I}}_{\gamma _2}^{-1}\left( {\hat{I}}_{\gamma _2}+\frac{n_2}{n_1}{\hat{I}}_{\beta _2}{\hat{\Sigma }}_{\beta _2}{\hat{I}}_{\beta _2}^\top \right) {\hat{I}}_{\gamma _2}^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\gamma _2}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _2,\beta _2)}{\partial \gamma _2}|_{\hat{\gamma _2},\hat{\beta _2}}\right)$, ${\hat{I}}_{\beta _2}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _2,\beta _2)}{\partial \beta _2}|_{\hat{\gamma _2},\hat{\beta _2}}\right)$ and ${\hat{\Sigma }}_{\beta _2}$ is a consistent estimator of $\Sigma _{\beta _2}$.

Proof

Define $\beta _2=\frac{\beta _1}{BF}$.

First note that,

$$\begin{aligned} Var({\tilde{X}}|{\varvec{V}},{\varvec{W}})=\Omega _1=\frac{1}{n_1}\sum _i\left\{ {\tilde{X}}_i-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top \beta _1\right\} ^2,\\ Var({\tilde{X}}|{\varvec{V}})=\Omega _2=\frac{1}{n_1}\sum _i\left\{ {\tilde{X}}_i-(1,{\varvec{V}}_i^\top ){\beta _t}\right\} ^2, \end{aligned}$$

where we let $\Omega _{1i}=\left\{ {\tilde{X}}_i-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )\beta _1\right\} ^2$ and $\Omega _{2i}=\left( {\tilde{X}}_i-(1,{\varvec{V}}_i^\top )\beta _t\right) ^2$.

Second, the estimating equations considered are

$$\begin{aligned} U_{121i}=(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top \Omega _{1}^{-1}({\tilde{X}}_i-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )\beta _1),\\ U_{122i}=(1,{\varvec{V}}_i^\top )^\top \Omega _{2}^{-1}({\tilde{X}}_i-(1,{\varvec{V}}_i^\top )\beta _t).\\ U_{123i}=({\tilde{X}}_i-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top {\beta })^2-\Omega _{1},\\ U_{124i}=({\tilde{X}}_i-(1,{\varvec{V}}_i^\top ){\beta _t})^2-\Omega _{2}, \end{aligned}$$

Third, we can derive the asymptotic normal distribution for $\beta$, $\beta _t$, $\Omega _1$ and $\Omega _2$ as

$$\begin{aligned} \sqrt{n_1}\left[ \left( \begin{array}{c} {\hat{\beta }}_1\\ \hat{\beta _t} \\ {\widehat{\Omega }}_1 \\ {\widehat{\Omega }}_2 \end{array}\right) -\left( \begin{array}{c} \beta \\ \beta _t \\ \Omega _1 \\ \Omega _2 \end{array}\right) \right] \rightarrow N(0, I^{-1}JI^{-\top }), \end{aligned}$$

where J is the variance covariance matrix of the above four estimating equations and I is a matrix composed by the expectation of derivatives of each estimating equation with respect to $\beta$, $\beta _t$, $\Omega _1$ and $\Omega _2$, respectively. Specifically,

$$\begin{aligned} I= & {} \left( \begin{array}{cccc} \frac{1}{n_1}\sum _i(\mathbb {X}_i)^\top (\mathbb {X}_i)&{}0&{}0&{}0\\ 0 &{}\frac{1}{n_1}\sum _i(\mathbb {X}_{ti})^\top (\mathbb {X}_{ti})&{}0&{}0\\ 0&{}0&{}1&{}0\\ 0&{}0&{}0&{}1\end{array}\right) \\ \end{aligned}$$

where $\mathbb {X}_i=(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top )^\top$ and $\mathbb {X}_{ti}=(1,{\varvec{V}}_i^\top )^\top$ Fourth, the asymptotic normal distribution for ${\hat{\beta }}_1$ and ${\widehat{BF}}$ jointly can be derived using delta method.

$$\begin{aligned} \sqrt{n_1} \left[ \left( \begin{array}{c} {\hat{\beta }}\\ {\widehat{BF}}\end{array}\right) -\left( \begin{array}{c} \beta \\ BF\end{array}\right) \right] \rightarrow N(0,CI^{-1}JI^{-\top }C^\top ), \end{aligned}$$

where C is a matrix derived by taking derivative of $\beta$ and BF each with respect to $\beta$, $\beta _t$, $V_1$ and $V_2$ respectively. For example,

$$\begin{aligned} \frac{\partial {BF}}{\partial \beta }=0, \frac{\partial {BF}}{\partial \beta _t}=0, \frac{\partial {BF}}{\partial \Omega _1}=\frac{-1}{\Omega _2-{\tilde{\sigma }}_{x}^2}, \frac{\partial {BF}}{\partial V_2}=\frac{\Omega _1-{\tilde{\sigma }}_{x}^2}{(\Omega _2-{\tilde{\sigma }}_{x}^2)^2} \end{aligned}$$

Fifth, ${\hat{\beta }}_2=\frac{{\hat{\beta }}_1}{{\hat{BF}}}$ can be derived using delta method.

$$\begin{aligned} \sqrt{n_1} \left( \begin{array}{c} {\hat{\beta }}_2 \end{array} -\begin{array}{c} \beta _2 \end{array} \right) \rightarrow N(0,C'(CI^{-1}JI^{-\top }C^\top )C'^\top ), \end{aligned}$$

where C’ is a matrix derived by taking derivative of $\beta _2$ each with respect to $\beta$ and BF, respectively. That is,

$$\begin{aligned} \frac{\partial {\beta _2}}{\partial \beta }=\frac{1}{BF}, \frac{\partial {\beta _2}}{\partial BF}=-\frac{\beta }{BF^2} \end{aligned}$$

Then we have estimating equation for $\beta _2$ as below:

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_1} U_i({\hat{\beta }}_2)\\= & {} \sum _{i=1}^{n_1} \left\{ U_i (\beta ^{*}_2) +\frac{\partial U}{\partial \beta _2}|_{\beta _2^{*}}({\hat{\beta }}_2-\beta ^{*}_2)+o(||{{\hat{\beta }}_2-\beta ^{*}_2}||)\right\} . \end{aligned}$$

With ULLN,

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _2)}{\partial \beta _2}|_{{\hat{\beta }}_2}=E\left( \frac{\partial U}{\partial \beta _2}|_{{\hat{\beta }}_2}\right) +o_p(1) \end{aligned}$$

and with continuous mapping theorem, we have

$$\begin{aligned} E\left( \frac{\partial U}{\partial \beta _1}|_{{\hat{\beta }}_2}\right) =E\left( \frac{\partial U}{\partial \beta _1}|_{\beta ^{*}_2}\right) +o_p(1) \end{aligned}$$

So we have

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _2)}{\partial \beta _1}|_{{\hat{\beta }}_2}=-I_{\beta _1}+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\beta ^{*}_2)=0$, we have

$$\begin{aligned} 0= & {} n_1^{-1}\sum _{i=1}^{n_1} \left\{ U_i(\beta ^{*}_2)-EU(\beta ^{*}_2)\right\} -I_{\beta _1}(\hat{\beta _2}-\beta ^{*}_2)\\&+o_p(1)||{{\hat{\beta }}_2-\beta ^{*}_2}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_1}({\hat{\beta }}_2-\beta ^{*}_2)=I_{\beta _2}^{-1}\left\{ n_1^{-1/2}\sum _{i=1}^{n_1} U_i (\beta ^{*}_2) \right\} +o_p(1). \end{aligned}$$

By CLT, we have

$$\begin{aligned} n_1^{-1/2}\sum _{i=1}^{n_2} \left\{ U_i(\beta ^{*}_2)-EU(\beta ^{*}_2)\right\} \Rightarrow _d N(0,Var(U_i(\beta ^{*}_2))) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\beta _2}=I_{\beta _2}^{-1}\left\{ Var(U(\beta _2)) \right\} I_{\beta _2}^{-\top }. \end{aligned}$$

By assumption, ${\hat{\Sigma }}_{\beta _2}=\Sigma _{\beta _2}+o_p(1)$ which is a consistent estimator of $\Sigma _{\beta _2}$. Then by applying Lemma 3, we have

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _2}={\hat{I}}_{\gamma _2}^{-1}\left( {\hat{I}}_{\gamma _2}+\frac{n_2}{n_1}{\hat{I}}_{\beta _2}{\hat{\Sigma }}_{\beta _2}{\hat{I}}_{\beta _2}^\top \right) {\hat{I}}_{\gamma _2}^{-\top } \end{aligned}$$

$\square$

Lemma 6

Assume $n_2/n_1\rightarrow C_{1}<\infty$ and $\gamma ^{*}_3$ is the unique solution to $EU(\gamma _3,\beta ^{*}_3)=0$ for an estimating function $U(\gamma _3,\beta _3)$. If ${\hat{\gamma }}_3$ solve the estimating equation $0=n_2^{-1}\sum _{i=1}^{n_2} U_i(\gamma _3,{\hat{\beta }}_3)$ where $\sqrt{n_1}({\hat{\beta }}_3-\beta ^{*}_3)\rightarrow N(0,\Sigma _{\beta _3})$, then we have $\sqrt{n_2}({\hat{\gamma }}_3-\gamma ^{*}_3)\rightarrow N(0, \Sigma _{\gamma _3})$ where

$$\begin{aligned} \Sigma _{\gamma _3}=I_{\gamma _3}^{-1}(I_{\gamma _3}+C_1I_{\beta _3}\Sigma _{\beta _3}I_{\beta _3}^\top )I_{\gamma _3}^{-\top }, \end{aligned}$$

where $I_{\gamma _3}=E\left( -\frac{\partial U(\gamma _3,\beta _3)}{\partial \gamma _3}|_{\gamma ^{*}_3,\beta ^{*}_3}\right)$ and $I_{\beta _3}=E\left( -\frac{\partial U(\gamma _3,\beta _3)}{\partial \beta _3}|_{\gamma ^{*}_3,\beta ^{*}_3}\right)$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _3}={\hat{I}}_{\gamma _3}^{-1}\left( {\hat{I}}_{\gamma _3}+\frac{n_2}{n_1}{\hat{I}}_{\beta _3}{\hat{\Sigma }}_{\beta _3}{\hat{I}}_{\beta _3}^\top \right) {\hat{I}}_{\gamma _3}^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\gamma _3}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _3,\beta _3)}{\partial \gamma _3}|_{\hat{\gamma _3},\hat{\beta _3}}\right)$, ${\hat{I}}_{\beta _3}=n_2^{-1}\sum _{i=1}^{n_2}\left( -\frac{\partial U_i(\gamma _3,\beta _3)}{\partial \beta _3}|_{\hat{\gamma _3},\hat{\beta _3}}\right)$ and ${\hat{\Sigma }}_{\beta _3}$ is a consistent estimator of $\Sigma _{\beta _3}$.

Proof

We can derive the asymptotic for ${\hat{\beta }}_3$ as below:

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_1} U_i({\hat{\beta }}_3)\\= & {} \sum _{i=1}^{n_1} \left\{ U_i (\beta ^{*}_3) +\frac{\partial U}{\partial \beta _3}|_{\beta _3^{*}}({\hat{\beta }}_3-\beta ^{*}_3)+o(||{{\hat{\beta }}_3-\beta ^{*}_3}||)\right\} . \end{aligned}$$

where $U_i(\beta _3^{*})=(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)^\top {\tilde{X}}_i-(1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)^\top (1,{\varvec{W}}_i^\top ,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)\beta _3^{*}.$ With ULLN,

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _3)}{\partial \beta _3}|_{{\hat{\beta }}_3}=E\left( \frac{\partial U}{\partial \beta _3}|_{{\hat{\beta }}_3}\right) +o_p(1) \end{aligned}$$

and with continuous mapping theorem, we have

$$\begin{aligned} E\left( \frac{\partial U}{\partial \beta _3}|_{{\hat{\beta }}_3}\right) =E\left( \frac{\partial U}{\partial \beta _3}|_{\beta ^{*}_3}\right) +o_p(1) \end{aligned}$$

So we have

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\beta _3)}{\partial \beta _3}|_{{\hat{\beta }}_3}=-I_{\beta _3}+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\beta ^{*}_3)=0$, we have

$$\begin{aligned} 0= & {} n_1^{-1}\sum _{i=1}^{n_1} \left\{ U_i(\beta ^{*}_3)-EU(\beta ^{*}_3)\right\} -I_{\beta _3}({\hat{\beta }}_3-\beta ^{*}_3)\\&+o_p(1)||{{\hat{\beta }}_3-\beta ^{*}_3}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_1}({\hat{\beta }}_3-\beta ^{*}_3)=I_{\beta _3}^{-1}\left\{ n_1^{-1/2}\sum _{i=1}^{n_1} U_i (\beta ^{*}_3) \right\} +o_p(1). \end{aligned}$$

By CLT, we have

$$\begin{aligned} n_1^{-1/2}\sum _{i=1}^{n_2} \left\{ U_i(\beta ^{*}_3)-EU(\beta ^{*}_3)\right\} \Rightarrow _d N(0,Var(U_i(\beta ^{*}_3))) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\beta _3}=I_{\beta _3}^{-1}\left\{ Var(U(\beta _3)) \right\} I_{\beta _3}^{-\top }. \end{aligned}$$

By assumption, ${\hat{\Sigma }}_{\beta _3}=\Sigma _{\beta _3}+o_p(1)$ which is a consistent estimator of $\Sigma _{\beta _3}$. By applying Lemma 3, we have

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _3}={\hat{I}}_{\gamma _3}^{-1}\left( {\hat{I}}_{\gamma _3}+\frac{n_2}{n_1}{\hat{I}}_{\beta _3}{\hat{\Sigma }}_{\beta _3}{\hat{I}}_{\beta _3}^\top \right) {\hat{I}}_{\gamma _3}^{-\top } \end{aligned}$$

$\square$

Lemma 7

Assume $\gamma ^{*}_4$ is the unique solution to $EU(\gamma _4)=0$ for an estimating function $U(\gamma _4)$. Solving the estimating equation $0=n_1^{-1}\sum _{i=1}^{n_1} U_i({\hat{\gamma }}_4)$, we have $\sqrt{n_1}({\hat{\gamma }}_4-\gamma ^{*}_4)\rightarrow N(0, \Sigma _{\gamma _4})$ where

$$\begin{aligned} \Sigma _{\gamma _4}=I_{\gamma _4}^{-1}(Var(U(\gamma _4)))I_{\gamma _4}^{-\top }, \end{aligned}$$

where $I_{\gamma _4}=E\left( -\frac{\partial U(\gamma _4)}{\partial \gamma _4}|_{\gamma ^{*}_4}\right)$. This variance can be consistently estimated by

$$\begin{aligned} {\hat{\Sigma }}_{\gamma _4}={\hat{I}}_{\gamma _4}^{-1}(Var(U(\gamma _4))){\hat{I}}_{\gamma _4}^{-\top }, \end{aligned}$$

where ${\hat{I}}_{\gamma _4}=n_1^{-1}\sum _{i=1}^{n_1}\left( -\frac{\partial U_i(\gamma _4)}{\partial \gamma _4}|_{\hat{\gamma _4}}\right)$ and ${\hat{\Sigma }}_{\gamma _4}$ is a consistent estimator of $\Sigma _{\gamma _4}$.

Proof

Derive asymptotic ${\hat{\gamma }}_4$ for method 4 directly

$$\begin{aligned} 0= & {} \sum _{i=1}^{n_1} U_i({\hat{\gamma }}_4)\\= & {} \sum _{i=1}^{n_1} \left\{ U_i (\gamma ^{*}_4) +\frac{\partial U}{\partial \gamma _4}|_{\gamma _4^{*}}({\hat{\gamma }}_4-\gamma ^{*}_4)+o(|| {{\hat{\gamma }}_4-\gamma ^{*}_4}||)\right\} . \end{aligned}$$

where $U_i(\gamma _4^{*})=(1,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)^\top {\tilde{X}}_i-(1,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)^T(1,{\varvec{V}}_i^\top ,{\varvec{Q}}_i)\gamma _4^{*}.$ With ULLN,

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\gamma _4)}{\partial \gamma _4}|_{{\hat{\gamma }}_4}=E\left( \frac{\partial U}{\partial \gamma _4}|_{{\hat{\gamma }}_4}\right) +o_p(1) \end{aligned}$$

and with continuous mapping theorem, we have

$$\begin{aligned} E\left( \frac{\partial U}{\partial \gamma _4}|_{{\hat{\gamma }}_4}\right) =E\left( \frac{\partial U}{\partial \gamma _4}|_{\gamma ^{*}_4}\right) +o_p(1) \end{aligned}$$

So we have

$$\begin{aligned} n_1^{-1}\sum _{i=1}^{n_1}\frac{\partial U_i(\gamma _4)}{\partial \gamma _4}|_{{\hat{\gamma }}_4}=-I_{\gamma _4}+o_p(1). \end{aligned}$$

Plug in the Taylor expansion, notice that $EU(\gamma ^{*}_4)=0$, we have

$$\begin{aligned} 0= & {} n_1^{-1}\sum _{i=1}^{n_1} \left\{ U_i(\gamma ^{*}_4)-EU(\gamma ^{*}_4)\right\} -I_{\gamma _4}(\hat{\gamma _4}-\gamma ^{*}_4)\\&+o_p(1)||{{\hat{\gamma }}_4-\gamma ^{*}_4}||. \end{aligned}$$

So we have

$$\begin{aligned} \sqrt{n_1}({\hat{\gamma }}_4-\gamma ^{*}_4)=I_{\gamma _4}^{-1}\left\{ n_1^{-1/2}\sum _{i=1}^{n_1} U_i (\gamma ^{*}_4) \right\} +o_p(1). \end{aligned}$$

By CLT, we have

$$\begin{aligned} n_1^{-1/2}\sum _{i=1}^{n_2} \left\{ U_i(\gamma ^{*}_4)-EU(\gamma ^{*}_4)\right\} \Rightarrow _d N(0,Var(U_i(\gamma ^{*}_4))) \end{aligned}$$

So we have

$$\begin{aligned} \Sigma _{\gamma _4}=I_{\gamma _4}^{-1}\left\{ Var(U(\gamma _4)) \right\} I_{\gamma _4}^{-\top }. \end{aligned}$$

By assumption, ${\hat{\Sigma }}_{\gamma _4}=\Sigma _{\gamma _4}+o_p(1)$ which is a consistent estimator of $\Sigma _{\gamma _4}$. $\square$

Combine different steps together, we obtain the final asymptotics result.

Theorem 2

Provide the asymptotic for method 1–4 under Cox model

With $\frac{n_3}{n_2}\rightarrow C_2$ and $\frac{n_2}{n_1}\rightarrow C_1$, we have $\sqrt{n_3}({\hat{\theta }}_1-\theta _1^{*})\rightarrow N(0, \Sigma _{\theta _ 1})$ where $\Sigma _{\theta _1}=I_{\theta _1}^{-1}(I_{\theta _1}+C_2I_{\gamma _1}\Sigma _{\gamma _1}I_{\gamma _1}^\top )I_{\theta _1}^{-\top }$ can be consistently estimated by ${\hat{\Sigma }}_{\theta _1}={\hat{I}}_{\theta _1}^{-1}({\hat{I}}_{\theta _ 1}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _1}{\hat{\Sigma }}_{\gamma _1}{\hat{I}}_{\gamma _1}^\top ){\hat{I}}_{\theta _1}^{-\top }$ for method 1.

With $\frac{n_3}{n_2}\rightarrow C_2$ and $\frac{n_2}{n_1}\rightarrow C_1$, we have $\sqrt{n_3}({\hat{\theta }}_2-\theta _2^{*})\rightarrow N(0, \Sigma _{\theta _ 2})$ where $\Sigma _{\theta _2}=I_{\theta _2}^{-1}(I_{\theta _2}+C_2I_{\gamma _2}\Sigma _{\gamma _2}I_{\gamma _2}^\top )I_{\theta _2}^{-\top }$ can be consistently estimated by ${\hat{\Sigma }}_{\theta _2}={\hat{I}}_{\theta _2}^{-1}({\hat{I}}_{\theta _ 2}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _2}{\hat{\Sigma }}_{\gamma _2}{\hat{I}}_{\gamma _2}^\top ){\hat{I}}_{\theta _2}^{-\top }$ for method 2.

With $\frac{n_3}{n_2}\rightarrow C_2$ and $\frac{n_2}{n_1}\rightarrow C_1$, we have $\sqrt{n_3}({\hat{\theta }}_3-\theta _3^{*})\rightarrow N(0, \Sigma _{\theta _ 3})$ where $\Sigma _{\theta _3}=I_{\theta _3}^{-1}(I_{\theta _3}+C_2I_{\gamma _3}\Sigma _{\gamma _3}I_{\gamma _3}^\top )I_{\theta _3}^{-\top }$ can be consistently estimated by ${\hat{\Sigma }}_{\theta _3}={\hat{I}}_{\theta _3}^{-1}({\hat{I}}_{\theta _ 3}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _3}{\hat{\Sigma }}_{\gamma _3}{\hat{I}}_{\gamma _3}^\top ){\hat{I}}_{\theta _3}^{-\top }$ for method 3.

With $\frac{n_3}{n_2}\rightarrow C_2$ and $\frac{n_2}{n_1}\rightarrow C_1$, we have $\sqrt{n_3}({\hat{\theta }}_4-\theta _4^{*})\rightarrow N(0, \Sigma _{\theta _ 4})$ where $\Sigma _{\theta _4}=I_{\theta _4}^{-1}(I_{\theta _4}+C_2I_{\gamma _4}\Sigma _{\gamma _4}I_{\gamma _4}^\top )I_{\theta _4}^{-\top }$ can be consistently estimated by ${\hat{\Sigma }}_{\theta _4}={\hat{I}}_{\theta _4}^{-1}({\hat{I}}_{\theta _ 4}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _4}{\hat{\Sigma }}_{\gamma _4}{\hat{I}}_{\gamma _4}^\top ){\hat{I}}_{\theta _4}^{-\top }$ for method 4.

Proof

By applying Lemmas 2 and  4, the asymptotic ${\Sigma }_{\theta _1}$ can be derived as ${\hat{\Sigma }}_{\theta _1}={\hat{I}}_{\theta _1}^{-1}({\hat{I}}_{\theta _ 1}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _1}{\hat{\Sigma }}_{\gamma _1}{\hat{I}}_{\gamma _1}^\top ){\hat{I}}_{\theta _1}^{-\top }$.

By applying Lemmas 2 and  5, the asymptotic ${\Sigma }_{\theta _2}$ can be derived as ${\hat{\Sigma }}_{\theta _2}={\hat{I}}_{\theta _2}^{-1}({\hat{I}}_{\theta _ 2}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _2}{\hat{\Sigma }}_{\gamma _2}{\hat{I}}_{\gamma _2}^\top ){\hat{I}}_{\theta _2}^{-\top }$.

By applying Lemmas 2 and  6, the asymptotic ${\Sigma }_{\theta _3}$ can be derived as ${\hat{\Sigma }}_{\theta _3}={\hat{I}}_{\theta _3}^{-1}({\hat{I}}_{\theta _ 3}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _3}{\hat{\Sigma }}_{\gamma _3}{\hat{I}}_{\gamma _3}^\top ){\hat{I}}_{\theta _3}^{-\top }$.

By applying Lemmas 2 and  7, the asymptotic ${\Sigma }_{\theta _4}$ can be derived as ${\hat{\Sigma }}_{\theta _4}={\hat{I}}_{\theta _4}^{-1}({\hat{I}}_{\theta _ 4}+\frac{n_3}{n_2}{\hat{I}}_{\gamma _4}{\hat{\Sigma }}_{\gamma _4}{\hat{I}}_{\gamma _4}^\top ){\hat{I}}_{\theta _4}^{-\top }$. $\square$

1.3 A.3: Efficiency Comparison

In this part, we heuristically compare the efficiency of method 3 and method 4 by comparing the average variation in the estimation of Z. For simplicity, we can assume without loss of generality that all variables are centered and only compare the efficiency in estimating ${\widehat{Z}}$ because the variance of ${\widehat{\theta }}$ is a monotone function of the variance of ${\widehat{Z}}$. We compare the expected variance under fixed design. For method 4,

$$\begin{aligned} E[Var({\widehat{Z}}_4|Q,{\varvec{V}})]=E[n_1^{-1}(1,Q,{\varvec{V}}^{\top })\left\{ (1,Q,{\varvec{V}}^{\top })^{\top }(1,Q,{\varvec{V}}^{\top })\right\} ^{-1}(1,Q,{\varvec{V}}^{\top })^{\top }Var({\tilde{X}}|Q,{\varvec{V}})], \end{aligned}$$

and for method 3,

$$\begin{aligned}&E[Var({\widehat{Z}}_3|Q,{\varvec{V}})]\\= & {} E[n_2^{-1}(1,Q,{\varvec{V}}^{\top })\left\{ (1,Q,{\varvec{V}}^{\top })^{\top }(1,Q,{\varvec{V}}^{\top })\right\} ^{-1}(1,Q,{\varvec{V}}^{\top })^{\top }Var({\tilde{X}}|Q,{\varvec{V}})]\\&+(n_1^{-1}-n_2^{-1})(1,Q,{\varvec{V}}^{\top })\left\{ (1,Q,{\varvec{V}}^{\top })^{\top }(1,Q,{\varvec{V}}^{\top })\right\} ^{-1}(1,Q,{\varvec{V}}^{\top })^{\top }Var({\tilde{X}}|Q,{\varvec{V}},{\varvec{W}})\\= & {} n_1^{-1}(1,Q,{\varvec{V}}^{\top })\left\{ (1,Q,{\varvec{V}}^{\top })^{\top }(1,Q,{\varvec{V}}^{\top })\right\} ^{-1}(1,Q,{\varvec{V}}^{\top })^{\top }Var({\tilde{X}}|Q,{\varvec{V}})\\&\times \left\{ \frac{n_1}{n_2}+(1-\frac{n_1}{n_2})(1-R^2_{({\tilde{X}},{\varvec{W}})|Q,{\varvec{V}}})\right\} . \end{aligned}$$

Appendix B: Details of Simulation Settings and Additional Results

1.1 B.1: Details of Simulation Settings

For each setting, we evaluate the strength of biomarker, self-reported data as well as observed prediction strength from each stage. Specifically, we consider the following quantities, coefficient and partial coefficient of multiple determination on long-term dietary intake by biomarker, mathematically, $R_{ZWV}^2=1-\frac{Var(Z|W,V)}{Var(Z)}$, $R_{ZW|V}^2=1-\frac{Var(Z|W,V)}{Var(Z|V)}$; coefficient and partial coefficient of multiple determination on long-term dietary intake by self-reported data, mathematically, $R_{ZQV}^2=1-\frac{Var(Z|Q,V)}{Var(Z)}$, $R_{ZQ|V}^2=1-\frac{Var(Z|Q,V)}{Var(Z|V)}$; coefficient and partial coefficient of multiple determination on long-term dietary intake by self-reported data and biomarker, mathematically, $R_{ZQWV}^2=1-\frac{Var(Z|W,Q,V)}{Var(Z)}$, $R_{ZQW|V}^2=1-\frac{Var(Z|W,Q,V)}{Var(Z|V)}$; coefficient and partial coefficient of multiple determination on consumed dietary intake by biomaker, mathematically, $R_{{{\tilde{X}}}WV}^2=1-\frac{Var({{\tilde{X}}}|W,V)}{Var({{\tilde{X}}})}$, $R_{{{\tilde{X}}}W|V}^2=1-\frac{Var({{\tilde{X}}}|W,V)}{Var({{\tilde{X}}}|V)}$; coefficient and partial coefficient of multiple determination on consumed dietary intake by biomaker and self-reported data, mathematically, $R_{{{\tilde{X}}}WQV}^2=1-\frac{Var({{\tilde{X}}}|W,Q,V)}{Var({{\tilde{X}}})}$, $R_{{{\tilde{X}}}WQ|V}^2=1-\frac{Var({{\tilde{X}}}|W,Q,V)}{Var({{\tilde{X}}}|V)}$, coefficient and partial coefficient of multiple determination on estimated dietary intake with Method 2 by self-reported data, mathematically, $R_{{\hat{X}}_2QV}^2=1-\frac{Var({\hat{X}}_2|Q,V)}{Var({\hat{X}}_2)}$ and $R_{{\hat{X}}_2Q|V}^2=1-\frac{Var({\hat{X}}_2|Q,V)}{Var({\hat{X}}_2|V)}$.

In settings 1, 2 and 3, we fixed the effect on Q by setting $a_0=4$, $a_1=1.5$ and $\sigma _q=3$. In setting 4, 5 and 6, we set $a_0=0.4$, $a_1=2$ and $\sigma _q=4$. In addition, we decrease the coefficient of X on W from 1.3 to 0.8 in the first three settings while we decrease the coefficient of X on W from 1.1 to 0.5 in the last three settings. Table 5 displays all different types of $R^2$ mentioned above for all six settings. To be more specific, by fixing the strength of self-reported data and correlation between true dietary intake and personal characteristic, we gradually decreased the strength of biomarker in the first three settings. In the last three settings, the correlation between true dietary intake and personal characteristic is set to be 0. The strength of self-reported data is again fixed but at a different level compared to the first three settings. We again decreased the strength of biomarker gradually in the last three settings. Below is the list of the six settings with varying parameters.

$b_1=1.3$, $\rho =0.6$, $a_0=4$, $a_1=1.5$, $\sigma _q=3$ (Setting 1);

$b_1=1.1$, $\rho =0.6$, $a_0=4$, $a_1=1.5$, $\sigma _q=3$ (Setting 2);

$b_1=0.8$, $\rho =0.6$, $a_0=4$, $a_1=1.5$, $\sigma _q=3$ (Setting 3);

$b_1=1.1$, $\rho =0$, $a_0=0.4$, $a_1=2$, $\sigma _q=4$ (Setting 4);

$b_1=0.8$, $\rho =0$, $a_0=0.4$, $a_1=2$, $\sigma _q=4$ (Setting 5);

$b_1=0.5$, $\rho =0$, $a_0=0.4$, $a_1=2$, $\sigma _q=4$ (Setting 6); The values of $R^2$s are listed in Table 5.

Table 5 List of $R^2$ and censoring rates (CR) under mixed censoring (Mix), uniform censoring (Unif) and exponential censoring (Exp) among different measurements

1.2 B.2: Additional Simulation Results

To better compare the robustness of these methods, we further conduct simulation under the setting where the relationships between the self-reported dietary intake and the short-term dietary intake are different before and after the controlled feeding study. Tables 6 and 7 summarize simulation results when the correlation structures between the FFQ and the true dietary intakes of the controlled feeding study and the full cohort are different under settings 1 and 4. With the correlation unequal between controlled feeding study and full cohort, method 2 with BF added has shown more robustness in controlling bias compared with method 3 and method 4. With the increase of difference from 10 to 50% in association between Q and Z, the performance of both method 3 and method 4 become worse (larger bias) while the performance in controlling bias of the estimator from method 2 is consistently good in most cases. In addition, in most settings, we observe that method 2 has smaller SD compared with method 3–4. The performance of method 2 is adequate even when partial $R^2$s (i.e., Setting 3: $R^2_{{\tilde{X}}W|V}=0.21;$ Setting 6: $R^2_{{\tilde{X}}W|V}=0.16$) from the biomarker construction step and the calibration equation building step are both low, which suggests that in the real application of method 2, one may not need to be too stringent on the threshold of partial $R^2$.

Table 6 Simulation results comparing different methods’ performance when the correlation structures between the short-term dietary intake and true dietary intakes of the controlled feeding study and the full cohort are different when $\lambda _0(t)= {0.002}$

Table 7 Simulation results comparing different methods’ performance when the correlation structures between the short-term dietary intake and true dietary intakes of the controlled feeding study and the full cohort are different when $\lambda _0(t)= {0.004}$