Appendix A: Proofs of technical results and other important results
Proposition A.1
Suppose that Assumptions 1–4 hold, we have
$$\begin{aligned}{} & {} \frac{1}{T}\left( \sum \limits _{{i}\in {\mathbb {N}}[0,sT]}({Z_{1i}^{\top }(p_{0})Z_{1i}(p_{0}))}/(t_{i+1}-t_{i}), \sum \limits _{{i}\in {\mathbb {N}}[sT,T]}{Z_{2i}^{\top }(p_{0})Z_{2i}(p_{0})}/(t_{i+1}-t_{i})\right) \nonumber \\{} & {} \quad \xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}} \left( s\varvec{{\Sigma }}_{1},(1-s)\varvec{{\Sigma }}_{2}\right) . \end{aligned}$$
(A.1)
Proof
The proof follows from the triangle inequality and Proposition 2.8 along with Proposition 4.1 in Lyu and Nkurunziza (2023). This completes the proof. \(\square \)
Let \(\gamma _{1}(s,T)\) be the smallest eigenvalue of matrix \(\displaystyle \sum \nolimits _{{i}\in {\mathbb {N}}[0,sT]}{Z_{1i}^{\top }(p)Z_{1i}(p)}/((t_{i+1}-t_{i})T)\) and \(\gamma _{2}(s,T)\) be the smallest eigenvalue of matrix \(\sum \nolimits _{{i}\in {\mathbb {N}}[sT,T]}{Z_{2i}^{\top }(p)Z_{2i}(p)}/((t_{i+1}-t_{i})T)\) and let \(\gamma _{k}\) be the smallest eigenvalue of \(\Sigma _{k}\), for \(k=1,2\).
Lemma A.1
If Assumptions 1–4 hold, then \(\gamma _{1}(s,T)\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}s\gamma _{1} \text{ and } \gamma _{2}(s,T)\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}(1-s)\gamma _{2}\).
Proof
From Proposition A.2 of Lyu and Nkurunziza (2023) and Proposition A.1 along with the fact that \(\varvec{{\Sigma }}_{1}\), \(\varvec{{\Sigma }}_{2}\) are positive definite matrices, we get \(\gamma _{1}(s,T)\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}\gamma _{1}(s)=s\gamma _{1}\) with \(\gamma _{1}>0\), and \(\gamma _{2}(s,T)\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}(1-s)\gamma _{2}\), with \(\gamma _{2}>0\). \(\square \)
Let \(\hat{\varvec{{\mu }}}_{j}(p_{*}+)=(\hat{\mu _{j1}},{\hat{\mu }}_{j2},\ldots ,{\hat{\mu }}_{jp_{*}},0_{p_{*}+1},\) \(\ldots ,0_{p},{\hat{\alpha }}_{j})^{\top },\) \(j=1,2\). We derive the following lemma, which is useful in proving that the information criterion \(\textrm{IC}(c,p)\) reaches its minimum value at the exact dimension \((c^{0},p_{0})\).
Lemma A.2
Suppose that Assumption 1–4 hold and suppose that at least one of the parameters, say \(\mu _{ji},~(\mu _{ji}\ne 0),~ p_{*}<i\leqslant p,j=1,2\), cannot be consistently estimated, then for large T,
$$\begin{aligned} \sum \limits _{i\in {\mathbb {N}}[0,sT]}\left( {Z_{1i}(p)}(\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+))\right) ^{2}/((t_{i+1}-t_{i})T)\geqslant s\gamma _{1}\left| \mu _{1i}\right| ^{2}>0,\\ \sum \limits _{i\in {\mathbb {N}}[sT,T]}\left( {Z_{2i}(p)}(\varvec{{\mu }}_{2}(p)-\hat{\varvec{{\mu }}}_{2}(p_{*}+))\right) ^{2}/((t_{i+1}-t_{i})T)\geqslant (1-s)\gamma _{2}\left| \varvec{{\mu }}_{2i}\right| ^{2}>0, \end{aligned}$$
with positive probability.
Proof
For the process on the observed interval [0, sT], if there is at least one parameter was not consistently estimated, say \(\mu _{1i},~(\mu _{1i}\ne 0),~ p_{*}<i\leqslant p\). We have
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,sT]}\left( {Z_{1i}(p)}(\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+))\right) ^{2}/(t_{i+1}-t_{i}) \\{} & {} \quad =(\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+))^{\top }\left[ \frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,sT]}{Z_{1i}^{\top }(p)Z_{1i}(p)}/(t_{i+1}-t_{i})\right] (\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+)) \\{} & {} \quad \geqslant \gamma _{1}(s,T)||\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+)||^{2}\\{} & {} \quad =\gamma _{1}(s,T)\left( \sum \limits _{i=1}^{p_{*}}({\hat{\mu }}_{ji}^{(1)} -\mu _{ji}^{(1)})^{2} +\sum \limits _{i=p_{*}+1}^{p_{0}}(\mu _{ji}^{(1)}-0)^{2}+({\hat{\alpha }}_{1}-\alpha _{1})^{2}\right) . \end{aligned}$$
Then, \(\sum \nolimits _{i\in {\mathbb {N}}[0,sT]}\left( {Z_{1i}(p)}(\varvec{{\mu }}_{1}(p)-\hat{\varvec{{\mu }}}_{1}(p_{*}+))\right) ^{2}/T \geqslant \gamma _{1}(s,T)\left| \mu _{1i}\right| ^{2},~~p_{*}\leqslant i\leqslant p_{0}\). By the proof of Lemma A.1, \(\gamma _{1}(s,T)\left| \mu _{1i}\right| ^{2}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{P}s\gamma _{1}\left| \mu _{1i}\right| ^{2}>0.\) By using the same techniques, we prove the second inequality. This completes the proof. \(\square \)
Before presenting this important result, we derive first the following propositions and lemmas, which play a crucial role in establishing this result. Note that Lemmas A.1 and A.2 are based on the exact rate of the change-point \(s=s^{0}\), while Propositions A.3–A.6 are based on \({\hat{s}}\) which is an estimator of s. Next, to emphasize the parameter dimension p in our notations, let W(s, T, p) be the vector W(s, T) with dimension p and Q(s, T, p) be the matrix Q(s, T, p) with size \(2(p+1)\times 2(p+1)\). Let \(W_{[0,sT]}(p)\) be the vector \(W_{[0,sT]}(s)\) with dimension p and \(Q_{[0,sT]}(p)\) be the matrix \(Q_{[0,sT]}\) with size \((p+1)\times (p+1)\). We derive the following result which is crucial in proving the consistency of \({\hat{s}}\).
Proposition A.2
If Assumptions 1–4 hold, then, (i) \(\left| \left| W(s,T,p)\right| \right| =O_{\mathrm{{P}}}(\sqrt{T}\textrm{log}^{a^{*}}(T))\) and (ii)
$$\begin{aligned} \left( \left| \left| \sum \limits _{i\in {\mathbb {N}}[0,sT]}\varepsilon _{i}{Z_{1i}(p)}/(t_{i+1}-t_{i})\right| \right| , \left| \left| \sum \limits _{i\in {\mathbb {N}}[sT,T]}\!\!\varepsilon _{i}{Z_{2i}(p)}/(t_{i+1}-t_{i})\right| \right| \right) \!=\!O_{\mathrm{{P}}}(\sqrt{T}\,\textrm{log}^{\textit{a}^{*}}(\textit{T})),\nonumber \\ \end{aligned}$$
(A.2)
for some \(0<a^{*}<{\varvec{a}}/2\).
Proof
The proof Part (i) is similar to that Proposition 4.3 in Lyu and Nkurunziza (2023) and the proof of Part (ii) is similar to that of Proposition 4.4 in Lyu and Nkurunziza (2023). \(\square \)
Proof of Proposition 3.1
Let \(\textrm{SSE}_{1}=\displaystyle \textrm{SSE}({\hat{s}},{\hat{p}},\hat{\varvec{{\theta }}})/N\), and \(\textrm{SSE}_{2}=\displaystyle \textrm{SSE}(s^{0},p_{0},\hat{\varvec{{\theta }}}_{0})/N\), where \(\hat{\varvec{{\theta }}}\) is the estimator based on the estimation of the change point and \(\hat{\varvec{{\theta }}}_{0}\) is the estimator based on the true change point, denoted by \(s^{0}\). Since we have \(\textrm{SSE}_{1}\leqslant \textrm{SSE}_{2}\) with probability 1, it remains to show that if the change point is not consistently estimated, \(\textrm{SSE}_{1}> \textrm{SSE}_{2}\) with positive probability yielding a contradiction. Indeed, if the change point is not consistently estimated, we have \(|{\hat{s}}T-s^{0}T|>\eta T\), for some constant \(0<\eta <1\). Without loss of generality, we assume that \(0<s^{0}T<{\hat{s}}T<T\). Let \({\hat{Y}}_{i}(p,s)\) be the predicted value of \(Y_{i}\) based on the parameter p and s and \(\hat{\varvec{{\theta }}}(p,s)\) be the estimator of \(\varvec{\theta }\) based on the parameter p and s. Then,
$$\begin{aligned}{} & {} \textrm{SSE}_{1}-\textrm{SSE}_{2} =\left( \frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}({\hat{p}},{\hat{s}})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}\right. \nonumber \\{} & {} \quad \left. -\frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}(p_{0},s^{0})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}\right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} \end{aligned}$$
(A.3)
$$\begin{aligned}{} & {} \quad +\left( \frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}({\hat{p}},{\hat{s}})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}-\frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}(p_{0},s^{0})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\nonumber \\ \end{aligned}$$
(A.4)
$$\begin{aligned}{} & {} \quad +\left( \frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}({\hat{p}},{\hat{s}})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}-\frac{1}{N}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{\hat{Y}}_{i}(p_{0},s^{0})}{\Delta _{N}}-\frac{Y_{i}}{\Delta _{N}}\right) ^{2}\right) {\mathbb {I}}_{\{{\hat{p}}<p_{0}\}}.\nonumber \\ \end{aligned}$$
(A.5)
Note that \(T=N\Delta _{N},Y_{i}=Z_{i}(p_{0})\varvec{{\theta }}(p_{0})+\varepsilon _{i}\), we have
$$\begin{aligned}{} & {} (A.3) =(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+))^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{{\mathcal {Z}}_{i}^{\top }({\hat{p}}){\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}}(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} \\{} & {} \quad -(\hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0}))^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{{\mathcal {Z}}_{i}^{\top }(p_{0}){\mathcal {Z}}_{i}(p_{0})}}{\Delta _{N}}(\hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})){\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} \\{} & {} \quad -\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+))\right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}\\{} & {} \quad +\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \varepsilon _{i}{{\mathcal {Z}}_{i}(p_{0})}(\hat{\varvec{{\theta }}}(p_{0},s_{0}) -\varvec{{\theta }}(p_{0}))\right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}. \end{aligned}$$
Since s is not consistently estimated, and from \(\left( {{\mathcal {Z}}_{i}({\hat{p}})}(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) ^{2}{\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}\geqslant 0\), we have
$$\begin{aligned} \frac{1}{T}\left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) ^{\top }\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{{\mathcal {Z}}_{i}^{\top }({\hat{p}}){\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}}\left( \hat{\varvec{{\theta }}} ({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} \end{aligned}$$
$$\begin{aligned} \begin{array}{ll} \leqslant \eta \left( \left( \varvec{{\mu }}_{1}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) ^{\top }\displaystyle \frac{1}{\eta T}\sum _{{i}\in {\mathbb {N}}[s^{0}T-\eta T,s^{0}T]}{\frac{{Z}_{1i}^{\top }({\hat{p}}){Z}_{1i}({\hat{p}})}{\Delta _{N}}} \left( \varvec{{\mu }}_{1}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) \right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}\\ +\eta \left( \left( \varvec{{\mu }}_{2}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}})\right) ^{\top }\displaystyle \frac{1}{\eta T}\sum _{{i}\in {\mathbb {N}}[s^{0}T,s^{0}T+\eta T]}{\frac{{Z}_{2i}^{\top }({\hat{p}}){Z}_{2i}({\hat{p}})}{\Delta _{N}}}\left( \varvec{{\mu }}_{2}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) \right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}. \end{array} \end{aligned}$$
(A.6)
Let \(\gamma _{11}(T), \gamma _{12}(T)\) be the smallest eigenvalues of the matrices \(\displaystyle \sum \limits _{{i}\in {\mathbb {N}}[s^{0}T-\eta T,s^{0}T]}{{Z}_{1i}^{\top }({\hat{p}}){Z}_{1i}({\hat{p}})}/(\Delta _{N}\eta T)\) and \(\displaystyle \sum \limits _{{i}\in {\mathbb {N}}[s^{0}T,s^{0}T+\eta T]}{{Z}_{2i}^{\top }({\hat{p}}){Z}_{2i}({\hat{p}})}/(\Delta _{N}\eta T)\), respectively, while \({\hat{p}}>p_{0}\). Then,
$$\begin{aligned} (A.6)\geqslant & {} \eta \min (\gamma _{11}(T),\gamma _{12}(T))\\{} & {} \left( \left| \left| \varvec{{\mu }}_{1}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2} +\left| \left| \varvec{{\mu }}_{2}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}\right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}. \end{aligned}$$
Using the convexity of a quadratic function, we have \(\left| \left| \varvec{{\mu }}_{1}(p_{0}+)-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}+\left| \left| \varvec{{\mu }}_{2}(p_{0}+) -\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}\geqslant \left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}\big /2\). Hence,
$$\begin{aligned}{} & {} \frac{\Delta _{N}}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{{\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}} (\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+))\right) ^{2}\\{} & {} \qquad \geqslant \eta \frac{\min (\gamma _{11}(T),\gamma _{12}(T))}{2} \left| \left| \varvec{{\theta }}_{1}(p_{0})-\varvec{{\theta }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}. \end{aligned}$$
By Proposition A.7, \(\gamma _{11}(T)\) and \(\gamma _{12}(T)\) are both bounded away from 0 and \(\eta \min \{\gamma _{11}(T),\gamma _{12}(T)\}\) is also bounded away from 0. Therefore, the right-hand side of the inequality \(\displaystyle \eta \min \{\gamma _{11}(T),\gamma _{12}(T)\}/2\) \(\left| \left| \varvec{{\mu }}_{1}(p_{0}) -\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}\) is strictly positive. Further, since \(\hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\xrightarrow [T\rightarrow \infty ]{a.s.}0\), we have
$$\begin{aligned} -\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) ^{\top }\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{{\mathcal {Z}}_{i}^{\top }(p_{0}){\mathcal {Z}}_{i}(p_{0})}}{\Delta _{N}}\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) \xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}] {\mathrm{{P}}}0. \end{aligned}$$
Propositions A.2 and A.6 imply that \(\frac{2}{T}\left| \left| \sum \limits _{i\in {\mathbb {N}}[0,T]}\varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}\right| \right| {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} =o_{\mathrm{{P}}}(1)\) and
\(\frac{2}{T}\left| \left| \sum \limits _{i\in {\mathbb {N}}[0, T]}\varepsilon _{i}{{\mathcal {Z}}_{i}(p_{0})}\right| \right| {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}=o_{\mathrm{{P}}}(1)\). So, we get
$$\begin{aligned}{} & {} -\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}\left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) \right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}\\{} & {} \quad +\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \varepsilon _{i}{{\mathcal {Z}}_{i}(p_{0})}\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) \right) {\mathbb {I}}_{\{{\hat{p}}>p_{0}\}}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}0. \end{aligned}$$
This is for large T,
$$\begin{aligned} (A.3)\geqslant C_{1}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} \end{aligned}$$
(A.7)
with positive probability, where \(C_{1}= \lim \nolimits _{T\rightarrow \infty }\displaystyle \min (s\gamma _{11}(T),(1-s)\gamma _{12}(T))/2>0\). Further, we have
$$\begin{aligned}{} & {} (A.4) =\left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) ^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\frac{{{\mathcal {Z}}_{i}^{\top }({\hat{p}}){\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}}\left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}} \\{} & {} \qquad -\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) ^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\frac{{{\mathcal {Z}}_{i}^{\top }(p_{0}){\mathcal {Z}}_{i}(p_{0})}}{\Delta _{N}}\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}} \\{} & {} \qquad -\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}\left( \hat{\varvec{{\theta }}} ({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) \right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\\{} & {} \qquad +\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]} \left( \varepsilon _{i}{{\mathcal {Z}}_{i}(p_{0})}\left( \hat{\varvec{{\theta }}}(p_{0},s_{0})-\varvec{{\theta }}(p_{0})\right) \right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}. \end{aligned}$$
Since s is not consistently estimated, and from \(\left( {{\mathcal {Z}}_{i}({\hat{p}})}({\hat{\theta }}\left( {\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0})\right) \right) ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\geqslant 0\), we have
$$\begin{aligned} \left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0})\right) ^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{{\mathcal {Z}}_{i}^{\top }({\hat{p}}){\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}}\left( \hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}+)\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}} \end{aligned}$$
$$\begin{aligned} \begin{array}{ll} \geqslant \eta \left( \left( \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) ^{\top }\displaystyle \frac{1}{\eta T}\sum _{{i}\in {\mathbb {N}}[s^{0}T-\eta T,s^{0}T]}{\frac{{Z}_{1i}^{\top }({\hat{p}}){Z}_{1i}({\hat{p}})}{\Delta _{N}}} \left( \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) \right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\\ ~+\eta \left( \left( \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) ^{\top }\displaystyle \frac{1}{\eta T}\sum _{{i}\in {\mathbb {N}}[s^{0}T,s^{0}T+\eta T]}{\frac{{Z}_{2i}^{\top }({\hat{p}}){Z}_{2i}({\hat{p}})}{\Delta _{N}}}\left( \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right) \right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}. \end{array} \end{aligned}$$
(A.8)
Let \(\gamma _{21}(T), \gamma _{22}(T)\) be the smallest eigenvalues of the matrices \(\sum \limits _{{i}\in {\mathbb {N}}[s^{0}T-\eta T,s^{0}T]}{{Z}_{1i}^{\top }(p_{0}){Z}_{1i}(p_{0})}/(\Delta _{N}\eta T)\) and \(\sum \limits _{{i}\in {\mathbb {N}}[s^{0}T,s^{0}T+\eta T]}{{Z}_{2i}^{\top }(p_{0}){Z}_{2i}(p_{0})}/(\Delta _{N}\eta T)\), respectively. Then,
$$\begin{aligned} (A.8)&\geqslant \eta \gamma _{21}(T)\left| \left| \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2} +\eta \gamma _{22}(T)\left| \left| \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\\&\geqslant \eta \min (\gamma _{21}(T),\gamma _{22}(T))\left( \left| \left| \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2} +\left| \left| \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}. \end{aligned}$$
Using the convexity of a quadratic function, we have
$$\begin{aligned} \left| \left| \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2}+ \left| \left| \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{1}({\hat{p}},{\hat{s}})\right| \right| ^{2} \geqslant \left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}\big /2. \end{aligned}$$
Hence,
$$\begin{aligned}{} & {} \frac{\Delta _{N}}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left( \frac{{{\mathcal {Z}}_{i}({\hat{p}})}}{\Delta _{N}}(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}})-\varvec{{\theta }}(p_{0}))\right) ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\\{} & {} \geqslant \frac{\eta \min (\gamma _{21}(T),\gamma _{22}(T))}{2}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}. \end{aligned}$$
By Proposition A.7, \(\gamma _{21}(T)\) and \(\gamma _{22}(T)\) are both bounded away from 0 and \(\eta \min (\gamma _{21}(T),\gamma _{22}(T))\) is also bounded away from 0. Therefore, the right-hand side of the inequality \(\displaystyle \eta \min \left( \gamma _{21}(T),\gamma _{22}(T)\right) /2\)
\(\left| \left| \varvec{{\mu }}_{1}(p_{0}) -\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}\) is positive. Similarly, we have
$$\begin{aligned}{} & {} ({\hat{\theta }}({\hat{p}},s^{0})-\varvec{{\theta }}(p_{0})^{\top }\frac{1}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]} \frac{{Z_{i}^{\top }({\hat{p}})Z_{i}({\hat{p}})}}{\Delta _{N}}({\hat{\theta }}({\hat{p}},s^{0})-\varvec{{\theta }}(p_{0}) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}0, \\{} & {} \quad -\frac{2}{T}\sum \limits _{i\in {\mathbb {N}}[0,T]}\left[ \left( \varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}(\hat{\varvec{{\theta }}}({\hat{p}},{\hat{s}}) -\varvec{{\theta }}(p_{0}))\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\right. \\{} & {} \left. +\left( \varepsilon _{i}{{\mathcal {Z}}_{i}({\hat{p}})}({\hat{\theta }}({\hat{p}},s^{0}) -\varvec{{\theta }}(p_{0}))\right) {\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\right] \xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}0, \end{aligned}$$
which implies that for large T,
$$\begin{aligned} (A.4)\geqslant C_{2}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}. \end{aligned}$$
(A.9)
with a positive probability, where \(C_{2}= \lim \limits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\eta \min (\gamma _{12}(T),\gamma _{22}(T))/2>0\). Similarly to the proof of (A.7) and (A.9), for large T, we have
$$\begin{aligned} (A.5)\geqslant C_{3}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}<p_{0}\}} \end{aligned}$$
(A.10)
with positive probability, where \(C_{3}= \lim \nolimits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\displaystyle \eta \min (\gamma _{31}(T),\gamma _{32}(T))/2>0\), and \(\gamma _{31}(T),\gamma _{32}(T)\) are the smallest eigenvalues of the matrices \(\displaystyle \sum \nolimits _{{i}\in {\mathbb {N}}[s^{0}T-\eta T,s^{0}T]}{Z_{1i}^{\top }({\hat{p}}){Z}_{1i}({\hat{p}})}/(\eta T\Delta _{N})\) and
\(\displaystyle \sum \nolimits _{{i}\in {\mathbb {N}}[s^{0}T,s^{0}T+\eta T]}{{Z}_{2i}^{\top }({\hat{p}}){Z}_{2i}({\hat{p}})}/(\eta T\Delta _{N})\), respectively. Finally, for large T, from (A.7), (A.9) and (A.10) we get, \(\textrm{SSE}_{1}-\textrm{SSE}_{2}\geqslant C_{1}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}>p_{0}\}} +C_{2}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}=p_{0}\}}\)
$$\begin{aligned} +C_{3}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}{\mathbb {I}}_{\{{\hat{p}}<p_{0}\}} \geqslant \min \{C_{1},C_{2},C_{3}\}\left| \left| \varvec{{\mu }}_{1}(p_{0})-\varvec{{\mu }}_{2}(p_{0})\right| \right| ^{2}>0, \end{aligned}$$
with a positive probability. This completes the proof. \(\square \)
Proposition A.3
Suppose that Assumptions 1–4 hold. Then,
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}/(t_{i+1}-t_{i})\nonumber \\{} & {} \qquad -\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}{Z_{1i}^{\top }(p_{0})Z_{1i}(p_{0})}/t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}\varvec{{0}}, \end{aligned}$$
(A.11)
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [{\hat{s}}T,T]}{{\mathcal {Z}}_{2i}^{\top }(p_{0}){\mathcal {Z}}_{2i}(p_{0})}/(t_{i+1}-t_{i})\nonumber \\{} & {} \qquad -\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,T]}{Z_{2i}^{\top }(p_{0})Z_{2i}(p_{0})}/(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}\varvec{{0}}. \end{aligned}$$
(A.12)
Proof
By the definition of \(Z_{i}(p_{0})\), we have
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{{i}\in {\mathbb {N}}[0,{\hat{s}}T]}\frac{{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}}{t_{i+1}-t_{i}}\\{} & {} \quad =\frac{1}{T}\begin{bmatrix} \sum \limits _{{i}\in {\mathbb {N}}[0,{\hat{s}}T]}\varvec{{b}}^{\top }(i)b(i)(t_{i+1}-t_{i})&{}~-\sum \limits _{{i}\in {\mathbb {N}}[0,{\hat{s}}T]}\varvec{{b}}^{\top }(i)({\ln X(t_{i})})(t_{i+1}-t_{i})\\ -\sum \limits _{{i}\in {\mathbb {N}}[0,{\hat{s}}T]}({\ln X(t_{i})})b(i)(t_{i+1}-t_{i})&{}\sum \limits _{{i}\in {\mathbb {N}}[0,{\hat{s}}T]}({\ln X(t_{i})})^{2}(t_{i+1}-t_{i}) \end{bmatrix}, \end{aligned}$$
Note that \({\ln X(t_{i})} ={\left\{ \begin{array}{ll} \ln X_{1}(t_{i}), &{}~~\textrm{if}~~ i\in {\mathbb {N}}[0,{\hat{s}}T]~~~\textrm{and}~~~{\hat{s}}<s\\ \ln X_{1}(t_{i}){\mathbb {I}}_{\{i\in {\mathbb {N}}[0,{s}T]\}}+\ln X_{2}(t_{i}){\mathbb {I}}_{\{i\in {\mathbb {N}}[{s}T,{\hat{s}}T]\}},&{} ~~\textrm{if}~~ i\in {\mathbb {N}}[0,{\hat{s}}T]~~~\textrm{and}~~~{\hat{s}}>s \end{array}\right. }\). First, since \({\hat{s}}\) is a consistent estimator of s, \(\forall ~\varepsilon >0\), \(\forall ~0<\delta <s/2\), we have
$$\begin{aligned} \textrm{P}\left( |{\hat{s}}-s|>\delta \right) <\varepsilon . \end{aligned}$$
(A.13)
for sufficiently large T. Then, \({\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}\right] \)
$$\begin{aligned}{} & {} ={\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,{\hat{s}}T]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s>0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \end{aligned}$$
(A.14)
$$\begin{aligned}{} & {} +{\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [{\hat{s}}T, sT]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s\leqslant 0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \end{aligned}$$
(A.15)
$$\begin{aligned}{} & {} +{\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,{\hat{s}}T]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s>0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \delta \}}\right] \end{aligned}$$
(A.16)
$$\begin{aligned}{} & {} +{\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [{\hat{s}}T, sT]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s\leqslant 0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \delta \}}\right] . \end{aligned}$$
(A.17)
$$\begin{aligned} (A.14)\leqslant {\mathbb {E}}\left[ \left( \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,{\hat{s}}T]}\left| \left| \varvec{{b}}^{\top }(i){\varvec{{b}}(i)}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s>0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \\ \leqslant (pK_{b})^{m/2}{\mathbb {E}}\left[ \left( {\hat{s}}-s\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}-s>0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \leqslant (pK_{b})^{m/2}\left( \delta \right) ^{m/2}. \end{aligned}$$
Similarly, we get (A.15)\(\leqslant (pK_{b})^{m/2}\left( \delta \right) ^{m/2}.\) Further, by Cauchy–Schwartz inequality,
$$\begin{aligned} (A.16)\leqslant \left\{ {\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,{\hat{s}}T]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m}\right] {\mathbb {E}}\left[ {\mathbb {I}}_{\{{\hat{s}}-s>0\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \delta \}} \right] \right\} ^{1/2} \\ \leqslant (pK_{b})^{m/2}\left\{ {\mathbb {E}}\left[ \left( {\hat{s}}-s\right) ^{m}\right] \textrm{P}\left( |{\hat{s}}-s|\geqslant \delta \right) \right\} ^{1/2} <(pK_{b})^{m/2}2^{m/2}\sqrt{\varepsilon }, \end{aligned}$$
where we used the fact that \(|{\hat{s}}-s|\leqslant 2\) a.s. Following the same techniques, we get (A.17)\(<(pK_{b})^{m/2}2^{m/2}\sqrt{\varepsilon }\). This implies that
$$\begin{aligned} \begin{array}{ll} {\mathbb {E}}\left[ \left| \left| \displaystyle \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}\varvec{{b}}^{\top }(i){\varvec{{b}}(i)}(t_{i+1}-t_{i})\right| \right| ^{m/2}\right] \\ <2(pK_{b})^{m/2}\left( \delta \right) ^{m/2}+2(pK_{b})^{m/2}2^{m/2}\sqrt{\varepsilon }=2(pK_{b})^{m/2}\left( \left( \delta \right) ^{m/2}+\sqrt{\varepsilon }\right) . \end{array} \end{aligned}$$
(A.18)
Second, from (A.13),
$$\begin{aligned} {\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i})){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\right| \right| ^{m/2}\right] \nonumber \\ \end{aligned}$$
$$\begin{aligned}{} & {} ={\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i})){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|>\delta \}}\right] \nonumber \\ \end{aligned}$$
(A.19)
$$\begin{aligned}{} & {} +{\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i})){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] . \nonumber \\ \end{aligned}$$
(A.20)
Next, we have
$$\begin{aligned} (A.19) \leqslant {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,T]}\left| \left| (\ln X_{2}(t_{i}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}>s+\delta \}}\right] \\ +{\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}\left| \left| (\ln X_{1}(t_{i}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}<s-\delta \}}\right] . \end{aligned}$$
Further, by Cauchy–Schwartz inequality,
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,T]}\left| \left| (\ln X_{2}(t_{i}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}>s+\delta \}}\right] \\{} & {} \quad \leqslant \frac{1}{T^{m/2}}\left\{ {\mathbb {E}}\left[ \left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left| \left| (\ln X_{2}(t_{i}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m}\right] \textrm{P}\left( {\hat{s}}>s+\delta \right) \right\} ^{1/2} \\{} & {} \quad \leqslant \frac{1}{T^{m/2}}(p_{0}K_{b})^{m/2}\left\{ ((1-s)T)^{m-1}\left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}{\mathbb {E}}\left[ \left| \ln X_{2}(t_{i})\right| ^{m}\right] (t_{i+1}-t_{i})\right) \textrm{P}\left( {\hat{s}}>s+\delta \right) \right\} ^{1/2} \\{} & {} \quad \leqslant \left( \frac{p_{0}K_{b}}{T}\right) ^{m/2}((1-s)T)^{m/2}\sqrt{\sup \limits _{t\geqslant 0}{\mathbb {E}}\left[ \left| \ln X(t)\right| ^{m}\right] }\left\{ \textrm{P}\left( {\hat{s}}>s+\delta \right) \right\} ^{1/2} \\ {}{} & {} \quad = (p_{0}K_{b}(1-s))^{m/2}\sqrt{\sup \limits _{t\geqslant 0}{\mathbb {E}}\left[ \left| \ln X(t)\right| ^{m}\right] }\sqrt{\varepsilon }. \end{aligned}$$
Similarly, we get
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}\left| \left| (\ln X_{1}(t_{i})){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{{\hat{s}}<s-\delta \}}\right] \\{} & {} <(p_{0}K_{b}s)^{m/2}\sqrt{\sup \limits _{t\geqslant 0}{\mathbb {E}}\left[ \left| \ln X(t)\right| ^{m}\right] }\sqrt{\varepsilon }. \end{aligned}$$
This implies that (A.19)\(<2(p_{0}K_{b}(1-s))^{m/2}\sqrt{\sup \limits _{t\geqslant 0}{\mathbb {E}}\left[ \left| \ln X(t)\right| ^{m}\right] }\sqrt{\varepsilon }.\) For (A.20), we have
$$\begin{aligned} {\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X(t_{i})){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \\ \leqslant {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,sT]}\left| \left| (\ln X_{1}(t_{i}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{s-\delta<{\hat{s}}<s\}}\right] \\ +{\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,(s+\delta )T]}\left| \left| (\ln X_{2}(i){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{s<{\hat{s}}<s+\delta \}}\right] \\ \leqslant 2{\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,(s+\delta )T]}\left| \left| ({\ln X(t_{i})}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m/2}{\mathbb {I}}_{\{s-\delta<{\hat{s}}<s+\delta \}}\right] . \end{aligned}$$
By Cauchy–Schwartz inequality and the fact that \(\textrm{P}(s-\delta<{\hat{s}}<s+\delta )\leqslant 1\),
$$\begin{aligned} {\mathbb {E}}\left[ \left| \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X(t_{i})){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\right| \right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \delta \}}\right] \\ \leqslant 2\left\{ {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,(s+\delta )T]}\left| \left| ({\ln X(t_{i})}){\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m}\right] \right\} ^{1/2}. \end{aligned}$$
Furthermore, by Jensen’s inequality and Proposition 2.2, we get
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left( \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,(s+\delta )T]}(|{\ln X(t_{i})}|)\left| \left| {\varvec{{b}}(t_{i})}\right| \right| (t_{i+1}-t_{i})\right) ^{m}\right] \\{} & {} \quad \leqslant (2\delta )^{m}K_{b}^{m}\sup \limits _{t\geqslant 0}{\mathbb {E}}[|\ln X(t)|^{m}]\frac{1}{2\delta T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,(s+\delta )T]}(t_{i+1}-t_{i}) \\{} & {} \quad = (2\delta )^{m}K_{b}^{m}\sup \limits _{t\geqslant 0}{\mathbb {E}}[|\ln X(t)|^{m}]. \end{aligned}$$
Since \(\delta \) and \(\varepsilon \) can be arbitrary small, we have
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\nonumber \\{} & {} \quad -\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}({\ln X_{1}(t_{i})}){\varvec{{b}}(t_{i})}(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}\varvec{{0}}_{1\times P_{0}}. \end{aligned}$$
(A.21)
Finally, for the last term, we have
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i}}))^{2}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}\right] \nonumber \\{} & {} \quad ={\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})})^{2}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \delta \}}\right] \nonumber \\ \end{aligned}$$
(A.22)
$$\begin{aligned}{} & {} \qquad +{\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}({\ln X(t_{i})}))^{2}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i})^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|<\delta \}}\right] . \nonumber \\ \nonumber \\ \end{aligned}$$
(A.23)
Further,
$$\begin{aligned} (A.22) \leqslant&{\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,T]}({\ln X_{2}(t_{i})})^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\&+{\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{1}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}<s- \delta \}}\right] . \end{aligned}$$
Then, from Jensen’s inequality,
$$\begin{aligned} (A.22)\leqslant&\frac{1}{T}{\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}|{\ln X_{2}(t_{i})}|^{m}(t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\&+\frac{1}{T}{\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [0,sT]}|{\ln X_{2}(t_{i})}|^{m}(t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}<s- \delta \}}\right] . \end{aligned}$$
Then, from (2.5), we get
$$\begin{aligned} \begin{array}{ll} (A.22)&{}\leqslant 3^{m-1}{\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |e^{-\alpha _{2} (t_{i}-\phi )}\ln {X_{0}^{\phi }}|^{m}+|r_{2}^{\phi }(t_{i}-\phi )|^{m}+|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\ &{}~+ 3^{m-1}{\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [0,sT]}\left( |e^{-\alpha _{2} (t_{i}-\phi )}\ln {X_{0}^{\phi }}|^{m}+|r_{2}^{\phi }(t_{i}-\phi )|^{m}+|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}<s- \delta \}}\right] . \end{array}\nonumber \\ \end{aligned}$$
(A.24)
First, by Assumption 2, we get
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |e^{-\alpha _{2} (t_{i}-\phi )}\ln {X_{0}^{\phi }}|^{m}+|r_{2}^{\phi }(t_{i}-\phi )|^{m}+|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\{} & {} \quad ={\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] \left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |e^{-m\alpha _{2} (t_{i}-\phi )}|\right) (t_{i+1}-t_{i})\right) \textrm{P}{\{{\hat{s}}>s+ \delta \}} \\{} & {} \quad +\left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |r_{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i})\right) \textrm{P}{\{{\hat{s}}>s+ \delta \}}\\{} & {} \quad +{\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+\delta \}}\right] . \end{aligned}$$
Since \({\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] <\infty \), we get
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] \left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |e^{-m\alpha _{2} (t_{i}-\phi )}|\right) (t_{i+1}-t_{i})\right) \textrm{P}{\{{\hat{s}}>s+ \delta \}}\nonumber \\{} & {} \quad < {\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] (1-s)T\textrm{P}{\{{\hat{s}}>s+ \delta \}}. \end{aligned}$$
(A.25)
We also get
$$\begin{aligned} \left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |r_{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i})\right) \textrm{P}{\{{\hat{s}}>s+ \delta \}} \leqslant \left( \displaystyle \frac{K_{\theta } K_{b}+\frac{1}{2}\sigma ^{2}}{\alpha _{2}}\right) ^{m}\! (1-s)T\textrm{P}{\{{\hat{s}}>s+ \delta \}}.\qquad \end{aligned}$$
(A.26)
From Cauchy–Schwartz inequality and Jensen’s inequality, we get
$$\begin{aligned} \begin{array}{ll} {\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\ \quad \leqslant \left\{ {\mathbb {E}}\left[ \left( \sum \limits _{i\in {\mathbb {N}} [sT,T]}|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}(t_{i+1}-t_{i})\right) ^{2}\right] \textrm{P}{\{{\hat{s}}>s+ \delta \}}\right\} ^{1/2}\\ \quad \leqslant (1-s)T\sigma ^{m}\left( \displaystyle \frac{1}{2\alpha _{2}}\right) ^{m/2}\sqrt{C_{m}\textrm{P}{\{{\hat{s}}>s+ \delta \}}}. \end{array} \end{aligned}$$
(A.27)
(A.25), (A.26), (A.27), and (A.13) imply that
$$\begin{aligned} \begin{array}{ll} {\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [sT,T]}\left( |e^{-\alpha _{2} (t_{i}-\phi )}\ln {X_{0}^{\phi }}|^{m}+|r_{2}^{\phi }(t_{i}-\phi )|^{m}+|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}>s+ \delta \}}\right] \\ \leqslant \left( {\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] +\left( \displaystyle \frac{K_{\theta } K_{b}+\frac{1}{2}\sigma ^{2}}{\alpha _{2}}\right) ^{m}\right) \varepsilon +\sigma ^{m}\left( \displaystyle \frac{1}{2\alpha _{2}}\right) ^{m/2}\sqrt{C_{m}\varepsilon }. \end{array}\nonumber \\ \end{aligned}$$
(A.28)
Similar to the proof of (A.25), (A.26), and (A.27), we get
$$\begin{aligned} \begin{array}{ll} {\mathbb {E}}\left[ \sum \limits _{i\in {\mathbb {N}} [0,sT]}\left( |e^{-\alpha _{2} (t_{i}-\phi )}\ln {X_{0}^{\phi }}|^{m}+|r_{2}^{\phi }(t_{i}-\phi )|^{m}+|\tau _{2}^{\phi }(t_{i}-\phi )|^{m}\right) (t_{i+1}-t_{i}){\mathbb {I}}_{\{{\hat{s}}<s- \delta \}}\right] \\ \quad \leqslant \left( {\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] +\left( \displaystyle \frac{K_{\theta } K_{b}+\frac{1}{2}\sigma ^{2}}{\alpha _{2}}\right) ^{m}\right) \varepsilon +\sigma ^{m}\left( \displaystyle \frac{1}{2\alpha _{2}}\right) ^{m/2}\sqrt{C_{m}\varepsilon }. \end{array}\nonumber \\ \end{aligned}$$
(A.29)
(A.24), (A.28) and (A.29) imply that
$$\begin{aligned} (A.22)\leqslant 2\left( \left( {\mathbb {E}}\left[ |\ln {X_{0}^{\phi }}|^{m}\right] +\left( \displaystyle \frac{K_{\theta } K_{b}+\frac{1}{2}\sigma ^{2}}{\alpha _{2}}\right) ^{m}\right) \varepsilon +\sigma ^{m}\left( \displaystyle \frac{1}{2\alpha _{2}}\right) ^{m/2}\sqrt{C_{m}\varepsilon }\right) .\nonumber \\ \end{aligned}$$
(A.30)
Following the same technique, we have
$$\begin{aligned} (A.23)&\leqslant {\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,sT]}(\ln X_{1}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{s-\delta<{\hat{s}}<s\}}\right] \\&\quad +{\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,(s+\delta )T]}(\ln X_{2}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}{\mathbb {I}}_{\{s<{\hat{s}}<s+\delta \}}\right] \\&\qquad \leqslant {\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,sT]}(\ln X_{1}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}\right] \\&\quad +{\mathbb {E}}\left[ \left| \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [sT,(s+\delta )T]}(\ln X_{2}(t_{i}))^{2}(t_{i+1}-t_{i})\right| ^{m/2}\right] . \end{aligned}$$
From Jensen’s inequality, this gives
$$\begin{aligned} (A.23){} & {} \leqslant (\delta )^{m/2}\frac{1}{\delta T}\sum \limits _{i\in {\mathbb {N}} [(s-\delta )T,sT]}{\mathbb {E}}\left[ |\ln X_{1}(t_{i})|^{m}\right] (t_{i+1}-t_{i})+(\delta )^{m/2}\frac{1}{\delta T}\\{} & {} \sum \limits _{i\in {\mathbb {N}} [sT,(s+\delta )T]}{\mathbb {E}}\left[ |\ln X_{2}(t_{i})|^{m}\right] (t_{i+1}-t_{i}). \end{aligned}$$
Proposition 2.2 gives that
$$\begin{aligned} (A.23)\leqslant \sup \limits _{t\geqslant 0}{\mathbb {E}}[|\ln X(t)|^{m}]\left( \delta \right) ^{m}. \end{aligned}$$
(A.31)
Since \(\delta \) and \(\varepsilon \) can be arbitrary small, together with (A.30) and (A.31), we get
$$\begin{aligned} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}(\ln X_{i})^{2}(t_{i+1}-t_{i})-\frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}(\ln X_{i})^{2}(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}0. \end{aligned}$$
(A.32)
(A.18), (A.21) and (A.32) imply that
$$\begin{aligned} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}/(t_{i+1}-t_{i})- \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,sT]}{Z_{1i}^{\top }(p_{0})Z_{1i}(p_{0})}/(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}} \varvec{{0}}. \end{aligned}$$
By using similar techniques, one proves the second statement. This completes the proof. \(\square \)
Proposition A.4
Suppose that Assumptions 1–4 hold. Then, we have
$$\begin{aligned}{} & {} \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\frac{{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}}{t_{i+1}-t_{i}}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}s\varvec{{\Sigma }}_{1}, \quad { } \text{ and } \quad { } \nonumber \\{} & {} \quad \frac{1}{T}\sum \limits _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\frac{{{\mathcal {Z}}_{2i}^{\top }(p_{0}){\mathcal {Z}}_{2i}(p_{0})}}{t_{i+1}-t_{i}}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}(1-s)\varvec{{\Sigma }}_{2}. \end{aligned}$$
(A.33)
Proof
To prove the first statement, it suffices to combines Propositions A.3, 2.8 along with some algebraic computations. The proof of the second statement in (A.33) is similar. This completes the proof. \(\square \)
Proposition A.5
If Assumptions 1–4 hold, then,
$$\begin{aligned}{} & {} \frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,{\hat{s}}T]}\epsilon _{i}{{\mathcal {Z}}_{1i}(p_{0})}/(t_{i+1}-t_{i})\nonumber \\{} & {} \quad - \frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,sT]}\epsilon _{i}{Z_{1i}(p_{0})}/(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}{\textbf{0}}_{1\times (p_{0}+1)}; \end{aligned}$$
(A.34)
$$\begin{aligned}{} & {} \frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[{\hat{s}}T,T]}\epsilon _{i}{{\mathcal {Z}}_{2i}(p_{0})}/(t_{i+1}-t_{i})\nonumber \\{} & {} \quad -\frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[sT,T]}\epsilon _{i}{Z_{2i}(p_{0})}/(t_{i+1}-t_{i})\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m/2}}{\textbf{0}}_{1\times (p_{0}+1)}. \end{aligned}$$
(A.35)
Proof
To prove (A.34), we use the inequality
$$\begin{aligned} \begin{array}{ll} \left| \left| \displaystyle \frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,{\hat{s}}T]}\epsilon _{i}{{\mathcal {Z}}_{1i}(p_{0})}/(t_{i+1}-t_{i}) -\frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,sT]}\epsilon _{i}{Z_{1i}}(p_{0})/(t_{i+1}-t_{i})\right| \right| ^{m/2} \\ \quad \leqslant 3^{m/2-1}\left( \left| \left| \displaystyle \frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,{\hat{s}}T]}\epsilon _{i}{{\mathcal {Z}}_{1i}(p_{0})}/(t_{i+1}-t_{i}) -\frac{\sigma }{\sqrt{T}}\int _{0}^{{\hat{s}}T}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}\right| \right| ^{m/2}\right. \\ \qquad +\left| \left| \displaystyle \frac{\sigma }{\sqrt{T}}\int _{0}^{{\hat{s}}T}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}-\frac{\sigma }{\sqrt{T}}\int _{0}^{{s}T}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}\right| \right| ^{m/2} \\ \qquad \left. +\left| \left| \displaystyle \frac{\sigma }{\sqrt{T}}\int _{0}^{{s}T}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}-\frac{1}{\sqrt{T}}\sum \limits _{i\in {\mathbb {N}}[0,sT]}\epsilon _{i}{{Z}_{1i}}(p_{0})/(t_{i+1}-t_{i})\right| \right| ^{m/2} \right) . \end{array} \end{aligned}$$
One can prove that, \(\displaystyle \sum \limits _{i\in {\mathbb {N}}[0,{\hat{s}}T]}\epsilon _{i}{{\mathcal {Z}}_{1i}}(p_{0})/((t_{i+1}-t_{i})\sqrt{T}) -\sigma \int _{0}^{{\hat{s}}T}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}/\sqrt{T}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m}}\varvec{{0}}_{1\times (p+1)}\), \(\displaystyle \sum \limits _{i\in {\mathbb {N}}[0,sT]}\epsilon _{i}{{Z}_{1i}}(p_{0})/((t_{i+1}-t_{i})\sqrt{T})-\sigma \int _{0}^{sT}\left( \varvec{{b}}(t),-\ln X(t)\right) dB_{t}/\sqrt{T}\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{L^{m}}{\varvec{{0}}}\), and by Lemmas A.3–A.4, \(\displaystyle \sigma \int _{0}^{{\hat{s}}T}\left( \varvec{{b}}(t),\right. \left. -\ln X(t)\right) dB_{t}/\sqrt{T}-\sigma \int _{0}^{{s}T}\left( \varvec{{b}}(t),\right. \left. -\ln X(t)\right) dB_{t}/\sqrt{T}\xrightarrow [T\rightarrow \infty ]{L^{m/2}}{\varvec{{0}}}\). This completes the proof of (A.34), and (A.35) is proved by similar techniques. \(\square \)
Proposition A.6
If Assumptions 1–4 hold, then, for some \(0<a^{*}<{\varvec{a}}/2\),
$$\begin{aligned} \left( \left| \left| \sum \limits _{i\in {\mathbb {N}}[0,{\hat{s}}T]}\varepsilon _{i}{{\mathcal {Z}}_{1i}(p)}/(t_{i+1}-t_{i})\right| \right| , \left| \left| \sum \limits _{i\in {\mathbb {N}}[{\hat{s}}T,T]}\varepsilon _{i}{{\mathcal {Z}}_{2i}(p)}/(t_{i+1}-t_{i})\right| \right| \right) =O_{\mathrm{{P}}}(\sqrt{T}\textrm{log}^{\textit{a}^{*}}(\textit{T})).\nonumber \\ \end{aligned}$$
(A.36)
Proof
The proof follows from the triangle inequality and Proposition A.2. \(\square \)
Propositions A.3–A.6 imply that \({\hat{\theta }}_{T}({\hat{s}})\) obtained from discretized version is consistent.
Proof of Proposition 3.2
The proof of the second claim follows directly from the first statement. Thus, only prove the first statement. From the log-likelihood function defined in (2.6) and SIC information criterion in (3.4), along with the fact that, by (3.2), \(Y_{i}={Z_{ki}(p_{0})\varvec{{\mu }}_{k}(p_{0})+\epsilon _{i},k=1,2}\),
$$\begin{aligned} \textrm{IC}(1,p)&=-2\left( \frac{1}{2\sigma ^{2}}\left( \sum _{i\in {\mathbb {N}}[0,T]}Y^{2}_{i}/(t_{i+1}-t_{i})-\sum _{i\in {\mathbb {N}} [0,{{\hat{s}}T}]}\left( {{\mathcal {Z}}_{1i}(p_{0})}\varvec{{\mu }}_{1}(p_{0})\right. \right. \right. \\&\left. \left. \left. +\varepsilon _{i}-{{\mathcal {Z}}_{1i}(p)}\hat{\varvec{{\mu }}}_{1}(p)\right) ^{2}/(t_{i+1}-t_{i})\right. \right. \\&\quad \left. \left. -\sum _{i\in {\mathbb {N}} [{{\hat{s}}T},T]}\left( {{\mathcal {Z}}_{2i}(p_{0})}\varvec{{\mu }}_{2}(p_{0})+\varepsilon _{i}-{{\mathcal {Z}}_{2i}(p)}\hat{\varvec{{\mu }}}_{2}(p)\right) ^{2}/(t_{i+1}-t_{i})\right) \right) \\&+2(p+1)\textrm{log}(N), \end{aligned}$$
and
$$\begin{aligned}{} & {} \textrm{IC}(0,p) =\\{} & {} -2\left( \frac{1}{2\sigma ^{2}}\left( \sum _{i\in {\mathbb {N}}[0,T]}\frac{Y^{2}_{i}}{t_{i+1}-t_{i}}-\sum _{i\in {\mathbb {N}} [0,T]}\frac{\left( {Z_{1i}(p_{0})}\varvec{{\mu }}_{1}(p_{0})+\varepsilon _{i}-{Z_{1i}(p)}\hat{\varvec{{\mu }}}_{1}(p)\right) ^{2}}{t_{i+1}-t_{i}}\right) \right) \\{} & {} +(p+1)\textrm{log}(N). \end{aligned}$$
First, suppose that \(c^{0}=0\), we need to compare \(\textrm{IC}(0,p_{0})\), \(\textrm{IC}(c,p)\), for \(c=1,~\textrm{or}~p\ne p_{0}\).
-
(1)
\(c=1, p=p_{0}\): In this case, we have \(\varvec{{\mu }}_{1}(p_{0})=\varvec{{\mu }}_{2}(p_{0})=\varvec{{\mu }}(p_{0})\) and one verifies that
$$\begin{aligned}{} & {} \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0}) =\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,T]}\frac{\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) ^{2}}{t_{i+1}-t_{i}} \\{} & {} \quad -\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,{{\hat{s}}T}]}\frac{\left( {{\mathcal {Z}}_{1i}(p_{0})\varvec{{\mu }}_{1}(p_{0}) -{\mathcal {Z}}_{1i}(p_{0})\hat{\varvec{{\mu }}}_{1}(p_{0})}\right) ^{2}}{t_{i+1}-t_{i}}\\{} & {} \quad -\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [{{\hat{s}}T},T]}\frac{\left( {{\mathcal {Z}}_{2i}(p_{0})\varvec{{\mu }}_{2}(p_{0}) -{\mathcal {Z}}_{2i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) ^{2}}{t_{i+1}-t_{i}} \\{} & {} \quad +\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,T]}\frac{2\varepsilon _{i}\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) }{t_{i+1}-t_{i}}\\{} & {} \quad -\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,{{\hat{s}}T}]}\frac{2\varepsilon _{i}\left( {{\mathcal {Z}}_{1i}(p_{0})\varvec{{\mu }}_{1}(p_{0}) -{\mathcal {Z}}_{1i}(p_{0})\hat{\varvec{{\mu }}}_{1}(p_{0})}\right) }{t_{i+1}-t_{i}} \\{} & {} \quad -\frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [{{\hat{s}}T},T]}\frac{2\varepsilon _{i}\left( {{\mathcal {Z}}_{2i}(p_{0})\varvec{{\mu }}_{2}(p_{0}) -{\mathcal {Z}}_{2i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) }{t_{i+1}-t_{i}}-(p_{0}+1)\textrm{log}(N). \end{aligned}$$
From Proposition 2.9, \({\hat{\theta }}(p_{0})=\varvec{{\theta }}(p_{0})+\sigma Q^{-1}(s,T,p_{0})W(s,T,p_{0}),\) we have
$$\begin{aligned}{} & {} \frac{1}{\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,T]}\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) ^{2}/(t_{i+1}-t_{i}) \\{} & {} \quad =\frac{1}{T}(TQ^{-1}(s,T,p_{0}))\frac{1}{\sqrt{T}}W(s,T,p_{0}))^{\top }\\{} & {} \quad \frac{1}{T}\sum _{i\in {\mathbb {N}} [0,T]}\frac{Z_{i}^{\top }(p_{0})Z_{i}(p_{0})}{t_{i+1}-t_{i}}(TQ^{-1}(s,T,p_{0}))\frac{1}{\sqrt{T}}W(s,T,p_{0})). \end{aligned}$$
By combining Propositions 2.8, A.2, A.3, A.4 and A.8, we get
$$\begin{aligned} \sum _{i\in {\mathbb {N}} [0,T]}\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) ^{2}/(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)), \end{aligned}$$
(A.37)
$$\begin{aligned} \sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\left( {{\mathcal {Z}}_{1i}(p_{0})\varvec{{\mu }}_{2}(p_{0})-{\mathcal {Z}}_{1i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) ^{2}/(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)), \\ \sum _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\left( {{\mathcal {Z}}_{2i}(p_{0})\varvec{{\mu }}_{2}(p_{0})-{\mathcal {Z}}_{2i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) ^{2}/(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)). \end{aligned}$$
Further, from Propositions A.2 and A.6, we have
$$\begin{aligned} \sum _{i\in {\mathbb {N}} [0,T]}2\varepsilon _{i}\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) /(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)), \end{aligned}$$
(A.38)
$$\begin{aligned} \sum _{i\in {\mathbb {N}} [0,{{\hat{s}}T}]}2\varepsilon _{i}\left( {{\mathcal {Z}}_{1i}(p_{0})\varvec{{\mu }}_{2}(p_{0})-{\mathcal {Z}}_{1i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) /(t_{i+1}-t_{i}) =O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)), \\ \sum _{i\in {\mathbb {N}} [{{\hat{s}}T},T]}2\varepsilon _{i}\left( {{\mathcal {Z}}_{2i}(p_{0})\varvec{{\mu }}_{2}(p_{0})-{\mathcal {Z}}_{2i}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})}\right) /(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{2a^{*}}(T)). \end{aligned}$$
Then, together with Assumption 4, \((p_{0}+1)\textrm{log}(N)=O(\textrm{log}^{a}(T))\), we have \(\textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})<0,\) whenever T tends to infinity and \(\Delta _{N}\) tends to 0 i.e. \(\lim \limits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\textrm{P}\left( \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})>0\right) =0.\)
-
(2)-(5)
Similarly, we prove that \(\lim \nolimits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\textrm{P}\left( \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})>0\right) =1\) for the cases where (2) \(c=1, p=p^{*}>p_{0}\); (3) \(c=1, p=p_{*}<p_{0}\); (4) \(c=0, p=p^{*}>p_{0}\) and (5) \(c=0, p=p_{*}<p_{0}\).
This completes the proof of first part.
Second, suppose that \(c^{0}=1\), we need to compare \(\textrm{IC}(1,p_{0})\), \(\textrm{IC}(c,p)\), for \(c=0,~\textrm{or}~p\ne p_{0}\).
-
(1).
\(c=0, p=p_{0}\):
$$\begin{aligned}{} & {} \frac{ \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})}{T} =\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,T]}\frac{\left( Z_{i}(p_{0})\varvec{{\mu }}_{1}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) ^{2}}{t_{i+1}-t_{i}} \\{} & {} \quad -\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\frac{{{\mathcal {Z}}_{1i}(p_{0})\left( \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}(p_{0})\right) ^{2}}}{t_{i+1}-t_{i}}\\{} & {} \quad -\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\frac{{{\mathcal {Z}}_{2i}(p_{0})\left( \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{2}(p_{0})\right) ^{2}}}{t_{i+1}-t_{i}} \\{} & {} \quad +\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,T]}\frac{2\varepsilon _{i}Z_{i}(p_{0})\left( \varvec{{\theta }}(p_{0})-{\hat{\theta }}(p_{0})\right) }{t_{i+1}-t_{i}}\\{} & {} \quad -\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\frac{2\varepsilon _{i}\left( {{\mathcal {Z}}_{1i}}(p_{0})\varvec{{\mu }}_{1}(p_{0})-{{\mathcal {Z}}_{1i}}(p_{0})\hat{\varvec{{\mu }}}_{1}(p_{0})\right) }{t_{i+1}-t_{i}} \\{} & {} \quad -\frac{1}{T\sigma ^{2}}\sum _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\frac{2\varepsilon _{i}\left( {{\mathcal {Z}}_{2i}}(p_{0})\varvec{{\mu }}_{2}(p_{0}) -{{\mathcal {Z}}_{2i}}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})\right) }{t_{i+1}-t_{i}} -\frac{(p_{0}+1)\textrm{log}(N)}{T}. \end{aligned}$$
From some algebraic computations, one can proves that
$$\begin{aligned} \sum _{i\in {\mathbb {N}} [0,T]}\left( Z_{i}(p_{0})\varvec{{\theta }}(p_{0})-Z_{i}(p_{0}){\hat{\theta }}(p_{0})\right) ^{2}/((t_{i+1}-t_{i})T\sigma ^{2})\geqslant C_{0}||\varvec{{\mu }}_{1}-\varvec{{\mu }}_{2}||>0 \end{aligned}$$
for some positive constant \(C_{0}\). Further, since \(\hat{\varvec{{\mu }}}_{1}(p_{0})-\varvec{{\mu }}_{1}(p_{0})=\sigma Q_{T}({\hat{s}})W_{T}({\hat{s}})\)
$$\begin{aligned}{} & {} \sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\left( {{\mathcal {Z}}_{1i}}(p_{0})\varvec{{\mu }}_{1}(p_{0})-{{\mathcal {Z}}_{1i}}(p_{0})\hat{\varvec{{\mu }}}_{1}(p_{0})\right) ^{2}/((t_{i+1}-t_{i})T\sigma ^{2}) \\{} & {} \quad =\frac{1}{T}(TQ_{T}^{-1}({\hat{s}},p_{0})\frac{1}{\sqrt{T}}W_{T}({\hat{s}},p_{0}))^{\top } \frac{1}{T}\sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\frac{{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}}{t_{i+1}-t_{i}}(TQ_{T}^{-1}({\hat{s}},p_{0})\\{} & {} \quad \frac{1}{\sqrt{T}}W_{T}({\hat{s}},p_{0})). \end{aligned}$$
By Propositions A.3, A.4 and A.8, we get \(\left| \left| \sum \nolimits _{i\in {\mathbb {N}} [0,{\hat{s}}T]}{{\mathcal {Z}}_{1i}^{\top }(p_{0}){\mathcal {Z}}_{1i}(p_{0})}/((t_{i+1}-t_{i})T)\right| \right| =O_{\mathrm{{P}}}(1)\), \(\Vert TQ_{T}^{-1}({\hat{s}},p_{0})\Vert =O_{\mathrm{{P}}}(1)\), and then, together with Proposition A.2, we have \(\Vert W_{T}(s,p)\Vert /T=o_{\mathrm{{P}}}(1)\), \(\Vert W_{T}(1,p)-W_{T}(s,p)\Vert /T=o_{\mathrm{{P}}}(1)\), \(\displaystyle \sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\left( {{\mathcal {Z}}_{1i}}(p_{0})\varvec{{\mu }}_{1}(p_{0})-{{\mathcal {Z}}_{1i}}(p_{0})\hat{\varvec{{\mu }}}_{1}(p_{0})\right) ^{2}/(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{a^{*}}(T))\), and \(\displaystyle \sum _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\left( {{\mathcal {Z}}_{2i}}(p_{0})\varvec{{\mu }}_{2}(p_{0})-{{\mathcal {Z}}_{2i}}(p_{0})\hat{\varvec{{\mu }}}_{2}(p_{0})\right) ^{2}/(t_{i+1}-t_{i})=O_{\mathrm{{P}}}(\textrm{log}^{a^{*}}(T))\). Hence, by using Proposition A.6, we get \(\displaystyle \sum _{i\in {\mathbb {N}} [0,T]}2\varepsilon _{i}Z_{i}(p_{0})\left( \varvec{{\theta }}(p_{0})-{\hat{\theta }}(p_{0})\right) /( (t_{i+1}-t_{i})T)\xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}]{\mathrm{{P}}}0\),
$$\begin{aligned}{} & {} \left( \frac{1}{T}\sum _{i\in {\mathbb {N}} [0,{\hat{s}}T]}\frac{2\varepsilon _{i}{{\mathcal {Z}}_{1i}}(p_{0})\left( \varvec{{\mu }}_{1}(p_{0})-\hat{\varvec{{\mu }}}_{1}(p_{0})\right) }{t_{i+1}-t_{i}},\right. \\{} & {} \left. \frac{1}{T}\sum _{i\in {\mathbb {N}} [{\hat{s}}T,T]}\frac{2\varepsilon _{i}{{\mathcal {Z}}_{2i}}(p_{0})\left( \varvec{{\mu }}_{2}(p_{0})-\hat{\varvec{{\mu }}}_{2}(p_{0})\right) }{t_{i+1}-t_{i}}\right) \\{} & {} \quad \xrightarrow [\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}] {\mathrm{{P}}}\varvec{{0}}. \end{aligned}$$
Therefore, by Assumption 4, we get \( \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})/T>0,\) whenever T is large and \(\Delta _{N}\) is arbitrary small, i.e. \(\lim \limits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\textrm{P}\left( \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})>0\right) =1.\)
-
(2)-(5).
Similarly, we prove that \(\lim \limits _{\begin{array}{c} T\rightarrow \infty \\ \Delta _{N}\rightarrow 0 \end{array}}\textrm{P}\left( \textrm{IC}(0,p_{0})-\textrm{IC}(1,p_{0})>0\right) =1\) for the cases where (2) \(c=0, p=p^{*}>p_{0}\); (3) \(c=0, p=p_{*}<p_{0}\); (4) \(c=1, p=p^{*}>p_{0}\) and (5) \(c=1, p=p_{*}<p_{0}\).
\(\square \)
We establish the following proposition, which plays important role in proving that \({\hat{s}}\) is a consistent estimator of the parameter s.
Proposition A.7
Let \(\phi =sT\). There exists an \(L_{0}>0\), such that for all \(L>L_{0}\), the minimum eigenvalues of the matrices \(\sum \nolimits _{{i}\in {\mathbb {N}}[\phi ,\phi +L]}{Z_{2i}^{\top }(p_{0})Z_{2i}}(p_{0})/((t_{i+1}-t_{i})L)\) and of \(\sum \nolimits _{{i}\in {\mathbb {N}}[\phi -L,\phi ]}{Z_{1i}^{\top }(p_{0})Z_{1i}}(p_{0})/((t_{i+1}-t_{i})L)\) and their respective continuous time versions \(Q_{[\phi ,\phi +L]}(p_{0})/L\) and \(Q_{[\phi -L,\phi ]}(p_{0})/L\) are all bounded away from 0.
The proof follows from Lemma A.1. Note that, by Proposition A.7, we weaken the assumption in Chen et al. (2020), Chen et al. (2018). More precisely, the established result shows that the assumptions in the quoted paper are stronger than needed.
Lemma A.3
Let \({\hat{s}}\) be \({\mathscr {F}}_{T}-\)measurable and a consistent estimator of s, with \(0\leqslant {\hat{s}}\leqslant 1\) a.s. and let \(\{\varvec{{b}}(t),t\geqslant 0\}\) be deterministic and bounded function. Then,
$$\begin{aligned} \frac{1}{\sqrt{T}}\int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\frac{1}{\sqrt{T}}\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\xrightarrow [T\rightarrow \infty ]{L^{m}}0. \end{aligned}$$
Proof
Let \(G(T)=T^{-m/2}{\mathbb {E}}\left[ \left| \displaystyle \int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\right| ^{m}\right] \). We need to prove that \(\lim \nolimits _{T\rightarrow \infty }G(T)=0\). Let \(||\varvec{{b}}(t)||\leqslant K_{b}\) and \(\epsilon >0\), we have \(\lim \limits _{T\rightarrow \infty }\textrm{P}\left( |{\hat{s}}-s|\geqslant \epsilon /(4K_{b}^{2})\right) =0\). We have
\(G(T)=G_{11}(T)+G_{12}(T)+G_{21}(T)+G_{22}(T)\) where
$$\begin{aligned} \begin{array}{l} G_{11}(T)=T^{-m/2}{\mathbb {E}}\left[ \left| \int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\right| ^{m}{\mathbb {I}}_{\{{\hat{s}}>s\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \frac{\epsilon }{4K_{b}^{2}}\}}\right] \\ G_{12}(T)=T^{-m/2}{\mathbb {E}}\left[ \left| \int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\right| ^{m}{\mathbb {I}}_{\{{\hat{s}}>s\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \frac{\epsilon }{4K_{b}^{2}}\}}\right] \\ G_{21}(T)=T^{-m/2}{\mathbb {E}}\left[ \left| \int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\right| ^{m}{\mathbb {I}}_{\{{\hat{s}}<s\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\leqslant \frac{\epsilon }{4K_{b}^{2}}\}}\right] \\ G_{22}(T)=T^{-m/2}{\mathbb {E}}\left[ \left| \int _{0}^{{\hat{s}}T}\varvec{{b}}(t)dB_{t}-\int _{0}^{sT}\varvec{{b}}(t)dB_{t}\right| ^{m}{\mathbb {I}}_{\{{\hat{s}}<s\}}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \frac{\epsilon }{4K_{b}^{2}}\}}\right] . \end{array} \end{aligned}$$
(A.39)
By using Cauchy–Schwarz’s inequality and Burkholder–Davis–Gundy’s inequality, we prove that \(\lim \limits _{T\rightarrow \infty }(G_{11}(T)=\lim \limits _{T\rightarrow \infty }G_{12}(T)=\lim \limits _{T\rightarrow \infty }G_{21}(T) =\lim \limits _{T\rightarrow \infty }G_{22}(T))=0.\) This completes the proof. \(\square \)
Lemma A.4
Suppose that Assumptions 1–4 hold and let \({\hat{s}}\) be a consistent estimator of s, with \(0<{\hat{s}}<1\) a.s. Then, \(\frac{1}{\sqrt{T}}\int _{0}^{{\hat{s}}T}(\ln X(t))dB_{t}-\frac{1}{\sqrt{T}}\int _{0}^{sT}(\ln X(t))dB_{t}\xrightarrow [T\rightarrow \infty ]{L^{m/2}}0\).
Proof
Since \({\hat{s}}\) is \({\mathscr {F}}_{T}-\)measurable and a consistent estimator of s, let \(\epsilon >0\), for every \(0<\delta <\min \{s,1-s\}\), we have \(\textrm{P}(|{\hat{s}}-s|\geqslant \delta )<\epsilon \), for sufficiently large T. Then,
$$\begin{aligned}{} & {} {\mathbb {E}}\left[ \left| \frac{1}{\sqrt{T}}\int _{0}^{{\hat{s}}T}(\ln X(t))dB_{t}-\frac{1}{\sqrt{T}}\int _{0}^{sT}(\ln X(t))dB_{t}\right| ^{m/2}\right] \end{aligned}$$
$$\begin{aligned}{} & {} \quad ={\mathbb {E}}\left[ \left| \frac{1}{\sqrt{T}}\int _{0}^{{\hat{s}}T}(\ln X(t))dB_{t}-\frac{1}{\sqrt{T}}\int _{0}^{sT}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|\geqslant \delta \}}\right] \end{aligned}$$
(A.40)
$$\begin{aligned}{} & {} \qquad +{\mathbb {E}}\left[ \left| \frac{1}{\sqrt{T}}\int _{0}^{{\hat{s}}T}(\ln X(t))dB_{t}-\frac{1}{\sqrt{T}}\int _{0}^{sT}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{|{\hat{s}}-s|<\delta \}}\right] . \end{aligned}$$
(A.41)
Further, one proves that
$$\begin{aligned}{} & {} (A.40) \leqslant T^{-m/4}{\mathbb {E}}\left[ \sup \limits _{0\leqslant u\leqslant sT}\left| \int _{u}^{sT}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}\leqslant s-\delta \}}\right] \nonumber \\{} & {} \qquad +T^{-m/4} {\mathbb {E}}\left[ \left| \sup \limits _{sT\leqslant u\leqslant T}\int _{sT}^{u}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{{\hat{s}}\geqslant s+\delta \}}\right] . \\{} & {} \quad (A.41) \leqslant T^{-m/4}{\mathbb {E}}\left[ \sup \limits _{(s-\delta )T<u<sT}\left| \int _{u}^{sT}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{s-\delta<{\hat{s}}\leqslant s\}}\right] \\{} & {} \quad +T^{-m/4}{\mathbb {E}}\left[ \sup \limits _{sT<u<(s+\delta )T}\left| \int _{sT}^{u}(\ln X(t))dB_{t}\right| ^{m/2}{\mathbb {I}}_{\{s<{\hat{s}}\leqslant s+\delta \}}\right] \end{aligned}$$
The proof is completed by combining Cauchy–Schwartz’s inequality, Burkholder-Davis-Gundy’s inequality and Jensen’s inequality. \(\square \)
The following two propositions show that \(\frac{1}{T}Q({\hat{s}},T)-\frac{1}{T}Q({s},T)\xrightarrow [T\rightarrow \infty ]{L^{m/2}}0\).
Proposition A.8
Suppose that Assumptions 1–4 hold and \({\hat{s}}\) is a consistent estimator of s, with \(0<{\hat{s}}<1\) a.s. Then, (1). \(\displaystyle \frac{1}{T}\int _{0}^{{\hat{s}}T}b^{\top }(t)\varvec{{b}}(t)dt-\frac{1}{T}\int _{0}^{sT}b^{\top }(t)\varvec{{b}}(t)dt\xrightarrow [T\rightarrow \infty ]{L^{m}}\textbf{0}\),
(2) \(\displaystyle \frac{1}{T}\int _{0}^{{\hat{s}}T}(\varvec{{b}}(t),-\ln X(t))^{\top }(\ln X(t))dt-\frac{1}{T}\int _{0}^{sT}(\varvec{{b}}(t),-\ln X(t))^{\top }(\ln X(t))dt\xrightarrow [T\rightarrow \infty ]{L^{m/2}}\textbf{0}\).
Proof
The proof follows by combining the consistency of \({\hat{s}}\) and Cauchy–Schwarz’s inequality along with some algebraic computations. \(\square \)
