Appendix A: Lemmas
1.1 A.1 Lemmas for the Multivariate Tuning Parameter Model
Regarding the multivariate tuning parameter model, we can establish the following lemmas. Under the model (i) - (iii), the posterior density of \(\kappa _i\) is given by
$$\begin{aligned} \begin{aligned} \Pi (\kappa _{i}\vert X_i)&\propto {\kappa _{i}}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) }. \end{aligned} \end{aligned}$$
In the following, we will use \(K (> 0)\) to denote a generic constant.
Lemma 1
Under the multivariate global-local prior model with treating the global parameter as a tuning parameter and under the regularity assumption (I), assuming \(\tau _n \rightarrow 0\) as \(n \rightarrow \infty \), for arbitrary \(\eta \) such that \(0< \eta < \min (1,a)\), when n is sufficiently large,
$$\begin{aligned} E(1-\kappa _i\vert \varvec{X}_i) \le K \tau _n^\eta \exp \left( \frac{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2}\right) . \end{aligned}$$
(A1)
Proof
For an arbitrary constant \(\xi >0\),
$$\begin{aligned} \begin{aligned}&E\left( 1-\kappa _i \mid \varvec{X}_i\right) \\ =&\tau _n E\left[ \lambda _i^2 \left( 1 + \lambda _i^2 \tau _n\right) ^{-1} \mid \varvec{X}_i\right] \\ =&\tau _n \frac{\int _0^\infty \left( 1+\lambda _{i}^2 \tau _n\right) ^{-\left( \frac{k}{2}+1\right) } \left( \lambda _{i}^2\right) ^{-a} L\left( \lambda _{i}^2\right) \exp \left( -\frac{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2\left( 1+\lambda _{i}^2 \tau _n \right) }\right) d\lambda _{i}^2}{\int _0^\infty (1+\lambda _{i}^2 \tau _n)^{-\frac{k}{2}} (\lambda _{i}^2)^{-a-1} L(\lambda _{i}^2) \exp \left( -\frac{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2(1+\lambda _{i}^2 \tau _n)}\right) d\lambda _{i}^2} \\ \le&\tau _n exp(\frac{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2}) I_1 / \left( I_2 L(\xi )\right) \end{aligned} \end{aligned}$$
(A2)
where
$$\begin{aligned} I_1 = \int _0^\infty \left( 1+\lambda _{i}^2 \tau _n\right) ^{-\left( \frac{k}{2}+1\right) } \left( \lambda _{i}^2\right) ^{-a} L\left( \lambda _{i}^2\right) d\lambda _{i}^2 \end{aligned}$$
(A3)
and
$$\begin{aligned} I_2 = \int _{\xi }^\infty \left( 1+\lambda _{i}^2 \tau _n\right) ^{-\frac{k}{2}} \left( \lambda _{i}^2\right) ^{-a-1} d\lambda _{i}^2 \end{aligned}$$
(A4)
But
$$\begin{aligned} I_2 \ge \int _{\xi }^\infty \left( 1+\lambda _{i}^2\right) ^{-\frac{k}{2}} \left( \lambda _{i}^2\right) ^{-a-1} d\lambda _{i}^2 = K. \end{aligned}$$
(A5)
Next, by choosing \(0< \eta < \min (1,a)\),
$$\begin{aligned} \begin{aligned} I_1&\!\le \! \int _0^1 (\lambda _{i}^2)(\lambda _{i}^2)^{-a-1} L(\lambda _{i}^2)d\lambda _{i}^2 \!+\! M \!\int _1^\infty (\lambda _{i}^2 \tau _n) ^{-1+\eta } (1\!+\! \lambda _{i}^2 \tau _n)^{-\frac{k}{2}-\eta } (\lambda _{i}^2)^{-a} d\lambda _{i}^2 \\&\!\le \! K^{-1} + M \tau _n^{\eta -1} (a-\eta )^{-1} \end{aligned} \end{aligned}$$
(A6)
Combining (A2)-(A6), \(E\left( 1-\kappa _i \mid \varvec{X}_i\right) \le K \tau _n^\eta \exp \left( \frac{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2}\right) \), when n is sufficiently large.\(\square \)
Lemma 2
Under the multivariate global-local prior model with treating the global parameter as a tuning parameter and under the regularity assumption (I), for arbitrary constants \(0< \xi < 1\) and \(0< \delta < 1\),
$$\begin{aligned} E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \le K \tau _n^{-a} \exp {\left[ -\frac{\xi (1-\delta )}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right] }. \end{aligned}$$
(A7)
Proof
For an arbitrary constant \(0< \xi < 1\),
$$\begin{aligned} \begin{aligned}&E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \\ =&\frac{\int _{\xi }^1 \kappa _{i}^{k/2+a} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}}{\int _0^{1} \kappa _{i}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}} \\ \le&\frac{\int _{\xi }^1 \kappa _{i}^{k/2+a} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}}{\int _0^{\xi \delta } \kappa _{i}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}} \\ \le&\exp {\left[ -\frac{\xi (1-\delta )}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right] } \frac{ \int _{\xi }^1 \kappa _{i}^{k/2+a} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) d\kappa _{i}}{\int _0^{\xi \delta } \kappa _{i}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) d\kappa _{i}}. \end{aligned} \end{aligned}$$
(A8)
Now observe that
$$\begin{aligned} \begin{aligned}&\int _{\xi }^1 \kappa _{i}^{k/2+a} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) d\kappa _{i} \\ =&\int _{0}^{(1-\xi )/(\xi \tau _n)} (1+\lambda _{i}^2 \tau _n)^{-(k/2+a)} (\lambda _{i}^2 \tau _n (1+\lambda _{i}^2 \tau _n)^{-1})^{-a-1} L(\lambda _{i}^2) \frac{\tau _n d \lambda _{i}^2}{(1+\lambda _{i}^2 \tau _n)^{2}} \\ =&\tau _n^{-a} \int _{0}^{(1-\xi )/(\xi \tau _n)} (1+\lambda _{i}^2 \tau _n)^{-(k/2+1)} (\lambda _{i}^2)^{-a-1} L(\lambda _{i}^2) d \lambda _{i}^2 \\ \le&\tau _n^{-a} \int _0^\infty (\lambda _{i}^2)^{-a-1} L(\lambda _{i}^2) d \lambda _{i}^2 = \tau _n^{-a} K^{-1}. \end{aligned} \end{aligned}$$
(A9)
By assumption (I) and \(\tau _n<1\),
$$\begin{aligned} \begin{aligned}&\int _0^{\xi \delta } \kappa _{i}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) d\kappa _{i} \\ \ge&\int _{0}^{\xi \delta } \kappa _{i}^{k/2+a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) d\kappa _{i} \\ \ge&L\left( \frac{1-\xi \delta }{\xi \delta }\right) \int _{0}^{\xi \delta } \kappa _{i}^{k/2+a-1} d \kappa _{i} \\ =&L\left( \frac{1-\xi \delta }{\xi \delta }\right) \frac{(\xi \delta )^{k/2+a}}{k/2+a}. \end{aligned} \end{aligned}$$
(A10)
Combining (A8) - (A10),
$$\begin{aligned} E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \le K \tau _n^{-a} \exp {\left[ -\frac{\xi (1-\delta )}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right] }. \end{aligned}$$
(A11)
\(\square \)
Lemma 3
Under the multivariate global-local prior model with treating the global parameter as a tuning parameter and under the regularity assumption (I), for an arbitrary constant \(0< \xi < 1\),
$$\begin{aligned} E(\kappa _i \mathbbm {1}_{[\kappa _i \le \xi ]} \vert \varvec{X}_i) \le K / \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i. \end{aligned}$$
(A12)
Proof
For an arbitrary constant \(0< \xi < 1\),
$$\begin{aligned} \begin{aligned}&E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i \le \xi \right] } \mid \varvec{X}_i\right) \\ =&\frac{\int _0^{\xi } \kappa _{i}^{k/2+a} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}}{\int _0^{1} \kappa _{i}^{k/2+a-1} (1-\kappa _{i})^{-a-1} L\left( \frac{1-\kappa _{i}}{\kappa _{i} \tau _n}\right) \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}} \\ \le&\frac{(1-\xi )^{-a-1}M}{L\left( \frac{1-\xi }{\xi \tau _n}\right) } \frac{\int _0^{\xi } \kappa _{i}^{k/2+a} \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}}{\int _0^{\xi } \kappa _{i}^{k/2+a-1} \exp {\left( -\frac{\kappa _{i}}{2} \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i \right) } d\kappa _{i}} \\ =&\frac{(1-\xi )^{-a-1}M}{L\left( \frac{1-\xi }{\xi \tau _n}\right) } \frac{\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} \left( \frac{2t}{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }\right) ^{k/2+a} \exp {\left( -t\right) } dt}{\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} \left( \frac{2t}{ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }\right) ^{k/2+a-1} \exp {\left( -t\right) } dt} \\ =&\frac{(1-\xi )^{-a-1}M}{L\left( \frac{1-\xi }{\xi \tau _n}\right) \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i } \frac{\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} t^{k/2+a} \exp {\left( -t\right) } dt}{\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} t^{k/2+a-1} \exp {\left( -t\right) } dt}. \end{aligned} \end{aligned}$$
(A13)
Integrating the numerator by parts,
$$\begin{aligned} \begin{aligned}&\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} t^{k/2+a} \exp {\left( -t\right) } dt = (k/2 + a) \times \\&\int _0^{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2} t^{k/2+a-1} \exp {\left( -t\right) } dt - \left( \frac{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2} \right) ^{\frac{k}{2}+a} \exp {\left( -\frac{\xi \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i }{2} \right) }. \end{aligned} \end{aligned}$$
(A14)
Combining (A13) and (A14),
$$\begin{aligned} E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i \le \xi \right] } \mid \varvec{X}_i\right) \le K/ \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i . \end{aligned}$$
(A15)
\(\square \)
1.2 A.2 Lemmas for the Multivariate Exponential-Inverse-Gamma Model
Under an Exponential-Inverse-Gamma model (i), (ii’) and (iii’) the posterior density of \(\kappa _i\) is given by
$$\begin{aligned} \begin{aligned} \Pi (\kappa _i\vert \varvec{X}_i) \propto \kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) . \end{aligned} \end{aligned}$$
Now, we give three basic inequalities involving the Exponential-Inverse-Gamma prior.
Lemma 4
Under the multivariate Exponential-Inverse-Gamma model, assuming \(c_n \rightarrow 0\) as \(n \rightarrow \infty \), for large n,
$$\begin{aligned} E(1-\kappa _i\vert \varvec{X}_i) \le K c_n^{d} \exp \left( \frac{\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) . \end{aligned}$$
(A16)
Proof
Firstly,
$$\begin{aligned} \begin{aligned} E(1-\kappa _i\vert \varvec{X}_i)&= \frac{\int _0^1 (1-\kappa _i) \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^1 \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i} \\&\le \exp (\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) \frac{\int _0^1 (1-\kappa _i)\kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^1 \kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}\\&:= \exp (\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) \frac{N}{D}. \end{aligned} \end{aligned}$$
Now,
$$\begin{aligned} \begin{aligned} D&\ge \int _{1-c_n}^1 \kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i\\&\ge \int _{1-c_n}^1 \kappa _i^{d+\frac{k}{2}-1} (c_n + c_n )^{-d-1} d\kappa _i\\&= \left( 2 c_n\right) ^{-(d+1)} \int _{1-c_n}^1 \kappa _i^{d+\frac{k}{2}-1} d \kappa _i\\&= \left( 2 c_n\right) ^{-(d+1)}\frac{ 1-(1-c_n)^{d+\frac{k}{2}}}{d+\frac{k}{2}}\\&\ge K c_n^{-(d+1)} \left[ c_n- \frac{d+\frac{k}{2}-1}{2}c_n^2\right] \\&= K c_n^{-d} \left[ 1- \frac{d+\frac{k}{2}-1}{2}c_n\right] \\&\ge K c_n^{-d}. \end{aligned} \end{aligned}$$
Next,
$$\begin{aligned} \begin{aligned} N&= \int _0^1 (1-\kappa _i)\kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i \\&\le \int _0^1 (1-\kappa _i)\kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i)^{-d-1} d\kappa _i \\&= \int _0^1 \kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i)^{-d} d\kappa _i \\&= \text {Beta}(1-d, \frac{k}{2}+d). \end{aligned} \end{aligned}$$
Therefore, when n is sufficiently large,
$$\begin{aligned} E(1-\kappa _i \vert \varvec{X}_i) \le K c_n^{d} \exp (\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) . \end{aligned}$$
\(\square \)
Lemma 5
Under the multivariate Exponential-Inverse-Gamma model, for arbitrary constants \(0< \xi < 1\) and \(0< \delta < 1\),
$$\begin{aligned} E(\kappa _i \mathbbm {1}_{[\kappa _i > \xi ]} \vert \varvec{X}_i) \le K c_n^{-d} \exp \left( -\frac{1}{2}\xi (1-\delta )\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i\right) . \end{aligned}$$
(A17)
Proof
For arbitrary constants \(0< \xi < 1\) and \(0< \delta < 1\),
$$\begin{aligned}{} & {} E(\kappa _i \mathbbm {1}_{ [\kappa _i > \xi ] }\vert \varvec{X}_i)\\= & {} \frac{\int _{\xi }^1 \kappa _i \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^1 \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i} \\\le & {} \frac{\int _{\xi }^1 \kappa _i^{d+\frac{k}{2}} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^{\xi \delta } \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}\\\le & {} \exp \left( -\frac{1}{2} \xi (1-\delta ) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i \right) \frac{\int _{\xi }^1 \kappa _i^{d+\frac{k}{2}} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^{\xi \delta } \kappa _i^{d+\frac{k}{2}-1} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}\\\le & {} \exp \left( -\frac{1}{2} \xi (1-\delta ) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i \right) \frac{\int _{\xi }^1 \kappa _i^{d+\frac{k}{2}} (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^{\xi \delta } \kappa _i^{d+\frac{k}{2}-1} (1 + c_n)^{-d-1} d\kappa _i}\\\le & {} \exp \left( -\frac{1}{2} \xi (1-\delta ) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i \right) (1+c_n)^{d+1} \frac{d+k/2}{(\xi \delta )^{d+k/2}}\int _{\xi }^1 (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i\\\le & {} K \exp \left( -\frac{1}{2} \xi (1-\delta ) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i \right) \left[ \frac{(1-\kappa _i + \xi c_n)^{-d}}{d}\right] ^1_\xi \\\le & {} K \exp \left( -\frac{1}{2} \xi (1-\delta ) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i \right) c_n^{-d}. \end{aligned}$$
\(\square \)
Lemma 6
Under the multivariate Exponential-Inverse-Gamma model, for an arbitrary constant \(0< \xi < 1\),
$$\begin{aligned} E(\kappa _i \mathbbm {1}_{[\kappa _i \le \xi ]} \vert \varvec{X}_i) \le K / \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i. \end{aligned}$$
(A18)
Proof
For an arbitrary constant \(0< \xi < 1\),
$$\begin{aligned}{} & {} E(\kappa _i \mathbbm {1}_{ [\kappa _i \le \xi ] }\vert \varvec{X}_i)\\= & {} \frac{\int _0^{\xi } \kappa _i^{d+\frac{k}{2}} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^1 \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i} \\\le & {} \frac{\int _0^{\xi } \kappa _i^{d+\frac{k}{2}} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i}{\int _0^\xi \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) (1-\kappa _i + \kappa _i c_n)^{-d-1} d\kappa _i} \\\le & {} \frac{(1-\xi )^{-d-1}}{(1+\xi c_n)^{-d-1}}\frac{\int _0^{\xi } \kappa _i^{d+\frac{k}{2}} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) d\kappa _i}{\int _0^\xi \kappa _i^{d+\frac{k}{2}-1} \exp \left( -\frac{\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2}\right) d\kappa _i} \\\le & {} \frac{(1-\xi )^{-d-1}}{(1+\xi c_n)^{-d-1} \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}\frac{\int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) t^{d+\frac{k}{2}} dt}{\int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) t^{d+\frac{k}{2}-1}dt}. \\ \end{aligned}$$
Integrating by parts,
$$\begin{aligned} \begin{aligned}&\int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) t^{d+\frac{k}{2}}dt\\ =&\left[ \!-\!\exp \left( -t\right) t^{d+\frac{k}{2}}\right] ^{t=\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2}_{t=0} \!+\! (d\!+\!\frac{k}{2}) \int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) t^{d+\frac{k}{2}-1} dt. \end{aligned} \end{aligned}$$
Therefore,
$$\begin{aligned} \frac{\int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) ^{d+\frac{k}{2}} dt}{\int _0^{\xi \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2} \exp \left( -t\right) t^{d+\frac{k}{2}-1}dt} \le d + \frac{k}{2}. \end{aligned}$$
Hence,
$$\begin{aligned} E(\kappa _i \mathbbm {1}_{ [\kappa _i \le \xi ] }\vert \varvec{X}_i) \le K/\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i. \end{aligned}$$
\(\square \)
Appendix B: Proofs of the Main Theorems
1.1 B.1 Proofs of Theorems in Section 2
Proof
of Theorem 1 For each \(\varvec{X}_{i} \overset{ind}{\sim }\ N_k(\varvec{\theta }_{i}, \varvec{\Sigma })\), define \(\varvec{Y}_{i} = (Y_{i1}, \dots , Y_{ik})^\top := \varvec{\Sigma }^{-1/2}\varvec{X}_{i}\) and \(\varvec{\lambda }_{i} = (\lambda _{i1}, \dots , \lambda _{ik})^\top := \varvec{\Sigma }^{-1/2}\varvec{\theta }_{i}\). So, all the components of \(\varvec{Y}_{i}\)’s are independently normally distributed, i.e., \({Y}_{ij} \overset{ind}{\sim }\ N({\lambda }_{ij}, 1), i = 1, \dots , n, j=1,\dots ,k\).
Given estimators \(\widehat{\varvec{\lambda }}_{i}\), if we use \(\widehat{\varvec{\theta }}_{i} = \varvec{\Sigma }^{1/2}\widehat{\varvec{\lambda }}_{i}\) to estimate \(\varvec{\theta }_{i}\)’s, then
$$E \Vert \widehat{\varvec{\theta }}_{i} - \varvec{\theta }_{i} \Vert _{\Sigma }^2 = E \Vert \widehat{\varvec{\lambda }}_{i} - \varvec{\lambda }_{i} \Vert ^2 = \sum _{j=1}^{k} E ( \widehat{{\lambda }}_{ij} - {\lambda }_{ij} )^2. $$
Also, since \(\{ \lambda _{ij}\} \in l_0[q'_{nk}]\) would imply that \(\{ \varvec{\theta }_{i}\} \in L_0[q_n]\), when we let \(q'_{nk} = q_n\), we have
$$\sup _{\{ \lambda _{ij}\} \in l_0[q'_{nk}]} \sum _{i=1}^n \sum _{j=1}^{k} E ( \widehat{{\lambda }}_{ij} - {\lambda }_{ij} )^2 \le \sup _{\{ \varvec{\theta }_{i}\} \in L_0[q_n]} \sum _{i=1}^n E \Vert \widehat{\varvec{\theta }}_{i} - \varvec{\theta }_{i} \Vert _{\Sigma }^2. $$
Finally, since \(\varvec{\Sigma }\) is positive definite, there is a one-to-one correspondence between \(\{\{\widehat{\varvec{\theta }}_{i}\}_{i=1}^n\}\) and \(\{\{\widehat{{\lambda }}_{ij}\}_{i=1,j=1}^{n,\quad k}\}\). So the above inequality still hold, if we further take the infinum over all possible estimators:
$$\inf _{\{\widehat{{\lambda }}_{ij}\}} \sup _{\{ \lambda _{ij}\} \in l_0[q'_{nk}]} \sum _{i=1}^n \sum _{j=1}^{k} E ( \widehat{{\lambda }}_{ij} - {\lambda }_{ij} )^2 \le \inf _{\{\widehat{\varvec{\theta }}_{i}\}} \sup _{\{ \varvec{\theta }_{i}\} \in L_0[q_n]} \sum _{i=1}^n E \Vert \widehat{\varvec{\theta }}_{i} - \varvec{\theta }_{i} \Vert _{\Sigma }^2. $$
Donoho et al. (1992) provides the result for the left hand side, which is, as \(n \rightarrow \infty \),
$$\begin{aligned} \begin{aligned} \inf _{\{\widehat{{\lambda }}_{ij}\}} \sup _{\{ \lambda _{ij}\} \in l_0[q'_{nk}]} \sum _{i=1}^n \sum _{j=1}^{k} E ( \widehat{{\lambda }}_{ij} - {\lambda }_{ij} )^2&= 2q'_{nk} \log (nk/q'_{nk}) (1+o(1)) \\&= 2q_{n} \log (n/q_{n}) (1+o(1)). \end{aligned} \end{aligned}$$
The last equality holds because we let \(q'_{nk} = q_n\).
So we have a minimax lower bound now,
$$\inf _{\{\widehat{\varvec{\theta }}_{i}\}} \sup _{\{ \varvec{\theta }_{i}\} \in L_0[q_n]} \sum _{i=1}^n E \Vert \widehat{\varvec{\theta }}_{i} - \varvec{\theta }_{i} \Vert _{\Sigma }^2 \ge 2q_{n} \log (n/q_{n}) (1+o(1)). $$
As we will see in (6) and (12), when using some particular priors, the error of the Bayes estimate of \(\varvec{\theta }_{i}\)’s will be at most \(2q_{n} \log (n/q_{n}) (1+o(1))\). This fact provides an upper bound for the minimax error, which coincides with the lower bound, and finishes this proof. \(\square \)
Proof
of Theorem 2 We first prove (4). Observe that
$$\begin{aligned} \Vert \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 = E\left( 1-\kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le E\left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2. \end{aligned}$$
Denote the \(l_2\)-norm by \(\Vert \cdot \Vert _2\). Then, making use of Lemma 1, for a sequence of positive constants \(\{a_n,n\ge 1\}\) to be specified later with \(a_n \rightarrow \infty \) as \(n \rightarrow \infty \),
$$\begin{aligned}{} & {} E_0\left[ E \left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] } \right] \nonumber \\\le & {} K \tau _n^\eta E_0\left[ \exp \left( \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i/ 2\right) \Vert \varvec{X}_i\Vert _2^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] } \right] \nonumber \\= & {} K \tau _n^\eta \int _{ \Vert \varvec{X}_i\Vert _2^2 \le a_n} \Vert \varvec{X}_i\Vert _2^2 \exp \left( \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i/2 - \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i / 2\right) d\varvec{X}_i \nonumber \\= & {} K \tau _n^\eta \int _{ \Vert \varvec{X}_i\Vert _2^2 \le a_n} \Vert \varvec{X}_i\Vert _2^2 d\varvec{X}_i \nonumber \\\le & {} K \tau _n^\eta a_n V_k(a_n)\nonumber \\= & {} K \tau _n^\eta a_n^{k+1}, \end{aligned}$$
(B19)
where \(V_k(r) = \frac{\pi ^{k/2}}{\Gamma (k/2+1)} r^{k}\) is the volume of a Euclidean ball of radius r in k-dimensional Euclidean space.
Moreover, under assumption (II),
$$\begin{aligned} \begin{aligned}&E_0\left[ E \left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2> a_n\right] } \right] \\ \le&E_0 \left( \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > \lambda _{\min } \left( \Sigma ^{-1}\right) a_n \right] } \right) \end{aligned} \end{aligned}$$
Noting that under \(\varvec{\theta }_{0i} = \varvec{0} \), \(\Vert \varvec{X}_i\Vert _{\Sigma }^2 \sim \chi _k^2\), one gets
$$\begin{aligned} \begin{aligned}&E_0\left( \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > \lambda _{\min } \left( \Sigma ^{-1}\right) a_n\right] } \right) \\ =&\int _{\lambda _{\min } \left( \Sigma ^{-1}\right) a_n}^\infty x \exp \left( -\frac{x}{2}\right) \frac{x^{k/2-1}}{\Gamma (k/2) 2^{k/2}} dx\\ =&2 \int _{\left( \lambda _{\min } \left( \Sigma ^{-1}\right) a_n \right) /2}^\infty \exp \left( -x\right) \frac{x^{k/2}}{\Gamma (k/2)} dx\\ \le&2 \exp \left( - \lambda _{\min } \left( \Sigma ^{-1}\right) a_n / 4 \right) \int _{\left( \lambda _{\min } \left( \Sigma ^{-1}\right) a_n \right) /2}^\infty \exp \left( -\frac{x}{2}\right) \frac{x^{k/2}}{\Gamma (k/2)} dx\\ \le&\exp \left( - \lambda _{\min } \left( \Sigma ^{-1}\right) a_n / 4\right) \left( \frac{k}{2}\right) 2 ^{\frac{k}{2}+2}. \end{aligned} \end{aligned}$$
Now choosing \(a_n = 4 \lambda _{\min }^{-1} \left( \Sigma ^{-1}\right) (1+\epsilon )\log (n/q_n)\), with \(\epsilon > 0\), one gets
$$\begin{aligned} E_0\left[ E \left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 > a_n\right] } \right] \le K (q_n/n)^{1+\epsilon }. \end{aligned}$$
(B20)
Finally, choosing \(\tau _n = \left( q_n/n\right) ^{\frac{1+\epsilon }{\eta }}\), the theorem follows from (B19) and (B20),
$$\begin{aligned} \begin{aligned} E_0 \left[ E\left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right]&\le K \left[ (q_n/n)^{1+\epsilon } \left( \log (n/q_n)\right) ^{k+1} + (q_n/n)^{1+\epsilon } \right] \\&\lesssim (q_n/n)^{1+\epsilon } \left( \log (n/q_n)\right) ^{k+1}, \text { as } n \rightarrow \infty . \end{aligned} \end{aligned}$$
Summing over all i’s for which \(\varvec{\theta }_{0i} = \varvec{0}\), one gets
$$ { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} {\sum \limits _{i: \varvec{\theta }_{0i} = \varvec{0}} E_0 \Vert \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 } / {\left( q_n \log \left( \frac{n}{q_n}\right) \right) } = o(1), \text { as } n \rightarrow \infty .$$
Now we prove (5). Use the inequality
$$\begin{aligned} \begin{aligned}&E_{\varvec{\theta }_{0i}} \Vert \widehat{\varvec{\theta }}^R_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 \\ =&E_{\varvec{\theta }_{0i}} \Vert (1-E\left( \kappa _i \mid \varvec{X}_i\right) )\varvec{X}_i - \varvec{X}_i + \varvec{X}_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 \\ =&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 + \Vert \varvec{X}_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 -2 \langle \varvec{\Sigma }^{-1/2}(\varvec{X}_i - \varvec{\theta }_{0i}), E\left( \kappa _i \mid \varvec{X}_i\right) \varvec{\Sigma }^{-1/2}\varvec{X}_i \rangle \right] \\ \le&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] + E_{\varvec{\theta }_{0i}} \Vert \varvec{X}_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 \\ +&2 \left( E_{\varvec{\theta }_{0i}} \Vert \varvec{X}_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] \right) ^\frac{1}{2} \end{aligned} \end{aligned}$$
But \(E_{\varvec{\theta }_{0i}} \Vert \varvec{X}_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 = k\), the above becomes
$$\begin{aligned} \begin{aligned} E_{\varvec{\theta }_{0i}} \Vert \widehat{\varvec{\theta }}^R_i - \varvec{\theta }_{0i}\Vert _{\Sigma }^2 \le&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] \\ +&2 k^{\frac{1}{2}} E_{\varvec{\theta }_{0i}}^\frac{1}{2} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] + k \\ \le&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] \\ +&2 k^{\frac{1}{2}} E_{\varvec{\theta }_{0i}}^\frac{1}{2} \left[ E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] + k \end{aligned} \end{aligned}$$
Since \(q_n k / q_n \log (n/q_n) \rightarrow 0\), as \(n \rightarrow \infty \), it suffices to show that
$$\begin{aligned} \limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2]}{4a q_n \log (n/q_n)} \le 1/(2\min (1,a)). \end{aligned}$$
(B21)
and
$$\begin{aligned} \limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}}^{\frac{1}{2}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2]}{q_n \log (n/q_n)} = 0. \end{aligned}$$
(B22)
In view of Lemma 2, for sufficiently large \(b_n > 0\), uniformly in \(\varvec{\theta }_{0i} \ne \varvec{0}\),
$$\begin{aligned} \begin{aligned}&E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i> \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2> b_n \right] } \\ \le&K \tau _n^{-a} \Vert \varvec{X}_i\Vert _{\Sigma }^2 \exp {\left[ -\frac{\xi (1-\delta )}{2} \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] } \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > b_n \right] } \\ \le&K \tau _n^{-a} b_n \exp {\left[ -\frac{\xi (1-\delta )}{2} b_n \right] } \\ \le&K(q_n/n)^{a(\rho -\epsilon )/\eta } \log \left( \frac{n}{q_n}\right) , \end{aligned} \end{aligned}$$
(B23)
by choosing \(b_n = \frac{1+\rho }{\eta } \frac{2a}{\xi (1-\delta )} \log \left( \frac{n}{q_n}\right) \), with \(\rho > \epsilon \), and recalling that \(\tau _n = \left( q_n/n\right) ^{\frac{1+\epsilon }{\eta }}\).
Then summing over all i’s for which \(\varvec{\theta }_{0i} \ne \varvec{0}\), one gets
$$\begin{aligned} \limsup \limits _{n \rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} \left[ E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i> \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > b_n \right] }\right] }{q_n \log (n/q_n)} = 0. \end{aligned}$$
(B24)
Finally,
$$\begin{aligned} \begin{aligned}&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] \\ \le&E_{\varvec{\theta }_{0i}} \left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] \\ \le&b_n \\ \le&\frac{1+\rho }{\eta } \frac{2a}{\xi (1-\delta )} \log \left( \frac{n}{q_n}\right) . \end{aligned} \end{aligned}$$
(B25)
Since the result above is independent of any specific choice of the parameters, by making \(\epsilon \rightarrow 0\), \(\rho \rightarrow 0\), \(\eta \rightarrow \min (1,a)\), \(\xi \rightarrow 1\) and \(\delta \rightarrow 0\), and summing over all i’s for which \(\varvec{\theta }_{0i} \ne \varvec{0}\), one gets
$$\begin{aligned} \begin{aligned}&\limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] }{4a q_n \log (n/q_n)} \\&\le 1/(2\min (1,a)). \end{aligned} \end{aligned}$$
(B26)
Together with Lemma 3 and (B24), this leads to (B21).
Altogether Lemma 2, (B23) and (B25) also imply that
$$\begin{aligned} E_{\varvec{\theta }_{0i}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2] \le K \log \left( \frac{n}{q_n}\right) (1+o(1)), \text { as } n \rightarrow \infty . \end{aligned}$$
(B27)
Again summing over all i’s for which \(\varvec{\theta }_{0i} \ne \varvec{0}\), one gets (B22). This completes the proof of (5).
(6) follows (4) and (5), immediately. In particular, when \(0 < a \le 1\), Theorem 1 provides a lower bound for the estimation error and (7) follows.\(\square \)
Proof
of Theorem 3 By Markov’s inequality and the independence of samples,
$$\begin{aligned} \begin{aligned}&E_{\{\varvec{\theta }_{0i}\}} \Pi (\sum _{i=1}^{n} \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 > q_n \log (\frac{n}{q_n}) \mid \{\varvec{X}_i\}) \\ \le&E_{\{\varvec{\theta }_{0i}\}} E(\sum _{i=1}^{n} \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \{\varvec{X}_i\}) / q_n \log (\frac{n}{q_n}) \\ =&\sum _{i=1}^{n} E_{\varvec{\theta }_{0i}} E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) / q_n \log (\frac{n}{q_n}). \end{aligned} \end{aligned}$$
Since
$$\varvec{\theta }_{i} \mid \varvec{X}_i, \kappa _i \sim N_k(\widehat{\varvec{\theta }}^R_i, (1-\kappa _i)\varvec{\Sigma }),$$
we have
$$\begin{aligned} \begin{aligned}&E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) \\ =&E\{ E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \kappa _i) \mid \varvec{X}_i \}\\ =&E\{ (1-\kappa _i) E( \frac{ \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 }{1-\kappa _i} \mid \varvec{X}_i, \kappa _i) \mid \varvec{X}_i \} \\ =&k E (1-\kappa _i \mid \varvec{X}_i ). \end{aligned} \end{aligned}$$
Now we only need to find a bound for \(E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i )\). When \(\varvec{\theta }_{0i} \ne 0\),
$$E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i ) \le 1.$$
When \(\varvec{\theta }_{0i} = 0\), letting \(\{a_n\}\) be a sequence of positive numbers that will be chosen later, using Lemma 1, for \(0< \eta < \min (0,a)\),
$$\begin{aligned}{} & {} E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i ) \\= & {} E_{\varvec{0}} \{E (1-\kappa _i \mid \varvec{X}_i ) \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] }\} + E_{\varvec{0}} \{E (1-\kappa _i \mid \varvec{X}_i ) \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2> a_n\right] }\} \\\le & {} E_{\varvec{0}} \{ K \tau _n^\eta \exp \left( \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i / 2 \right) \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] }\} + E_{\varvec{0}} \{\mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2> a_n\right] }\} \\= & {} K \tau _n^\eta V_k(a_n) + K Pr (\chi ^2_k > a_n) \\\le & {} K \tau _n^\eta a_n^k + K a_n^{k/2} \exp (-a_n/2), \end{aligned}$$
where \(V_k(r)\) is the volume of a Euclidean ball of radius r in k-dimensional Euclidean space, and the probability is bounded using the Chernoff bound for the \(\chi ^2\) random variables.
For some \(\epsilon > 0\), choose \(a_n = 2(1+\epsilon ) \log (n/q_n)\) and \(\tau _n = (q_n/n)^{(1+\epsilon )/\eta }\). Then,
$$\begin{aligned} \begin{aligned}&\sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]} \sum _{i=1}^{n} E_{\varvec{\theta }_{0i}} E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) / q_n \log (\frac{n}{q_n}) \\ \le&\frac{q_n + (n-q_n)(K \tau _n^\eta a_n^k + K a_n^{k/2} \exp (-a_n/2))}{q_n \log (\frac{n}{q_n})} \rightarrow 0, \end{aligned} \end{aligned}$$
as \(n \rightarrow \infty \). This proves (8).
Next, since
$$E( \Vert \varvec{\theta }_i - {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \mid \varvec{X}_i) \le 2E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^R_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) + 2E( \Vert \widehat{\varvec{\theta }}^R_i - {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \mid \varvec{X}_i),$$
by Markov’s inequality, (6) and (8) immediately lead to (9).\(\square \)
Proof
of Theorem 4 The proof of (10) is similar to the proof of (4), but now using Lemma 4. We start from
$$\begin{aligned} \Vert \widehat{\varvec{\theta }}^{EIG}_i \Vert _{\Sigma }^2 = E\left( 1-\kappa _i \mid \varvec{X}_i\right) ^2 \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le E\left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2. \end{aligned}$$
Then, for a sequence of positive constants \(\{a_n,n\ge 1\}\) to be specified later with \(a_n \rightarrow \infty \) as \(n \rightarrow \infty \),
$$\begin{aligned} \begin{aligned}&E_0\left[ E \left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] } \right] \\ \le&K c_n^d E_0\left[ \exp \left( \varvec{X}_i^T \varvec{\Sigma }^{-1} \varvec{X}_i /2 \right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] } \right] \\ \le&K c_n^d a_n^{k+1}, \end{aligned} \end{aligned}$$
and,
$$\begin{aligned} \begin{aligned}&E_0\left[ E \left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2> a_n\right] } \right] \\ \le&E_0 \left( \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > \lambda _{\min } \left( \Sigma ^{-1}\right) a_n \right] } \right) \\ \le&\exp \left( -\lambda _{\min } \left( \Sigma ^{-1}\right) a_n/4\right) . \end{aligned} \end{aligned}$$
By choosing \(a_n = 4 \lambda _{\min }^{-1} \left( \Sigma ^{-1}\right) (1+\epsilon )\log (n/q_n)\) and \(\tau _n = \left( q_n/n\right) ^{\frac{1+\epsilon }{d}}\), with \(\epsilon > 0\), one gets
$$\begin{aligned} \begin{aligned} E_0 \left[ E\left( 1-\kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right]&\le K \left[ (q_n/n)^{1+\epsilon } \left( \log (n/q_n)\right) ^{k+1} + (q_n/n)^{1+\epsilon } \right] \\&\lesssim (q_n/n)^{1+\epsilon } \left( \log (n/q_n)\right) ^{k+1}, \text { as } n \rightarrow \infty . \end{aligned} \end{aligned}$$
and hence
$${ \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} {\sum \limits _{i: \varvec{\theta }_{0i} = \varvec{0}} E_0 \Vert \widehat{\varvec{\theta }}^{EIG}_i \Vert _{\Sigma }^2} / {\left( q_n \log \left( \frac{n}{q_n}\right) \right) } = o(1), \text { as } n \rightarrow \infty .$$
The proof of (11) is similar to the proof of (5), but now using Lemmas 5 and 6. Again, it suffices to show that
$$\begin{aligned} \limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2]}{2\lambda _{\max }(\Sigma ) q_n \log (n/q_n)} \le 1. \end{aligned}$$
(B28)
and
$$\begin{aligned} \limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}}^{\frac{1}{2}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2]}{q_n \log (n/q_n)} = 0. \end{aligned}$$
(B29)
By Lemma 5, for sufficiently large \(b_n > 0\), uniformly in \(\varvec{\theta }_{0i} \ne \varvec{0}\),
$$\begin{aligned} \begin{aligned}&E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i> \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2> b_n \right] } \\ \le&K c_n^{-d} \Vert \varvec{X}_i\Vert _{\Sigma }^2 \exp {\left[ -\frac{\xi (1-\delta )}{2} \Vert \varvec{X}_i\Vert _{\Sigma }^2 \right] } \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > b_n \right] } \\ \le&K c_n^{-d} b_n \exp {\left[ -\frac{\xi (1-\delta )}{2} b_n \right] } \\ \le&K(q_n/n)^{\rho -\epsilon } \log \left( \frac{n}{q_n}\right) , \end{aligned} \end{aligned}$$
by choosing \(b_n = \frac{2(1+\rho )}{\xi (1-\delta )} \log \left( \frac{n}{q_n}\right) \), with \(\rho > \epsilon \), and recalling that \(c_n = \left( q_n/n\right) ^{\frac{1+\epsilon }{d}}\). So,
$$\begin{aligned} \limsup \limits _{n \rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} \left[ E \left( \kappa _i \mathbbm {1}_{\left[ \kappa _i> \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 > b_n \right] }\right] }{q_n \log (n/q_n)} = 0. \end{aligned}$$
(B30)
Also,
$$\begin{aligned} \begin{aligned}&E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] \\ \le&E_{\varvec{\theta }_{0i}} \left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] \\ \le&b_n \\ \le&\frac{2(1+\rho )}{\xi (1-\delta )} \log \left( \frac{n}{q_n}\right) . \end{aligned} \end{aligned}$$
(B31)
By making \(\epsilon \rightarrow 0\), \(\rho \rightarrow 0\), \(\eta \rightarrow \min (1,a)\), \(\xi \rightarrow 1\) and \(\delta \rightarrow 0\), one gets
$$\begin{aligned} \limsup \limits _{n\rightarrow \infty } { \sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]}} \frac{\sum \limits _{i: \varvec{\theta }_{0i} \ne \varvec{0}} E_{\varvec{\theta }_{0i}} \left[ E\left( \kappa _i \mathbbm {1}_{\left[ \kappa _i > \xi \right] } \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2 \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _{\Sigma }^2 \le b_n \right] } \right] }{2 q_n \log (n/q_n)} \le 1. \end{aligned}$$
(B32)
Together with Lemma 6 and (B30), this leads to (B28).
Altogether Lemma 5, (B30) and (B31) also imply that
$$\begin{aligned} E_{\varvec{\theta }_{0i}} [E\left( \kappa _i \mid \varvec{X}_i\right) \Vert \varvec{X}_i\Vert _{\Sigma }^2] \le K \log \left( \frac{n}{q_n}\right) (1+o(1)), \text { as } n \rightarrow \infty . \end{aligned}$$
(B33)
Consequently, one gets (B29). This completes the proof of (11).
Finally, (12) follows the previous two results and Theorem 1. \(\square \)
Proof
of Theorem 5 Similar to the Proof of Theorem 3, we only need to find a bound for \(E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i )\). When \(\varvec{\theta }_{0i} \ne 0\),
$$E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i ) \le 1.$$
When \(\varvec{\theta }_{0i} = 0\), letting \(\{a_n\}\) be a sequence of positive numbers that will be chosen later, using Lemma 4,
$$\begin{aligned} \begin{aligned}&E_{\varvec{\theta }_{0i}} E (1-\kappa _i \mid \varvec{X}_i ) \\ =&E_{\varvec{0}} \{E (1-\kappa _i \mid \varvec{X}_i ) \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 \le a_n\right] }\} + E_{\varvec{0}} \{E (1-\kappa _i \mid \varvec{X}_i ) \mathbbm {1}_{\left[ \Vert \varvec{X}_i\Vert _2^2 > a_n\right] }\} \\ \le&K c_n^d a_n^k + K a_n^{k/2} \exp (-a_n/2). \end{aligned} \end{aligned}$$
For some \(\epsilon > 0\), choose \(a_n = 2(1+\epsilon ) \log (n/q_n)\) and \(c_n = (q_n/n)^{(1+\epsilon )/d}\). Then,
$$\begin{aligned} \begin{aligned}&\sup _{\{\varvec{\theta }_{0i}\} \in L_0[q_n]} \sum _{i=1}^{n} E_{\varvec{\theta }_{0i}} E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^{EIG}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) / q_n \log (\frac{n}{q_n}) \\ \le&\frac{q_n + (n-q_n)(K c_n^d a_n^k + K a_n^{k/2} \exp (-a_n/2))}{q_n \log (\frac{n}{q_n})} \rightarrow 0, \end{aligned} \end{aligned}$$
as \(n \rightarrow \infty \). This proves (13).
Next, since
$$E( \Vert \varvec{\theta }_i - {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \mid \varvec{X}_i) \le 2E( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}^{EIG}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i) + 2E( \Vert \widehat{\varvec{\theta }}^{EIG}_i - {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \mid \varvec{X}_i),$$
by Markov’s inequality, (12) and (13) immediately lead to (14). \(\square \)
1.2 B.2 Proofs of Theorems in Section 3
Proof
of Theorem 6 We use \(\widehat{\varvec{\theta }}_{i}\) for \(\widehat{\varvec{\theta }}_{i}^R\) in this proof.
Proof of (16). Look at the case where \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \le K_S \tau _n\). Note that we could write \(\varvec{X}_i = \varvec{\theta }_{0i} + \varvec{\epsilon }_i\), where \(\varvec{\epsilon }_i \sim \varvec{N}_{k}(\varvec{0}, \varvec{\Sigma })\) and hence \(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \sim \chi ^2_k\). So,
$$\begin{aligned} \begin{aligned} \Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2&= \Vert \varvec{\theta }_{0i} - E(1-\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma }^2 \\&=\Vert E(\kappa _i \mid \varvec{X}_i)\varvec{\theta }_{0i} - E(1-\kappa _i \mid \varvec{X}_i) \varvec{\epsilon }_i \Vert _{\Sigma }^2 \\&\le 2 E(\kappa _i \mid \varvec{X}_i)^2 \Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 + 2 E(1-\kappa _i \mid \varvec{X}_i)^2 \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \\&\le 2 K_s \tau _n + 2 E(1-\kappa _i \mid \varvec{X}_i) \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2. \end{aligned} \end{aligned}$$
By Lemma 1, for any fixed \(\eta < a\),
$$\begin{aligned} \begin{aligned} E(1-\kappa _i \mid \varvec{X}_i)&\le K \tau _n^\eta \exp \left( \frac{\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2} \right) \\&\le \tau _n^\eta \exp \left( \Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \right) \exp \left( \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2\right) . \end{aligned} \end{aligned}$$
We will show that, for small enough \(\tau _n\),
$$\begin{aligned} \widehat{r}_i(\alpha , \tau _n) \ge \chi ^2_{k, A\alpha } c \tau _n (1+o(1)) , \end{aligned}$$
(B34)
for some \(c>0\) which we will fix later and any fixed \(A > K m / (a c^a)\) with, specifically, \(K = \int _0^\infty u^{-a-1} L(u) du\) here. Given this, for fixed \(a/(1+\rho )< \eta < a\) and for any fixed L,
$$\begin{aligned} \begin{aligned} P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \in \widehat{C}^{R}_{i})&= P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n)) \\&\ge P_{\varvec{\theta }_{0i}} (K \tau _n + K \tau _n^\eta e^{\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2} \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le K \tau _n^{a/(1+\rho )} (1+o(1))) \\&\ge P_{\varvec{\theta }_{0i}} (K \tau _n + K \tau _n^\eta e^{\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2} \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le K \tau _n^{a/(1+\rho )} (1+o(1)) \mid \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }) \\&\times P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\&\rightarrow 1 \times (1-\alpha ) = 1-\alpha , \end{aligned} \end{aligned}$$
as \(\tau _n \rightarrow 0\), since the left hand side of the inequality in the conditional probability is of a higher order of infinitesimal.
Now, it remains to prove (B34). Due to the normality of the posterior
$$\begin{aligned} \varvec{\theta }_i \mid \varvec{X}_i, \lambda _i^2 \sim \varvec{N}_k ((1-\kappa _i)\varvec{X}_i, (1-\kappa _i) \varvec{\Sigma }), \end{aligned}$$
applying Anderson’s lemma, we have
$$\begin{aligned} \begin{aligned} \Pi (\Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \!>\! \widehat{r}_i(\alpha , \tau _n) \!\mid \! \varvec{X}_i, \lambda _i^2) \!\ge \! \Pi ( \Vert \varvec{\theta }_i \!-\! (1\!-\!\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 \!>\! \widehat{r}_i(\alpha , \tau _n) \!\mid \! \varvec{X}_i, \lambda _i^2). \end{aligned} \end{aligned}$$
Thus,
$$\begin{aligned} \begin{aligned} \alpha&= \int _0^\infty \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2> \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\&\ge \int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2. \end{aligned} \end{aligned}$$
Recall that \(\pi (\lambda _i^2 \mid \varvec{X}_i) \propto (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp (-\frac{\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i}{2 (1+\lambda _i^2 \tau _n)})\). Let \(\widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) \propto (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2)\) be another density. Then, since \(\pi (\lambda _i^2 \mid \varvec{X}_i)/\widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i)\) and \( \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2)\) are both increasing in \(\lambda _i^2\)
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2. \end{aligned} \end{aligned}$$
On the other hand, since \(\Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 / (1-\kappa _i) \mid \varvec{X}_i, \lambda _i^2 \sim \chi ^2_k\), and \(1-\kappa _i \ge c \tau _n (1+o(1))\) when \(\lambda _i^2 \ge c\) for any \(c > 0\). Thus, for some fixed \(A(>0)\) to be determined later,
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } c \tau _n (1+o(1)) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\int _{c}^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } c \tau _n (1+o(1)) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\int _{c}^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \chi ^2_{k, A\alpha } (1-\kappa _i) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ =&A \alpha \widetilde{\Pi }(\lambda _i^2 \ge c \mid \varvec{X}_i). \end{aligned} \end{aligned}$$
Since, by the dominated convergence theorem,
$$\begin{aligned} \begin{aligned} \widetilde{\Pi }(\lambda _i^2 \ge c \mid \varvec{X}_i) =&\frac{\int _{c}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2}{\int _{0}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2} \\ \rightarrow&\frac{\int _{c}^{\infty } (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2}{\int _{0}^{\infty } (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2} \\ \ge&K m \int _{c}^\infty (\lambda _i^2)^{-a-1} d \lambda _i^2 = K m / (a c^a). \end{aligned} \end{aligned}$$
First fixing c to be such that \( K m / (a c^a) > \alpha \), if we further fix \(A > a c^a/(K m)\) with \(K = \int _0^\infty u^{-a-1} L(u) du\), then
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } \tau _n/2 (1+o(1)) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\alpha A K m / (a c^a) \\>&\alpha \\ \ge&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \widetilde{\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2. \end{aligned} \end{aligned}$$
This implies that \(\widehat{r}_i(\alpha , \tau _n) \ge \chi ^2_{k, A\alpha } c \tau _n (1+o(1))\) which completes the proof of the first part.
Proof of (17) For the case where \(f_{\tau _n} \tau _n \le \Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \le K_M \log \frac{1}{\tau _n}\), we start with the inequality
$$\begin{aligned} \Vert \varvec{X}_{i} \Vert _{\Sigma } \le \Vert \varvec{\theta }_{0i} \Vert _{\Sigma } + \Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }. \end{aligned}$$
For \(K_0 > K_M\),
$$\begin{aligned} \begin{aligned} \Vert \varvec{X}_{i} \Vert _{\Sigma } - \left( K_0 \log \frac{1}{\tau _n} \right) ^{1/2} \le \Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma } + \left( \sqrt{K_M} - \sqrt{K_0}\right) \left( \log \frac{1}{\tau _n} \right) ^{1/2} \le 0, \end{aligned} \end{aligned}$$
if
$$\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma } \le \left( \sqrt{K_0} - \sqrt{K_M}\right) \left( \log \frac{1}{\tau _n} \right) ^{1/2},$$
the probability of which converges to 1 as \(\tau _n \rightarrow 0\).
We now study \(\widehat{r}_i(\alpha , \tau _n)\) in this case. We will find an upper bound for \(\widehat{r}_i(\alpha , \tau _n)\). As the first ingredient, for \(B > 0\),
$$\begin{aligned} \begin{aligned}&{\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i) \\ =&\frac{\int _{B}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2}{\int _{0}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2}. \end{aligned} \end{aligned}$$
The denominator
$$\begin{aligned} \begin{aligned}&\int _{0}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2 \\ \ge&\exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2}\right) \int _{1}^{2} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2 \\ \ge&\exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2}\right) (3)^{-k/2} L(1) \int _{1}^{2} (\lambda _i^2)^{-a-1} d\lambda _i^2\\ =&K \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2}\right) . \end{aligned} \end{aligned}$$
Hence,
$$\begin{aligned} \begin{aligned} {\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i) \le&K \exp \left( \frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2}\right) \int _{B}^{\infty } (\lambda _i^2)^{-a-1} d\lambda _i^2 \\ =&K \exp \left( \frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2}\right) B^{-a}. \end{aligned} \end{aligned}$$
(B35)
Next, by the inequality
$$\begin{aligned} \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \le 2 \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 + 2 \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2, \end{aligned}$$
for \(r > 0\),
$$\begin{aligned} \frac{1}{2} \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 - \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge r \end{aligned}$$
would imply that
$$\begin{aligned} \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge r. \end{aligned}$$
Now,
$$\begin{aligned} \begin{aligned}&\Pi ( \frac{1}{2} \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 - \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge r \mid \varvec{X}_i, \lambda _i^2 ) \\ \le&\Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge r \mid \varvec{X}_i, \lambda _i^2 ). \end{aligned} \end{aligned}$$
(B36)
Using (B36),
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \!\ge \! 2r \!+\! 2 \sup _{\lambda _i^2 \le B} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 \\ \le&\int _0^B \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \!\ge \! 2r \!+\! 2 \sup _{\lambda _i^2 \le B} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 \\ +&{\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i) \\ \le&\int _0^B \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge 2r + 2 \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 \\ +&{\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i) \\ \le&\int _0^B \Pi ( \Vert \varvec{\theta }_i \!-\! (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge r \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 + {\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i). \end{aligned} \end{aligned}$$
Since \(1-\kappa _i \le \frac{B\tau _n}{1+B\tau _n}\) when \(\lambda _i^2 \le B\), we can bound the first term above by \(\alpha /2\) via
$$\begin{aligned} \begin{aligned}&\int _0^B \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge \chi ^2_{k,\alpha /2} \frac{B\tau _n}{1+B\tau _n} \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 \\ \le&\int _0^B \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge \chi ^2_{k,\alpha /2} (1-\kappa _i) \mid \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2 \\ \le&\alpha /2. \end{aligned} \end{aligned}$$
As for the second term, by (B35), when \(\varvec{X}_i\) is fixed, for large enough B,
$$\begin{aligned} {\Pi }(\lambda _i^2 \ge B \mid \varvec{X}_i) \le \alpha / 2. \end{aligned}$$
(B37)
This leads to
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2\\ =&\alpha \\ \ge \!&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge 2r \!+\! 2 \sup _{\lambda _i^2 \le B} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \!\mid \! \varvec{X}_i, \lambda _i^2 ) \pi (\lambda _i^2 \mid \varvec{X}_i) d \lambda _i^2, \end{aligned} \end{aligned}$$
if \(r = \chi ^2_{k,\alpha /2} B\tau _n / (1+B\tau _n)\) and B is large enough. Thus, when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0 \log \frac{1}{\tau _n}\), for small enough \(\tau _n\),
$$\begin{aligned} \begin{aligned} \widehat{r}_i(\alpha , \tau _n) \le&2\chi ^2_{k,\alpha /2} \frac{B\tau _n}{1+B\tau _n} + 2 \sup _{\lambda _i^2 \le B} \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&2\chi ^2_{k,\alpha /2} \frac{B\tau _n}{1\!+\!B\tau _n} \!+\! 4 \!\sup _{\lambda _i^2 \le B} \Vert E(1\!-\!\kappa _i \!\mid \! \varvec{X}_i) {\varvec{X}}_{i} \Vert _{\Sigma }^2 \!+\! 4 \!\sup _{\lambda _i^2 \le B} \Vert (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&2\chi ^2_{k,\alpha /2} B\tau _n + 8 B \tau _n \Vert \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&K B \tau _n \log \frac{1}{\tau _n}. \end{aligned} \end{aligned}$$
Again, applying Lemma 1, when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0 \log \frac{1}{\tau _n}\), for fixed \(\eta < a\),
$$\begin{aligned} \begin{aligned} \Vert \widehat{\varvec{\theta }}_{i}\Vert _{\Sigma }^2 \le&E(1-\kappa _i \mid \varvec{X}_i) \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \\ \le&K \tau _n^{\eta } \exp \left( \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 / 2 \right) \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \\ \le&K \tau _n^{\eta } \exp \left( K_0 \log \frac{1}{\tau _n} / 2 \right) K_0 \log \frac{1}{\tau _n}\\ =&K \tau _n^{\eta - K_0 / 2} \log \frac{1}{\tau _n}. \end{aligned} \end{aligned}$$
If we choose B to be such that \(B^a = K\tau _n^{-K_0/2}\), in which the factor K makes (B37) hold, we then have, as \(\tau _n \rightarrow 0\),
$$\begin{aligned} \begin{aligned} 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n) \le K \tau _n^{\eta - K_0 / 2} \log \frac{1}{\tau _n} + K \left( \tau _n^{1 - K_0/2a} \log \frac{1}{\tau _n} \right) ^{a/(1+\rho )} = o(1), \end{aligned} \end{aligned}$$
if we require \(K_M < 2a \) and fix \(\eta \) and \(K_0\) to be such that \(K_M< K_0< 2\eta < 2a\). This leads to that, when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0 \log \frac{1}{\tau _n}\), for some \(f_{\tau _n}\) such that \(f_{\tau _n} \rightarrow \infty \) and \(f_{\tau _n} \tau _n \rightarrow 0\) as \(\tau _n \rightarrow 0\),
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \ge f_{\tau _n} \tau _n \ge 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n). \end{aligned}$$
Finally, by the inequality
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \le 2 \Vert {\varvec{\theta }}_{0i} - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2, \end{aligned}$$
the fact that
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \ge 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n), \end{aligned}$$
would imply that
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n). \end{aligned}$$
So,
$$\begin{aligned} \begin{aligned}&P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \notin \widehat{C}^{R}_{i}) \\ =&P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2> L \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n)) \\ \ge&P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} \!-\! \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \!>\! L \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n) \!\mid \! \Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \!\le \! (\sqrt{K_0} \!-\! \sqrt{K_M})^2 \log 1/\tau _n ) \\ \times&P(\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \le (\sqrt{K_0} - \sqrt{K_M})^2 \log 1/\tau _n)\\ =&P(\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \le (\sqrt{K_0} - \sqrt{K_M})^2 \log 1/\tau _n)\\ \rightarrow&1, \end{aligned} \end{aligned}$$
as \(\tau _n \rightarrow 0\), for any fixed \(L > 0\).
Proof of (18). Consider the case where \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n}\). We write
$$\begin{aligned} \begin{aligned} \Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }&= \Vert \varvec{\theta }_{0i} - \varvec{X}_i - E(\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma } \\&\le \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma } + \Vert E(\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma }. \end{aligned} \end{aligned}$$
Applying both Lemmas 2 and 3, for any fixed \(\xi , \delta \in (0,1)\),
$$\begin{aligned} \begin{aligned} \Vert E(\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma }^2&\le 2 \left( \frac{K}{\Vert \varvec{X}_i \Vert _{\Sigma }^2}\right) ^2 \Vert \varvec{X}_i \Vert _{\Sigma }^2 \\&+ 2 \left( K \tau _n^{-a} \exp \left( -\frac{\xi (1-\delta )}{2} \Vert \varvec{X}_i \Vert _{\Sigma }^2 \right) \right) ^2 \Vert \varvec{X}_i \Vert _{\Sigma }^2, \end{aligned} \end{aligned}$$
Further,
$$\Vert \varvec{X}_i \Vert _{\Sigma } = \Vert \varvec{\theta }_{0i} + \varvec{\epsilon }_i \Vert _{\Sigma } \ge |\Vert \varvec{\theta }_{0i} \Vert _{\Sigma } - \Vert \varvec{\epsilon }_i \Vert _{\Sigma } |,$$
in which \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n}\), by assumption, and \(\Vert \varvec{\epsilon }_i \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }\) with probability \(1-\alpha \). Since \(\tau _n \rightarrow 0\), when \(\Vert \varvec{\epsilon }_i \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }\), we will have
$$\Vert \varvec{X}_i \Vert _{\Sigma } \ge \left( K_L \log \frac{1}{\tau _n} \right) ^{1/2} - \left( \chi ^2_{k, \alpha }\right) ^{1/2}$$
and, consequently,
$$\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1)).$$
By choosing \(K_L\), \(\xi \), and \(\delta \) to be such that \(K_L (1+o(1)) > \frac{2a}{\xi (1-\delta )}\), e.g., \(K_L = 3a\) and \(\xi (1-\delta ) = 3/4\), when \(\Vert \varvec{\epsilon }_i \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }\), we will have, as \(\tau _n \rightarrow 0\),
$$\begin{aligned} \Vert E(\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma }^2 \le o(1), \end{aligned}$$
hence,
$$\begin{aligned} \begin{aligned} \Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2&\le \left( \sqrt{\chi ^2_{k,\alpha }} + o(1) \right) ^2 = \chi ^2_{k,\alpha } (1 + o(1)). \end{aligned} \end{aligned}$$
(B38)
Then, we find a lower bound for \(\widehat{r}_i(\alpha , \tau _n)\) in this case. Making use of the posterior normality and Anderson’s lemma again,
$$\begin{aligned} \begin{aligned} \alpha&= \int _0^\infty \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2> \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\&\ge \int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2. \end{aligned} \end{aligned}$$
On the other hand, since \(1-\kappa _i \ge g_{\tau _n} / (1+g_{\tau _n}) = 1 + o(1),\) if \(\lambda _i^2 \ge g_{\tau _n}/\tau _n\) with \(g_{\tau _n} = (\log 1/\tau _n)^{1/3}\), for some fixed A,
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } g_{\tau _n} / (1+g_{\tau _n}) \mid \varvec{X}_i, \lambda _i^2) {\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\int _{g_{\tau _n}/\tau _n}^\infty \Pi ( \Vert \varvec{\theta }_i \!-\! (1\!-\!\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } g_{\tau _n} / (1+g_{\tau _n}) \mid \varvec{X}_i, \lambda _i^2) {\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\int _{g_{\tau _n}/\tau _n}^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \chi ^2_{k, A\alpha } (1-\kappa _i) \mid \varvec{X}_i, \lambda _i^2) {\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ =&A \alpha {\Pi }(\lambda _i^2 \ge g_{\tau _n}/\tau _n \mid \varvec{X}_i). \end{aligned} \end{aligned}$$
We now study the posterior probability in a situation where \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\):
$$\begin{aligned} \begin{aligned}&{\Pi }(\lambda _i^2 < g_{\tau _n}/\tau _n \mid \varvec{X}_i) \\ =&\frac{\int ^{g_{\tau _n}/\tau _n}_{0} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2}{\int _{0}^{\infty } (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2} \\ :=&N/D. \end{aligned} \end{aligned}$$
Next,
$$\begin{aligned} \begin{aligned} D \ge&\int _{(g_{\tau _n}+1)/\tau _n}^{2(g_{\tau _n}+1)/\tau _n} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2 \\ \ge&m (2 g_{\tau _n}+3)^{-k/2} \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (2+g_{\tau _n})}\right) \int _{(g_{\tau _n}+1)/\tau _n}^{2(g_{\tau _n}+1)/\tau _n} (\lambda _i^2)^{-a-1} d\lambda _i^2 \\ =&K (2 g_{\tau _n}+3)^{-k/2} \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (2+g_{\tau _n})}\right) \left( \frac{\tau _n}{g_{\tau _n}+1}\right) ^a. \end{aligned} \end{aligned}$$
Next, fixing a constant \(c > 0\), when \(g_{\tau _n}\) is large enough,
$$\begin{aligned} \begin{aligned} N =&\int ^{g_{\tau _n}/\tau _n}_{0} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2\\ =&\left( \int ^{c/\tau _n}_{0} + \int ^{g_{\tau _n}/\tau _n}_{c/\tau _n} \right) (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2 \\ =:&N_1 + N_2. \end{aligned} \end{aligned}$$
The first term
$$\begin{aligned} \begin{aligned} N_1 =&\int ^{c/\tau _n}_{0} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2 \\ \le&\int ^{c/\tau _n}_{0} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+c)}\right) d\lambda _i^2 \\ \le&\exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+c)}\right) \int ^{\infty }_{0} (\lambda _i^2)^{-a-1} L(\lambda _i^2) d\lambda _i^2 \\ =&K \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+c)}\right) . \end{aligned} \end{aligned}$$
When \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\),
$$\begin{aligned} \begin{aligned} N_1/D \le&K (2 g_{\tau _n}+3)^{k/2} \left( \frac{g_{\tau _n}+1}{\tau _n}\right) ^a \exp \left( \frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (2+g_{\tau _n})}-\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+c)}\right) \\ \le&K (2 g_{\tau _n}+3)^{k/2} \left( \frac{g_{\tau _n}+1}{\tau _n}\right) ^a \exp \left( -\frac{K_L}{2} \log \frac{1}{\tau _n} \frac{1}{(1+c)} (1+o(1)) \right) \\ =&K (2 g_{\tau _n}+3)^{k/2} \left( g_{\tau _n}+1\right) ^a \tau _n^{\frac{K_L}{2(1+c)}(1+o(1))-a} = o(1), \end{aligned} \end{aligned}$$
(B39)
as \(\tau _n \rightarrow 0\), if c is chosen properly, e.g., \(c = 1/3\). Our choice of \(K_L\) makes sure that the above is o(1) as \(\tau _n \rightarrow 0\), given that \(g_{\tau _n} = (\log 1/\tau _n)^{1/3}\). Moreover,
$$\begin{aligned} \begin{aligned} N_2 =&\int _{c/\tau _n}^{g_{\tau _n}/\tau _n} (1+\lambda _i^2 \tau _n)^{-k/2} (\lambda _i^2)^{-a-1} L(\lambda _i^2) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+\lambda _i^2 \tau _n)}\right) d\lambda _i^2 \\ \le&M (1+c)^{-k/2} \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+g_{\tau _n})}\right) \int _{c/\tau _n}^{g_{\tau _n}/\tau _n} (\lambda _i^2)^{-a-1} d\lambda _i^2 \\ \le&M (1+c)^{-k/2} \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+g_{\tau _n})}\right) (\tau _n / c)^{a+1} \frac{g_{\tau _n} - c}{\tau _n}. \end{aligned} \end{aligned}$$
It follows that
$$\begin{aligned} \begin{aligned} N_2/D \le&K (2 g_{\tau _n}+3)^{k/2} \left( g_{\tau _n}+1\right) ^a (g_{\tau _n} - c) \exp \left( -\frac{\Vert \varvec{X}_i \Vert _{\Sigma }^2}{2 (1+g_{\tau _n})(2+g_{\tau _n})}\right) . \end{aligned} \end{aligned}$$
Since we have chosen \(g_{\tau _n} = (\log 1/\tau _n)^{1/3}\), when \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\), we will have, as \(\tau _n \rightarrow 0\),
$$\begin{aligned} N_2/D = o(1). \end{aligned}$$
(B40)
Combining (B39) and (B40), when \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\), as \(\tau _n \rightarrow 0\),
$$\begin{aligned} \begin{aligned} {\Pi }(\lambda _i^2 < g_{\tau _n}/\tau _n \mid \varvec{X}_i) = o(1), \end{aligned} \end{aligned}$$
and hence
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \chi ^2_{k, A\alpha } g_{\tau _n} / (1+g_{\tau _n}) \mid \varvec{X}_i, \lambda _i^2) {\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&A \alpha {\Pi }(\lambda _i^2 \ge g_{\tau _n}/\tau _n \mid \varvec{X}_i) \\ =&A \alpha (1-o(1)). \end{aligned} \end{aligned}$$
Here, let \(A = \beta /\alpha \) for some fixed \(\beta > \alpha \). For small enough \(\tau _n\), we will have
$$\begin{aligned} \begin{aligned}&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, \beta } g_{\tau _n} / (1+g_{\tau _n}) \mid \varvec{X}_i, \lambda _i^2) {\pi }(\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2 \\ \ge&\beta (1-o(1)) \\>&\alpha \\ \ge&\int _0^\infty \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , \tau _n) \mid \varvec{X}_i, \lambda _i^2) \pi (\lambda _i^2 \mid \varvec{X}_i) d\lambda _i^2. \end{aligned} \end{aligned}$$
It follows that
$$\begin{aligned} \widehat{r}_i(\alpha , \tau _n) \ge \chi ^2_{k, \beta } g_{\tau _n} / (1+g_{\tau _n}) = \chi ^2_{k, \beta } (1+o(1)), \end{aligned}$$
(B41)
when \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\).
As a reminder, when \(\tau _n\) is small enough, \(\Vert \varvec{\epsilon }_i \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }\) implies that \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n} (1+o(1))\).
So, by (B38) along with (B41), in the case where \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \ge K_L \log \frac{1}{\tau _n}\), for any fixed \(\beta > \alpha \), as \(\tau _n \rightarrow 0\),
$$\begin{aligned} \begin{aligned} P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \in \widehat{C}^{R}_{i})&= P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n)) \\&\!\ge P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{a/(1+\rho )}(\alpha , \tau _n) \mid \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }) \\&\times P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\&\!\ge P_{\varvec{\theta }_{0i}} (\chi ^2_{k,\alpha }(1 \!+\! o(1)) \!\le \! L (\chi ^2_{k, \beta })^{a/(1+\rho )} (1\!+\!o(1)) \!\mid \! \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \!\le \! \chi ^2_{k, \alpha }) \\&\!\times P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\&\!\rightarrow 1 \times (1-\alpha ) = 1-\alpha , \end{aligned} \end{aligned}$$
if we choose \(L > \chi ^2_{k,\alpha }(\chi ^2_{k,\beta })^{-a/(1+\rho )}\), e.g., \(L = 2\chi ^2_{k,\alpha }(\chi ^2_{k,\beta })^{-a/(1+\rho )}\).\(\square \)
Proof
of Theorem 7 We use \(\widehat{\varvec{\theta }}_{i}\) for \(\widehat{\varvec{\theta }}_{i}^{EIG}\) in this proof.
Proof of (20). We first find a lower bound for \(\widehat{r}_i(\alpha , c_n)\). Recall that \(\pi (\kappa _i \mid \varvec{X}_i) \propto \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} \exp (-\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2)\). Similar to the proof of (16), let \(\widetilde{\pi }(\kappa _i \mid \varvec{X}_i) \propto \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1}\) be another density. Also, under the setup of this theorem,
$$\begin{aligned} \varvec{\theta }_i \mid \varvec{X}_i, \kappa _i \sim \varvec{N}_k ((1-\kappa _i)\varvec{X}_i, (1-\kappa _i) \varvec{\Sigma }), \end{aligned}$$
Now we can proceed similarly as the proof of (B34). For some \(A(>0)\) and some \(v (> 1)\) to be determined later,
$$\begin{aligned} \begin{aligned}&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } c_n^v \mid \varvec{X}_i, \kappa _i) \widetilde{\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ \ge&\int _0^{ 1-c_n^v} \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } c_n^v \mid \varvec{X}_i, \kappa _i) \widetilde{\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ \ge&\int _0^{ 1-c_n^v} \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \chi ^2_{k, A\alpha } (1-\kappa _i) \mid \varvec{X}_i, \kappa _i) \widetilde{\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ =&A \alpha \widetilde{\Pi }(\kappa _i \le 1-c_n^v \mid \varvec{X}_i). \end{aligned} \end{aligned}$$
Since
$$\begin{aligned} \begin{aligned} \widetilde{\Pi }(\kappa _i > 1-c_n^v \mid \varvec{X}_i)&= \frac{\int _{ 1-c_n^v}^1 \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} d\kappa _i}{\int _0^1 \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} d\kappa _i} \\&\le \frac{\int _{1-c_n^v}^1 c_n^{-d-1} d\kappa _i}{\int _{1-c_n}^1 (1-c_n)^{d+k/2-1} (c_n + c_n (1-c_n))^{-d-1} d\kappa _i} \\&= c_n^{v-1} \frac{(2-c_n)^{d+1}}{(1-c_n)^{d+k/2-1}}, \end{aligned} \end{aligned}$$
suppose \(\alpha < 1/2\), when \(c_n\) is small enough, if we fix \(A = 2\), for example,
$$\begin{aligned} A \alpha \widetilde{\Pi }(\kappa _i \le 1-c_n^v \mid \varvec{X}_i) > \alpha . \end{aligned}$$
On the other hand, by Anderson’s lemma,
$$\begin{aligned} \begin{aligned} \alpha&= \int _0^1 \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2> \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i \\&\ge \int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i. \end{aligned} \end{aligned}$$
So,
$$\begin{aligned} \begin{aligned}&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } c_n^v \mid \varvec{X}_i, \kappa _i) \widetilde{\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i\\> \alpha \ge&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \widetilde{\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i. \end{aligned} \end{aligned}$$
This implies that
$$\begin{aligned} \widehat{r}_i(\alpha , c_n) \ge \chi ^2_{k, A\alpha } c_n^v. \end{aligned}$$
(B42)
For the case where \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \le K_S' c_n\). Again,
$$\begin{aligned} \begin{aligned} \Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2&= \Vert \varvec{\theta }_{0i} - E(1-\kappa _i \mid \varvec{X}_i)\varvec{X}_i \Vert _{\Sigma }^2 \\&=\Vert E(\kappa _i \mid \varvec{X}_i)\varvec{\theta }_{0i} - E(1-\kappa _i \mid \varvec{X}_i) \varvec{\epsilon }_i \Vert _{\Sigma }^2 \\&\le 2 E(\kappa _i \mid \varvec{X}_i)^2 \Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 + 2 E(1-\kappa _i \mid \varvec{X}_i)^2 \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \\&\le 2 K_s \tau _n + 2 E(1-\kappa _i \mid \varvec{X}_i) \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2. \end{aligned} \end{aligned}$$
Using Lemma 4 and (B42), when we choose v to be such that \(v{d/(1+\rho )}< d < 1\), e.g., \(v = 1+\rho /2\),
$$\begin{aligned} \begin{aligned} P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \in \widehat{C}^{EIG}_{i})&= P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{d/(1+\rho )} (\alpha , c_n)) \\&\ge P_{\varvec{\theta }_{0i}} (K c_n + K c_n^d e^{\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2} \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le (\chi ^2_{k, A\alpha } c_n^v)^{d/(1+\rho )} ) \\&\ge P_{\varvec{\theta }_{0i}} (K c_n + K c_n^d e^{\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2} \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le (\chi ^2_{k, A\alpha } c_n^v)^{d/(1+\rho )} \mid \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }) \\&\times P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\&\rightarrow 1 \times (1-\alpha ) = 1-\alpha , \end{aligned} \end{aligned}$$
as \(c_n \rightarrow 0\), since the left hand side of the inequality in the conditional probability is of a higher order of infinitesimal.
Proof of (21) For the case where \( f'_{c_n} c_n \le \Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \le K'_M \log \frac{1}{c_n}\), similar to the proof of (17), for \(K_0' > K_M'\),
$$\begin{aligned} \begin{aligned} \Vert \varvec{X}_{i} \Vert _{\Sigma } - \left( K_0' \log \frac{1}{c_n} \right) ^{1/2} \le \Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma } + \left( \sqrt{K_M'} - \sqrt{K_0'}\right) \left( \log \frac{1}{c_n} \right) ^{1/2} \le 0, \end{aligned} \end{aligned}$$
if
$$\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma } \le \left( \sqrt{K_0'} - \sqrt{K_M'}\right) \left( \log \frac{1}{c_n} \right) ^{1/2},$$
the probability of which converges to 1 as \(c_n \rightarrow 0\).
Using (B36) again,
$$\begin{aligned}{} & {} \int _0^1 \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge 2r \!+\! 2 \sup _{\kappa _i \ge {B_n}} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \!\mid \! \varvec{X}_i, \kappa _i ) \pi (\kappa _i \!\mid \! \varvec{X}_i) d \kappa _i \\\le & {} \int _{B_n}^1 \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \!\ge \! 2r \!+\! 2 \sup _{\kappa _i \ge {B_n}} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1\!-\!\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \kappa _i ) \pi (\kappa _i \!\mid \! \varvec{X}_i) d \kappa _i \\+ & {} {\Pi }(\kappa _i< {B_n} \mid \varvec{X}_i) \\\le & {} \int _{B_n}^1 \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge 2r + 2 \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \mid \varvec{X}_i, \kappa _i ) \pi (\kappa _i \mid \varvec{X}_i) d \kappa _i \\+ & {} {\Pi }(\kappa _i< {B_n} \mid \varvec{X}_i) \\\le & {} \int _{B_n}^1 \Pi ( \Vert \varvec{\theta }_i \!-\! (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \!\ge \! r \mid \varvec{X}_i, \kappa _i ) \pi (\kappa _i \mid \varvec{X}_i) d \kappa _i \!+\! {\Pi }(\kappa _i < {B_n} \mid \varvec{X}_i). \end{aligned}$$
For the first term, let \(r = \chi ^2_{k,\alpha /2} (1-{B_n})\),
$$\begin{aligned} \begin{aligned}&\int _{B_n}^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge \chi ^2_{k,\alpha /2} (1-{B_n}) \mid \varvec{X}_i, \kappa _i ) \pi (\kappa _i \mid \varvec{X}_i) d \kappa _i \\ \le&\int _{B_n}^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \ge \chi ^2_{k,\alpha /2} (1-\kappa _i) \mid \varvec{X}_i, \kappa _i ) \pi (\kappa _i \mid \varvec{X}_i) d \kappa _i \\ \le&\alpha /2. \end{aligned} \end{aligned}$$
For the other term, when \(\varvec{X}_i\) is fixed, if \({B_n} = 1 - c_n^{d/(2d+2)}\), as \(c_n \rightarrow 0\),
$$\begin{aligned} \begin{aligned}&{\Pi }(\kappa _i < {B_n} \mid \varvec{X}_i) \\ =&\frac{\int _{0}^{{1-c_n^{d/(2d+2)}}} \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} \exp (-\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d \kappa _i }{\int _{0}^{1}\kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} \exp (-\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d \kappa _i } \\ \le&\frac{\int _{0}^{{1-c_n^{d/(2d+2)}}} (c_n^{d/(2d+2)} + c_n (1-c_n^{d/(2d+2)}))^{-d-1} d \kappa _i }{\exp (-\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) \int _{1-c_n}^{1} (1-c_n)^{d+k/2-1} (c_n + c_n (1-c_n))^{-d-1} d \kappa _i } \\ =&\frac{c_n^{-d/2} (1+c_n^{(d+2)/(2d+2)}-c_n)^{-d-1}}{\exp (-\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) c_n^{-d} (1-c_n)^{d+k/2-1} (2-c_n)^{-d-1} } \\ \le&K \exp (\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) {c_n}^{d/2} \le \alpha /2. \end{aligned} \end{aligned}$$
So,
$$\begin{aligned} \begin{aligned}&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i\\ =&\alpha \\ \ge&\! \int _0^1 \Pi ( \Vert \varvec{\theta }_i \!-\! \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge 2r \!+\! 2 \sup _{\kappa _i \ge {B_n}} \Vert \widehat{\varvec{\theta }}_{i} \!-\! (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \!\mid \! \varvec{X}_i, \kappa _i ) \pi (\kappa _i \mid \varvec{X}_i) d \kappa _i, \end{aligned} \end{aligned}$$
if \(r = \chi ^2_{k,\alpha /2} (1-{B_n})\) and \({B_n} = 1 - c_n^{d/(2d+2)}\). Thus, when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0' \log \frac{1}{c_n}\), for small enough \(c_n\),
$$\begin{aligned} \begin{aligned} \widehat{r}_i(\alpha , c_n) \le&2\chi ^2_{k,\alpha /2} (1-{B_n}) + 2 \sup _{\kappa _i \ge {B_n}} \Vert \widehat{\varvec{\theta }}_{i} - (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&2\chi ^2_{k,\alpha /2} c_n^{d/(2d+2)} + 4 \sup _{\kappa _i \ge {B_n}} \Vert E(1-\kappa _i \mid \varvec{X}_i) {\varvec{X}}_{i} \Vert _{\Sigma }^2 + 4 \sup _{\kappa _i \ge {B_n}} \Vert (1-\kappa _i) \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&2\chi ^2_{k,\alpha /2} c_n^{d/(2d+2)} + 8 c_n^{d/(2d+2)} \Vert \varvec{X}_i \Vert _{\Sigma }^2 \\ \le&K c_n^{d/(2d+2)} \log \frac{1}{c_n}. \end{aligned} \end{aligned}$$
Finally, by the inequality
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \le 2 \Vert {\varvec{\theta }}_{0i} - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2, \end{aligned}$$
the fact that
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \ge 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{d/(1+\rho )}(\alpha , \tau _n), \end{aligned}$$
would imply that
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 \ge \widehat{r}_i^{d/(1+\rho )}(\alpha , \tau _n). \end{aligned}$$
Applying Lemma 4, when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0' \log \frac{1}{c_n}\),
$$\begin{aligned} \begin{aligned} \Vert \widehat{\varvec{\theta }}_{i}\Vert _{\Sigma }^2 \le&E(1-\kappa _i \mid \varvec{X}_i) \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \\ \le&K c_n^d \exp \left( \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 / 2 \right) \Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \\ \le&K c_n^d \exp \left( K_0' \log \frac{1}{c_n} / 2 \right) K_0' \log \frac{1}{c_n}\\ =&K c_n^{d - K_0' / 2} \log \frac{1}{c_n}. \end{aligned} \end{aligned}$$
We then have, as \(c_n \rightarrow 0\),
$$\begin{aligned} \begin{aligned} 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{d/(1+\rho )}(\alpha , c_n) \le K c_n^{d - K_0' / 2} \log \frac{1}{c_n} + K \left( c_n^{d/(2d+2)} \log \frac{1}{c_n}\right) ^{d/(1+\rho )} = o(1), \end{aligned} \end{aligned}$$
if we require \(K_M < 2d \) and fix \(K_0'\) to be such that \(K_M< K_0' < 2d\). This implies that when \(\Vert \varvec{X}_{i} \Vert _{\Sigma }^2 \le K_0' \log \frac{1}{c_n}\), for some \(f_{\tau _n}\) such that \(f_{c_n}' \rightarrow \infty \) and \(f_{c_n}' c_n \rightarrow 0\) as \(c_n \rightarrow 0\),
$$\begin{aligned} \Vert {\varvec{\theta }}_{0i} \Vert _{\Sigma }^2 \ge f_{c_n}' c_n \ge 2 \Vert \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2 + 2 \widehat{r}_i^{d/(1+\rho )}(\alpha , c_n). \end{aligned}$$
Finally,
$$\begin{aligned} \begin{aligned}&P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \notin \widehat{C}^{EIG}_{i}) \\ =&P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2> L \widehat{r}_i^{d/(1+\rho )}(\alpha , \tau _n)) \\ \ge&P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} \!-\! \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \!>\! L \widehat{r}_i^{d/(1+\rho )}(\alpha , \tau _n) \mid \Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \le (\sqrt{K_0} \!-\! \sqrt{K_M})^2 \log 1/\tau _n ) \\ \times&P(\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \le (\sqrt{K_0} - \sqrt{K_M})^2 \log 1/\tau _n)\\ =&P(\Vert \varvec{\epsilon }_{0i} \Vert _{\Sigma }^2 \le (\sqrt{K_0} - \sqrt{K_M})^2 \log 1/\tau _n)\\ \rightarrow&1, \end{aligned} \end{aligned}$$
as \(c_n \rightarrow 0\), for any fixed \(L > 0\).
Proof of (22). Similar to the proof of (18), using Lemmas 5 and 6, when \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \ge K_L' \log \frac{1}{c_n}\) and \(\Vert \varvec{\epsilon }_i \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }\), we have, as \(c_n \rightarrow 0\),
$$\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L' \log \frac{1}{c_n} (1+o(1)),$$
and, for \(K_L' (1+o(1)) > \frac{2d}{\xi (1-\delta )}\),
$$\begin{aligned} \begin{aligned} \Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2&\le \chi ^2_{k,\alpha } (1 + o(1)). \end{aligned} \end{aligned}$$
(B43)
Then, we find a lower bound for \(\widehat{r}_i(\alpha , c_n)\) in this case. Making use of the posterior normality and Anderson’s lemma again,
$$\begin{aligned} \begin{aligned} \alpha&= \int _0^1 \Pi ( \Vert \varvec{\theta }_i - \widehat{\varvec{\theta }}_{i} \Vert _{\Sigma }^2> \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i \\&\ge \int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i. \end{aligned} \end{aligned}$$
On the other hand, since \(1-\kappa _i \ge 1 - c_n\) if \(\kappa _i \le c_n\), for some fixed A,
$$\begin{aligned} \begin{aligned}&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } (1 - c_n) \mid \varvec{X}_i, \kappa _i) {\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ \ge&\int _{0}^{c_n} \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } (1 - c_n) \mid \varvec{X}_i, \kappa _i) {\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ \ge&\int _{0}^{c_n} \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \chi ^2_{k, A\alpha } (1-\kappa _i) \mid \varvec{X}_i, \kappa _i) {\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ =&A \alpha {\Pi }( \kappa _i \le c_n \mid \varvec{X}_i). \end{aligned} \end{aligned}$$
When \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L' \log \frac{1}{c_n} (1+o(1)),\) as \(n \rightarrow \infty \),
$$\begin{aligned} \begin{aligned}&{\Pi }(\kappa _i \ge c_n \mid \varvec{X}_i) \\ \le&\frac{\int _{c_n}^{1} \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} \exp (-\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d\kappa _i }{\int _{1-c_n^{1/d}}^{1} \kappa _i^{d+k/2-1} (1 - \kappa _i + c_n \kappa _i)^{-d-1} \exp (-\kappa _i \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d\kappa _i} \\ \le&\frac{\int _{c_n}^{1} c_n^{-d-1} \exp (-c_n \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d\kappa _i }{\int _{1-c_n^{1/d}}^1 (1-c_n^{1/d})^{d+k/2-1} (c_n^{1/d} + c_n (1-c_n^{1/d}))^{-d-1} \exp (-\varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) d\kappa _i} \\ =&c_n^{-d} \exp (- (1-c_n) \varvec{X}_i^T\varvec{\Sigma }^{-1}\varvec{X}_i/2) (1+o(1)) \\ \le&c_n^{-d} \exp \left( - (1-c_n) K_L' \log \frac{1}{c_n} (1+o(1)) /2 \right) (1+o(1)) \\ =&c_n^{(1-c_n) K_L' (1+o(1)) /2 - d} = o(1). \end{aligned} \end{aligned}$$
since we have already chosen \(K_L' (1+o(1))> \frac{2d}{\xi (1-\delta )} > 2d\).
Let \(A = \beta /\alpha \) for some fixed \(\beta > \alpha \). For small enough \(c_n\), we will have
$$\begin{aligned} \begin{aligned}&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2> \chi ^2_{k, A\alpha } (1 - c_n) \mid \varvec{X}_i, \kappa _i) {\pi }(\kappa _i \mid \varvec{X}_i) d\kappa _i \\ \ge&\beta (1-o(1)) \\>&\alpha \\ \ge&\int _0^1 \Pi ( \Vert \varvec{\theta }_i - (1-\kappa _i) \varvec{X}_{i} \Vert _{\Sigma }^2 > \widehat{r}_i(\alpha , c_n) \mid \varvec{X}_i, \kappa _i) \pi (\kappa _i \mid \varvec{X}_i) d\kappa _i. \end{aligned} \end{aligned}$$
It follows that, as \(c_n \rightarrow 0\),
$$\begin{aligned} \widehat{r}_i(\alpha , c_n) \ge \chi ^2_{k, \beta }(1 - c_n) = \chi ^2_{k, \beta } (1+o(1)), \end{aligned}$$
(B44)
when \(\Vert \varvec{X}_i \Vert _{\Sigma }^2 \ge K_L' \log \frac{1}{c_n} (1+o(1))\).
So, by (B43) along with (B44), in the case where \(\Vert \varvec{\theta }_{0i} \Vert _{\Sigma }^2 \ge K_L' \log \frac{1}{c_n}\), for any fixed \(\beta > \alpha \), as \(c_n \rightarrow 0\),
$$\begin{aligned} P_{\varvec{\theta }_{0i}} (\varvec{\theta }_{0i} \in \widehat{C}^{EIG}_{i})= & {} P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{d/(1+\rho )}(\alpha , c_n)) \\\ge & {} P_{\varvec{\theta }_{0i}} (\Vert \varvec{\theta }_{0i} - \widehat{\varvec{\theta }}_i \Vert _{\Sigma }^2 \le L \widehat{r}_i^{d/(1+\rho )}(\alpha , c_n) \mid \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }) \\\times & {} P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\\ge & {} P_{\varvec{\theta }_{0i}} (\chi ^2_{k,\alpha }(1 + o(1)) \le L (\chi ^2_{k, \beta })^{d/(1+\rho )} (1+o(1)) \mid \Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha }) \\\times & {} P(\Vert \varvec{\epsilon }_{i} \Vert _{\Sigma }^2 \le \chi ^2_{k, \alpha })\\\rightarrow & {} 1 \times (1-\alpha ) = 1-\alpha , \end{aligned}$$
if we choose \(L > \chi ^2_{k,\alpha }(\chi ^2_{k,\beta })^{-{d/(1+\rho )}}\), e.g., \(L = 2\chi ^2_{k,\alpha }(\chi ^2_{k,\beta })^{-{d/(1+\rho )}}\).\(\square \)