Appendix
Derivation of Eq. (32) First, we define
$$\begin{aligned} \alpha _{ij}=\delta _{ij}-\Delta _1 \hat{a}-{\Delta }_2 \hat{a},\qquad \beta _{ij}={\epsilon }_{ij}-\Delta _1 \hat{b}-{\Delta }_2\hat{b}, \end{aligned}$$
and
$$\begin{aligned} \zeta _{ij}= \frac{2}{ \tilde{R}_i}\tilde{u}_{ij} \alpha _{ij} +\frac{2}{ \tilde{R}_i}\tilde{v}_{ij}\beta _{ij} + \frac{\alpha _{ij}^2}{\tilde{R}_i^2}+\frac{\beta _{ij}^2}{\tilde{R}_i^2}. \end{aligned}$$
Now, for each \(i=1\), 2 and \(j=1,\ldots ,n_i\), expanding the distance \(d_{ij}\) to the second-order terms gives
$$\begin{aligned} d_{ij} =\tilde{R}_i \sqrt{1+\zeta _{ij}} - \tilde{R}_i- \Delta _1 \hat{R}_i-\Delta _2 \hat{R}_i. \end{aligned}$$
This expression can be simplified further using the approximation
$$\begin{aligned} \sqrt{1+\zeta _{ij}}&\approx 1+ \frac{1}{2} \zeta _{ij}-\frac{1}{8} \zeta _{ij}^2\\&\approx 1+ \frac{2}{ 2\tilde{R}_i}\tilde{u}_{ij} \alpha _{ij} +\frac{2}{ 2\tilde{R}_i}\tilde{v}_{ij}\beta _{ij} + \frac{\alpha _{ij}^2}{2\tilde{R}_i^2}+\frac{ \beta _{ij}^2}{2\tilde{R}_i^2}\\&\qquad -\frac{1}{8}\left( \frac{2}{\tilde{R}_i}\tilde{u}_{ij} \alpha _{ij} +\frac{2}{ \tilde{R}_i}\tilde{v}_{ij}\beta _{ij}\right) ^2. \end{aligned}$$
Keeping all terms of order \(\mathscr {O}_\mathrm{P}(\sigma ^2)\) and discarding the less significant terms in \(\sqrt{1+\zeta _{ij}}\), the above expression becomes
$$\begin{aligned}&\approx 1+ \frac{2}{ 2\tilde{R}_i}\tilde{u}_{ij} \alpha _{ij} +\frac{2}{ 2\tilde{R}_i}\tilde{v}_{ij}\beta _{ij} + \frac{(\delta _{ij}-\Delta _1 \hat{a})^2}{2\tilde{R}_i^2}\\&\qquad +\frac{ ({\epsilon }_{ij}-\Delta _1 \hat{b})^2}{2\tilde{R}_i^2} \\&\qquad -\frac{1}{2\tilde{R}_i^2}\left( \tilde{u}_{ij} (\delta _{ij}-\Delta _1 \hat{a}) +\tilde{v}_{ij}({\epsilon }_{ij}-\Delta _1 \hat{b})\right) ^2 \\&\approx 1+ \frac{1}{\tilde{R}_i}\tilde{u}_{ij} \alpha _{ij} +\frac{1}{ \tilde{R}_i}\tilde{v}_{ij}\beta _{ij} + \frac{\left( 1-\tilde{u}_{ij}^2\right) (\delta _{ij}-\Delta _1 \hat{a})^2}{2\tilde{R}_i^2}\\&\qquad +\frac{ \left( 1-\tilde{v}_{ij}^2\right) ({\epsilon }_{ij}-\Delta _1 \hat{b})^2}{2\tilde{R}_i^2}\\&\qquad -\frac{2}{2\tilde{R}_i^2}\, \tilde{u}_{ij}\tilde{v}_{ij} (\delta _{ij}-\Delta _1 \hat{a}) ({\epsilon }_{ij}-\Delta _1 \hat{b}). \end{aligned}$$
Since \(1-\tilde{u}_{ij}^2=\tilde{v}_{ij}^2\), we conclude
$$\begin{aligned} d_{ij}&=\tilde{R}_i\sqrt{1+\zeta _{ij}}-\hat{R}_i\\&\approx \tilde{u}_{ij} \alpha _{ij} +\tilde{v}_{ij}\beta _{ij} + \frac{\tilde{v}_{ij}^2(\delta _{ij}-\Delta _1 \hat{a})^2}{2\tilde{R}_i}\\&\qquad +\frac{ \tilde{u}_{ij}^2({\epsilon }_{ij}-\Delta _1 \hat{b})^2}{2\tilde{R}_i}\\&\qquad -\frac{2}{2\tilde{R}_i}\, \tilde{u}_{ij}\tilde{v}_{ij} (\delta _{ij}-\Delta _1 \hat{a}) ({\epsilon }_{ij}-\Delta _1 \hat{b})\\&\approx q_{ij}-\tilde{u}_{ij}{\Delta }_2 \hat{a}-\tilde{v}_{ij}{\Delta }_2 \hat{b}-{\Delta }_2 \hat{R}_i. \end{aligned}$$
Proof of Lemma 1
The first three assertions will be proven if we show
$$\begin{aligned} {\mathbb E}\left( \rho _{ij}^2\right) =\sigma ^2, \quad {\mathbb E}\left( \tau _{ij}^2\right) =\sigma ^2\tilde{h}_{ij},\quad {\mathbb E}(\tau _{ij}\rho _{ij})=0. \end{aligned}$$
Firstly, observe that for any \(j=1,\ldots ,n_i\)
$$\begin{aligned} {\mathbb E}\left( \rho _{ij}^2\right) =\tilde{\mathbf {t}}_{ij}^\mathrm{T}{\mathbb E}\left( \check{\mathbf {n}}_{ij}\check{\mathbf {n}}_{ij}^\mathrm{T}\right) \tilde{\mathbf {t}}_{ij}=\hat{\delta }_{ij} \sigma ^2 \tilde{\mathbf {t}}_{ij}^\mathrm{T}\tilde{\mathbf {t}}_{ij}= \sigma ^2, \end{aligned}$$
which comes from the identity \(\Vert \tilde{\mathbf {t}}_{ij}\Vert _2^2=\tilde{u}_{ij}^2+\tilde{v}_{ij}^2=1\). Next, we compute \({\mathbb E}(\tau _{ij}^2)\). Since \(\mathrm{cov}({\Delta }_1\hat{{\varvec{\theta }}}_\mathrm{m})=\sigma ^2\tilde{\mathbf {R}}\), then, for any \(j=1,\ldots ,n_i\), one has
$$\begin{aligned} {\mathbb E}(\tau _{ij}^2)=\tilde{\mathbf {t}}_{ij}^\mathrm{T}{\mathbb E}\left( {\Delta }_1\hat{{\varvec{\theta }}}_\mathrm{m}{\Delta }_1\hat{{\varvec{\theta }}}_\mathrm{m}^\mathrm{T}\right) \tilde{\mathbf {t}}_{ij}=\sigma ^2\tilde{h}_{ij}. \end{aligned}$$
Finally, we compute \({\mathbb E}(\tau _{ij}\rho _{ij})\). Recall the definition of \({\Delta }_1\hat{{\varvec{\theta }}}_m=\tilde{\mathbf {K}}\mathbf {f}_1=\tilde{\mathbf {R}}\tilde{\mathbf {W}}^\mathrm{T}\mathbf {f}_1\), which can be rewritten as \(\tilde{\mathbf {R}}\mathbf {g}\) with
$$\begin{aligned} \mathbf {g}=\left[ \sum _{i=1}^2\sum _{t=1}^{n_i}\tilde{u}_{it} f_{1it},\, \sum _{i=1}^2\sum _{t=1}^{n_i} \tilde{v}_{it} f_{1it}, \sum _{n=1}^{n_1} f_{11t},\sum _{t=1}^{n_2} f_{12t} \right] ^\mathrm{T}, \end{aligned}$$
then for each \(t=1,\ldots ,n_i\), \({\mathbb E}(f_{1it}\rho _{ij})={\mathbb E}(f_{1it}(\check{\mathbf {n}}_{ij}^\mathrm{T}\tilde{\mathbf {t}}_{ij}))=0 \) for all \(j=1,\cdots ,n_i\). Thus, \({\mathbb E}(\tau _{ij}\rho _{ij})=0\).
Next, we compute the expectations of the outer products of \(\mathbf {a}_i\), \(\mathbf {b}_i\), and \(\mathbf {c}_i\) starting with \({\mathbb E}(\mathbf {a}_i\mathbf {a}_k^\mathrm{T})\). Since
$$\begin{aligned} {\mathbb E}(a_{ij} a_{kl})={\mathbb E}(\rho _{ij}^2\rho _{kl}^2)={\mathbb E}\left( \left( \tilde{\mathbf {t}}_{ij}^\mathrm{T}\check{\mathbf {n}}_{ij}\right) ^2\left( \check{\mathbf {n}}_{kl}^\mathrm{T}\tilde{\mathbf {t}}_{kl}\right) ^2\right) . \end{aligned}$$
Using the Isserlis’ Theorem [11] gives
$$\begin{aligned} {\mathbb E}(a_{ij} a_{kl})= & {} {\mathbb E}\left( \left[ \tilde{\mathbf {t}}_{ij}^\mathrm{T}\check{\mathbf {n}}_{ij}\right] ^2\right) \, {\mathbb E}\left( \left[ \check{\mathbf {n}}_{kl}^\mathrm{T}\tilde{\mathbf {t}}_{kl}\right] ^2\right) \\&+\,2\,{\mathbb E}\left( \left( \tilde{\mathbf {t}}_{ij}^\mathrm{T}\check{\mathbf {n}}_{ij}\right) \left( \check{\mathbf {n}}_{kl}^\mathrm{T}\tilde{\mathbf {t}}_{kl}\right) \right) . \end{aligned}$$
But it is easy to show that \({\mathbb E}((\tilde{\mathbf {t}}_{ij}^\mathrm{T}\check{\mathbf {n}}_{ij})^2)=\sigma ^2\) and
$$\begin{aligned} {\mathbb E}\left( \left( \tilde{\mathbf {t}}_{ij}^\mathrm{T}\check{\mathbf {n}}_{ij}\right) \left( \check{\mathbf {n}}_{kl}^\mathrm{T}\tilde{\mathbf {t}}_{kl}\right) \right) =\hat{\delta }_{ik}\hat{\delta }_{jl}\sigma ^2. \end{aligned}$$
Therefore,
$$\begin{aligned} {\mathbb E}(a_{ij} a_{kl})=\sigma ^4+2 \hat{\delta }_{ik}\hat{\delta }_{jl}\sigma ^4. \end{aligned}$$
Next, we compute \({\mathbb E}(b_{ij}b_{kl})=4{\mathbb E}\left( \tau _{ij}\rho _{ij}\tau _{kl} \rho _{kl}\right) \). The pair \((\tau _{ij},\rho _{kl})\) are uncorrelated for any choice of i and k. Besides,
$$\begin{aligned} {\mathbb E}(\rho _{ij} \rho _{kl})=\hat{\delta }_{ik}\hat{\delta }_{jl}\sigma ^2 , \text { and } {\mathbb E}(\tau _{ij}\tau _{kl})=\sigma ^2\tilde{\mathbf {t}}_{ij}^\mathrm{T}\tilde{\mathbf {R}}\tilde{\mathbf {t}}_{kl}=\sigma ^2\tilde{h}_{ijkl}. \end{aligned}$$
Therefore,
$$\begin{aligned} {\mathbb E}(b_{ij}b_{kl})=4\,{\mathbb E}\left( \tau _{ij}\tau _{kl}\right) {\mathbb E}\left( \rho _{ij} \rho _{kl}\right) =4 \sigma ^4\tilde{h}_{ijkl}\hat{\delta }_{ik}\hat{\delta }_{jl}. \end{aligned}$$
Next we evaluate \({\mathbb E}(c_{ij}c_{kl})={\mathbb E}(\tau _{ij}^2\tau _{kl}^2)\). If \(\tilde{h}_{ijij}\) is expressed by \(\tilde{h}_{ij}\), then using the general formulas of the expected value of the product of two quadratic forms of random variables [11], we obtain
$$\begin{aligned} {\mathbb E}(c_{ij}c_{kl})= & {} {\mathbb E}\left( \tau _{ij}^2\right) {\mathbb E}\left( \tau _{kl}^2\right) +2[{\mathbb E}(\tau _{ij}\tau _{kl} )]^2\\ {}= & {} \sigma ^4\left( \tilde{h}_{ij}\tilde{h}_{kl}+2\tilde{h}_{ijkl}\right) . \end{aligned}$$
In the same analog, we find
$$\begin{aligned} {\mathbb E}(a_{ij}c_{kl})={\mathbb E}\left( \tau _{ij}^2\rho _{kl}^2\right) ={\mathbb E}\left( \tau _{ij}^2\right) {\mathbb E}\left( \rho _{kl}^2\right) =\sigma ^4\tilde{h}_{ij}, \end{aligned}$$
because \(\tau _{ij}\) and \(\rho _{kl})\) are independent random variables. Finally, following the same reason, we observe that \({\mathbb E}(a_{ij}b_{kl})=-2\,{\mathbb E}(\rho _{ij}^2\tau _{kl} \rho _{kl})=0\) and also \( {\mathbb E}(b_{ij}c_{kl})=0\). From these results, we find the desired identities. This completes the proof of the lemma. \(\square \)
Derivation of Eqs. (41)–(43) The MSE of \({\Delta }_1\hat{{\varvec{\theta }}}_\mathrm{m}\) is equal to \(\sigma ^2\tilde{\mathbf {R}}\). We shall next evaluate \(\mathrm{MSE}({\Delta }_2\hat{{\varvec{\theta }}}_\mathrm{m})\). Let us define \(\check{\mathbf {a}}=(\check{\mathbf {a}}_1^\mathrm{T} , \, \check{\mathbf {a}}_2^\mathrm{T})^\mathrm{T}\) and other variables in the same manner. Then
$$\begin{aligned} {\mathbb E}(\check{\mathbf {a}}\check{\mathbf {a}}^\mathrm{T})&=\left[ \begin{array}{cc} {\mathbb E}\left( \check{\mathbf {a}}_1\check{\mathbf {a}}_1^\mathrm{T}\right) &{}\quad {\mathbb E}\left( \check{\mathbf {a}}_1\check{\mathbf {a}}_2^\mathrm{T}\right) \\ {\mathbb E}\left( \check{\mathbf {a}}_2\check{\mathbf {a}}_1^\mathrm{T}\right) &{} \quad {\mathbb E}\left( \check{\mathbf {a}}_2\check{\mathbf {a}}_2^\mathrm{T}\right) \end{array}\right] \\&=\sigma ^4\left[ \begin{array}{cc}\check{{\mathbf {1}}}_{n_1}\check{{\mathbf {1}}}_{n_1}^\mathrm{T}&{}\quad \check{{\mathbf {1}}}_{n_1}\check{{\mathbf {1}}}_{n_2}^\mathrm{T}\\ \check{{\mathbf {1}}}_{n_2}\check{{\mathbf {1}}}_{n_1}^\mathrm{T} &{}\quad \check{{\mathbf {1}}}_{n_2}\check{{\mathbf {1}}}_{n_2}^\mathrm{T} \end{array}\right] +2\sigma ^4\check{\mathbf {I}}_{n}, \end{aligned}$$
where
$$\begin{aligned} \check{\mathbf {I}}_{n}=\left[ \begin{array}{cc}\frac{1}{4\tilde{R}_1^2}\mathbf {I}_{n_1}&{}\quad {\mathbf {0}}_{n_1\times n_2}\\ {\mathbf {0}}_{n_2\times n_1} &{} \quad \frac{1}{4\tilde{R}_2^2}\mathbf {I}_{n_2} \end{array}\right] =\left[ \begin{array}{cc}\check{\mathbf {I}}_{n_1}&{}\quad {\mathbf {0}}_{n_1\times n_2}\\ {\mathbf {0}}_{n_2\times n_1} &{}\quad \check{\mathbf {I}}_{n_2} \end{array}\right] . \end{aligned}$$
Premultiplying and then postmultiplying \({\mathbb E}(\check{\mathbf {a}}\check{\mathbf {a}}^\mathrm{T})\) by \(\tilde{\mathbf {K}}\) and \(\tilde{\mathbf {K}}^\mathrm{T}\), respectively, lead to
$$\begin{aligned} \tilde{\mathbf {K}}{\mathbb E}\left( \check{\mathbf {a}}\check{\mathbf {a}}^\mathrm{T}\right) \tilde{\mathbf {K}}^\mathrm{T}=\sigma ^4\tilde{\mathbf {b}}_1\tilde{\mathbf {b}}_1^\mathrm{T} +2\sigma ^4 \tilde{\mathbf {K}}\check{\mathbf {I}}_n \tilde{\mathbf {K}}^\mathrm{T}. \end{aligned}$$
(82)
Now we compute \({\mathbb E}\left( \check{\mathbf {a}}\check{\mathbf {c}}^\mathrm{T} \right) \). It is easy to show that
$$\begin{aligned} {\mathbb E}\left( \check{\mathbf {a}}\check{\mathbf {c}}^\mathrm{T}\right)&\!=\!\left[ \! \begin{array}{cc}{\mathbb E}\left( \check{\mathbf {a}}_1\check{\mathbf {c}}_1^\mathrm{T}\right) &{}\quad {\mathbb E}\left( \check{\mathbf {a}}_1\check{\mathbf {c}}_2^\mathrm{T}\right) \\ {\mathbb E}\left( \check{\mathbf {a}}_2\check{\mathbf {c}}_1^\mathrm{T}\right) &{}\quad {\mathbb E}\left( \check{\mathbf {a}}_2\check{\mathbf {c}}_2^\mathrm{T}\right) \end{array}\!\right] \!=\!\sigma ^4\left[ \begin{array}{cc}\check{\tilde{\mathbf {h}}}_1\check{{\mathbf {1}}}_{n_1}^\mathrm{T}&{}\quad \check{\tilde{\mathbf {h}}}_1\check{{\mathbf {1}}}_{n_2}^\mathrm{T}\\ \check{\tilde{\mathbf {h}}}_2\check{{\mathbf {1}}}_{n_1}^\mathrm{T} &{}\quad \check{\tilde{\mathbf {h}}}_2\check{{\mathbf {1}}}_{n_2}^\mathrm{T} \end{array}\right] , \end{aligned}$$
and as such, if it is premultiplied and then postmultiplied by \(\tilde{\mathbf {K}}\) and \(\tilde{\mathbf {K}}^\mathrm{T}\), respectively, we obtain
$$\begin{aligned} 2\mathscr {S}\left[ \tilde{\mathbf {K}}{\mathbb E}\left( \check{\mathbf {a}}\check{\mathbf {c}}^\mathrm{T}\right) \tilde{\mathbf {K}}^\mathrm{T}\right]&= 2\mathscr {S}\left[ \sigma ^4 \tilde{\mathbf {K}}\tilde{\mathbf {h}}\left( \check{\hat{\mathbf {e}}}_3+ \check{\hat{\mathbf {e}}}_4\right) ^\mathrm{T}\right] \\&=2\sigma ^4\mathscr {S}\left[ \tilde{\mathbf {b}}_2\tilde{\mathbf {b}}_1^\mathrm{T}\right] . \end{aligned}$$
Finally, the less important but the most complicated expressions in the MSE of \({\Delta }_2 \hat{{\varvec{\theta }}}_\mathrm{m}\) come from \({\mathbb E}(\check{\mathbf {b}}\check{\mathbf {b}}^\mathrm{T})\) and \({\mathbb E}(\check{\mathbf {c}}\check{\mathbf {c}}^\mathrm{T})\). After lengthy but direct calculations, we have
$$\begin{aligned} {\mathbb E}\left[ \left( (\check{\mathbf {b}}+\check{\mathbf {c}})(\check{\mathbf {b}}+\check{\mathbf {c}})\right) ^\mathrm{T}\right] =\sigma ^4\left( \check{\tilde{\mathbf {h}}}\check{\tilde{\mathbf {h}}}^\mathrm{T}+ 2\check{\tilde{\mathbf {H}}}+4\check{\tilde{\mathbf {D}}}_{\tilde{\mathbf {h}}_1,\tilde{\mathbf {h}}_2}\right) , \end{aligned}$$
where
$$\begin{aligned} \check{\tilde{\mathbf {H}}} =\left[ \begin{array}{cc}\check{\tilde{\mathbf {H}}}_{11} &{}\quad \check{\tilde{\mathbf {H}}}_{12} \\ \check{\tilde{\mathbf {H}}}_{21} &{}\quad \check{\tilde{\mathbf {H}}}_{22} \end{array}\right] , \end{aligned}$$
and \(\check{\tilde{\mathbf {D}}}_{\tilde{\mathbf {h}}_1,\tilde{\mathbf {h}}_2}=\mathrm{diag}(\check{\tilde{\mathbf {h}}}_1, \check{\tilde{\mathbf {h}}}_2 )\). Premultiplying and postmultiplying this expression by \(\tilde{\mathbf {K}}\) and \(\tilde{\mathbf {K}}^\mathrm{T}\) give
$$\begin{aligned}&\tilde{\mathbf {K}}{\mathbb E}\left[ \left( (\check{\mathbf {b}}+\check{\mathbf {c}})(\check{\mathbf {b}}+\check{\mathbf {c})}\right) ^\mathrm{T}\right] \tilde{\mathbf {K}}^\mathrm{T}\\&\quad =\sigma ^4\tilde{\mathbf {b}}_2\tilde{\mathbf {b}}_2^\mathrm{T}+2\sigma ^4 \tilde{\mathbf {K}}\left( \check{\tilde{\mathbf {H}}}+2\check{\tilde{\mathbf {D}}}_{\tilde{\mathbf {h}}_1,\tilde{\mathbf {h}}_2}\right) \tilde{\mathbf {K}}^\mathrm{T}. \end{aligned}$$
Finally, combining all expressions together gives us the MSE of \({\Delta }_2\hat{{\varvec{\theta }}}_\mathrm{m}\).
Proof of Eq. (70)
Let \(\varvec{\varLambda }_{ij}=({\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1 ^1\mathbf {\mathscr {M}}{\tilde{{\varvec{\phi }}}})\Delta _1^1\mathbf {\mathscr {M}}\). Then
$$\begin{aligned} {\mathbb E}\left( \varvec{\varLambda }_{ij}\right)&={\sum _{k=1}^2\sum _{l=1}^{n_k} \frac{2}{\tilde{\zeta }_k^{2}}{\mathbb E}}\nonumber \\&\quad {\left[ \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\left( \Delta _1\mathbf {z}_{kl}^\mathrm{T}{\tilde{{\varvec{\phi }}}}\right) \tilde{\mathbf {z}}_{kl}\right) \mathscr {S}\left[ \Delta _1 \mathbf {z}_{kl}\tilde{\mathbf {z}}_{kl}^\mathrm{T}\right] \right] }\nonumber \\&= 2\sigma ^2\sum _{k=1}^2\sum _{l=1}^{n_k}\frac{1}{ \tilde{\zeta }_k^{2}} \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {z}}_{kl}\right) \mathscr {S}\left[ \tilde{\mathbf {V}}_{kl}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_{kl}^\mathrm{T}\right] . \end{aligned}$$
(83)
Here again \(\mathscr {S}(\bullet )= (\bullet + \bullet ^\mathrm{T})/2\) is the symmetrization operator. Recall that \(\gamma _{ij}={\hat{{\varvec{\phi }}}}^\mathrm{T}\mathbf {V}_{ij}{\hat{{\varvec{\phi }}}}\) and \({\Delta }_1\gamma _{ij}\) is given in Eq. (63). We shall evaluate Eq. (69) term by term. First, consider
$$\begin{aligned} \mathbf {I}_1&:={\mathbb E}\left( \Delta _1^1 \mathbf {\mathscr {M}}\tilde{\mathbf {\mathscr {M}}}^{-}\Delta _1^1 \mathbf {\mathscr {M}}\right) \\&=4\sum _{i,k=1}^2\sum _{j=1}^{n_i} \sum _{l=1}^{n_k}\tilde{\zeta }_i^{-1}\tilde{\zeta }_k^{-1} {\mathbb E}\left( \mathscr {S}\left[ \tilde{\mathbf {z}}_{ij}\Delta _1 \mathbf {z}_{ij}^\mathrm{T}\right] \right) \tilde{\mathbf {\mathscr {M}}}^- \\&\qquad \mathscr {S}\left[ \Delta _1\, \mathbf {z}_{kl}\tilde{\mathbf {z}}_{kl}^\mathrm{T}]\right) \\&=4\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-2}\times {\mathbb E}\left( \mathscr {S}\left[ \tilde{\mathbf {z}}_{ij}\Delta _1 \mathbf {z}_{ij}^\mathrm{T}\right] \right) \tilde{\mathbf {\mathscr {M}}}^-\\&\qquad \left( \mathscr {S}\left[ \tilde{\mathbf {z}}_{ij}\Delta _1 \mathbf {z}_{ij}^\mathrm{T}\right] \right) . \end{aligned}$$
By direct inspection, \({\mathbb E}\bigl (\Delta _1\, \mathbf {z}_{ij}^\mathrm{T}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1\,\mathbf {z}_{ij}\bigr )=\sigma ^2\mathrm{tr}(\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {V}}_{ij})\), hence if we using the definition of \(\tilde{\psi }_{ij}=\tilde{\mathbf {z}}_{ij}^\mathrm{T}\tilde{\mathbf {\mathscr {M}}}^{-}\tilde{\mathbf {z}}_{ij}\), then
$$\begin{aligned} \mathbf {I}_1= & {} \sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-2}\left[ 2\mathscr {S}\left[ \tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {M}}_{ij}\right] \right. \nonumber \\&\left. +\, \mathrm{tr}\,\left( \tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {V}}_{ij}\right) \tilde{\mathbf {M}}_{ij}+\tilde{\psi }_{ij}\tilde{\mathbf {V}}_{ij} \right] . \end{aligned}$$
(84)
We shall make use of an auxiliary formula below that follows from Eq. (67):
$$\begin{aligned} {\mathbb E}\left( \Delta _1\mathbf {A}\Delta _1\mathbf {z}_{ij}^\mathrm{T}\right) =-\sigma ^2\tilde{\zeta }_i^{-1} \tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {z}}_{ij}{\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij} \end{aligned}$$
(85)
and hence \({\mathbb E}(\Delta _1\mathbf {z}_{ij}\Delta _1{\hat{{\varvec{\phi }}}}^\mathrm{T}) =-\sigma ^2\tilde{\zeta }_i^{-1}\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_i^\mathrm{T}\tilde{\mathbf {\mathscr {M}}}^-\). Next we compute
$$\begin{aligned} \mathbf {I}_2:= & {} {\mathbb E}\left( \Delta _1^2 \mathbf {\mathscr {M}}\tilde{\mathbf {\mathscr {M}}}^{-}\Delta _1^1\mathbf {\mathscr {M}}\right) \nonumber \\= & {} -\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-2}{\mathbb E}\left( \Delta _1\gamma _{ij}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^{-}\Delta _1^1 \mathbf {\mathscr {M}}\right) . \end{aligned}$$
(86)
Using \({\Delta }_1^1\mathbf {\mathscr {M}}=\sum _{i=1}^2 \sum _{j=1}^{n_i}\tilde{\zeta }_i^{-1}{\Delta }_1\mathbf {M}_{ij}\), and then Eqs. (63) and (85) give
$$\begin{aligned} \mathbf {I}_2&=-\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}{\mathbb E}\left[ \left( 2\left( \Delta _1{\hat{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\right) \right. \right. \\&\left. \left. \qquad +\,{\tilde{{\varvec{\phi }}}}^\mathrm{T}\Delta _1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1^1 \mathbf {\mathscr {M}}\right] \\&=-\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}{\mathbb E}\left( 2\left( \Delta _1{\hat{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1^1\mathbf {\mathscr {M}}\right) \\&\qquad -\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}{\mathbb E}\left( \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\Delta _1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1^1\mathbf {\mathscr {M}}\right) \\&=-2\sum _{i,k=1}^2\sum _{j=1}^{n_i} \sum _{l=1}^{n_k} \tilde{\zeta }_i^{-2}\tilde{\zeta }_k^{-1} \tilde{\mathbf {M}}_{ij} \mathbf {\mathscr {M}}^-{\mathbb E}\\&\qquad \left( {\Delta }_1\mathbf {M}_{kl}{\Delta }_1{\hat{{\varvec{\phi }}}}^\mathrm{T} \tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\right) \\&\qquad \,-\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-2}{\mathbb E}\left( \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\Delta _1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\Delta _1^1\mathbf {\mathscr {M}}\right) \\&=4\sigma ^2\sum _{i,k=1}^2\sum _{j=1}^{n_i}\sum _{l=1}^{n_k}\tilde{\zeta }_i^{-2}\tilde{\zeta }_k^{-2}\left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {z}}_{kl}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\\&\qquad \mathscr {S}\left[ \tilde{\mathbf {V}}_{kl}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_{kl}^\mathrm{T}\right] -\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-3}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-{\mathbb E}\\&\qquad \left( \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\Delta _1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}\right) \Delta _1\mathbf {M}_{ij}\right) . \end{aligned}$$
Now recall that \(\tilde{\mathbf {V}}_{ij}=\tilde{\mathbf {a}}_{ij}\tilde{\mathbf {a}}_{ij}^\mathrm{T}+\tilde{\mathbf {b}}_{ij}\tilde{\mathbf {b}}_{ij}^\mathrm{T}\); thus, the noisy version of \(\tilde{\mathbf {V}}_{ij}\) has the first-order error term expressed as
$$\begin{aligned} {\Delta }_1\mathbf {V}_{ij}=2\mathscr {S}\left[ \tilde{\mathbf {a}}_{ij}{\Delta }_1 \mathbf {a}_{ij}^\mathrm{T}+\tilde{\mathbf {b}}_{ij}{\Delta }_1\mathbf {b}_{ij}^\mathrm{T}\right] . \end{aligned}$$
Thus, \({\tilde{{\varvec{\phi }}}}^\mathrm{T}{\Delta }_1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}=({\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {T}}_{ij}{\tilde{{\varvec{\phi }}}})\delta _{ij} +({\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {S}}_{ij}{\tilde{{\varvec{\phi }}}}){\epsilon }_{ij}\). Also \({\mathbb E}(\delta _{ij}{\Delta }_1\mathbf {M}_{ij})=2 \sigma ^2\mathscr {S}[\tilde{\mathbf {a}}_{ij}\tilde{\mathbf {z}}_{ij}^\mathrm{T}]\) and \({\mathbb E}({\epsilon }_{ij}{\Delta }_1\mathbf {M}_{ij})=2\sigma ^2\mathscr {S}[\tilde{\mathbf {b}}_{ij}\tilde{\mathbf {z}}_{ij}^\mathrm{T}]\), where \(\tilde{\mathbf {a}}_{ij}\) and \(\tilde{\mathbf {b}}_{ij}\) denote the first and second columns of \(\nabla \mathbf {z}_{ij}\). Therefore, \({\mathbb E}[({\tilde{{\varvec{\phi }}}}^\mathrm{T}\Delta _1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}})\Delta _1\mathbf {M}_{ij}]\)
$$\begin{aligned}= & {} 2\sigma ^2\left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {T}}_{ij}{\tilde{{\varvec{\phi }}}}\right) \mathscr {S}\left[ \tilde{\mathbf {a}}_{ij}\tilde{\mathbf {z}}_{ij}^\mathrm{T}\right] +2\sigma ^2 \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {S}}_{ij}{\tilde{{\varvec{\phi }}}}\right) \mathscr {S}\left[ \tilde{\mathbf {b}}_{ij}\tilde{\mathbf {z}}_{ij}^\mathrm{T}\right] \nonumber \\= & {} 2\sigma ^2\mathscr {S}[\tilde{\varvec{\varGamma }}_{ij}], \end{aligned}$$
(87)
where \(\tilde{\varvec{\varGamma }}_{ij}\) is defined in Eq. (73). They imply that
$$\begin{aligned} \mathbf {I}_2= & {} 4\sigma ^2\sum _{i,k=1}^2\sum _{j=1}^{n_i} \sum _{l=1}^{n_k}\frac{1}{\tilde{\zeta }_i^{2}\tilde{\zeta }_k^{2}} \left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {z}}_{kl}\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\nonumber \\&\quad \mathscr {S}\left[ \tilde{\mathbf {V}}_{kl}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_{kl}^\mathrm{T}\right] \nonumber \\&-2\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\frac{1}{\tilde{\zeta }_i^{3}}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\mathscr {S}[\tilde{\varvec{\varGamma }}_{ij}]. \end{aligned}$$
(88)
Next, using \({\Delta }_2\mathbf {M}_{ij}=2\mathscr {S}[{\Delta }_2\mathbf {z}_{ij}\tilde{\mathbf {z}}_{ij}^\mathrm{T}]+{\Delta }_1\mathbf {z}_{ij}{\Delta }_1\mathbf {z}_{ij}^\mathrm{T}\) and \({\Delta }_2\mathbf {z}_{ij}=(\delta _{ij}^2+{\epsilon }_{ij}^2)\mathbf {e}_1'\) give
$$\begin{aligned} {\mathbb E}(\Delta _2^1\mathbf {\mathscr {M}})= & {} \sum _{i=1}^2\sum _{i=1}^{n_i}\tilde{\zeta }_i^{-1}{\mathbb E}(\Delta _2\mathbf {M}_{ij})\nonumber \\= & {} \sum _{i=1}^2\sum _{i=1}^{n_i}\tilde{\zeta }_i^{-1}{\mathbb E}\left( \Delta _1\mathbf {z}_{ij}\Delta _1\mathbf {z}_{ij}^\mathrm{T}+2\mathscr {S}\left[ \tilde{\mathbf {z}}_{ij}\Delta _2\mathbf {z}_{ij}^\mathrm{T}\right] \right) ,\nonumber \\ \end{aligned}$$
(89)
which simply becomes
$$\begin{aligned} {\mathbb E}\left( \Delta _2^1\mathbf {\mathscr {M}}\right)= & {} \sigma ^2\sum _{i=1}^2 \sum _{j=1}^{n_i}\frac{1}{\tilde{\zeta }_i}\left( \tilde{\mathbf {V}}_{ij}+4\mathscr {S}\left[ \mathbf {e}_1'\tilde{\mathbf {z}}_{ij}^\mathrm{T} \right] \right) . \end{aligned}$$
(90)
Lastly, we evaluate
$$\begin{aligned} {\mathbb E}\left( \Delta _2^2 \mathbf {\mathscr {M}}\right)&=-\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}{\mathbb E}(\Delta _1\gamma _{ij}\Delta _1\mathbf {M}_{ij})\\&=-\sum _{i=1}^2\sum _{j=1}^{n_i}\frac{1}{\tilde{\zeta }_i^{2}}{\mathbb E}\left( \left( 2\left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\Delta _1{\hat{{\varvec{\phi }}}}\right) \right. \right. \\&\qquad \left. \left. +\,{\tilde{{\varvec{\phi }}}}^\mathrm{T}{\Delta }_1\mathbf {V}_{ij}{\tilde{{\varvec{\phi }}}}\right) \Delta _1\mathbf {M}_{ij}\right) . \end{aligned}$$
The first term comes from Eq. (85), and the second comes from Eq. (87). Thus,
$$\begin{aligned} {\mathbb E}\left( \Delta _2^2\mathbf {\mathscr {M}}\right)= & {} 2\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\left( \frac{2}{\tilde{\zeta }_i^{3}}\left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^{-}\tilde{\mathbf {z}}_{ij}\right) \right. \nonumber \\&\left. \mathscr {S}\left[ \tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_{ij}^\mathrm{T}\right] -\frac{1}{\tilde{\zeta }_i^{2}}\mathscr {S}[\tilde{\varvec{\varGamma }}_{ij}]\right) \end{aligned}$$
(91)
This completes the integration of Eq. (69).
To prove Eq. (70), we multiply the identities in Eqs. (84), (88), (90), and (91) by \({\tilde{{\varvec{\phi }}}}\). First,
$$\begin{aligned} \mathbf {I}_1{\tilde{{\varvec{\phi }}}}=\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-2}\bigl (\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-+ \tilde{\psi }_{ij}\mathbf {I}_5 \bigr )\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}. \end{aligned}$$
(92)
Using the following useful identities
$$\begin{aligned} 2\mathscr {S}\left[ \tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\tilde{\mathbf {z}}_{ij}^\mathrm{T}\right] {\tilde{{\varvec{\phi }}}}=\tilde{\mathbf {z}}_{ij}\left( {\tilde{{\varvec{\phi }}}}^\mathrm{T}\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\right) =\zeta _i\tilde{\mathbf {z}}_{ij},\,\, \text { and }\, \tilde{\varvec{\varGamma }}_i^\mathrm{T}{\tilde{{\varvec{\phi }}}}= {\mathbf {0}}. \end{aligned}$$
We have \(\mathbf {I}_2{\tilde{{\varvec{\phi }}}}\)
$$\begin{aligned} \mathbf {I}_2{\tilde{{\varvec{\phi }}}}&=\!2\sigma ^2\sum _{i,k=1}^2\sum _{j=1}^{n_i} \sum _{l=1}^{n_k}\tilde{\zeta }_i^{-2}\tilde{\zeta }_k^{-1}\left( \!{\tilde{{\varvec{\phi }}}}^\mathrm{T} \tilde{\mathbf {V}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {z}}_{kl}\!\right) \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^- \tilde{\mathbf {z}}_{kl}\\&-\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-3} \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\varvec{\varGamma }}_{ij}{\tilde{{\varvec{\phi }}}}, \end{aligned}$$
and further,
$$\begin{aligned} \mathbf {I}_2{\tilde{{\varvec{\phi }}}}&=\!2\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\! \left[ \!\sum _{k=1}^2\sum _{l=1}^{n_k}\tilde{\zeta }_k^{-1}\tilde{\mathbf {M}}_{kl} \!\!\right] \tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\\&\quad - \sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-3}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\varvec{\varGamma }}_{ij}{\tilde{{\varvec{\phi }}}}. \end{aligned}$$
Applying the definition of \(\tilde{\mathbf {\mathscr {M}}}\) and the identity \(\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {\mathscr {M}}}\tilde{\mathbf {\mathscr {M}}}^-=\tilde{\mathbf {\mathscr {M}}}^-\) yields
$$\begin{aligned} \mathbf {I}_2{\tilde{{\varvec{\phi }}}}= & {} 2\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-2}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^- \tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}\nonumber \\&- \sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\tilde{\zeta }_i^{-3} \tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\varvec{\varGamma }}_{ij}{\tilde{{\varvec{\phi }}}}. \end{aligned}$$
(93)
Next
$$\begin{aligned} {\mathbb E}\left( \Delta _2^1 \mathbf {\mathscr {M}}\right) {\tilde{{\varvec{\phi }}}}=\sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i} \tilde{\zeta }_i^{-1}\left( \tilde{\mathbf {V}}_{ij}+2\tilde{\mathbf {z}}_{ij}{\mathbf {e}_1'} ^\mathrm{T} \right) {\tilde{{\varvec{\phi }}}}. \end{aligned}$$
(94)
Lastly, we apply the above identities again to evaluate
$$\begin{aligned} {\mathbb E}\left( {\Delta }_2^2\mathbf {\mathscr {M}}\right) {\tilde{{\varvec{\phi }}}}= & {} \sigma ^2\sum _{i=1}^2\sum _{j=1}^{n_i}\nonumber \\&\left( 2\tilde{\zeta }_i^{-2}\tilde{\mathbf {M}}_{ij}\tilde{\mathbf {\mathscr {M}}}^-\tilde{\mathbf {V}}_{ij}{\tilde{{\varvec{\phi }}}}-\tilde{\zeta }_i^{-2}\tilde{\varvec{\varGamma }}_{ij}\tilde{\mathbf {A}}\right) . \end{aligned}$$
(95)
Now, observing that \(\tilde{\mathbf {G}}_2^\mathrm{T} {\tilde{{\varvec{\phi }}}}={\mathbf {0}}\) and combining Eqs. (92), (93), and (94), (95) complete the derivation of Eq. (70). Note that some terms in Eq. (93) and Eq. (95) cancel each other. \(\square \)
