Proof of Theorem 1

Part (a). We prove the result by contradiction. That is, suppose (3.1) fails. Then, ${\lim \sup }_{N \to \infty } (P_{N}({\theta }_{N}\in CI_{\alpha }^{1})-P_{N}({\theta }_{N}\in CI_{\alpha }^{2}))\neq 0$ or ${\lim \inf }_{N \to \infty } (P_{N}({\theta }_{N}\in CI_{\alpha }^{1})-P_{N}({\theta }_{N}\in CI_{\alpha }^{2}))\neq 0$ . In either case, we can find a subsequence $\{k_{N}\}_{N\in \mathbb {N}}$ s.t.

(A-1) $$ \begin{align} \lim_{N \to \infty} (P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{1})-P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{2})) ~ \neq ~0. \end{align} $$

The proof is completed by showing that (A-1) cannot hold.

By possibly taking a further subsequence,

(A-2) $$ \begin{align} &\left( \begin{array}{@{}c@{}} {\theta}_{l}(P_{k_{N}}),{\theta}_{u}(P_{k_{N}}),\sigma_{l}(P_{k_{N}}),\sigma_{u}(P_{k_{N}}),\rho (P_{k_{N}}),\\ \sqrt{k_{N}}({\theta}_{u}(P_{k_{N}})-{\theta}_{l}(P_{k_{N}})), \sqrt{k_{N}}({\theta}_{l}(P_{k_{N}})-{\theta}_{k_{N}}),\sqrt{k_{N}}({\theta}_{k_{N}}-{\theta}_{u}(P_{k_{N}})) \end{array} \right) \nonumber \\ & \to ({\theta}_{l},{\theta}_{u},\sigma_{l},\sigma_{u},\rho ,\mu ,\Psi_{l},\Psi_{u})\in \bar{\mathbb{R}}\times \bar{\mathbb{R}}\times [\underline{\sigma },\overline{\sigma }]\times [\underline{ \sigma },\overline{\sigma }]\times [-1,1]\times \bar{\mathbb{R}}_{+}\times \bar{\mathbb{R}}\times \bar{\mathbb{R}}. \end{align} $$

We then divide the argument into four exhaustive cases, depending on the possible values of $(\mu ,\Psi _{l},\Psi _{u})$ . In this regard, note that ${\theta }_{N}\in \Theta _{I}(P_{N})^{c}$ implies that either (i) $\sqrt {k_{N}}({\theta }_{l}(P_{k_{N}})-{\theta }_{k_{N}})>0$ or (ii) $\sqrt {k_{N}}({\theta }_{k_{N}}-{\theta }_{u}(P_{k_{N}}))>0$ . By taking limits, we conclude that either (i) $\Psi _{l}\geq 0$ or (ii) $\Psi _{u}\geq 0$ . The proof is completed by showing that none of the following exhaustive cases satisfy (A-1).

Case 1: $\mu =\infty $ and $\Psi _{l}\geq 0$ . Then, consider the following derivation:

$$ \begin{align*} \lim_{N \to \infty} P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{1})~\overset{(1)}{=}~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/{\sigma}_{l}) ~\overset{(2)}{=}~\lim_{N \to \infty} P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{2}), \end{align*} $$

where (1) holds by (A-2), $F_1(\mu ,{\sigma }_l,{\sigma }_u) = \Phi ^{-1}(1-\alpha )$ , and part (a) of Lemma A.1, and (2) by part (a) of Lemma A.2. This equation implies that (A-1) fails.

Case 2: $\mu =\infty $ and $\Psi _{u}\geq 0$ . This case is analogous to case 1 except that we replace parts (a) of Lemmas A.1 and A.2 with parts (b) of these results.

Case 3: $\mu \in \mathbb {R}_{+}$ and $\Psi _{l}\geq 0$ . By $\mu \in \mathbb {R}_{+}$ , it follows that ${\theta }_{u}(P_{k_{N}})-{\theta }_{l}(P_{k_{N}}) \to 0$ . By this and Lemma A.4, we get $\rho =1$ and $\sigma _{l}=\sigma _{u}$ . We set $\sigma =\sigma _{l}=\sigma _{u}$ . Then, consider the following derivation:

$$ \begin{align*} \lim_{N \to \infty} P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{1})~&\overset{(1)}{=}~\Phi \left( (\Psi_{l} +\mu )/\sigma +G(\mu /\sigma )\right) -\Phi \left( \Psi_{l} /\sigma -G(\mu /\sigma )\right) \\ ~&\overset{(2)}{=}~\lim_{N \to \infty} P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{2}), \end{align*} $$

where (1) holds by (A-2), $F_1(\mu , \sigma ,\sigma ) = G(\mu /\sigma )$ , and part (a) of Lemma A.1, and (2) by part (c) of Lemma A.2. This equation implies that (A-1) fails.

Case 4: $\mu \in \mathbb {R}_{+}$ and $\Psi _{u}\geq 0$ . This case is analogous to case 3 except that we replace part (a) of Lemma A.1 with part (b), and part (c) of Lemma A.2 with part (d).

Part (b). We prove the result by contradiction. That is, suppose that ${\lim \inf }_{N \to \infty }(P_{N}({\theta }_{N}\in CI_{\alpha }^{3})-P_{N}({\theta }_{N}\in CI_{\alpha }^{j}))<0$ for some $j=1,2$ . Then, we can find a subsequence $\{k_{N}\}_{N\in \mathbb {N}}$ s.t.

(A-3) $$ \begin{align} \lim_{N \to \infty} (P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{3})-P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{j})) ~<~0~~\text{ for some }j=1,2. \end{align} $$

The proof is completed by showing that (A-3) cannot hold.

By possibly taking a further subsequence,

(A-4) $$ \begin{align} & \left( \begin{array}{@{}c@{}} {\theta}_{l}(P_{k_{N}}),{\theta}_{u}(P_{k_{N}}),\sigma_{l}(P_{k_{N}}),\sigma_{u}(P_{k_{N}}),\rho (P_{k_{N}}), \\ \sqrt{k_{N}}({\theta}_{u}(P_{k_{N}})-{\theta}_{l}(P_{k_{N}})),\sqrt{k_{N}} ({\theta}_{l}(P_{k_{N}})-{\theta}_{k_{N}}),\sqrt{k_{N}}({\theta}_{k_{N}}-{\theta}_{u}(P_{k_{N}})) \end{array} \right) \nonumber \\ & \to ({\theta}_{l},{\theta}_{u},\sigma_{l},\sigma_{u},\rho ,\mu ,\Psi_{l},\Psi_{u})\in \bar{\mathbb{R}}\times \bar{\mathbb{R}}\times [ \underline{\sigma },\overline{\sigma }]\times [ \underline{ \sigma },\overline{\sigma }]\times [ -1,1]\times \bar{\mathbb{R}}_{+}\times \bar{\mathbb{R}}\times \bar{\mathbb{R}}. \end{align} $$

We then divide the argument into four exhaustive cases, depending on the possible values of $(\mu ,\Psi _{l},\Psi _{u})$ . As before, we note that ${\theta }_{N}\in \Theta _{I}(P_{N})^{c}$ implies that either (i) $\Psi _{l}\geq 0$ or (ii) $\Psi _{u}\geq 0$ . The proof is completed by showing that none of the following exhaustive cases satisfy (A-3).

Case 1: $\mu =\infty $ and $\Psi _{l}\geq 0$ . Then, consider the following derivation for $j=1,2$ ,

$$ \begin{align*} \lim_{N \to \infty }P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{j}) ~\overset{(1)}{=} ~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/\sigma_{l})~\overset{(2)}{\leq} \lim_{N \to \infty }P_{k_{N}}({\theta}_{k_{N}}\in CI_{\alpha}^{3}), \end{align*} $$

where (1) holds by (A-4), $F_1(\mu ,{\sigma }_l,{\sigma }_u) = \Phi ^{-1}(1-\alpha )$ , and part (a) of Lemma A.1 and Lemma A.2, and (2) by part (a) of Lemma A.3. This equation implies that (A-3) fails.

Case 2: $\mu =\infty $ and $\Psi _{u}\geq 0$ . This case is analogous to case 1 except that we replace parts (a) of Lemmas A.1 and A.2 with parts (b) of these results, and part (a) of Lemma A.3 with part (f).

Case 3: $\mu \in \mathbb {R}_{+}$ and $\Psi _{l}\geq 0$ . By $\mu \in \mathbb {R}_{+}$ and Lemma A.4, we get $\rho =1$ and $\sigma _{l}=\sigma _{u}$ . We set $\sigma =\sigma _{l}=\sigma _{u}$ .

As a preliminary derivation, note that

(A-5) $$ \begin{align} \Phi ^{-1}(1-\alpha /2)~\geq ~G(\mu /\sigma ). \end{align} $$

To see this, note that $\Phi (c+\mu /\sigma )-\Phi (-c)$ is strictly increasing in $c$ , and

$$ \begin{align*}& \Phi (\Phi ^{-1}(1-\alpha /2)+\mu /\sigma )-\Phi (-\Phi ^{-1}(1-\alpha /2))\\&\quad\overset{(1)}{\geq }~\Phi (\Phi ^{-1}(1-\alpha /2))-\Phi (-\Phi^{-1}(1-\alpha /2)) \\ &\quad=~1-\alpha \\ & \quad\overset{(2)}{=}~\Phi (G(\mu /\sigma )+\mu /\sigma )-\Phi (-G(\mu /\sigma)), \end{align*} $$

where (1) holds by $\mu /\sigma \geq 0$ , and (2) by definition of $G(y)$ .

For $j=1,2$ , we then have the following derivation:

(A-6) $$ \begin{align} \lim_{N\to \infty }P_{k_{N}}({\theta}_{k_{N}} \in CI_{\alpha}^{3}) &~\overset{(1)}{=}~\Phi \left( (\Psi_{l}+\mu )/\sigma +\Phi ^{-1}(1-\alpha /2)\right) -\Phi \left( \Psi_{l}/\sigma -\Phi ^{-1}(1-\alpha /2)\right) \nonumber \\ &~\overset{(2)}{\geq }~\Phi \left( (\Psi_{l}+\mu )/\sigma +G(\mu /\sigma )\right) -\Phi \left( \Psi_{l}/\sigma -G(\mu /\sigma )\right) \nonumber \\ &~\overset{(3)}{=}~\lim_{N\to \infty }P_{k_{N}}({\theta}_{k_{N}} \in CI_{\alpha}^{j}), \end{align} $$

where (1) holds by part (e) of Lemma A.3, (2) by (A-5) and that $\Phi ( (\Psi _{l}+\mu )/\sigma +c) -\Phi ( \Psi _{l}/\sigma -c) $ is increasing in $c$ , and (3) by part (a) of Lemma A.1, part (c) of Lemma A.2, and $F_{1}(\mu ,\sigma ,\sigma )=G(\mu /\sigma )$ . Note that this equation implies that (A-3) fails.

Case 4: $\mu \in \mathbb {R}_{+}$ and $\Psi _{u}\geq 0$ . This case is analogous to case 3 except that we replace part (a) of Lemma A.1 with part (b), part (c) of Lemma A.2 with part (d), and part (e) of Lemma A.3 with part (j).

To conclude the proof of this part, it suffices to verify that (3.2) holds strictly for a suitably chosen sequence $\{(P_{N},{\theta }_{N})\in \mathcal {P}\times \Theta _{I}(P_{N})^{c}\}_{N\in \mathbb {N}}$ . We can consider two cases: ${\lim \sup }_{N \to \infty }\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}))<\infty $ or ${\lim \sup }_{N \to \infty }\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}))=\infty $ . For brevity, we focus on the first case.

To this end, consider Example 1 with $\underline {Y}=0$ , $\overline {Y}=1$ , $\{Y_{i}|Z_{i}=1\}\sim Be(1/2)$ , and $Z_{i}\sim Be(1-\pi (P_{N}))$ , with $\pi (P_{N}) = a_1 /\sqrt {N} \downarrow 0$ for any $a_1 \in (0,\infty )$ , which leads to ${\lim \sup }_{N \to \infty }\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})) <\infty $ . Consider coverage of ${\theta }_{N}={\theta }_{l}(P_{N})-a_2/ \sqrt {N}\in \Theta _{I}(P_{N})^{c}$ for any $a_2>0$ using $CI_{\alpha }^{3}$ implemented with $b_{N}=(\ln N)/\sqrt {N}$ . Then, (A-2) and (A-4) hold with $({\theta }_{l},{\theta }_{u},\sigma ,\sigma ,\rho ,\mu ,\Psi _{l},\Psi _{u}) = (1/2, 1/2, 1/2, 1/2, 1, a_1, a_2, - a_1 -a_2)$ . For $j=1,2$ , we then obtain

(A-7) $$ \begin{align} \lim_{N \to \infty} P_{N}({\theta}_{N}\in CI_{\alpha}^{j})~&\overset{(1)}{=}~\Phi \left( 2(a_2 +a_1 )+G(2a_1 )\right) -\Phi \left( 2 a_2 -G(2a_1 )\right) \nonumber \\ ~&\overset{(2)}{<}~\Phi \left( 2( a_2 +a_1 )+\Phi ^{-1}(1-\alpha /2)\right) -\Phi \left( 2 a_2 -\Phi ^{-1}(1-\alpha /2)\right) \nonumber \\ ~&\overset{(3)}{=}~ \lim_{N \to \infty} P_{N}({\theta}_{N}\in CI_{\alpha}^{3}), \end{align} $$

where (1) holds by part (a) of Lemmas A.1 and A.2, (2) by repeating the derivation in (A-5), and (3) by part (e) of Lemma A.3. For concreteness, $a_1=1$ , $a_2=1$ , and $\alpha =0.05$ give $\lim _{N\to \infty }P_N({\theta }_N\in CI_\alpha ^{j})=0.36<0.48=\lim _{N\to \infty }P_N({\theta }_N\in CI_\alpha ^{3})$ for $j=1,2$ . Note that (A-7) implies that (3.2) holds strictly.

Part (c). These results are shown in the Supplementary Material.

Proof of Theorem 2

We prove (3.5) by contradiction. To this end, assume (3.5) fails, that is,

(A-8) $$ \begin{align} \underset{N \to \infty }{\text{lim inf} }~~\big(P_{N}({\theta}_{N}\in CI_{\alpha}^{1,I})-P_{N}({\theta}_{N}\in CI_{\alpha}^{1,E})\big)~<~0. \end{align} $$

By possibly taking a subsequence $\{k_{N}\}_{N\geq 1}$ , the following sequence converges:

(A-9) $$ \begin{align} &\left( \begin{array}{@{}c@{}} {\theta}_{l}(P_{k_{N}}),{\theta}_{u}(P_{k_{N}}),\sigma_{l}^{E}(P_{{ k_{N}}}),\sigma_{u}^{E}(P_{{k_{N}}}),\rho ^{E}(P_{{k_{N}}}),\sigma_{l}^{I}(P_{{k_{N}}}),\sigma_{u}^{I}(P_{{k_{N}}}),\rho ^{I}(P_{{k_{N}} }),\\ \sqrt{k_{N}}({\theta}_{u}(P_{k_{N}})-{\theta}_{l}(P_{k_{N}})), \sqrt{k_{N}}({\theta}_{l}(P_{k_{N}})-{\theta}_{k_{N}}),\sqrt{k_{N}}({\theta}_{k_{N}}-{\theta}_{u}(P_{k_{N}})) \end{array} \right) \nonumber \\ & \to ({\theta}_{l},{\theta}_{u},\sigma^{E}_{l},\sigma^{E}_{u},\rho^{E},\sigma^{I}_{l},\sigma^{I}_{u},\rho^{I} ,\mu ,\Psi_{l},\Psi_{u}) \nonumber \\ &\in \bar{\mathbb{R}}\times \bar{\mathbb{R}}\times [\underline{\sigma },\overline{\sigma }]\times [\underline{ \sigma },\overline{\sigma }]\times [-1,1]\times [\underline{\sigma },\overline{\sigma }]\times [\underline{ \sigma },\overline{\sigma }]\times [-1,1]\times \bar{\mathbb{R}}_{+}\times \bar{\mathbb{R}}\times \bar{\mathbb{R}}. \end{align} $$

By Assumption 1, $\sigma _{l}^{E}\leq \sigma _{l}^{I}$ and $\sigma _{u}^{E}\leq \sigma _{u}^{I}$ . We then divide the argument into four exhaustive cases, depending on the possible values of $(\mu ,\Psi _{l},\Psi _{u})$ . As in the proof of Theorem 1, we note that ${\theta }_{N}\in \Theta _{I}(P_{N})^{c}$ implies that either (i) $\Psi _{l}\geq 0$ or (ii) $\Psi _{u}\geq 0$ . The proof is completed by showing that none of the following exhaustive cases satisfy (A-8).

Case 1: $\mu =\infty $ and $\Psi _{l}\geq 0$ . Then,

$$ \begin{align*} \lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,E})&~\overset{(1)}{=}~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/\sigma_{l}^{E}) \\ &~\overset{(2)}{\leq }~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/\sigma_{l}^{I}) \\ &~\overset{(3)}{=}~\lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,I}), \end{align*} $$

as desired, where (1) and (3) hold by part (a) of Lemma A.1, and (2) by $\Psi _{l}\geq 0$ and $\sigma _{l}^{E}\leq \sigma _{l}^{I}$ .

Case 2: $\mu =\infty $ and $\Psi _{u}\geq 0$ . This case is analogous to case 1 except that we replace part (a) of Lemma A.1 with part (b).

Case 3: $\mu \in \mathbb {R}_{+}$ and $\Psi _{l}\geq 0$ . By $\mu \in \mathbb {R}_{+}$ and Lemma A.4, $\sigma ^{E}_{l}=\sigma ^{E}_{u}$ , $\sigma ^{I}_{l}=\sigma ^{I}_{u}$ , $\rho ^{E} =1$ , and $\rho ^{I} =1$ . We set $\sigma ^{I}=\sigma _{l}^{I}=\sigma _{u}^{I}$ and $\sigma ^{E}=\sigma _{l}^{E}=\sigma _{u}^{E}$ . Then,

$$ \begin{align*} \lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,E})& ~\overset{(1)}{=}~\Phi \left( (\Psi_{l}+\mu )/\sigma ^{E}+G(\mu /\sigma ^{E})\right) -\Phi \left( \Psi_{l}/\sigma ^{E}-G(\mu /\sigma ^{E})\right) \\ &~\overset{(2)}{=}~H(\sigma ^{E},\mu ,\Psi_{l}) \\ &~\overset{(3)}{\leq }~H(\sigma ^{I},\mu ,\Psi_{l}) \\ &~\overset{(4)}{=}~\Phi \left( (\Psi_{l}+\mu )/\sigma ^{I}+G(\mu /\sigma ^{I})\right) -\Phi \left( \Psi_{l}/\sigma ^{I}-G(\mu /\sigma ^{I})\right) \\ &~\overset{(5)}{=}~\lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,I}), \end{align*} $$

as desired, where (1) and (5) hold by part (c) of Lemma A.1, (2) and (4) by $H:[\underline { \sigma },\overline {\sigma }]\times \mathbb {R}_{+}\times \bar {\mathbb {R}}_{+} \to \mathbb {R}$ defined in Lemma A.5, and (3) by part (a) of Lemma A.5 and $\sigma _{l}^{E}\leq \sigma _{l}^{I}$ .

Case 4: $\mu \in \mathbb {R}_{+}$ and $\Psi _{u}\geq 0$ . This case is analogous to case 3 except that we replace part (c) of Lemma A.1 with part (d).

To conclude the proof, it suffices to show that the inequality in (3.5) holds strictly for suitable sequences of $\{(P_{N},{\theta }_{N})\in \mathcal {P}\times \Theta _{I}(P_{N})^{c}\}_{N\in \mathbb {N}}$ . To this end, consider $\{(P_{N},{\theta }_{N})\}_{N\in \mathbb {N}}$ that satisfies (A-9) with $\sigma ^{E}<\sigma ^{I}$ , and either (i) $ \Psi _{l}>0$ or (ii) $\Psi _{u}>0$ . For brevity, focus on (i), but analogous results hold for (ii).

We now have two cases. If $\mu =\infty $ , then the desired strict inequality follows from the derivation in case 1. The difference with this derivation is that the weak inequality now becomes strict under $\Psi _{l}>0$ and $\sigma ^{E}<\sigma ^{I}$ . In turn, if $\mu \in \mathbb {R}_{+}$ , the desired strict inequality follows from the derivations in case 3. The difference with this derivation is that the weak inequality now becomes strict under $\Psi _{l}>0$ , $\sigma ^{E}<\sigma ^{I}$ , and part (b) of Lemma A.5.

We now illustrate the case with $\mu =\infty $ in the context of Example 1. Within this example, set $\underline {Y}=0$ , $\overline {Y}=1$ , $\{Y_{i}|Z_{i}=1\}\sim Be(1/2)$ , and $Z_{i}\sim Be(1-\pi (P_{N}))$ , with $\pi (P_{N}) =a_1 \in (0,1)$ , which leads to ${\lim \sup }_{N \to \infty }\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})) =\infty $ . Our goal is to cover ${\theta }_{N}={\theta }_{l}(P_{N})- a_2/ \sqrt {N}\in \Theta _{I}(P_{N})^{c}$ for some $ a_2>0$ . In this context, consider two estimators for the bounds. The first one is (2.3), which serves as the efficient estimator. The second one is as in (2.3) but using only the first $\lfloor a_3 N\rfloor $ observations of the sample for any $ a_3 \in (0,1)$ . This will be the inefficient estimator, as it uses a fraction of the available sample. Both bounds estimators can be represented as

(A-10) $$ \begin{align} ( {\hat{\theta}}_{l}( \lambda ) ,{\hat{\theta}}_{u}( \lambda ) ) ~=~\Big( \frac{1}{\lfloor \lambda N\rfloor } \sum_{i=1}^{\lfloor \lambda N\rfloor }( Y_{i}Z_{i}+\underline{Y}( 1-Z_{i}) ) ,\frac{1}{\lfloor \lambda N\rfloor } \sum_{i=1}^{\lfloor \lambda N\rfloor }( Y_{i}Z_{i}+\overline{Y}( 1-Z_{i}) ) \Big). \end{align} $$

For $\lambda =1$ , we get $( {\hat {\theta }}_{l}^{E}, {\hat {\theta }}_{u}^{E}) =( {\hat {\theta }}_{l}( 1) ,\hat { \theta }_{u}( 1) ) $ , and for $\lambda = a_3\in (0,1) $ , we get $( {\hat {\theta }}_{l}^{I},{\hat {\theta }}_{u}^{I}) =( {\hat {\theta }}_{l}( a_3) ,\hat { \theta }_{u}( a_3))$ . Finally, we assume that the estimators for $( \sigma _{l},\sigma _{u},\rho )$ are all obtained by their corresponding sample analogs: $( {\hat {\sigma }}_{l}^{E},{\hat {\sigma }}_{u}^{E},\hat {\rho } ^{E}) $ uses the entire sample (i.e., $\lambda =1$ ), and $( \hat { \sigma }_{l}^{I},{\hat {\sigma }}_{u}^{I},\hat {\rho }^{I}) $ relies only on the first $\lfloor a_3 N\rfloor $ observations of the sample, with the standard-deviation estimators scaled by $\sqrt {N/\lfloor a_3N\rfloor }$ . By repeating arguments in Example 1, we can verify that $ ( {\hat {\theta }}_{l}^{E},{\hat {\theta }}_{u}^{E},{\hat {\sigma }}_{l}^{E},\hat { \sigma }_{u}^{E},\hat {\rho }^{E}) $ and $( {\hat {\theta }}_{l}^{I}, {\hat {\theta }}_{u}^{I},{\hat {\sigma }}_{l}^{I},{\hat {\sigma }}_{u}^{I},\hat {\rho } ^{I}) $ satisfy Assumption 1. Moreover, we can show that (A-9) holds with

$$ \begin{align*} &({\theta}_{l},{\theta}_{u},\sigma^{E}_{l},\sigma^{E}_{u},\rho^{E},\sigma^{I}_{l},\sigma^{I}_{u},\rho^{I} ,\mu ,\Psi_{l},\Psi_{u}) \\ &= \Big(\frac{1-a_1}{2},\frac{1+a_1}{2}, \frac{\sqrt{1-a_1^{2}}}{2},\frac{\sqrt{1-a_1^{2}}}{2},\frac{1-a_1}{1+a_1}, \frac{\sqrt{1-a_1^{2}}}{2\sqrt{ a_3}},\frac{\sqrt{1-a_1^{2}}}{2\sqrt{ a_3}},\frac{1-a_1}{1+a_1}, +\infty,a_{2},-\infty\Big). \end{align*} $$

At this point, we can repeat the derivation in case 1 to get the following:

$$ \begin{align*} \lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,E})&~\overset{(1)}{=}~\Phi \Big(\Phi ^{-1}(1-\alpha )-2 a_2 /\sqrt{1-a_1^2}\Big) \\ &~\overset{(2)}{<}~\Phi \Big(\Phi ^{-1}(1-\alpha )-2\sqrt{ a_3} a_2 /\sqrt{1-a_1^2}\Big) \\ &~\overset{(3)}{=}~\lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{1,I}), \end{align*} $$

as desired, where (1) and (3) hold by part (a) of Lemma A.1, and (2) by $a_3 \in (0,1)$ . For concreteness, $a_1=1/2$ , $a_2=1$ , $a_3=0.3$ , and $\alpha =0.05$ yield $\lim _{N\to \infty }P_N({\theta }_N\in CI_{\alpha }^{1,I})=0.69>0.25=\lim _{N\to \infty }P_N({\theta }_N\in CI_{\alpha }^{1,E})$ .

Proof of Theorem 3

This result follows from part (a) of Theorem 1 and Theorem 2.

Proof of Theorem 4

To show this result, we construct specific sequences where (3.7) and (3.8) can occur. We focus on sequences $\{(P_{N},{\theta }_{N})\in \mathcal {P}\times \Theta _{I}(P_{N})^{c}\}_{N\in \mathbb {N}}$ s.t.

(A-11) $$ \begin{align} &\Bigg( \begin{array}{@{}c@{}} {\theta}_{l}( P_{N}) ,{\theta}_{u}( P_{N}) ,\sigma_{l}^{E}( P_{N}) ,\sigma_{u}^{E}( P_{N}) ,\rho ^{E}( P_{N}) ,\sigma_{l}^{I}( P_{N}) ,\sigma_{u}^{I}( P_{N}) ,\rho ^{I}( P_{N}) \\ \sqrt{N}( {\theta}_{u}( P_{N}) -{\theta}_{l}( P_{N}) ) ,\sqrt{N}( {\theta}_{l}( P_{N}) -{\theta}_{N}) ,\sqrt{N}( {\theta}_{N}-{\theta}_{u}( P_{N}) ) \end{array} \Bigg) \nonumber \\ &\to ( {\theta}_{l},{\theta}_{u},\sigma_{l}^{E},\sigma_{u}^{E},\rho ^{E},\sigma_{l}^{I},\sigma_{u}^{I},\rho ^{I}, \mu ,\Psi_{l},\Psi_{u}), \end{align} $$

with $\Psi _{l}\geq 0$ and $\mu \in \mathbb {R}_{+}$ . By Lemma A.4, $\rho ^{E}=\rho ^{I} =1$ and $\sigma _{l}^E=\sigma _{u}^E$ , and $\sigma _{l}^{I}=\sigma _{u}^{I}$ .

We can construct concrete sequences in the context of Example 1. In particular, we use Example 1 with $\underline {Y}=0$ , $\overline {Y}=1$ , $\{Y_{i}|Z_{i}=1\}\sim Be(1/2)$ , and $Z_{i}\sim Be(1-\pi (P_{N}))$ , where $\pi (P_{N})~=~a_1/\sqrt {N}~\downarrow ~0$ for some $a_1>0$ . We consider coverage of ${\theta }_{N}={\theta }_{l}(P_{N}) - a_2/\sqrt {N}\in \Theta _{I}(P_{N})^{c}$ for some $a_2>0$ and $CI_{\alpha }^{3}$ implemented with the subsample of the data $ \{(Y_{i},Z_{i})\}_{i=1}^{\lfloor \lambda N\rfloor }$ for $\lambda \in (0,1]$ . In this context, we consider two estimators for the bounds. The efficient estimator uses the full sample, while the inefficient estimator uses only a fraction $\lambda = a_3 \in (0,1)$ of the sample. By repeating arguments in Theorem 2, we deduce that (A-11) holds with

$$ \begin{align*} ( {\theta}_{l},{\theta}_{u},\sigma_{l}^{E},\sigma_{u}^{E},\rho ^{E},\sigma_{l}^{I},\sigma_{u}^{I},\rho ^{I}, \mu ,\Psi_{l}) ~=~ \Big(\frac{1}{2},\frac{1}{2}, \frac{1}{2},\frac{1}{2},1, \frac{1}{2\sqrt{a_3}},\frac{1}{2\sqrt{a_3}},1,a_1,a_2\Big). \end{align*} $$

Part (e) of Lemma A.3 then yields

$$ \begin{align*} \lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,E})& =\Phi \big( 2( a_2+a_1)+\Phi ^{-1}( 1-\alpha /2) \big)- \Phi \big( 2a_2-\Phi ^{-1}( 1-\alpha /2) \big), \\ \lim_{N \to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,I})&=\Phi \big( 2\sqrt{a_3}(a_2+a_1 ) +\Phi ^{-1}( 1-\alpha /2) \big)\\&\quad -\Phi \big( 2\sqrt{a_3}a_2-\Phi ^{-1}( 1-\alpha /2) \big). \end{align*} $$

We now verify the strict inequalities. To obtain (3.7), set $a_1=a_2=1$ , $a_3=0.3$ , and $\alpha =0.05$ , which give

$$ \begin{align*} \lim_{N\to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,E})~=~0.48~<~0.81~=~\lim_{N\to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,I}). \end{align*} $$

To obtain (3.8), set $a_1=0.15$ , $a_2=0.01$ , $a_3=0.3$ , and $\alpha =0.05$ , which give

$$ \begin{align*} \lim_{N\to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,E})~=~0.963 ~>~ 0.958~=~\lim_{N\to \infty} P_{N}({\theta}_{N} \in CI_{\alpha}^{3,I}). \end{align*} $$

This completes the proof.

Proof of Theorem 5

This result is shown in the Supplementary Material.

Proof. We begin by confirming that $F_{1}(\delta ,\sigma _{l},\sigma _{u})$ is unique and positive. To this end, note that the left-hand side of (A-12) is strictly increasing and continuous in $c$ . Moreover, at $c=0$ , it is equal to $\Phi (\delta /\max \{\sigma _{l},\sigma _{u}\})-\Phi (0)\leq 0.5<1-\alpha $ , whereas, as $c\to \infty $ , $\lim _{c\to \infty }{\Phi ( c+{\delta }/{\max \{\sigma _{l},\sigma _{u}}\}) -\Phi (-c)}=1>1-\alpha $ . This establishes existence, uniqueness, and positivity.

Next, we show that $F_{1}:\bar {\mathbb {R}}_{+}\times [\underline { \sigma },\overline {\sigma }]\times [\underline {\sigma },\overline { \sigma }] \to \bar {\mathbb {R}}$ is a continuous function of its arguments. To this end, consider any sequence $\{(\delta _{M},\sigma _{M,l},\sigma _{M,u})\}_{M\in \mathbb {N}}$ s.t. $\lim _{M\to \infty } (\delta _{M},\sigma _{M,l},\sigma _{M,u}) =(\delta ,\sigma _{l},\sigma _{u})\in \bar { \mathbb {R}}_{+}\times [\underline {\sigma },\overline {\sigma }]\times [\underline {\sigma },\overline {\sigma }]$ . In search of a contradiction, assume that $\{F_{1}(\delta _{M},\sigma _{M,l},\sigma _{M,u})\}_{M \in \mathbb {N}}$ does not converge to $F_{1}(\delta ,\sigma _{l},\sigma _{u})$ . Then, there exist $\Delta>0$ and a subsequence $\{k_M\}_{M\in {\mathbb {N}}}$ such that either $F_1(\delta _{k_M},{\sigma }_{k_M,l},{\sigma }_{k_M,u})-F_1(\delta ,{\sigma }_l,{\sigma }_u)>\Delta $ for all $M\in {\mathbb {N}}$ or $F_1(\delta _{k_M},{\sigma }_{k_M,l},{\sigma }_{k_M,u})-F_1(\delta ,{\sigma }_l,{\sigma }_u)<-\Delta $ for all $M\in {\mathbb {N}}$ . We assume the first case, but the argument is analogous to the other case. Since $\lim _{M\to \infty } (\delta _{M},\sigma _{M,l},\sigma _{M,u}) = (\delta ,\sigma _{l},\sigma _{u})\in \bar { \mathbb {R}}_{+}\times [\underline {\sigma } ,\bar {\sigma }]\times [\underline {\sigma } , \bar {\sigma }]$ , we have that $\varphi _{k_{M}}=\delta _{k_{M}}/\max \{\sigma _{k_{M},l},\sigma _{k_{M},u}\} \to \varphi =\delta /\max \{\sigma _{l},\sigma _{u}\}$ as $M\to \infty $ . Here, there are two cases: case 1: $\delta <\infty $ , and so $ \varphi <\infty $ , or case 2: $\delta =\varphi =\infty $ .

Case 1: $\delta <\infty $ . Then, $\exists M_{1}\in \mathbb {N}$ s.t. $|\varphi _{k_{M}}-\varphi |<\Delta /2$ for all $M>M_{1}$ . Then, for all $M\geq M_{1}$ , we reach the following contradiction:

(A-14) $$ \begin{align} 1-\alpha & ~\overset{(1)}{=}~\Phi (F_{1}(\delta_{k_{M}},\sigma_{k_{M},l},\sigma_{k_{M},u})+\varphi_{k_{M}})-\Phi (-F_{1}(\delta_{k_{M}},\sigma_{k_{M},l},\sigma_{k_{M},u})) \nonumber \\ &~ \overset{(2)}{\geq }~\Phi (\Delta /2+F_{1}(\delta ,\sigma_{l},\sigma_{u})+\varphi )-\Phi (-\Delta -F_{1}(\delta ,\sigma_{l},\sigma_{u})) \nonumber \\ &~ \overset{(3)}{>}~\Phi (F_{1}(\delta ,\sigma_{l},\sigma_{u})+\varphi )-\Phi (-F_{1}(\delta ,\sigma_{l},\sigma_{u})) \nonumber \\ &~ \overset{(4)}{=}~1-\alpha , \end{align} $$

where (1) holds by (A-12) with $c=F_{1}(\delta _{k_{M}},\sigma _{k_{M},l},\sigma _{k_{M},u})$ , (2) by $F_{1}(\delta _{k_{M}},\sigma _{k_{M},l},\sigma _{k_{M},u})-F_{1}(\delta ,\sigma _{l},\sigma _{u})>\Delta $ and $|\varphi _{k_{M}}-\varphi |<\Delta /2$ , and (3) by $\Delta>0$ , and (4) by (A-12) with $c=F_{1}(\delta ,\sigma _{l},\sigma _{u})$ .

Case 2: $\delta =\varphi =\infty $ . In this case, $F_{1}(\delta ,\sigma _{l},\sigma _{u})=\Phi ^{-1}(1-\alpha )$ and $\varphi _{k_{M}} \to \infty $ . By this and $\Delta>0$ , $\exists M_{1}\in \mathbb {N}$ s.t. for all $M>M_{1}$ ,

(A-15) $$ \begin{align} \Phi (\Delta +\Phi ^{-1}(1-\alpha )+\varphi_{k_{M}})-\Phi (-\Delta -\Phi ^{-1}(1-\alpha ))~>~1-\alpha. \end{align} $$

Then, for all $M\geq M_{1}$ , we have the following contradiction:

$$ \begin{align*} 1-\alpha & ~\overset{(1)}{=}~\Phi (F_{1}(\delta_{k_{M}},\sigma_{k_{M},l},\sigma_{k_{M},u})+\varphi_{k_{M}})-\Phi (-F_{1}(\delta_{k_{M}},\sigma_{k_{M},l},\sigma_{k_{M},u})) \nonumber \\ & ~\overset{(2)}{\geq }~\Phi (\Delta +F_{1}(\delta ,\sigma_{l},\sigma_{u})+\varphi_{k_{M}})-\Phi (-\Delta -F_{1}(\delta ,\sigma_{l},\sigma_{u})) \nonumber \\ & ~\overset{(3)}{=}~\Phi (\Delta +\Phi ^{-1}(1-\alpha )+\varphi_{k_{M}})-\Phi (-\Delta -\Phi ^{-1}(1-\alpha ))\\ & ~\overset{(4)}{>}~1-\alpha , \end{align*} $$

as desired, where (1) holds by (A-12) with $c=F_{1}(\delta _{k_{M}},\sigma _{k_{M},l},\sigma _{k_{M},u})$ , (2) by $F_{1}(\delta _{k_{M}},\sigma _{k_{M},l},\sigma _{k_{M},u})-F_{1}(\delta ,\sigma _{l},\sigma _{u})>\Delta $ , (3) by $F_{1}(\delta ,\sigma _{l},\sigma _{u})=\Phi ^{-1}(1-\alpha )$ , and (4) by (A-15).

We now divide the rest of the argument into its parts.

Part (a). Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-13),

(A-16) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u})~\overset{d}{ \to }~(z_{1}\sigma_{l},(z_{2}\sqrt{1-\rho ^{2}}+z_{1}\rho )\sigma_{u},\sigma_{l},\sigma_{u}), \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Then,

$$ \begin{align*} &P_{N}({\theta}_{N} \in CI_{\alpha}^{1}) \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{1}(\sqrt{N}( {\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u})\leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \\ \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+{\hat{\sigma}}_{u}F_{1}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}) \end{array} \right) \\ & \overset{(2)}{\to}~P\left( \begin{array}{@{}c@{}} z_{1}\sigma_{l}-\sigma_{l}F_{1}(z_{2}\sigma_{u}\sqrt{1-\rho ^{2}} +z_{1}(\rho \sigma_{u}-\sigma_{l})+\mu ,\sigma_{l},\sigma_{u})\leq -\Psi_{l} \leq \\ (z_{2}\sqrt{1-\rho ^{2}}+z_{1}\rho )\sigma_{u}+\mu +\sigma_{u}F_{1}(z_{2}\sigma_{u}\sqrt{1-\rho ^{2}}+z_{1}(\rho \sigma_{u}-\sigma_{l})+\mu ,\sigma_{l},\sigma_{u}) \end{array} \right) \\ & \overset{(3)}{=}~P\left( \begin{array}{@{}c@{}} \left\{ z_{1}-F_{1}(z_{2}\sigma_{u}\sqrt{1-\rho ^{2}}+z_{1}(\rho \sigma_{u}-\sigma_{l})+\mu ,\sigma_{l},\sigma_{u})\leq -\frac{\Psi_{l} }{\sigma_{l} }\right\} \cap \\ \left\{ -\frac{\Psi_{l} +\mu }{\sigma_{u}}\leq z_{2}\sqrt{1-\rho ^{2}} +z_{1}\rho +F_{1}(z_{2}\sigma_{u}\sqrt{1-\rho ^{2}}+z_{1}(\rho \sigma_{u}-\sigma_{l})+\mu ,\sigma_{l},\sigma_{u})\right\} \end{array} \right) , \end{align*} $$

as desired, where (1) holds by (2.4) and (2.5), which give ${c}^1=F_{1}( \sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u})$ , (2) by (A-13), (A-16), the continuity of $F_{1}(\delta ,\sigma _{l},\sigma _{u})$ , and the CMT, and (3) by $\sigma _{l},\sigma _{u}\geq \underline {\sigma }>0$ .

Part (b). This argument is analogous to part (a), and it is thus omitted.

Proof. Let $C_{2}:\bar {\mathbb {R}}_{+}\times [\underline {\sigma },\overline { \sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1] \rightrightarrows {\bar {\mathbb {R}}^{2}}$ be the following correspondence:

(A-18) $$ \begin{align} C_{2}(\delta ,\sigma_{l},\sigma_{u},\rho )~=~\left\{\! \begin{array}{l} ( c_{l},c_{u}) \in {\bar{\mathbb{R}}^{2}}~~~\text{ s.t.} \\ P(\{-c_{l}\leq z_{1}\}\cap \{\rho z_{1}\leq c_{u}+{\delta }/{\sigma_{u}}+\sqrt{1-\rho ^{2}}z_{2}\})~\geq ~1-\alpha ~\text{ and} \\ P(\{-c_{l}-{\delta }/{\sigma_{l}}+\sqrt{1-\rho ^{2}}z_{2}\leq \rho z_{1}\}\cap\{z_{1}\leq c_{u}\})~\geq ~1-\alpha , \end{array} \!\right\} , \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ , let $S_2:\bar {\mathbb {R}}_{+}\times [\underline {\sigma },\overline { \sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1] \rightrightarrows {\bar {\mathbb {R}}^{2}}$ be the following minimizer correspondence:

(A-19) $$ \begin{align} S_{2}(\delta ,\sigma_{l},\sigma_{u},\rho )~=~\underset{( c_{l},c_{u}) \in C_{2}(\delta ,\sigma_{l},\sigma_{u},\rho )}{\text {arg min} }(\sigma_{l}c_{l}+\sigma_{u}c_{u}), \end{align} $$

and let $F_{2}=(F_{2,l},F_{2,u}):\bar {\mathbb {R}}_{+}\times [ \underline {\sigma },\overline {\sigma }]\times [\underline {\sigma }, \overline {\sigma }]\times [-1,1] \to {\bar {\mathbb {R}}^{2}}$ be the choice in $S_{2}$ used to implement $CI_{\alpha }^2$ . By (2.6) and (2.7), we have that $c^2=F_{2}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ .

We now study the continuity properties of $F_{2}$ . For our purposes, it suffices to determine the limit of $\{F_{2}(\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N} }$ for two types of sequences $\{(\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N} }$ : (i) those converging to $(\infty ,\sigma _{l},\sigma _{u},\rho )$ with $ (\sigma _{l},\sigma _{u},\rho )\in [\underline {\sigma },\overline { \sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]$ and (ii) those converging to $(\delta ,\sigma ,\sigma ,1)$ with $(\delta ,\sigma )\in \bar {\mathbb {R}}_{+}\times [\underline { \sigma },\overline {\sigma }]$ .

First, consider $\{F_{2}(\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N} }$ for $\lim _{M\to \infty } (\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M}) = (\infty ,\sigma _{l},\sigma _{u},\rho )$ and $(\sigma _{l},\sigma _{u},\rho )\in [\underline {\sigma },\overline {\sigma } ]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]$ . It is not hard to verify that

$$ \begin{align*} C_{2}(\infty ,\sigma_{l},\sigma_{u},\rho )& ~=~\left\{ \left( c_{l},c_{u}\right) \in {\bar{\mathbb{R}}^{2}}~~\text{s.t.}~~~\Phi (c_{l})\geq 1-\alpha ~~\text{and}~~\Phi (c_{u})\geq 1-\alpha \right\}, \\ S_{2}(\infty ,\sigma_{l},\sigma_{u},\rho )& ~=~\{(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha ))\},\\ F_{2}(\infty ,\sigma_{l},\sigma_{u},\rho )& ~=~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align*} $$

Furthermore, part (a) of Lemma A.7 shows that $C_{2}$ is continuous at $(\delta ,\sigma _{l},\sigma _{u},\rho )=(\infty ,\sigma _{l},\sigma _{u},\rho )$ . Then, Lemma A.9 shows that $F_{2}$ is continuous at $(\delta ,\sigma _{l},\sigma _{u},\rho )=(\infty ,\sigma _{l},\sigma _{u},\rho )$ , that is,

(A-20) $$ \begin{align} \underset{(\delta_{M},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\infty ,\sigma_{l},\sigma_{u},\rho )}{\text{lim} } F_{2}(\delta_{M},\sigma_{M,l},\sigma_{M,u},\rho_{M})~=~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align} $$

Second, consider $\{F_{2}(\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N} }$ for $\lim _{M\to \infty } (\delta _{M},\sigma _{M,l},\sigma _{M,u},\rho _{M}) = (\delta ,\sigma ,\sigma ,1)$ and $(\delta ,\sigma )\in \bar {\mathbb {R}}_{+}\times [\underline {\sigma },\overline {\sigma }]$ . It is not hard to verify that

$$ \begin{align*} C_{2}(\delta ,\sigma ,\sigma ,1)=\left\{ \begin{array}{@{}l@{}} \left( c_{l},c_{u}\right) \in {\bar{\mathbb{R}}^{2}}~~~\text{ s.t.} \\ \Phi (c_{u}+{\delta }/{\sigma })-\Phi (-c_{l})\geq 1-\alpha ~\text{and}~\Phi (c_{u})-\Phi (-c_{l}-{\delta }/{\sigma })\geq 1-\alpha \end{array} \right\}. \end{align*} $$

In turn, Lemma A.6 shows that

$$ \begin{align*} S_{2}(\delta ,\sigma ,\sigma ,1)~=\{(G(\delta /\sigma ),G(\delta /\sigma ))\}~\text{ and }~F_{2}(\delta ,\sigma ,\sigma ,1)~=~(G(\delta /\sigma ),G(\delta /\sigma )). \end{align*} $$

Furthermore, part (b) of Lemma A.7 shows that $C_{2}$ is continuous at $(\delta ,\sigma _{l},\sigma _{u},\rho )=(\delta ,\sigma ,\sigma ,1)$ . Under these conditions, Lemma A.9 shows that $F_{2}$ is continuous at $(\delta ,\sigma _{l},\sigma _{u},\rho )=(\delta ,\sigma ,\sigma ,1)$ , that is,

(A-21) $$ \begin{align} \underset{(\delta_{M},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\delta ,\sigma ,\sigma ,1)}{\text{lim} }F_{2}(\delta_{M},\sigma_{M,l},\sigma_{M,u},\rho_{M})~=~(G(\delta /\sigma ),G(\delta /\sigma )). \end{align} $$

We now divide the rest of the argument into its parts.

Part (a). Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-17),

(A-22) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho})~\overset{d}{ \to }~(z_{1}\sigma_{l},(z_{2}\sqrt{1-\rho ^{2}}+z_{1}\rho )\sigma_{u},\sigma_{l},\sigma_{u},\rho), \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Then,

$$ \begin{align*} & P_{N}({\theta}_{N} \in CI_{\alpha}^{2}) \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l} F_{2,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \\ \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+{\hat{\sigma}}_{u}F_{2,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \end{array} \right) \\ & \overset{(2)}{\to}~P(z_{1}\sigma_{l}-\sigma_{l}\Phi ^{-1}(1-\alpha )\leq -\Psi_{l} ) \\ & \overset{(3)}{=}~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/\sigma_{l}), \end{align*} $$

as desired, where (1) holds by (2.6) and (2.7), which give ${c}^2=F_{2}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\hat { \sigma }_{l},{\hat {\sigma }}_{u},\hat {\rho })$ , (2) by (A-17), (A-20), (A-22), $\mu = \infty $ , and the CMT, and (3) by $\sigma _{l},\sigma _{u}\geq \underline {\sigma }>0$ and $z_{1}\sim \mathcal {N}(0,1)$ .

Part (b). This argument is analogous to part (a) and it is therefore omitted.

Part (c). Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-17), (A-22) holds. Moreover, by Lemma A.4, $\sigma _{l}=\sigma _{u}$ and $\rho =1$ . We set $\sigma =\sigma _{l}=\sigma _{u}$ . Then,

$$ \begin{align*} &P_{N}({\theta}_{N} \in CI_{\alpha}^{2}) \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{2,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{ \rho}) \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \\ \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+{\hat{\sigma}}_{u}F_{2,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \end{array} \right) \\ & \overset{(2)}{\to}~P\left( \sigma z_{1}-\sigma G(\mu /\sigma )\leq -\Psi_{l} \leq z_{1}\sigma +\mu +\sigma G(\mu /\sigma )\right) \\ & \overset{(3)}{=}~\Phi \left( (\Psi_{l} +\mu )/\sigma +G(\mu /\sigma )\right) -\Phi \left( \Psi_{l} /\sigma -G(\mu /\sigma )\right) , \end{align*} $$

as desired, where (1) holds by (2.6) and (2.7), which give ${c}^2=F_{2}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\hat { \sigma }_{l},{\hat {\sigma }}_{u},\hat {\rho })$ , (2) by $\sigma =\sigma _{l}=\sigma _{u}$ , $\rho =1$ , (A-17), (A-21), (A-22), and the CMT, and (3) by $\sigma \geq \underline {\sigma }>0$ and $z_{1}\sim \mathcal {N}(0,1)$ .

Part (d). This argument is analogous to part (c), and it is thus omitted.

Proof. Let $C_{3}:\bar {\mathbb {R}}_{+}\times \bar {\mathbb {R}}\times [ \underline {\sigma },\overline {\sigma }]\times [\underline {\sigma }, \overline {\sigma }]\times [-1,1] \rightrightarrows {\bar {\mathbb {R}}^{2}}$ be the following correspondence:

(A-24) $$ \begin{align} & C_{3}(\delta_{1},\delta_{2},\sigma_{l},\sigma_{u},\rho ) \nonumber \\ & =~\left\{\begin{array}{l} \left( c_{l},c_{u}\right) \in {\bar{\mathbb{R}}^{2}}~~~\text{ s.t.} \\ P(\{-c_{l}\leq z_{1}\}~\cap ~\{\rho z_{1}\leq c_{u}+{\delta_{1}1[\delta_{2}>0]}/{\sigma_{u}}+\sqrt{1-\rho ^{2}}z_{2}\})~\geq ~1-\alpha ~\text{and}\\ P(\{-c_{l}-{\delta_{1}1[\delta_{2}>0]}/{\sigma_{l}}+\sqrt{1-\rho ^{2}} z_{2}\leq \rho z_{1}\}~\cap ~\{z_{1}\leq c_{u}\})~\geq ~1-\alpha , \end{array}\right\} , \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ , let $S_{3}:\bar {\mathbb {R}}_{+}\times \bar {\mathbb {R}}\times [ \underline {\sigma },\overline {\sigma }]\times [\underline {\sigma }, \overline {\sigma }]\times [-1,1] \rightrightarrows {\bar {\mathbb {R}}^{2}}$ be the following minimizer correspondence:

(A-25) $$ \begin{align} S_{3}(\delta_{1},\delta_{2},\sigma_{l},\sigma_{u},\rho )~=~\underset{ \left( c_{l},c_{u}\right) \in C_{3}(\delta_{1},\delta_{2},\sigma_{l},\sigma_{u},\rho )}{\text{arg min}}(\sigma_{l}c_{l}+\sigma_{u}c_{u}), \end{align} $$

and let $F_{3}=(F_{3,l},F_{3,u}):\bar {\mathbb {R}}_{+}\times \bar {\mathbb {R}} \times [\underline {\sigma },\overline {\sigma }]\times [ \underline {\sigma },\overline {\sigma }]\times [-1,1] \to {\bar { \mathbb {R}}^{2}}$ be the choice in $S_{3}(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )$ used to implement $CI_{\alpha }^3$ . By (2.8), (2.9), and Lemma A.11, we have that $ c^{3}=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l})-\sqrt {N}b_{N},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat { \rho })$ a.s.

Part (a). We first show that $( c_{l},c_{u}) \in C_{3}(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )$ satisfies $c_{l},c_{u}\geq \Phi ^{-1}(1-\alpha )>0$ . To show this for $c_{l}$ , note that the first constraint in (A-24) implies that

$$ \begin{align*} P(-c_{l}\leq z_{1})~\geq~ P(\{-c_{l}\leq z_{1}\}~\cap ~\{\rho z_{1}\leq c_{u}+ {\delta_{1}1[\delta_{2}>0]}/{\sigma_{u}}+\sqrt{1-\rho ^{2}} z_{2}\})~\geq ~1-\alpha , \end{align*} $$

which implies $c_{l}\geq \Phi ^{-1}(1-\alpha )$ . In turn, note that $\Phi ^{-1}(1-\alpha )>0$ by $\alpha \in (0,0.5)$ . A similar argument using the second constraint in (A-24) implies that $c_{u}\geq \Phi ^{-1}(1-\alpha )>0$ . Since $F_{3}(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )\in S_{3}(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )\subseteq C_{3}(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )$ ,

(A-26) $$ \begin{align} F_{3,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}), \sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l})-\sqrt{N}b_{N},{\hat{\sigma}}_{l}, {\hat{\sigma}}_{u},\hat{\rho})~&\geq~ \Phi ^{-1}(1-\alpha ) \nonumber \\ F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}), \sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l})-\sqrt{N}b_{N},{\hat{\sigma}}_{l}, {\hat{\sigma}}_{u},\hat{\rho})~&\geq~ \Phi ^{-1}(1-\alpha ). \end{align} $$

Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-27) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho})~\overset{d}{\to}~(z_{1}\sigma_{l},(z_{2}\sqrt{1-\rho ^{2}}+z_{1}\rho )\sigma_{u},\sigma_{l},\sigma_{u},\rho), \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Then,

(A-28) $$ \begin{align} &P_{N}({\theta}_{N} \in CI_{\alpha}^{3}) \nonumber \\ & ~\overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+\\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \end{array} \right) \nonumber \\ & ~\overset{(2)}{\geq }~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}\Phi ^{-1}(1-\alpha )\leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \\ \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+\\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \end{array} \right) \nonumber \\ &\overset{(3)}{\to}~P\left( z_{1}\sigma_{l}-\sigma_{l}\Phi^{-1}(1-\alpha )\leq -\Psi_{l}\right) \nonumber \\ &\overset{(4)}{=}~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l} /\sigma_{l}), \end{align} $$

where (1) holds by Lemma A.11, (2.8), and (2.9), which give ${c}^3=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}) - \sqrt {N}b_{N},\hat { \sigma }_{l},{\hat {\sigma }}_{u},\hat {\rho })$ a.s., (2) by (A-26), (3) by (A-23), (A-26), (A-27), $\mu = \infty $ , and the CMT, and (4) by ${\sigma }_{l} \geq \underline {\sigma }>0$ and $z_{1}\sim \mathcal {N}(0,1)$ . The desired result is implied by (A-28).

Part (b). Note that $\sqrt {N}\left ( {\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})-b_{N}\right ) \to \eta =0$ and $b_{N}\to 0$ imply ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})\to 0$ . Lemma A.4 then implies that (A-23) holds with $(\sigma _{l},\sigma _{u},\rho )=(\sigma ,\sigma ,1)$ for some $\sigma \in [\underline {\sigma },\bar {\sigma }]$ .

We now derive the desired limiting result. It suffices to show that, for any $\varepsilon>0$ ,

(A-29) $$ \begin{align} \limsup_{N\to \infty }P_{N}({\theta}_{N}\in CI_{\alpha}^{3})~\leq~ \Phi ( \Phi ^{-1}(1-\alpha /2)+\varepsilon -\Psi_{l}/\sigma ). \end{align} $$

To see why, note that the desired limit follows from (A-29) as $\varepsilon \downarrow 0$ . To this end, we fix $\varepsilon>0$ arbitrarily for the remainder of this part.

Since $({\hat {\theta }}_{l}, {\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{ \sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-30) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) ~\overset{d}{ \to }~(z\sigma ,z\sigma ,\sigma ,\sigma ,1), \end{align} $$

where $z \sim \mathcal {N}(0,1)$ .

Next, we introduce some definitions. Define $A_{N}=\{({\hat {\theta }}_{u}-{\hat {\theta }}_{l})>b_{N}\}$ . Let $C_{0}:[-1,1]\rightrightarrows {\bar {\mathbb {R}}^{2}}$ be the following correspondence:

$$ \begin{align*} C_{0}(\rho )=~\left\{ \begin{array}{l} \left( c_{l},c_{u}\right) \in {\bar{\mathbb{R}}^{2}}~~~\text{ s.t.} \\ P(\{-c_{l}\leq z_{1}\}~\cap ~\{\rho z_{1}\leq c_{u}+\sqrt{1-\rho ^{2}} z_{2}\})~\geq ~1-\alpha ~\text{and} \\ P(\{-c_{l}+\sqrt{1-\rho ^{2}}z_{2}\leq \rho z_{1}\}~\cap ~\{z_{1}\leq c_{u}\})~\geq ~1-\alpha , \end{array} \right\} , \end{align*} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ , let $S_{0}:[\underline {\sigma },\overline {\sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]\rightrightarrows {\bar {\mathbb {R}} ^{2}}$ be the following minimizer correspondence:

(A-31) $$ \begin{align} S_{0}(\sigma_{l},\sigma_{u},\rho )~=~\underset{\left( c_{l},c_{u}\right) \in C_{0}(\rho )}{\text{arg min}}(\sigma_{l}c_{l}+\sigma_{u}c_{u}), \end{align} $$

and let $F_{0}=(F_{0,l},F_{0,u}):[\underline {\sigma },\overline {\sigma } ]\times [\underline {\sigma },\overline {\sigma }]\times [ -1,1]\to {\bar {\mathbb {R}}^{2}}$ be the choice in $S_{0}(\sigma _{l},\sigma _{u},\rho )$ used to implement $CI_{\alpha }^3$ when $A_{N}^c$ occurs. By repeating arguments in part (b) of Lemma A.8, it follows that $C_{0}$ is continuous at $\rho = 1$ . Also note that $S_{0}(\sigma ,\sigma ,1 ) = \{(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2))\}$ . Then, by repeating the arguments of Lemma A.10, we obtain that $F_{0}$ is continuous at $(\sigma ,\sigma ,1)$ and $F_{0}(\sigma ,\sigma ,1) = (\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2))$ . By this and (A-30),

(A-32) $$ \begin{align} F_{0}({\hat{\sigma}}_{l} ,{\hat{\sigma}}_{u} ,\hat{\rho})~\overset{p}{\to}~(\Phi^{-1}(1-\alpha/2),\Phi^{-1}(1-\alpha/2)). \end{align} $$

Let $C_2$ be defined as in (A-18), $S_2$ be defined as in (A-19), and $F_{2}=(F_{2,l},F_{2,u}):\bar {\mathbb {R}}_{+}\times [ \underline {\sigma },\overline {\sigma }]\times [\underline {\sigma }, \overline {\sigma }]\times [-1,1] \to {\bar {\mathbb {R}}^{2}}$ be the choice in $S_{2}$ used to implement $CI_{\alpha }^3$ when $A_{N}$ occurs. Recall from Lemma A.7 that $C_2$ in (A-18) is continuous at $(\infty ,\sigma ,\sigma ,1)$ . Also, note that from the definition in (A-19) and Lemmas A.5 and A.6, we obtain $S_{2}(\infty ,\sigma ,\sigma ,1) = \{(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha ))\}$ . From this and Lemma A.9, it follows that $F_{2}$ is continuous at $(\infty ,\sigma ,\sigma ,1)$ and $F_{2}(\infty ,\sigma ,\sigma ,1) = (\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha ))$ . By this and (A-30),

(A-33) $$ \begin{align} F_{2}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}), {\hat{\sigma}}_{l} ,{\hat{\sigma}}_{u} ,\hat{\rho})~\overset{p}{\to}~(\Phi^{-1}(1-\alpha),\Phi^{-1}(1-\alpha)). \end{align} $$

Next, define $E_{N}=\{ F_{3,l}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}-b_{N}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })>\Phi ^{-1}(1-\alpha /2)+\varepsilon \} $ . By construction,

(A-34) $$ \begin{align} &F_{3,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \nonumber \\ &=F_{2,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},\hat{ \sigma}_{u},\hat{\rho})\times I\left\{ A_{N}\right\} +F_{0,l}({\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho})\times I\left\{ A_{N}^{c}\right\}. \end{align} $$

Then,

(A-35) $$ \begin{align} P_{N}( E_{N}) &~=~P_{N}( E_{N}\cap A_{N}) +P_{N}( E_{N}\cap A_{N}^{c}) \nonumber \\ &~\overset{(1)}{=}~\left\{ \begin{array}{@{}c@{}} P_{N}( \{ F_{2,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\hat{ \sigma}_{l},{\hat{\sigma}}_{u},\hat{\rho})>\Phi ^{-1}(1-\alpha /2)+\varepsilon \} \cap A_{N}) \nonumber \\ +P_{N}( \{ F_{0,l}({\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho} )>\Phi ^{-1}(1-\alpha /2)+\varepsilon \} \cap A_{N}^{c}) \end{array} \right\} \nonumber \\ &~\leq~ \left\{ \begin{array}{@{}c@{}} P_{N}( F_{2,l}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho})>\Phi ^{-1}(1-\alpha /2)+\varepsilon ) \\ +P_{N}( F_{0,l}({\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho})>\Phi ^{-1}(1-\alpha /2)+\varepsilon ) \end{array} \right\} \nonumber \\ &~\overset{(2)}{ \to }~0, \end{align} $$

where (1) holds by (A-34) and (2) by (A-32) and (A-33). To complete the proof, note that

$$ \begin{align*} & P_{N}({\theta}_{N}\in CI_{\alpha}^{3}) \nonumber \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+ \\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho}) \end{array} \right) \nonumber \\ & \leq ~P_{N}(\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho})\\& \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N})) ) \nonumber \\ & \overset{(2)}{\leq} ~P_{N}( \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-\hat{ \sigma}_{l}(\Phi ^{-1}(1-\alpha /2)+\varepsilon) \leq \sqrt{N} ({\theta}_{N}-{\theta}_{l}(P_{N}))) +P_{N}( E_{N}) \\ &\overset{(3)}{\to}~ \Phi ( \Phi ^{-1}(1-\alpha /2)+\varepsilon -\Psi_{l}/\sigma ) , \end{align*} $$

where (1) holds by Lemma A.11, (2.8), and (2.9), (2) by definition of $E_N$ , and (3) by (A-30) and (A-35). This derivation implies (A-29) and concludes the proof of this part.

Part (c). We divide the analysis into two cases.

Case 1: $\eta \in (0,\infty )$ . By $\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})-b_{N}) \to \eta \in (0,\infty )$ and $b_{N} \to 0$ , we conclude that ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ . Lemma A.4 then implies $(\sigma _{l}(P_{N}),\sigma _{u}(P_{N}),\rho (P_{N})) \to (\sigma ,\sigma ,1)$ with $\sigma \in [\underline {\sigma },\bar {\sigma }]$ .

We now study the continuity properties of $F_{3}$ for any sequence $\{(\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N}}$ with $\lim _{M\to \infty } (\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M}) = (\infty ,\eta ,\sigma ,\sigma ,1)$ . By evaluating in (A-24), we get

$$ \begin{align*} C_{3}(\infty ,\eta ,\sigma ,\sigma ,1)~=~\left\{ ( c_{l},c_{u}) \in {\bar{\mathbb{R}}^{2}}~\text{s.t.}~\Phi (c_{l})~\geq ~1-\alpha ~\text{and}~\Phi (c_{u})~\geq ~1-\alpha \right\}. \end{align*} $$

In turn, Lemma A.6 shows that

$$ \begin{align*} S_{3}(\infty ,\eta ,\sigma ,\sigma ,1)~&=~\{(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha ))\}\\ F_{3}(\infty ,\eta ,\sigma ,\sigma ,1)~&=~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align*} $$

Furthermore, part (a) of Lemma A.8 shows that $C_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\eta ,\sigma ,\sigma ,1)$ . Under these conditions, Lemma A.10 shows that $F_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\eta ,\sigma ,\sigma ,1)$ , that is,

(A-36) $$ \begin{align} &\underset{(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\infty ,\eta ,\sigma ,\sigma ,1)}{\text{lim}}F_{3}(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M})\nonumber\\ & \quad =~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align} $$

Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u}, \hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-37) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) ~\overset{d}{ \to }~(z\sigma ,z\sigma ,\sigma ,\sigma ,1), \end{align} $$

where $z\sim \mathcal {N}(0,1)$ . Then,

(A-38) $$ \begin{align} &P_{N}({\theta}_{N}\in CI_{\alpha}^{3}) \nonumber \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+ \\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}( {\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho}) \end{array} \right) \nonumber \\ & \overset{(2)}{\to}~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/\sigma ), \end{align} $$

as desired, where (1) holds by Lemma A.11, (2.8), and (2.9), which give ${c}^{3}=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}-b_{N}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ a.s., and (2) by (A-23), (A-36), (A-37), and the CMT.

Case 2: $\eta =\infty $ . We now study continuity properties of $F_{3}$ for any convergent sequence $\{(\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N}}$ with $\lim _{M\to \infty } (\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M}) = (\infty ,\infty ,\sigma _{l},\sigma _{u},\rho )$ with $(\sigma _{l},\sigma _{u},\rho )\in [\underline {\sigma },\overline { \sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]$ . By evaluating in (A-24), we get

$$ \begin{align*} C_{3}(\infty ,\infty ,\sigma_{l},\sigma_{u},\rho )~=~\left\{ ( c_{l},c_{u}) \in {\bar{\mathbb{R}}^{2}}~\text{s.t.}~\Phi (c_{l})~\geq ~1-\alpha ~\text{and}~\Phi (c_{u})~\geq ~1-\alpha \right\}. \end{align*} $$

In turn, Lemma A.6 shows that

$$ \begin{align*} S_{3}(\infty ,\infty ,\sigma_{l},\sigma_{u},\rho )~&=~\{(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha ))\}\\ F_{3}(\infty ,\infty ,\sigma_{l},\sigma_{u},\rho )~&=~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align*} $$

Furthermore, part (a) of Lemma A.8 shows that $C_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\infty ,\sigma _{l},\sigma _{u},\rho )$ with $(\sigma _{l},\sigma _{u},\rho )\in [\underline {\sigma },\overline {\sigma }]\times [ \underline {\sigma },\overline {\sigma }]\times [-1,1]$ . Under these conditions, Lemma A.10 shows that $F_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\infty ,\sigma _{l},\sigma _{u},\rho )$ , that is,

(A-39) $$ \begin{align} &\underset{(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\infty ,\infty ,\sigma_{l},\sigma_{u},\rho )}{\text{lim}} F_{3}(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M})\nonumber\\&\quad=~(\Phi ^{-1}(1-\alpha ),\Phi ^{-1}(1-\alpha )). \end{align} $$

Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u}, \hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-40) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) ~\overset{d}{ \to }~(z_{1}\sigma_{l},(z_{2}\sqrt{1-\rho ^{2}} +z_{1}\rho )\sigma_{u},\sigma_{l},\sigma_{u},\rho ), \end{align} $$

where $(z_{1},z_{2})\sim \mathcal {N}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Then,

(A-41) $$ \begin{align} & P_{N}({\theta}_{N}\in CI_{\alpha}^{3}) \nonumber \\ & ~\overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+ \\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}( {\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho}) \end{array} \right) \nonumber \\ & ~\overset{(2)}{ \to }~\Phi (\Phi ^{-1}(1-\alpha )-\Psi_{l}/{\sigma}_{l} ), \end{align} $$

as desired, where (1) holds by Lemma A.11, (2.8), and (2.9), which give ${c}^{3}=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}-b_{N}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ a.s., and (2) by (A-23), (A-39), (A-40), and the CMT.

Part (d). We now show that ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ . To see this, note that we have two cases: $\eta \in (-\infty ,0)$ or $\eta =-\infty $ . If $\eta \in (-\infty ,0)$ , $\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})-b_{N}) \to \eta \in (-\infty ,0)$ . By this and $ b_{N} \to 0$ , we get that ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ . If $\eta =-\infty $ , ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})\leq -1/\sqrt {N}+b_{N}$ for all sufficiently large $N$ . By this, ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})\geq 0$ , and $ b_{N} \to 0$ , we get that ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ . Since ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ , Lemma A.4 then implies $(\sigma _{l}(P_{N}),\sigma _{u}(P_{N}),\rho (P_{N})) \to (\sigma ,\sigma ,1)$ with $\sigma \in [\underline { \sigma },\bar {\sigma }]$ .

We now study the continuity properties of $F_{3}$ for any sequence $\{(\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N}}$ with $\lim _{M\to \infty } (\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M}) = (\infty ,\eta ,\sigma ,\sigma ,1)$ . By evaluating in (A-24), we get

$$ \begin{align*} C_{3}(\infty ,\eta ,\sigma ,\sigma ,1)~=~\left\{ ( c_{l},c_{u}) \in {\bar{\mathbb{R}}^{2}}~\text{ s.t.}~P(-c_{l}\leq z_{1}\leq c_{u})~\geq ~1-\alpha \right\}. \end{align*} $$

In turn, Lemma A.6 shows that

$$ \begin{align*} S_{3}(\infty ,\eta ,\sigma ,\sigma ,1 )~=~\{(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2))\}\\ \!\!\!F_{3}(\infty ,\eta ,\sigma ,\sigma ,1 )~=~(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2)). \end{align*} $$

Furthermore, part (a) of Lemma A.8 shows that $C_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\eta ,\sigma ,\sigma ,1)$ . Under these conditions, Lemma A.10 shows that $F_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\infty ,\eta ,\sigma ,\sigma ,1)$ , that is,

(A-42) $$ \begin{align} &\underset{(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\infty ,\eta ,\sigma ,\sigma ,1)}{\text{lim}}F_{3}(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M})\nonumber\\&\quad=~(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2)). \end{align} $$

Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u}, \hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-43) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \overset{d}{ \to }(z\sigma ,z\sigma ,\sigma ,\sigma ,1), \end{align} $$

where $z\sim \mathcal {N}(0,1)$ . Then,

(A-44) $$ \begin{align} & P_{N}({\theta}_{N}\in CI_{\alpha}^{3}) \nonumber \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+ \\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho}) \end{array} \right) \nonumber \\ & \overset{(2)}{ \to }~\Phi (\Phi ^{-1}(1-\alpha /2)-\Psi_{l}/\sigma ), \end{align} $$

as desired, where (1) holds by Lemma A.11, (2.8), and (2.9), which give ${c}^{3}=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}-b_{N}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ a.s., and (2) by (A-23), (A-42), (A-43), and the CMT.

Part (e). By $\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})) \to \mu \in \mathbb {R}_{+}$ and $b_{N} \to 0$ , we conclude that ${\theta }_{u}(P_{N})-{\theta }_{l}(P_{N}) \to 0$ . Lemma A.4 then implies $(\sigma _{l}(P_{N}),\sigma _{u}(P_{N}),\rho (P_{N})) \to (\sigma ,\sigma ,1)$ with $\sigma \in [\underline {\sigma },\bar {\sigma }]$ . Also, by $\sqrt {N}b_{N} \to \infty $ , we conclude that $\sqrt {N}({\theta }_{u}(P_{N})-{\theta }_{l}(P_{N})-b_{N}) \to -\infty $ .

We now study the continuity properties of $F_{3}$ for any sequence $\{(\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M})\}_{M\in \mathbb {N}}$ with $\lim _{M\to \infty } (\delta _{M,1},\delta _{M,2},\sigma _{M,l},\sigma _{M,u},\rho _{M}) =(\mu ,-\infty ,\sigma ,\sigma ,1)$ with $(\mu ,\sigma )\in \mathbb {R}_{+}\times [\underline {\sigma },\overline {\sigma }]$ . By evaluating in (A-24), we get

$$ \begin{align*} C_{3}(\mu ,-\infty ,\sigma ,\sigma ,1)=\left\{ ( c_{l},c_{u}) \in {\bar{\mathbb{R}}^{2}}~\text{ s.t.}~P(-c_{l}\leq z_{1}\leq c_{u})~\geq ~1-\alpha \right\}. \end{align*} $$

In turn, Lemma A.6 shows that

$$ \begin{align*} S_{3}(\mu ,-\infty,\sigma ,\sigma ,1 )~=~\{(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2))\}\\ \!\!\!F_{3}(\mu ,-\infty ,\sigma ,\sigma ,1 )~=~(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2)). \end{align*} $$

Furthermore, part (a) of Lemma A.8 shows that $C_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\mu ,-\infty ,\sigma ,\sigma ,1)$ . Under these conditions, Lemma A.10 shows that $F_{3}$ is continuous at $(\delta _{1},\delta _{2},\sigma _{l},\sigma _{u},\rho )=(\mu ,-\infty ,\sigma ,\sigma ,1)$ , that is,

(A-45) $$ \begin{align} &\underset{(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M}) \to (\mu ,-\infty ,\sigma ,\sigma ,1)}{\text{lim}}F_{3}(\delta_{M,1},\delta_{M,2},\sigma_{M,l},\sigma_{M,u},\rho_{M})\nonumber\\&\quad=~(\Phi ^{-1}(1-\alpha /2),\Phi ^{-1}(1-\alpha /2)). \end{align} $$

Since $({\hat {\theta }}_{l},{\hat {\theta }}_{u},{\hat {\sigma }}_{l},{\hat {\sigma }}_{u}, \hat {\rho })$ satisfies OBS with parameter $({\theta }_{l}(P),{\theta }_{u}(P),{\sigma }_{l}(P),{\sigma }_{u}(P),{\rho }(P))$ and (A-23),

(A-46) $$ \begin{align} (\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N})),\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N})),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \overset{d}{ \to }(z\sigma ,z\sigma ,\sigma ,\sigma ,1), \end{align} $$

where $z\sim \mathcal {N}(0,1)$ . Then,

$$ \begin{align*} & P_{N}({\theta}_{N}\in CI_{\alpha}^{3}) \\ & \overset{(1)}{=}~P_{N}\left( \begin{array}{@{}c@{}} \sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(P_{N}))-{\hat{\sigma}}_{l}F_{3,l}(\sqrt{N }({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u},\hat{\rho}) \\ \leq \sqrt{N}({\theta}_{N}-{\theta}_{l}(P_{N}))\leq \sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(P_{N}))+\sqrt{N}({\theta}_{u}(P_{N})-{\theta}_{l}(P_{N}))+ \\ {\hat{\sigma}}_{u}F_{3,u}(\sqrt{N}({\hat{\theta}}_{u}-{\hat{\theta}}_{l}),\sqrt{N}( {\hat{\theta}}_{u}-{\hat{\theta}}_{l}-b_{N}),{\hat{\sigma}}_{l},{\hat{\sigma}}_{u}, \hat{\rho}) \end{array} \right) \\ & \overset{(2)}{ \to }~P\left( \sigma z-\sigma \Phi ^{-1}(1-\alpha /2)\leq -\Psi_{l}\leq z\sigma +\mu +\sigma \Phi ^{-1}(1-\alpha /2)\right) \\ & \overset{(3)}{=}~\Phi \left( (\Psi_{l}+\mu )/\sigma +\Phi ^{-1}(1-\alpha /2)\right) -\Phi \left( \Psi_{l}/\sigma -\Phi ^{-1}(1-\alpha /2)\right), \end{align*} $$

as desired, where (1) holds by Lemma A.11, (2.8), and (2.9), which give ${c}^{3}=F_{3}(\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}),\sqrt {N}({\hat {\theta }}_{u}-{\hat {\theta }}_{l}-b_{N}),{\hat {\sigma }}_{l},{\hat {\sigma }}_{u},\hat {\rho })$ a.s., (2) by (A-23), (A-45), (A-46), and the CMT, and (3) by $\sigma \geq \underline {\sigma }>0$ and $z\sim \mathcal {N}(0,1)$ .

Parts (f)–(j). These are analogous to those in parts (a)–(e), respectively, and are thus omitted.

Proof. By (A-47), we can construct a subsequence $\{k_{N}\}_{N\geq 1}$ of $\{N\}_{N\geq 1}$ s.t. $\sqrt {N}({\theta }_{u}(P_{k_{N}})-{\theta }_{l}(P_{k_{N}})) \to 0$ . Construct the subsequence recursively as follows. First, set $k_{1}=1$ , and so $P_{k_{1}}=P_{1}$ . Next, for any $N\in \mathbb {N}$ with $ N>1$ , find $k_{N}\in \mathbb {N}$ s.t. $k_{N}>k_{N-1}$ and $|\sqrt {N}({\theta }_{u}(P_{k_{N}})-{\theta }_{l}(P_{k_{N}}))|\leq 1/N$ . (This is always possible because of (A-47).) By repeating this construction iteratively, and taking limits as $N \to \infty $ , we get $\sqrt {N}({\theta }_{u}(P_{k_{N}})-{\theta }_{l}(P_{k_{N}})) \to 0$ . Next, consider the sequence $\{\tilde {P}_{N}\}_{N\geq 1}$ with $\tilde {P}_{N}=P_{k_{N}}$ . By construction, we get

(A-49) $$ \begin{align} \sqrt{N}({\theta}_{u}(\tilde{P}_{N})-{\theta}_{l}(\tilde{P}_{N}))~ \to ~0. \end{align} $$

Since $\{\tilde {P}_{N}\}_{N\geq 1}$ is a subsequence of $\{P_{N}\}_{N\geq 1}$ , (A-48) implies that

(A-50) $$ \begin{align} (\rho (\tilde{P}_{N}),\sigma_{l}(\tilde{P}_{N}),\sigma_{u}(\tilde{P}_{N}))~ \to ~(\rho ,\sigma_{l},\sigma_{u}). \end{align} $$

To complete the proof, it suffices to show that $\rho =1$ and $\sigma _{l}=\sigma _{u}$ .

First, note that OBS applied to $\{\tilde {P}_{N}\}_{N\geq 1}$ and (A-50) imply

(A-51) $$ \begin{align} \sqrt{N}\left( \begin{array}{@{}c@{}} {\hat{\theta}}_{l}-{\theta}_{l}(\tilde{P}_{N}), \\ {\hat{\theta}}_{u}-{\theta}_{u}(\tilde{P}_{N}) \end{array} \right) ~\overset{d}{ \to }~N\left( {\mathbf{0}}_{2\times 1},\left( \begin{array}{@{}cc@{}} \sigma_{l}^{2} & \rho \sigma_{l}\sigma_{u} \\ \rho \sigma_{l}\sigma_{u} & \sigma_{u}^{2} \end{array} \right) \right). \end{align} $$

Let $Z_{N}\equiv \sqrt {N}({\hat {\theta }}_{l}-{\hat {\theta }}_{u})$ . Then, consider the following derivation:

(A-52) $$ \begin{align} Z_{N}& ~=~\sqrt{N}({\hat{\theta}}_{l}-{\theta}_{l}(\tilde{P}_{N}))-\sqrt{N}({\hat{\theta}}_{u}-{\theta}_{u}(\tilde{P}_{N}))-\sqrt{N}({\theta}_{u}(\tilde{P}_{N})-{\theta}_{l}(\tilde{P}_{N})) \nonumber \\ & ~{\overset{d}{ \to }}~Z\sim \mathcal{N}(0,\sigma_{l}^{2}+\sigma_{u}^{2}-2\rho \sigma_{l}\sigma_{u}), \end{align} $$

where the convergence holds by (A-49) and (A-51). By OBS, $\tilde {P}_{N}(Z_{N}\leq 0)=\tilde {P}_{N}({\hat {\theta }}_{l}\leq {\hat {\theta }}_{u})=1$ . This and (A-52) then imply that $\sigma _{l}^{2}+\sigma _{u}^{2}-2\rho \sigma _{l}\sigma _{u}=0$ or, equivalently,

(A-53) $$ \begin{align} (\sigma_{l}-\sigma_{u})^{2}~=~2(\rho -1)\sigma_{l}\sigma_{u}. \end{align} $$

To see why, note that (A-52) and $\sigma _{l}^{2}+\sigma _{u}^{2}-2\rho \sigma _{l}\sigma _{u}>0$ imply $\tilde {P}_{N}(Z_{N}\leq 0) \to P(Z\leq 0)<1$ , which contradicts $\tilde {P}_{N}(Z_{N}\leq 0)=1$ .

We are now ready to conclude the proof. First, note that (A-53), $(\sigma _{l}-\sigma _{u})^{2}\geq 0$ , and $(\rho ,\sigma _{l},\sigma _{u})\in [-1,1]\times [\underline {\sigma },\overline {\sigma }]\times [\underline {\sigma },\overline {\sigma }]$ imply that $\rho =1$ . In turn, (A-53) and $\rho =1$ imply that $(\sigma _{l}-\sigma _{u})^{2}=0$ or, equivalently, $\sigma _{l}=\sigma _{u}$ .

Proof. We first verify that (A-55) has a unique and positive solution. For any $y\in \mathbb {R}_{+}$ , let $L(c) \equiv \Phi (c+y) -\Phi (-c) $ denote the left-hand side of (A-55). Note that $L$ is continuous and strictly increasing because $L'(c)=\phi (c+y)+\phi (-c)>0$ . Moreover, $L(0) =\Phi (y) -\Phi (0) \leq 0.5<1-\alpha $ and $\lim _{c\to \infty }L(c)=1>1-\alpha $ . Thus, by the intermediate value theorem, (A-55) has a solution. Strict monotonicity gives uniqueness, and $L(0)<1-\alpha $ implies that the solution is positive.

Note that $H( \sigma ,\mu ,\Psi )$ is differentiable in $\sigma $ . Therefore, part (a) follows from verifying that, for all $( \sigma ,\mu ,\Psi ) \in [ \underline {\sigma },\overline {\sigma }] \times {\mathbb {R}}_{+}\times \bar {\mathbb {R}}_{+}$ ,

(A-56) $$ \begin{align} \frac{\partial H( \sigma ,\mu ,\Psi ) }{\partial \sigma }~\geq~ 0. \end{align} $$

Moreover, part (b) follows from verifying that (A-56) becomes strict when $\Psi \in (0,\infty )$ .

We divide the proof into cases.

Case 1: $\Psi =\infty $ . This case implies that $H( \sigma ,\mu ,\infty ) =0$ , and so $\partial H( \sigma ,\mu ,\infty ) /\partial \sigma =0$ .

Case 2: $\Psi <\infty $ . Note that

(A-57) $$ \begin{align} \frac{\partial H( \sigma ,\mu ,\Psi ) }{\partial \sigma }=- \frac{\small\left( \phi ( \tfrac{\Psi +\mu }{\sigma }{+}G(\tfrac{\mu }{\sigma })) \left(\left( G^{\prime }(\tfrac{\mu }{\sigma }){+}1\right) \mu {+}\Psi \right) {+}\phi ( \tfrac{\Psi }{\sigma }-G(\tfrac{\mu }{\sigma })) \left(G^{\prime }(\tfrac{\mu }{\sigma })\mu -\Psi \right) \right)}{\sigma ^{2}}. \end{align} $$

We complete our proof by showing that the numerator of (A-57) is non-positive or, equivalently,

(A-58) $$ \begin{align} \exp \big( -{( \tfrac{\Psi }{\sigma }-G(\tfrac{\mu }{\sigma })) ^{2}}/{2}\big) ~( \Psi -G^{\prime }( \tfrac{\mu }{\sigma }) \mu ) &\geq \exp \big( -{( \tfrac{\Psi +\mu }{\sigma }+G(\tfrac{\mu }{\sigma })) ^{2}}/{2}\big) \nonumber\\&\quad ( ( G^{\prime }(\tfrac{\mu }{\sigma })+1) \mu +\Psi ) , \end{align} $$

and that this becomes strict when $\Psi>0$ .

We now divide the analysis into three sub-cases.

Case 2.1: $\mu =0$ and $\Psi =0$ . In this case, (A-58) is equivalent to $0\geq 0$ .

Case 2.2: $\mu =0$ and $\Psi \in (0,\infty )$ . In this case, the strict inequality in (A-58) is equivalent to $( \tfrac {\Psi }{\sigma }-G( 0)) ^{2}< ( \tfrac {\Psi }{\sigma }+G(0)) ^{2}$ which, in turn, is equivalent to $G(0)\tfrac {\Psi }{\sigma }>0$ . This follows from $G(0)>0$ and $\sigma \geq \underline {\sigma }>0$ .

Case 2.3: $\mu>0$ and $\Psi \in [0,\infty )$ . As a preliminary result, we now show that

(A-59) $$ \begin{align} G^{\prime }( \tfrac{\mu }{\sigma }) ~=~\frac{-\phi ( {G}( \tfrac{\mu }{\sigma }) +\tfrac{\mu }{\sigma }) }{\phi ( {G}( \tfrac{\mu }{\sigma }) +\tfrac{\mu }{\sigma }) +\phi ( -{G}( \tfrac{\mu }{\sigma })) }~<~0. \end{align} $$

To this end, note that for any $y\in \mathbb {R}_{+}$ , ${G}( y) $ satisfies $F(y)\equiv \Phi ( G(y)+y) -\Phi ( -G(y)) -(1-\alpha )=0$ . Since the relationship holds for all $y\in \mathbb {R}_{+}$ , we have that $F^{\prime }( y) =0$ , that is,

$$ \begin{align*} \phi ( G(y)+y) ( G^{\prime }( y) +1) +\phi ( -G(y)) G^{\prime }( y) ~=~0. \end{align*} $$

By evaluating at $y=\mu /\sigma $ with $( \sigma ,\mu ) \in [ \underline {\sigma },\overline {\sigma }] \times \bar {\mathbb {R}}_{+}$ and solving for $G^{\prime }( \mu /\sigma ) $ , (A-59) follows.

Since $G^{\prime }( \tfrac {\mu }{\sigma }) <0$ (by (A-59)), $\Psi \geq 0$ , and $\mu>0$ , we have that $\Psi -G^{\prime }( \tfrac {\mu }{\sigma })\mu>0$ , and so (A-58) is equivalent to

(A-60) $$ \begin{align} \exp \Big( \tfrac{1}{2}{( \tfrac{\Psi +\mu }{\sigma }+G( \tfrac{\mu }{ \sigma })) ^{2}}-\tfrac{1}{2}{( \tfrac{\Psi }{\sigma }-G( \tfrac{ \mu }{\sigma })) ^{2}}\Big) ~\geq~ \frac{( G^{\prime }( \tfrac{\mu }{\sigma })+1) \mu +\Psi }{\Psi -G^{\prime }( \tfrac{\mu }{\sigma })\mu }. \end{align} $$

By combining with (A-59), (A-60) is equivalent to

(A-61) $$ \begin{align} &\Big( \tfrac{\Psi }{\mu }+\tfrac{\phi ( G( \tfrac{\mu }{\sigma }) +\tfrac{\mu }{\sigma }) }{\phi ( G( \tfrac{\mu }{ \sigma }) +\tfrac{\mu }{\sigma }) +\phi ( -G( \tfrac{ \mu }{\sigma })) }\Big) ~ \exp \big( \tfrac{\Psi \mu }{\sigma ^{2}}+\tfrac{\mu ^{2}}{2\sigma ^{2}}+G( \tfrac{\mu }{\sigma }) \tfrac{2\Psi +\mu }{\sigma }\big) \nonumber\\& \quad\geq ~\tfrac{\phi ( -G( \tfrac{\mu }{\sigma })) }{ \phi ( G( \tfrac{\mu }{\sigma }) +\tfrac{\mu }{\sigma }) +\phi ( -G( \tfrac{\mu }{\sigma })) }+\tfrac{ \Psi }{\mu }. \end{align} $$

To verify (A-61), it suffices to check that:

(A-62) $$ \begin{align} ( \Psi /\mu )~ \exp \big( \tfrac{\Psi \mu }{\sigma ^{2}}+\tfrac{\mu ^{2}}{2\sigma ^{2}}+G( \tfrac{\mu }{\sigma }) \tfrac{2\Psi +\mu }{\sigma }\big) & ~\geq~ \Psi /\mu , \nonumber \\ \phi ( G(\mu /\sigma )+\mu /\sigma ) ~ \exp \big( \tfrac{\Psi \mu }{\sigma ^{2}}+\tfrac{\mu ^{2}}{2\sigma ^{2}}+G( \tfrac{\mu }{\sigma }) \tfrac{2\Psi +\mu }{\sigma }\big) & ~\geq~ \phi ( -G(\mu /\sigma )). \end{align} $$

Both of these can be shown by algebra and the fact that $\Psi \geq 0$ , $\mu>0$ , $G( \tfrac {\mu }{\sigma })>0$ , and $\sigma \geq \underline {\sigma }>0$ .

Finally, we note that $\Psi>0$ further implies that both inequalities in (A-62) hold strictly. By retracing our steps, this implies that (A-58), (A-60), and (A-61) all become strict in this case.

Proof. First, we show that $c_{l}^{\ast },c_{u}^{\ast }> 0$ . To see why, note that $c_{l}^{\ast }\leq 0$ implies a violation of the first constraint in (A-63) and $c_{u}^{\ast }\leq 0$ would imply a violation of the second constraint in (A-63). For brevity, we only show the first one. To see this, note that $c_{l}^{\ast }\leq 0$ implies

$$ \begin{align*} \Phi ( c_{u}^{\ast }+\delta /\sigma ) -\Phi ( -c_{l}^{\ast }) ~\leq~ 1-\Phi ( -c_{l}^{\ast })~ =~\Phi ( c_{l}^{\ast }) ~\leq~ \Phi ( 0) ~=~1/2~<~1-\alpha. \end{align*} $$

This is a violation of the first constraint, as desired.

By $c_{l}^{\ast },c_{u}^{\ast }> 0$ and $\sigma>0$ , we have that (A-63) is equivalent to

(A-64) $$ \begin{align} &\min_{c_{l},c_{u}\in \mathbb{R}}( c_{l}+c_{u})~~~\text{s.t.}~~ \nonumber \\ &\Phi( c_{u}+\delta /\sigma ) +\Phi ( c_{l}) \geq 2-\alpha ~~\text{and}~~\Phi ( c_{u}) +\Phi ( c_{l}+\delta /\sigma ) \geq 2-\alpha. \end{align} $$

We focus on (A-64) for the remainder of the proof.

We now show that at any solution $( c_{l}^{\ast },c_{u}^{\ast }) $ one of the two constraints must be binding. To see why, suppose that this is not the case and $( c_{l}^{\ast },c_{u}^{\ast }) $ minimizes $( c_{l}+c_{u}) $ with $\Phi ( c_{u}+\delta /\sigma ) +\Phi ( c_{l}) \geq 2-\alpha $ and $\Phi ( c_{u}) +\Phi ( c_{l}+\delta /\sigma ) \geq 2-\alpha $ . Since $\Phi $ is continuous and the constraints are slack, we can then find an alternative candidate solution $( {\tilde {c}}_{l}^{\ast },{\tilde {c}}_{u}^{\ast })$ with $0\leq {\tilde {c}}_{l}^{\ast } <c_{l}^{\ast }$ and $0\leq {\tilde {c}}_{u}^{\ast } <c_{u}^{\ast }$ that also satisfies the constraints. Since ${\tilde {c}}_{l}^{\ast }+{\tilde {c}}_{u}^{\ast }<c_{l}^{\ast }+c_{u}^{\ast }$ , this contradicts that $( c_{l}^{\ast },c_{u}^{\ast }) $ was a minimizer.

We now show that the solution is unique. Suppose that this is not the case, and that $( c_{l}^{\ast },c_{u}^{\ast }) $ and $( {\tilde {c}}_{l}^{\ast },{\tilde {c}}_{u}^{\ast }) $ are solutions to the problem and $( c_{l}^{\ast },c_{u}^{\ast }) \neq ( {\tilde {c}}_{l}^{\ast },{\tilde {c}}_{u}^{\ast }) $ . Let $V=c_{l}^{\ast }+c_{u}^{\ast }={\tilde {c}}_{l}^{\ast }+{\tilde {c}}_{u}^{\ast }$ denote their (common) minimized value.

Then, consider $( ( c_{l}^{\ast }+{\tilde {c}}_{l}^{\ast }) /2,( c_{u}^{\ast }+{\tilde {c}}_{u}^{\ast }) /2) $ as another candidate solution. Note that this achieves the same minimized value $( c_{l}^{\ast }+c_{u}^{\ast }) /2+( {\tilde {c}}_{l}^{\ast }+{\tilde {c}}_{u}^{\ast }) /2=V$ . Then, we have

(A-65) $$ \begin{align} \Phi ( \tfrac{c_{u}^{\ast }+{\tilde{c}}_{u}^{\ast }}{2}+\delta /\sigma ) +\Phi ( \tfrac{c_{l}^{\ast }+{\tilde{c}}_{l}^{\ast }}{2}) &\overset{(1)}{>}\frac{ \Phi ( c_{u}^{\ast }+\delta /\sigma ) +\Phi ( c_{l}^{\ast }) +\Phi ( {\tilde{c}}_{u}^{\ast }+\delta /\sigma ) +\Phi ( {\tilde{c}}_{l}^{\ast }) }{2} \overset{(3)}{\geq} 2-\alpha \nonumber \\ \Phi ( \tfrac{c_{u}^{\ast }+{\tilde{c}}_{u}^{\ast }}{2}) +\Phi( \tfrac{c_{l}^{\ast }+{\tilde{c}}_{l}^{\ast }}{2}+\delta /\sigma ) &\overset{(2)}{>}\frac{\Phi ( c_{u}^{\ast }) +\Phi ( c_{l}^{\ast }+\delta /\sigma) +\Phi ( {\tilde{c}}_{u}^{\ast }) +\Phi ( {\tilde{c}}_{l}^{\ast}+\delta /\sigma ) }{2} \overset{(4)}{\geq} 2-\alpha , \end{align} $$

where (1) and (2) hold by $c_{l}^{\ast },c_{u}^{\ast },{\tilde {c}}_{l}^{\ast },{\tilde {c}}_{u}^{\ast }> 0$ and $\delta /\sigma \geq 0$ , $\Phi ( x) $ is strictly concave for $x>0$ , and either $c_{l}^{\ast } \neq {\tilde {c}}_{l}^{\ast }$ or $c_{u}^{\ast } \neq {\tilde {c}}_{u}^{\ast }$ , and (3) and (4) hold because $( c_{l}^{\ast },c_{u}^{\ast })$ and $( {\tilde {c}}_{l}^{\ast },{\tilde {c}}_{u}^{\ast }) $ satisfy the constraints. Note that (A-65) implies that $( ( c_{l}^{\ast }+{\tilde {c}}_{l}^{\ast }) /2,( c_{u}^{\ast }+{\tilde {c}}_{u}^{\ast }) /2) $ is an optimal solution where neither constraint is binding (i.e., an interior solution), which is a contradiction.

Next, we show that the unique solution $( c_{l}^{\ast },c_{u}^{\ast }) $ satisfies $c_{l}^{\ast }=c_{u}^{\ast }$ . Suppose this is not the case, that is, $c_{l}^{\ast } \neq c_{u}^{\ast }$ . From here, it is easy to see that $( (c_{l}^{\ast }+c_{u}^{\ast })/2,(c_{l}^{\ast }+c_{u}^{\ast })/2) $ is a different candidate that is feasible and achieves the same minimized value. But this then implies that the solution is not unique, which is a contradiction.

By combining all previous arguments, we get that the unique solution is $( c_{l}^{\ast },c_{u}^{\ast }) =( c,c) $ with $c>0$ that satisfies $\Phi ( c+\delta /\sigma ) -\Phi ( -c) \geq 1-\alpha $ or $\Phi ( c) -\Phi ( -c-\delta /\sigma ) \geq 1-\alpha $ , one of the two constraints with equality. We now show that this implies that both hold with equality. To this end, suppose that the first one holds with equality, that is, $\Phi ( c+\delta /\sigma ) -\Phi ( -c) =1-\alpha $ . This is equivalent to $\Phi ( c+\delta /\sigma ) +\Phi ( c) =2-\alpha $ or, equivalently, $\Phi ( c) -\Phi ( -c-\delta /\sigma ) =1-\alpha $ , which is the second equality. Finally, we note that the first constraint with equality is exactly how we define $G(\delta /\sigma )$ , completing the proof.

Proof. Part (a). For any $({\sigma }_l,{\sigma }_u,\rho )\in [\underline {\sigma },\overline {\sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]$ , consider $(\tilde {\delta },{\tilde {\sigma }}_{l},{\tilde {\sigma }}_{u},\tilde {\rho })=(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ .

We first show UHC. Let $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)=(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ and $(c_{M,l},c_{M,u}) \in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ with $\lim _{M\to \infty } (c_{M,l},c_{M,u})= (c_l,c_u)$ . By (A-18), $(c_{M,l},c_{M,u})\in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ implies

(A-66) $$ \begin{align} 1-\alpha~&\leq~P(\{-c_{M,l}\leq z_1\}\cap \{\rho_M z_1\leq c_{M,u}+{\delta_M}/{{\sigma}_{M,u}}+\sqrt{1-\rho_M^2}z_2\})~\leq~1-\Phi(-c_{M,l}) \nonumber \\ 1-\alpha~&\leq~P(\{-c_{M,l}-{\delta_M}/{{\sigma}_{M,l}}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\}\cap\{z_1\leq c_{M,u}\})~\leq~\Phi(c_{M,u}). \end{align} $$

Taking limits in both equations in (A-66) as $M\to \infty $ yields $1-\Phi (-c_l)\geq 1-\alpha $ and $\Phi (c_u)\geq 1-\alpha $ . By (A-18), this is equivalent to $(c_l,c_u)\in C_2(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ , proving UHC.

We now show LHC. Let $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)=(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ and fix $(c_l,c_u)\in C_2(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ . For each $k\in {\mathbb {N}}$ , set ${\tilde {c}}_{k,l}=c_l+1/k$ and ${\tilde {c}}_{k,u}=c_u+1/k$ . Then $\lim _{k\to \infty } ({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})=(c_l,c_u)$ , $1-\Phi (-{\tilde {c}}_{k,l})>1-\alpha $ , and $\Phi ({\tilde {c}}_{k,u})>1-\alpha $ . Moreover, for each fixed $k\in {\mathbb {N}}$ , $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)=(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ and the dominated convergence theorem give

$$ \begin{align*} \begin{aligned} &\lim_{M \to \infty} P\big(\{-{\tilde{c}}_{k,l}\leq z_1\}\cap \big\{\rho_M z_1\leq {\tilde{c}}_{k,u}+{\delta_M}/{{\sigma}_{M,u}} +\sqrt{1-\rho_M^2}z_2\big\}\big)=1-\Phi(-{\tilde{c}}_{k,l})>1-\alpha,\\ &\lim_{M \to \infty} P\big(\big\{-{\tilde{c}}_{k,l}-{\delta_M}/{{\sigma}_{M,l}}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\big\}\cap\{z_1\leq {\tilde{c}}_{k,u}\}\big)\,\,\,\,\,~=~\Phi({\tilde{c}}_{k,u})~>~1-\alpha, \end{aligned} \end{align*} $$

where $(z_1,z_2)\sim {\mathcal {N}}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Hence, for each fixed $k\in \mathbb {N}$ , there exists $M_k\in \mathbb {N}$ such that $({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})\in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M\geq M_k$ . Without loss of generality, we can take $M_k \in \mathbb {N}$ strictly increasing in $k$ . Then, for any $M \in \mathbb {N}$ , proceed as follows. For $M <M_1$ , set $(c_{M,l},c_{M,u})= (\infty ,\infty )$ . For $M \in [M_k,M_{k+1})$ with $k\in \mathbb {N}$ , set $(c_{M,l},c_{M,u})=({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})$ . By construction, this means that $(c_{M,l},c_{M,u}) \in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M \in \mathbb {N}$ . Moreover, since $M_k$ is strictly increasing, the index $k$ corresponding to the interval $M\in [M_k,M_{k+1})$ satisfies $k\to \infty $ as $M\to \infty $ . Therefore, as $\lim _{M\to \infty } (c_{M,l},c_{M,u})=\lim _{k\to \infty } ({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})=(c_l,c_u)$ . Thus, we have constructed a sequence $(c_{M,l},c_{M,u})$ satisfying $(c_{M,l},c_{M,u}) \in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ with $\lim _{M\to \infty } (c_{M,l},c_{M,u})=(c_l,c_u)$ , which proves LHC.

Part (b). For any $(\delta ,\sigma )\in \bar {{\mathbb {R}}}_{+}\times [\underline {\sigma },\overline {\sigma }]$ , consider $(\tilde {\delta },{\tilde {\sigma }}_{l},{\tilde {\sigma }}_{u},\tilde {\rho })=(\delta ,\sigma ,\sigma ,1)$ . If $\delta =\infty $ , the result follows from part (a), so suppose that $\delta <\infty $ .

We first show UHC. Let $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M) = (\delta ,\sigma ,\sigma ,1)$ and $(c_{M,l},c_{M,u})\in \\C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ with $\lim _{M\to \infty } (c_{M,l},c_{M,u})=(c_l,c_u)$ . Since $\delta <\infty $ , we have $\delta _M<\infty $ for all sufficiently large $M$ . By (A-18), $(c_{M,l},c_{M,u})\in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ implies that

(A-67) $$ \begin{align} &P\big(\{-c_{M,l}\leq z_1\}\cap \big\{\rho_M z_1\leq c_{M,u}+{\delta_M}/{{\sigma}_{M,u}}+\sqrt{1-\rho_M^2}z_2\big\}\big)~\geq~1-\alpha, \nonumber \\ &P\big(\big\{-c_{M,l}-{\delta_M}/{{\sigma}_{M,l}}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\big\}\cap\{z_1\leq c_{M,u}\}\big)~\geq~1-\alpha. \end{align} $$

Moreover, $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)=(\delta ,\sigma ,\sigma ,1)$ and the dominated convergence theorem give

(A-68) $$ \begin{align} &\lim_{M \to \infty} P\big(\{-c_{M,l}\leq z_1\}\cap \big\{\rho_M z_1\leq c_{M,u}+{\delta_M}/{{\sigma}_{M,u}} +\sqrt{1-\rho_M^2}z_2\big\}\big)\nonumber\\&\quad =~\Phi(c_u+{\delta}/{\sigma})-\Phi(-c_l), \nonumber \\ &\lim_{M \to \infty} P\big(\big\{-c_{M,l}-{\delta_M}/{{\sigma}_{M,l}}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\big\}\cap\{z_1\leq c_{M,u}\}\big)\nonumber\\&\quad=~\Phi(c_u)-\Phi(-c_l-{\delta}/{\sigma}), \end{align} $$

where $(z_1,z_2)\sim {\mathcal {N}}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Combining (A-67) and (A-68) implies that $\Phi (c_u+{\delta }/{\sigma })-\Phi (-c_l)\geq 1-\alpha $ and $\Phi (c_u)-\Phi (-c_l-{\delta }/{\sigma })\geq 1-\alpha $ . By (A-18), this is equivalent to $(c_l,c_u)\in C_2(\delta ,\sigma ,\sigma ,1)$ , proving UHC.

We now show LHC. Let $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M) = (\delta ,\sigma ,\sigma ,1)$ and fix $(c_l,c_u)\in C_2(\delta ,\sigma ,\sigma ,1)$ . For each $k\in {\mathbb {N}}$ , let ${\tilde {c}}_{k,l}=c_l+1/k$ and ${\tilde {c}}_{k,u}=c_u+1/k$ . Then $\lim _{k\to \infty } ({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})=(c_l,c_u)$ , $\Phi ({\tilde {c}}_{k,u}+\tfrac {\delta }{\sigma })-\Phi (-{\tilde {c}}_{k,l})>1-\alpha $ , and $\Phi ({\tilde {c}}_{k,u})-\Phi (-{\tilde {c}}_{k,l}-\tfrac {\delta }{\sigma })>1-\alpha $ . Moreover, for each fixed $k\in {\mathbb {N}}$ , $\lim _{M\to \infty } (\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)= (\delta ,\sigma ,\sigma ,1)$ and the dominated convergence theorem give

$$ \begin{align*} \begin{aligned} &\lim_{M \to \infty} P\big(\{-{\tilde{c}}_{k,l}\leq z_1\}\cap \big\{\rho_M z_1\leq {\tilde{c}}_{k,u}+\tfrac{\delta_M}{{\sigma}_{M,u}} +\sqrt{1-\rho_M^2}z_2\big\}\big)\nonumber\\&\,\,\,\,\quad=~\Phi({\tilde{c}}_{k,u}+\tfrac{\delta}{\sigma})-\Phi(-{\tilde{c}}_{k,l})~>1-\alpha,\\ &\lim_{M \to \infty} P\big(\big\{-{\tilde{c}}_{k,l}-\tfrac{\delta_M}{{\sigma}_{M,l}}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\big\}\cap\{z_1\leq {\tilde{c}}_{k,u}\}\big)\nonumber\\&\,\,\,\,\quad=~\Phi({\tilde{c}}_{k,u})-\Phi(-{\tilde{c}}_{k,l}-\tfrac{\delta}{\sigma})~>1-\alpha, \end{aligned} \end{align*} $$

where $(z_1,z_2)\sim {\mathcal {N}}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Hence, for each fixed $k\in \mathbb {N}$ , there exists $M_k\in \mathbb {N}$ such that $({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})\in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M\geq M_k$ . Without loss of generality, we can take $M_k \in \mathbb {N}$ strictly increasing in $k$ . Then, proceed as follows for every $M \in \mathbb {N}$ . For $M <M_1$ , set $(c_{M,l},c_{M,u})= (\infty ,\infty )$ . For $M \in [M_k,M_{k+1})$ with $k\in \mathbb {N}$ , set $(c_{M,l},c_{M,u})=({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})$ . By construction, this means that $(c_{M,l},c_{M,u}) \in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M \in \mathbb {N}$ . Moreover, since $M_k$ is strictly increasing, the index $k$ corresponding to the interval $M\in [M_k,M_{k+1})$ satisfies $k\to \infty $ as $M\to \infty $ . Therefore, $\lim _{M\to \infty } (c_{M,l},c_{M,u})= \lim _{M\to \infty } ({\tilde {c}}_{M_k,l},{\tilde {c}}_{M_k,u})=(c_l,c_u)$ . Thus, we have constructed a sequence $(c_{M,l},c_{M,u})\in C_2(\delta _M,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ with $\lim _{M\to \infty } (c_{M,l},c_{M,u})=(c_l,c_u)$ , which proves LHC.

Proof. By (A-18) and (A-24), $C_3(\delta _1,\delta _2,{\sigma }_l,{\sigma }_u,\rho ) = C_2(\delta _1 1[\delta _2>0],{\sigma }_l,{\sigma }_u,\rho )$ for all $(\delta _1,\delta _2,{\sigma }_l,{\sigma }_u,\rho )$ . We divide the rest of the argument into parts.

Part (a). Consider $({\tilde {\delta }}_{1},{\tilde {\delta }}_{2},{\tilde {\sigma }}_{l},{\tilde {\sigma }}_{u},\tilde {\rho })=(\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho )$ with $(\eta ,{\sigma }_l,{\sigma }_u,\rho )\in (\bar {{\mathbb {R}}}\setminus \{0\})\times [\underline {\sigma },\overline {\sigma }]\times [\underline {\sigma },\overline {\sigma }]\times [-1,1]$ . Let $\{(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)\}_{M \in \mathbb {N}}$ be a convergence sequence that satisfies $ (\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M) \to (\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho )$ as $M\to \infty $ .

First, suppose that $\eta>0$ . Then $\delta _{M,2}>0$ for all sufficiently large $M$ , and so $\delta _{M,1}1[\delta _{M,2}>0]=\delta _{M,1}\to \infty $ . Therefore, for all sufficiently large $M$ , $C_3(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)= C_2(\delta _{M,1},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ , and $C_3(\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho )= C_2(\infty ,{\sigma }_l,{\sigma }_u,\rho )$ . Thus, the UHC and LHC of $C_3$ follow from part (a) of Lemma A.7.

Now suppose that $\eta <0$ . Then $\delta _{M,2}\leq 0$ for all sufficiently large $M$ , and so $\delta _{M,1}1[\delta _{M,2}>0]=0$ . Hence, for all sufficiently large $M$ , $C_3(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)=C_2(0,{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ and $C_3(\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho ) =C_2(0,{\sigma }_l,{\sigma }_u,\rho ).$

We first show UHC in this case. Let $(c_{M,l},c_{M,u})\in C_3(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ with $\lim _{M\to \infty } (c_{M,l},c_{M,u}) = (c_l,c_u)$ . By (A-24), membership implies that, for all sufficiently large $M$ ,

(A-69) $$ \begin{align} &P\big(\{-c_{M,l}\leq z_1\}\cap\{\rho_M z_1\leq c_{M,u}+\sqrt{1-\rho_M^2}z_2\}\big)~\geq~1-\alpha, \nonumber \\ &P\big(\{-c_{M,l}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\}\cap\{z_1\leq c_{M,u}\}\big)~\geq~1-\alpha. \end{align} $$

Moreover, the dominated convergence theorem gives

(A-70) $$ \begin{align} &\lim_{M\to\infty}P\big(\{-c_{M,l}\leq z_1\}\cap\{\rho_M z_1\leq c_{M,u}+\sqrt{1-\rho_M^2}z_2\}\big)\nonumber\\&\quad=~P\big(\{-c_l\leq z_1\}\cap\{\rho z_1\leq c_u+\sqrt{1-\rho^2}z_2\}\big), \nonumber \\ &\lim_{M\to\infty}P\big(\{-c_{M,l}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\}\cap\{z_1\leq c_{M,u}\}\big)\nonumber\\&\quad=~P\big(\{-c_l+\sqrt{1-\rho^2}z_2\leq \rho z_1\}\cap\{z_1\leq c_u\}\big), \end{align} $$

where $(z_1,z_2)\sim {\mathcal {N}}({\mathbf {0}}_{2\times 1},{\mathbf {I}}_{2\times 2})$ . Combining (A-69) and (A-70) implies that $(c_l,c_u)\in C_2(0,{\sigma }_l,{\sigma }_u,\rho )$ . Since $C_3(\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho )=C_2(0,{\sigma }_l,{\sigma }_u,\rho )$ , this proves UHC.

We now show LHC in the case with $\eta <0$ . Fix $(c_l,c_u)\in C_3(\infty ,\eta ,{\sigma }_l,{\sigma }_u,\rho )=C_2(0,{\sigma }_l,{\sigma }_u,\rho )$ . For each $k\in {\mathbb {N}}$ , set ${\tilde {c}}_{k,l}=c_l+1/k$ and ${\tilde {c}}_{k,u}=c_u+1/k$ . Then $\lim _{k\to \infty } ({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})=(c_l,c_u)$ , and the two defining inequalities for $C_2(0,{\sigma }_l,{\sigma }_u,\rho )$ hold strictly at $({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})$ . Moreover, for each fixed $k\in {\mathbb {N}}$ , the dominated convergence theorem gives

$$ \begin{align*} &\lim_{M\to\infty}P\big(\{-{\tilde{c}}_{k,l}\leq z_1\}\cap\{\rho_M z_1\leq {\tilde{c}}_{k,u}+\sqrt{1-\rho_M^2}z_2\}\big)\\ &~=~P\big(\{-{\tilde{c}}_{k,l}\leq z_1\}\cap\{\rho z_1\leq {\tilde{c}}_{k,u}+\sqrt{1-\rho^2}z_2\}\big)~>~1-\alpha, \end{align*} $$

and

$$ \begin{align*} &\lim_{M\to\infty}P\big(\{-{\tilde{c}}_{k,l}+\sqrt{1-\rho_M^2}z_2\leq \rho_M z_1\}\cap\{z_1\leq {\tilde{c}}_{k,u}\}\big)\\ &~=~P\big(\{-{\tilde{c}}_{k,l}+\sqrt{1-\rho^2}z_2\leq \rho z_1\}\cap\{z_1\leq {\tilde{c}}_{k,u}\}\big)~>~1-\alpha. \end{align*} $$

Hence, for each fixed $k\in {\mathbb {N}}$ , there exists $M_k\in {\mathbb {N}}$ such that $({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})\in C_3(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M\geq M_k$ . Without loss of generality, we can take $M_k\in {\mathbb {N}}$ strictly increasing in $k$ . For $M<M_1$ , set $(c_{M,l},c_{M,u})=(\infty ,\infty )$ . For $M\in [M_k,M_{k+1})$ with $k\in {\mathbb {N}}$ , set $(c_{M,l},c_{M,u})=({\tilde {c}}_{k,l},{\tilde {c}}_{k,u})$ . By construction, $(c_{M,l},c_{M,u})\in C_3(\delta _{M,1},\delta _{M,2},{\sigma }_{M,l},{\sigma }_{M,u},\rho _M)$ for all $M\in {\mathbb {N}}$ . Moreover, since $M_k$ is strictly increasing, the index $k$ corresponding to the interval $M\in [M_k,M_{k+1})$ satisfies $k\to \infty $ as $M\to \infty $ . Therefore, $\lim_{M \to \infty} (c_{M,l},c_{M,u}) = (c_l,c_u)$ . This proves LHC, completing the proof of part (a).