Estimation of the lead–lag parameter between two stochastic processes driven by fractional Brownian motions

Abstract

In this paper, we consider the problem of estimating the lead–lag parameter between two stochastic processes driven by fractional Brownian motions (fBMs) of the Hurst parameter greater than 1/2. First we propose a lead–lag model between two stochastic processes involving fBMs, and then construct a consistent estimator of the lead–lag parameter with possible convergence rate. Our estimator has the following two features. Firstly, we can construct the lead–lag estimator without using the Hurst parameters of the underlying fBMs. Secondly, our estimator can deal with some non-synchronous and irregular observations. We explicitly calculate possible convergence rate when the observation times are (1) synchronous and equidistant, and (2) given by the Poisson sampling scheme. We also present numerical simulations of our results using the R package YUIMA.

Appendix A

Proof of Lemma 5.2

First we define an auxiliary set $A_{n}(K)$ by

$$\begin{aligned} A_{n}(K) = \bigcap _{l=1,2} \left\{ \frac{r_{n}^{2H_{l}-1+\mu }v_{n}^{-\gamma (\mu )}}{\sum _{i:\overline{I_{i}^{l,n}} \le T }| I_{i}^{l,n} |^{2H_{l}}} \le K \right\} . \end{aligned}$$

Since $\sum _{i:\overline{I_{i}^{l,n}} \le T }| I_{i}^{l,n} |^{2H_{l}} \le T r_{n}^{2H_{l}-1}$ holds, we have

$$\begin{aligned} r_{n} \le (TK)^{1/\mu } v_{n}^{\gamma (\mu )/\mu }, \end{aligned}$$

on the set $A_{n}(K)$. In particular, it holds that $( r_{n}/v_{n} )1_{A_{n}(K)} \rightarrow 0$ as $n\rightarrow \infty$ (recall that $\gamma (\mu ) >\mu$). Thanks to (B3), it holds that $\sup _{n}{\mathbb {P}}(A_{n}(K)^{c}) < \epsilon$ for each $\epsilon >0$ if $K= K(\epsilon )$ is sufficiently large. Let $R^{l,n}_{1}$ denote $(\sum _{i : \overline{I_{i}^{l,n}} \in J}B^{l}((I_{i}^{l,n})_{a})^{2})/(\sum _{i:\overline{I_{i}^{l,n}} \in J}|I_{i}^{l,n}|^{2H_{l}} )$ for simplicity. For any positive number $\epsilon _{1} >0$, we have

$$\begin{aligned} {\mathbb {P}}\left\{ \left| R^{l,n}_{1} -1 \right| \ge \epsilon _{1} \right\}&\le {\mathbb {P}}\left\{ \left| R^{l,n}_{1} -1 \right| \ge \epsilon _{1} ,\ A_{n}(K) \right\} + {\mathbb {P}}\{ A_{n}(K)^{c} \}. \end{aligned}$$

Hence we obtain

$$\begin{aligned} {\mathbb {P}}\left\{ \left| R^{l,n}_{1} -1 \right| \ge \epsilon _{1} \right\}&\le {\mathbb {P}}\left\{ \left| R^{l,n}_{1} -1 \right| \ge \epsilon _{1},\ A_{n}(K(\epsilon _{2})) \right\} + \epsilon _{2} \nonumber \\&\le \epsilon _{1}^{-2} {\mathbb {E}}\left\{ \left| R^{l,n}_{1} -1 \right| ^{2} {\mathbf {1}}_{A_{n}(K(\epsilon _{2})) } \right\} + \epsilon _{2}. \end{aligned}$$

(A.1)

By using [B1] and conditioning, we can calculate the expectation in (A.1) as

$$\begin{aligned}&{\mathbb {E}}\left\{ \left| R^{l,n}_{1} -1 \right| ^{2} {\mathbf {1}}_{A_{n}(K(\epsilon _{2})) } \right\} \\&= {\mathbb {E}}\left\{ {\mathbb {E}}\left\{ \left( \frac{ \sum _{i : \overline{I_{i}^{l,n}} \in J}( B^{l}( (I_{i}^{l,n})_{a} )^{2} - | I_{i}^{l,n} |^{2H_{l}} )}{ \sum _{i : \overline{I_{i}^{l,n}} \in J}| I_{i}^{l,n} |^{2H_{l}} } \right) ^{2} {\mathbf {1}}_{A_{n}(K(\epsilon _{2})) } \bigg | \sigma ({\mathcal {T}}) \right\} \right\} \\&= {\mathbb {E}} \left\{ \frac{ {\mathbf {1}}_{A_{n}(K(\epsilon _{2}))} {\mathbb {E}} \left\{ \left( \sum _{i : \overline{I_{i}^{l,n}} \in J}\left( B^{l}((I_{i}^{l,n})_{a} )^{2} - | I_{i}^{l,n} |^{2H_{l}} \right) \right) ^{2} \bigg | \sigma ({\mathcal {T}}) \right\} }{ (\sum _{i : \overline{I_{i}^{l,n}} \in J}|I_{i}^{l,n} |^{2H_{l}})^{2} } \right\} . \end{aligned}$$

Since $B^{{l}}((I_{i}^{l,n})_{a} )^{2} - | (I_{i}^{l,n}) |^{2H_{l}} = B^{{l}}((I_{i}^{l,n})_{a} )^{2} - | (I_{i}^{l,n})_{a} |^{2H_{l}}$ coincides with the multiple Wiener integral ${\mathbb {I}}_{2}( h^{l}( (I_{i}^{l,n})_{a} )^{\otimes 2} )$ (recall that the definition of $h^{l}(I)$ is given in (2.5)) by Theorem 2.7, we have

$$\begin{aligned} {\mathbb {E}} \left\{ \left( \sum _{i : \overline{I_{i}^{l,n}} \in J}\left( B^{l}((I_{i}^{l,n})_{a})^{2} - | I_{i}^{l,n} |^{2H_{l}} \right) \right) ^{2} \bigg | \sigma ({\mathcal {T}}) \right\}&= {\mathbb {E}}\left\{ {\mathbb {I}}_{2}^{2} \left( \sum _{i : \overline{I_{i}^{l,n}} \in J} h^{l}( (I_{i}^{l,n})_{a} )^{\otimes 2} \right) \bigg | \sigma ({\mathcal {T}}) \right\} \\&= \sum _{i,j} \langle h^{l}( (I_{i}^{l,n})_{a} ) , h^{l}( (I_{j}^{l,n})_{a} ) \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})}^{2} \\&= \sum _{i,j} \langle {\mathbf {1}}_{ (I_{i}^{l,n})_{a} } , {\mathbf {1}}_{ (I_{j}^{l,n})_{a} } \rangle _{{\mathcal {H}}_{H_{l}} } \end{aligned}$$

Since $| \langle {\mathbf {1}}_{ (I_{i}^{l,n})_{a} } , {\mathbf {1}}_{ (I_{j}^{l,n})_{a} } \rangle _{{\mathcal {H}}_{H_{l}} } | \le r_{n}^{2H_{l}}$ because of Cauchy–Schwarz inequality, we have

$$\begin{aligned} {\mathbb {E}} \left\{ \left( \sum _{i : \overline{I_{i}^{l,n}} \in J}\left( B^{l}((I_{i}^{l,n})_{a})^{2} - | I_{i}^{l,n} |^{2H_{l}} \right) \right) ^{2} \bigg | \sigma ({\mathcal {T}}) \right\}&\le r_{n}^{2H_{l}} \sum _{i,j} \langle {\mathbf {1}}_{ (I_{i}^{l,n})_{a} } , {\mathbf {1}}_{ (I_{j}^{l,n})_{a} } \rangle _{{\mathcal {H}}_{H_{l}} } \\&\lesssim r_{n}^{2H_{l}}. \end{aligned}$$

Note that $\langle {\mathbf {1}}_{ (I_{i}^{l,n})_{a} } , {\mathbf {1}}_{ (I_{j}^{l,n})_{a} } \rangle _{{\mathcal {H}}_{H_{l}} }$ is non-negative. Plugging this into (A.1), we obtain

$$\begin{aligned} {\mathbb {P}}\left\{ \left| R^{l,n}_{1} -1 \right| \ge \epsilon _{1} \right\}&\lesssim \epsilon _{1}^{-2} {\mathbb {E}}\left\{ \left( \frac{ r_{n}^{H_{l} } }{ \sum _{i : \overline{I_{i}^{l,n}} \in J}| I_{i}^{l,n} |^{2H_{l}} } \right) ^{2} {\mathbf {1}}_{A_{n}(K(\epsilon _{2}))} \right\} + \epsilon _{2} \\&= \epsilon _{1}^{-2} {\mathbb {E}}\left\{ \left( \frac{ r_{n}^{2H_{l}-1+\mu } v_{n}^{-\gamma (\mu )} }{ \sum _{i : \overline{I_{i}^{l,n}} \in J}| I_{i}^{l,n} |^{2H_{l}} } r_{n}^{1-H_{l}-\mu } v_{n}^{\gamma (\mu )} \right) ^{2} {\mathbf {1}}_{A_{n}(K(\epsilon _{2}))} \right\} + \epsilon _{2} \\&\lesssim \epsilon _{1}^{-2} (TK(\epsilon _{2}))^{2/\mu } v_{n}^{2(1 -H_{l})} + \epsilon _{2}, \end{aligned}$$

if we choose sufficiently small $\mu >0$ such that $1-H-\mu >0$ holds. Since $\epsilon _{1}$ and $\epsilon _{2}$ are arbitrary, we obtain (5.2) by letting $n\rightarrow \infty$. $\square$

Proof of Proposition 5.3

The set ${\mathcal {G}}^{n}\cap \{{\tilde{\theta }}\in \Theta \mid |{\tilde{\theta }}-\theta |\ge \epsilon v_{n}\}$ can be decomposed as

$$\begin{aligned} {\mathcal {G}}^{n}\cap \{{\tilde{\theta }}\in \Theta \mid |{\tilde{\theta }}-\theta |\ge \epsilon v_{n}\}&= ( {\mathcal {G}}^{n}_{\ge 0}\cap \{{\tilde{\theta }}\in \Theta \mid {\tilde{\theta }} \ge \theta + \epsilon v_{n} \} ) \cup ( {\mathcal {G}}^{n}_{\ge 0}\cap \{{\tilde{\theta }}\in \Theta \mid {\tilde{\theta }} \le \theta - \epsilon v_{n} \} ) \\&\quad \cup ({\mathcal {G}}^{n}_{<0}\cap \{ {\tilde{\theta }}\in \Theta \mid {\tilde{\theta }} \le \theta - \epsilon v_{n} \}) \\&=: {\mathcal {G}}^{n}_{1} \cup {\mathcal {G}}^{n}_{2} \cup {\mathcal {G}}^{n}_{3}. \end{aligned}$$

Clearly (5.3) is equivalent to

$$\begin{aligned} \sup _{{\mathcal {G}}^{n}_{i}}\left| \frac{R^{n}_{2}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| \rightarrow ^{p} 0 \end{aligned}$$

(A.2)

for $i=1,2,3$. We only prove (A.2) when $i=1$. The other cases are similar.

Let us assume $H_{1} \le H_{2}$ for the moment. Since ${\tilde{\theta }} \in {\mathcal {G}}^{n}_{1}$, we have

$$\begin{aligned} R_{2}^{n}\big ({\tilde{\theta }}\big )&= \sum _{i,j:\overline{I_{i}^{1,n}}\le T} B^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big )B^{2}\big (\big (I_{j}^{2,n}\big )\big ){\mathbf {1}}_{ (I_{i}^{1,n})_{\theta }\cap (I_{j}^{2,n})_{\theta -{\tilde{\theta }}}\ne \emptyset } \\&= \sum _{i:\overline{I_{i}^{1,n}}\le T} B^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big )B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta }\big ) \\&= \sum _{i:\overline{I_{i}^{1,n}}\le T} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) , h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big ) \big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})} \\&\quad +\sum _{i:\overline{I_{i}^{1,n}}\le T} \Bigl ( B^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big )B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta }\big ) \\&\quad - \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) , h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big ) \big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})} \Bigr ) \\&=: R_{3}^{n}\big ({\tilde{\theta }}\big ) + R_{4}^{n}\big ({\tilde{\theta }}\big ). \end{aligned}$$

Hence it suffices to show

$$\begin{aligned} \sup _{{\tilde{\theta }}\in {\mathcal {G}}^{n}_{1}} \left| \frac{R_{3}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| \rightarrow ^{p} 0 \end{aligned}$$

(A.3)

and

$$\begin{aligned} \sup _{{\tilde{\theta }} \in {\mathcal {G}}_{1}^{n} }\left| \frac{R_{4}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| \rightarrow ^{p} 0 \end{aligned}$$

(A.4)

as $n\rightarrow \infty$.

The limit (A.3). It suffices to show

$$\begin{aligned} \sup _{{\tilde{\theta }}\in {\mathcal {G}}^{n}_{1}} \left| \frac{R_{3}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| {\mathbf {1}}_{A_{n}(K)} \rightarrow 0 \end{aligned}$$

(A.5)

as $n\rightarrow \infty$ for each $K>0$. By (2.6) in Proposition 2.9, it holds that

$$\begin{aligned} |R_{3}^{n}({\tilde{\theta }})|&= |\rho | c_{H_{1}}c_{H_{2}} \sum _{i:\overline{I_{i}^{1,n}}\le T} \int _{ (I_{i}^{1,n})_{\theta } }du\int _{{\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}((I_{i}^{1,n})_{\theta })_{{\tilde{\theta }}-\theta }}dv\ \beta (u,v) \\&\quad \times |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}}. \end{aligned}$$

Note that for each $K>0$ and $\epsilon >0$ we can choose $n_{0}=n_{0}(K,\epsilon )$ such that $n\ge n_{0}$ implies

$$\begin{aligned} (TK)^{1/\mu }v_{n}^{(\gamma (\mu )/\mu )-1} \le \frac{\epsilon }{3}. \end{aligned}$$

Hence if $n\ge n_{0}$ then $r_{n}{\mathbf {1}}_{A_{n}(K)} \le (\epsilon v_{n}/3)$ holds. We can always assume $n\ge n_{0}$ in this proof. Since ${\tilde{\theta }} \in {\mathcal {G}}^{n}_{1}$ and $r_{n}\le (\epsilon v_{n})/3$ on $A_{n}(K)$, we have $|u-v| \ge (\epsilon v_{n})/3$. Therefore

$$\begin{aligned} | R_{3}^{n}({\tilde{\theta }}) |\lesssim (\epsilon v_{n})^{H_{1}+H_{2}-2} \sum _{i:\overline{I_{i}^{1,n}}\le T} \int _{ (I_{i}^{1,n})_{\theta } }du\ u^{H_{1}-H_{2}} \int _{{\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}((I_{i}^{1,n})_{\theta })_{{\tilde{\theta }}-\theta }}dv\ v^{H_{2}-H_{1}} \end{aligned}$$

(A.6)

on $A_{n}(K)$. Since $H_{1}\le H_{2}$, we have

$$\begin{aligned} \int _{{\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}((I_{i}^{1,n})_{\theta })_{{\tilde{\theta }}-\theta }}dv\ v^{H_{2}-H_{1}} \lesssim r_{n}. \end{aligned}$$

(A.7)

Since the integral $\int _{ (I_{i}^{1,n})_{\theta } }du\ u^{H_{1}-H_{2}}$ is summable (note that $H_{1}-H_{2} \in (-1/2,1/2)$), we have

$$\begin{aligned} | R_{3}^{n}({\tilde{\theta }}) |\lesssim (\epsilon v_{n})^{H_{1}+H_{2}-2} r_{n}. \end{aligned}$$

(A.8)

By (A.8) and (B3), we have

$$\begin{aligned} \left| \frac{R_{3}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| {\mathbf {1}}_{A_{n}(K)} \lesssim v_{n}^{\gamma (\mu )-\mu } {\mathbf {1}}_{A_{n}(K)}. \end{aligned}$$

(A.9)

Since the right-hand-side of (A.9) does not depend on ${\tilde{\theta }}$, we obtain (A.5).

The limit (A.4). We define ${\tilde{h}}_{i}^{n}({\tilde{\theta }}) \in L^{2}([0,T+2\delta ];{\mathbb {R}}^{2}) {\tilde{\otimes }} L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})$ by

$$\begin{aligned} {\tilde{h}}_{i}^{n}({\tilde{\theta }})(\omega ) = h^{1}( (I_{i}^{1,n})_{\theta }(\omega ) ) {\tilde{\otimes }} h^{2}( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}((I_{i}^{1,n})_{\theta })_{{\tilde{\theta }}-\theta }(\omega ) ). \end{aligned}$$

Let $\epsilon _{1}$ and $\epsilon _{2}$ be positive numbers. By the same conditioning technique as in the proof of Lemma 5.2, we have

$$\begin{aligned} {\mathbb {P}}\left\{ \sup _{{\tilde{\theta }} \in {\mathcal {G}}^{n}_{1} }\left| \frac{R_{4}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| \ge \epsilon _{1} \right\} \le \epsilon _{2} + \epsilon _{1}^{-2p}\sum _{ {\tilde{\theta }} \in {\mathcal {G}}^{n}_{1} } {\mathbb {E}}\left\{ \frac{{\mathbf {1}}_{A_{n}(K(\epsilon _{2}))} {\mathbb {E}}\left\{ \left| {\mathbb {I}}_{2}(\sum _{i:\overline{I_{i}^{1,n}}\le T} {\tilde{h}}^{n}_{i}({\tilde{\theta }}) ) \right| ^{2p} \bigg | \sigma ({\mathcal {T}}) \right\} }{|D^{n}({\tilde{\theta }})|^{2p}} \right\} \end{aligned}$$

(A.10)

for any $p\ge 1$. Using Theorem 2.8, we obtain

$$\begin{aligned} {\mathbb {E}}\left\{ \left| {\mathbb {I}}_{2}\left( \sum _{i:\overline{I_{i}^{1,n}}\le T} {\tilde{h}}^{n}_{i}({\tilde{\theta }})\right) \right| ^{2p}\bigg | \sigma ({\mathcal {T}}) \right\} \le C_{p} {\mathbb {E}} \left\{ \left| {\mathbb {I}}_{2}\left( \sum _{i:\overline{I_{i}^{1,n}}\le T} {\tilde{h}}^{n}_{i}({\tilde{\theta }})\right) \right| ^{2} \bigg | \sigma ({\mathcal {T}}) \right\} ^{p} \end{aligned}$$

(A.11)

where $C_{p}$ is a positive constant depending only on $p\ge 1$. Let $R_{5}^{n}({\tilde{\theta }})$ denote the expectation in the right-hand-side of (A.11). A simple calculation using Proposition 2.6 yields

$$\begin{aligned} R_{5}^{n}({\tilde{\theta }}\big )&= \sum _{i,k:\overline{I_{i}^{1,n}}\le T,\ \overline{I_{k}^{2,n}}\le T} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta } \big ) \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\quad \quad \times \big \langle h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta }, h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\quad \quad + \sum _{i,k:\overline{I_{i}^{1,n}}\le T,\ \overline{I_{k}^{2,n}}\le T} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\quad \quad \times \big \langle h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\quad =: R_{5,1}^{n}\big ({\tilde{\theta }}\big ) + R_{5,2}^{n}\big ({\tilde{\theta }}\big ). \end{aligned}$$

(1) The first term can be estimated as follows. Since

$$\begin{aligned} \big \langle h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta }, h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \lesssim r_{n}^{2H_{2}}, \end{aligned}$$

we have

$$\begin{aligned} | R_{5,1}^{n}({\tilde{\theta }}) | \lesssim r_{n}^{2H_{2}} \end{aligned}$$

(A.12)

on $A_{n}(K(\epsilon _{2}))$. Note that $I \mapsto h^{1}(I)$ is linear so that

$$\begin{aligned} \sum _{i,k: \overline{I_{i}^{1,n}} \le T,\ \overline{I_{k}^{1,n}}\le T} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta }\big ) \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} < \infty . \end{aligned}$$

(2) The second term can be estimated as follows. We first recall that

$$\begin{aligned}&\big | \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \big | \\&\quad = |\rho | c_{H_{1}}c_{H_{2}} \int _{ (I_{i}^{1,n})_{\theta } }du\int _{{\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}((I_{k}^{1,n})_{\theta })_{{\tilde{\theta }}-\theta }}dv\ \beta (u,v) \\&\qquad \times |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}}. \end{aligned}$$

By the same reasoning as (A.6), we have

$$\begin{aligned} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \lesssim \left( \frac{\epsilon v_{n}}{3} \right) ^{H_{1}+H_{2}-2} r_{n} | I_{i}^{1,n} | \end{aligned}$$

if $i\le k$, and

$$\begin{aligned} \big \langle h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \lesssim \left( \frac{\epsilon v_{n}}{3} \right) ^{H_{1}+H_{2}-2} r_{n} | I_{k}^{1,n} | \end{aligned}$$

if $i>k$, on $A_{n}(K(\epsilon _{2}))$. Therefore, it holds that

$$\begin{aligned} \big |R_{5,2}^{n}\big ({\tilde{\theta }}\big )\big |&\lesssim r_{n}v_{n}^{H_{1}+H_{2}-2} \Biggl ( \sum _{i\le k } \big |I_{i}^{1,n}\big | \big |\big \langle h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \big | \nonumber \\&\quad + \sum _{k>i} \big |I_{k}^{1,n}\big | \big |\big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}-\theta } \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \big | \Biggr ) \nonumber \\&\lesssim r_{n}^{1+H_{2}} v_{n}^{H_{1}+H_{2}-2} \end{aligned}$$

(A.13)

on $A_{n}(K(\epsilon _{2}))$. Plugging (A.12) and (A.13) into (A.10), we obtain

$$\begin{aligned} {\mathbb {P}}\left\{ \sup _{{\tilde{\theta }} \in {\mathcal {G}}_{1}^{n} }\left| \frac{R_{4}^{n}({\tilde{\theta }})}{D^{n}({\tilde{\theta }})} \right| \ge \epsilon _{1} \right\}&\lesssim \epsilon _{2} + \epsilon _{1}^{-2p} \sum _{ {\tilde{\theta }} \in {\mathcal {G}}_{1}^{n} } {\mathbb {E}}\left\{ \frac{ {\mathbf {1}}_{A_{n}(K(\epsilon _{2}))} |r_{n}^{2H_{2}} + r_{n}^{1+H_{2}}v_{n}^{H_{1}+H_{2}-2} |^{p} }{ |D^{n}({\tilde{\theta }})|^{2p} } \right\} \\&\lesssim \epsilon _{2} + \epsilon _{1}^{-2p} (\# {\mathcal {G}}^{n}) v_{n}^{p(1-H_{1})} \end{aligned}$$

by (B3). Thanks to (C2), we have $\lim _{n\rightarrow \infty }(\# {\mathcal {G}}^{n}) v_{n}^{p(1-H_{1})}= 0$ if we choose sufficiently large $p\ge 1$. This gives (A.4).

Thanks to (A.3) and (A.4), we obtain (A.2) when $H_{1} \le H_{2}$. Now let us consider the case where $H_{1} > H_{2}$. In the case where $H_{1} > H_{2}$, we use an alternative expression of $R_{2}^{n}({\tilde{\theta }})$:

$$\begin{aligned} R_{2}^{n}({\tilde{\theta }})&= \sum _{i,j: \overline{I_{i}^{1,n}} \le T} B^{1}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )B^{2}\big ( I_{j}^{2,n} \big ) {\mathbf {1}}_{ (I_{i}^{1,n})_{{\tilde{\theta }}} \cap I_{j}^{2,n} \ne \emptyset } \nonumber \\&= \sum _{j} B^{2}\big (I_{j}^{2,n}\big ) B^{1} \big ( {\mathcal {I}}_{{\tilde{\theta }}}^{1,n,\le T} ( I_{j}^{2,n} \big )_{\theta -{\tilde{\theta }}} ), \end{aligned}$$

(A.14)

where the symbol ${\mathcal {I}}_{{\tilde{\theta }}}^{1,n,\le T}$ denotes the family of shifted intervals

$$\begin{aligned} {\mathcal {I}}_{{\tilde{\theta }}}^{1,n,\le T} =\big \{ \big (I_{i}^{1,n}\big )_{{\tilde{\theta }}} \mid \overline{I_{i}^{1,n}} \le T \big \}. \end{aligned}$$

Using the expression (A.14), we can prove (A.2) when $H_{1} > H_{2}$ in the same manner as we did in the case where $H_{1} \le H_{2}$. $\square$

Proof of Proposition 5.4

We formally extend $B^{1}$ and $B^{2}$ by setting $B^{1}_{t} = B^{2}_{t} =0$ for $t<0$. Let us first suppose that ${\tilde{\theta }}_{n} \in \Theta _{\ge 0}$. This is the case for sufficiently large n if $\theta \in (0,\delta )$. Then we decompose $R_{2}^{n}({\tilde{\theta }}_{n})$ into three terms:

$$\begin{aligned} R_{2}^{n}\big ({\tilde{\theta }}_{n}\big )&= \sum _{i:\overline{I_{i}^{1,n}}\le T} B^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big )\left( B^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}} \big ( \big (I_{i}^{1,n}\big )_{\theta } \big )_{{\tilde{\theta }}_{n}-\theta } \big )- B^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}} \big ( \big (I_{i}^{1,n}\big )_{\theta } \big )\cap [0,T+2\delta ]\big ) \right) \nonumber \\&\quad + \sum _{i:\overline{I_{i}^{1,n}}\le T} \Bigl (B^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) B^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}} \big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) \cap [0,T+2\delta ] \big ) \nonumber \\&\quad - \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) , h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )\cap [0,T+2\delta ] \big ) \big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})} \Bigr ) \nonumber \\&\quad + \sum _{i:\overline{I_{i}^{1,n}}\le T}\big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ) , h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )\cap [0,T+2\delta ] \big ) \big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})} \nonumber \\&=: R_{6}^{n}\big ({\tilde{\theta }}_{n}\big ) + R_{7}^{n}\big ({\tilde{\theta }}_{n}\big ) + R_{8}^{n}\big ({\tilde{\theta }}_{n}\big ). \end{aligned}$$

(A.15)

To simplify the notation, we set ${\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}((I_{i}^{1,n})_{\theta })^{+} = {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}((I_{i}^{1,n})_{\theta })\cap [0,T+2\delta ]$. Let us show

$$\begin{aligned} \left| \frac{R_{6}^{n}({\tilde{\theta }}_{n})}{D^{n}({\tilde{\theta }}_{n})} \right| \rightarrow ^{p} 0, \end{aligned}$$

(A.16)

$$\begin{aligned} \left| \frac{R_{7}^{n}({\tilde{\theta }}_{n})}{D^{n}({\tilde{\theta }}_{n})} \right| \rightarrow ^{p} 0 \end{aligned}$$

(A.17)

and that there exists $c_{*}>0$ such that

$$\begin{aligned} {\mathbb {P}} \left\{ \left| \frac{R_{8}^{n}({\tilde{\theta }}_{n})}{D^{n}({\tilde{\theta }}_{n})} \right| \ge c_{*}^{\prime } \right\} \rightarrow 1 \end{aligned}$$

(A.18)

as $n\rightarrow \infty$.

The limit (A.16). Thanks to (B3), it suffices to show

$$\begin{aligned} \left| \frac{R_{6}^{n}({\tilde{\theta }}_{n})}{D^{n}({\tilde{\theta }}_{n})} \right| {\mathbf {1}}_{A_{n}(K)} \rightarrow ^{p} 0 \end{aligned}$$

(A.19)

as $n\rightarrow \infty$. By the conditioning argument as in the proof of Lemma 5.2, we have

$$\begin{aligned} {\mathbb {E}}\left\{ \frac{|R_{6}^{n}(\tilde{\theta _{n}})|{\mathbf {1}}_{A_{n}(K)}}{|D^{n}({\tilde{\theta }}_{n})|} \right\}&\le {\mathbb {E}}\Biggl \{ \frac{{\mathbf {1}}_{A_{n}(K)} }{ | D^{n} | } \sum _{i} {\mathbb {E}}\big \{ \big | B^{1}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )\big | \nonumber \\&\quad \times \big | B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}_{n}-\theta }\big ) - B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )^{+}\big ) \big | \mid \sigma ({\mathcal {T}}) \big \} \Biggr \}. \end{aligned}$$

(A.20)

Let $p>1$ and $q>1$ be conjugate indices: $1/p + 1/q = 1$. Using (conditional) Hölder’s inequality and Theorem 2.8, we have

$$\begin{aligned}&\sum _{i} {\mathbb {E}}\big \{ \big | B^{1}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )\big | \big | B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}_{n}-\theta }\big ) - B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )^{+}\big ) \big | \mid \sigma ({\mathcal {T}}) \big \} \nonumber \\&\le \biggl ( \sum _{i}{\mathbb {E}}\big \{\big |B\big (\big (I_{i}^{1,n}\big )_{\theta }\big )\big |^{p} \mid \sigma ({\mathcal {T}}) \big \}\biggr )^{1/p}\nonumber \\&\quad \biggl (\sum _{i}{\mathbb {E}}\bigl \{\big |B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )_{{\tilde{\theta }}_{n}-\theta }\big ) - B^{2}\big ({\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta }\big )^{+}\big )\big |^{q} \mid \sigma ({\mathcal {T}}) \bigr \}\biggr )^{1/q} \nonumber \\&\lesssim r_{n}^{H_{1}-1/p} \big (\big (\#{\mathcal {I}}^{1,n}\big )\big |\theta -{\tilde{\theta }}_{n}\big |^{qH_{2}}\big )^{1/q}. \end{aligned}$$

(A.21)

Plugging (A.21) into (A.20) with $p=\bigl (1-\frac{H_{2}}{1-\varsigma }\bigr )^{-1}$ and $q=\bigl ( \frac{H_{2}}{1-\varsigma } \bigr )^{-1}$, we obtain

$$\begin{aligned} {\mathbb {E}}\left\{ \frac{|R_{6}^{n}(\tilde{\theta _{n}})|{\mathbf {1}}_{A_{n}(K)}}{|D^{n}({\tilde{\theta }}_{n})|} \right\} \le {\mathbb {E}}\Biggl \{ \frac{r_{n}^{H_{1}+H_{2}-1+\mu + \frac{H_{2}\varsigma }{1-\varsigma } - \mu }\bigl ( (\#{\mathcal {I}}^{1,n})|\theta -{\tilde{\theta }}_{n}|^{1-\varsigma } \bigr )^{\frac{H_{2}}{1-\varsigma }} }{|D^{n}({\tilde{\theta }}_{n})|}{\mathbf {1}}_{A_{n}(K)}\Biggr \}. \end{aligned}$$

(A.22)

Now we can choose sufficiently small $\mu >0$ satisfying $\frac{H_{2}\varsigma }{1-\varsigma } - \mu >0$. Applying (B3), (C3) and Jensen’s inequality to (A.22), we complete the proof of (A.16).

The limit (A.17). This can be proved in the same way as (A.4). Note that in this case the term corresponding to $R_{5,2}^{n}({\tilde{\theta }})$ is

$$\begin{aligned} R_{7,2}^{n}({\tilde{\theta }}_{n})&:=\sum _{i,k:\overline{I_{i}^{1,n}}\le T,\ \overline{I_{k}^{2,n}}\le T} \big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )^{+} \big ) \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\quad \times \big \langle h^{1}\big ( \big (I_{k}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{i}^{1,n}\big )_{\theta })^{+}\big ) \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})}. \end{aligned}$$

This is estimated as follows:

$$\begin{aligned} \big |R_{7,2}^{n}({\tilde{\theta }}_{n})\big |&\lesssim r_{n}^{H_{1}+H_{2}} \sum _{i,k:\overline{I_{i}^{1,n}}\le T,\ \overline{I_{k}^{2,n}}\le T}\big \langle h^{1}\big ( \big (I_{i}^{1,n}\big )_{\theta } \big ), h^{2}\big ( {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}\big (\big (I_{k}^{1,n}\big )_{\theta }\big )^{+}\big ) \big \rangle _{L^{2}([0,T+2\delta ] ; {\mathbb {R}}^{2})} \\&\lesssim r_{n}^{H_{1}+2H_{2}} \big (\#{\mathcal {I}}^{1,n}\big ) \\&= r_{n}^{2H_{1}-1+\mu } r_{n}^{2H_{2}-1+\mu } r_{n}^{2-H_{1}-2\mu }\big (\#{\mathcal {I}}^{1,n}\big ). \end{aligned}$$

This, combined with (B4), completes the proof of (A.17).

The limit (A.18). Finally we analyze the term $R_{8}^{n}({\tilde{\theta }}_{n})$. Recall that

$$\begin{aligned} \big |R_{8}^{n}({\tilde{\theta }}_{n})\big |&= |\rho | c_{H_{1}}c_{H_{2}} \sum _{i:\overline{I_{i}^{n}}\le T} \int _{ (I_{i}^{1,n})_{\theta } }du\int _{{\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}((I_{i}^{1,n})_{\theta })^{+}}dv\ \beta (u,v) \\&\quad \times |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}}. \end{aligned}$$

Since $(I_{i}^{1,n})_{\theta } \subset {\mathcal {I}}^{2,n}_{\theta -{\tilde{\theta }}_{n}}((I_{i}^{1,n})_{\theta })^{+}$ for all i with $\overline{I_{i}^{1,n}}\le T$, it holds that

$$\begin{aligned} \big |R_{8}^{n}({\tilde{\theta }}_{n})\big |&\ge |\rho | c_{H_{1}}c_{H_{2}} \sum _{i:\overline{I_{i}^{1,n}}\le T} \int _{ (I_{i}^{1,n})_{\theta } }du\int _{(I_{i}^{1,n})_{\theta }}dv\ \beta (u,v) \nonumber \\&\quad \times |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}}. \end{aligned}$$

(A.23)

To obtain a lower bound for (A.23), we note the following fact: for any $\epsilon >0$, it holds that

$$\begin{aligned}&\sum _{i:\overline{I_{i}^{1,n}}\le T}\int _{ (I_{i}^{1,n})_{\theta } }du\int _{(I_{i}^{1,n})_{\theta }}dv\ \beta (u,v) |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}} \nonumber \\&\quad \ge \sum _{i:\overline{I_{i}^{1,n}} \le T, \underline{I_{i}^{1,n}} \ge \epsilon }\int _{ (I_{i}^{1,n})_{\theta } }du\int _{(I_{i}^{1,n})_{\theta }}dv\ \beta (u,v) |u-v|^{H_{1}+H_{2}-2} u^{H_{1}-H_{2}} v^{H_{2}-H_{1}}\nonumber \\&\quad \gtrsim \sum _{i:\overline{I_{i}^{1,n}} \le T, \underline{I_{i}^{1,n}} \ge \epsilon }\int _{ (I_{i}^{1,n})_{\theta } }du\int _{(I_{i}^{1,n})_{\theta }}dv\ |u-v|^{H_{1}+H_{2}-2}. \end{aligned}$$

(A.24)

Combining (A.23) nad (A.24), we obtain

$$\begin{aligned} |R_{8}^{n}({\tilde{\theta }}_{n})|&\gtrsim \sum _{i:\overline{I_{i}^{1,n}} \le T, \underline{I_{i}^{1,n}} \ge \epsilon } | I_{i}^{1,n} |^{H_{1}+H_{2}} \end{aligned}$$

(A.25)

for any $\epsilon >0$. Therefore we obtain

$$\begin{aligned} \left| \frac{R_{8}^{n}({\tilde{\theta }}_{n})}{D^{n}({\tilde{\theta }}_{n})} \right| \gtrsim \frac{ \sum _{i:\overline{I_{i}^{1,n}} \le T, \underline{I_{i}^{1,n}} \ge \epsilon } | I_{i}^{1,n} |^{H_{1}+H_{2}} }{ \sqrt{\sum _{i} |I_{i}^{1,n}|^{2H_{1}} } \sqrt{ \sum _{j} |I_{j}^{2,n}|^{2H_{2}} } }. \end{aligned}$$

We complete the proof of (A.18) by (B2) (especially (3.2)).

As we noted above, if the true parameter value $\theta$ is positive, then ${\tilde{\theta }}_{n}$ is in $\Theta _{\ge 0}$ for sufficiently large n since $|{\tilde{\theta }}_{n} - \theta |\le \rho _{n}$. Therefore we can conclude (5.4) by (A.16)–(A.18) if $\theta >0$. However ${\tilde{\theta }}_{n}$ may be negative for any n when $\theta =0$. In order to obtain (5.4) when $\theta =0$, it suffices to show (A.16)–(A.18) hold for $\theta =0$ and ${\tilde{\theta }}_{n} \in [-\rho _{n},0)$. This is completely analogous to the case where ${\tilde{\theta }}_{n} \in \Theta _{\ge 0}$, so that we only add a few remarks and omit the proof.

If $\theta =0$ and ${\tilde{\theta }}_{n} \in [-\rho _{n},0)$, then (A.15) becomes

$$\begin{aligned} R_{2}^{n}\big ({\tilde{\theta }}_{n}\big )&= \sum _{i,j:\overline{I_{j}^{2,n}} \le T } B^{1}\big (I_{i}^{1,n}\big ) B^{2}\big (I_{j}^{2,n}\big ) {\mathbf {1}}_{ (I_{i}^{1,n})_{{\tilde{\theta }}_{n}} \cap I_{j}^{2,n} \ne \emptyset } \nonumber \\&= \sum _{j:\overline{I_{j}^{2,n}} \le T } \left( B^{1}\big ({\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}\big (I_{j}^{2,n}\big )_{-{\tilde{\theta }}_{n}}\big ) - B^{1}\big ({\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}\big (I_{j}^{2,n}\big )^{+} \big ) \right) B^{2}\big (I_{j}^{2,n}\big ) \nonumber \\&\quad +\sum _{j:\overline{I_{j}^{2,n}} \le T } \left( B^{1}({\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}\big (I_{j}^{2,n}\big )^{+}\big ) B^{2}\big (I_{j}^{2,n}\big ) - \big \langle h^{1}\big ( {\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}\big (I_{j}^{2,n}\big )^{+} \big ) , h^{2}\big ( I_{j}^{2,n} \big ) \big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})} \right) \nonumber \\&\quad + \sum _{j:\overline{I_{j}^{2,n}} \le T } \big \langle h^{1}\big ( {\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}\big (I_{j}^{2,n}\big )^{+} \big ) , h^{2}\big ( I_{j}^{2,n} \big )\big \rangle _{L^{2}([0,T+2\delta ];{\mathbb {R}}^{2})}. \end{aligned}$$

(A.26)

Here the symbol ${\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}(I_{j}^{2,n})^{+}$ denotes ${\mathcal {I}}_{{\tilde{\theta }}_{n}}^{1,n}(I_{j}^{2,n}) \cap [0,T+2\delta ]$ as before. Let ${\hat{R}}_{8}^{n}({\tilde{\theta }}_{n})$ denote the last term in (A.26). Then the inequality corresponding to (A.25) is

$$\begin{aligned} \big |{\hat{R}}_{8}^{n}({\tilde{\theta }}_{n})\big |&\gtrsim \sum _{j:\overline{I_{j}^{2,n}} \le T, \underline{I_{j}^{2,n}} \ge \epsilon } \big | I_{i}^{2,n} \big |^{H_{1} + H_{2}}. \end{aligned}$$

In particular, we use (3.3) of (B2) in this case. $\square$

Proof of Proposition 5.4

By the Hölder continuity of $B^{l}$, we have, for any $\epsilon >0$,

$$\begin{aligned} \big |X^{l}\big (I_{i}^{l,n}\big )^{2} - \sigma _{l}^{2} B^{l}\big (\big (I_{i}^{l,n}\big )_{a_{l}}\big )^{2}\big |&= \big |A^{l}\big (I_{i}^{l,n}\big )^{2} + 2\sigma _{l} A^{l}\big (I_{i}^{l,n}\big ) B^{l}\big (\big (I_{i}^{l,n}\big )_{a_{l}}\big ) \big |\\&\lesssim \big | I_{i}^{l,n} \big |^{2H_{l}} + \big | I_{i}^{l,n} \big |^{1+H_{l}-\epsilon } \\&\lesssim r_{n} \big | I_{i}^{l,n} \big | + r_{n}^{H_{l}-\epsilon } \big | I_{i}^{l,n} \big |. \end{aligned}$$

Since $H_{l}-\epsilon = (2H_{l}-1) +1 -H_{l} -\epsilon$, we obtain (5.6) by (B3) if we choose $\epsilon >0$ such that $1-H_{l}-\epsilon >0$. $\square$

Proof of Lemma 5.7

Note that

$$\begin{aligned} X^{1}\big (I_{i}^{1,n}\big ) X^{2}\big (I_{j}^{2,n}\big )&= A^{1}\big (I_{i}^{1,n}\big )A^{2}\big (I_{j}^{2,n}\big ) + \sigma _{2} A^{1}\big (I_{i}^{1,n}\big ) B^{2}\big (I_{j}^{2,n}\big ) \\&\quad + \sigma _{1} B^{1}\big (\big (I_{i}^{1,n}\big )_{\theta }\big ) A^{2}\big (I_{j}^{2,n}\big ) + \sigma _{1}\sigma _{2} B^{1}\big (\big (I_{i}^{1,n})_{\theta }\big ) B^{2}\big (I_{j}^{2,n}\big ). \end{aligned}$$

Obviously (5.7) is equivalent to

$$\begin{aligned} \sup _{{\tilde{\theta }} \in {\mathcal {G}}^{n} \cap \Theta _{\ge 0} } \left| \frac{ {\check{R}}_{2}^{n}({\tilde{\theta }}) - \sigma _{1}\sigma _{2}R_{2}^{n}({\tilde{\theta }}) }{D^{n}({\tilde{\theta }})} \right| \rightarrow ^{p} 0 \end{aligned}$$

(A.27)

and

$$\begin{aligned} \sup _{{\tilde{\theta }} \in {\mathcal {G}}^{n} \cap \Theta _{<0} } \left| \frac{ {\check{R}}_{2}^{n}({\tilde{\theta }}) - \sigma _{1}\sigma _{2}R_{2}^{n}({\tilde{\theta }}) }{D^{n}({\tilde{\theta }})} \right| \rightarrow ^{p} 0. \end{aligned}$$

(A.28)

We only prove (A.27). The proof of (A.28) is completely analogous. Since ${\tilde{\theta }} \in \Theta _{\ge 0}$, we have

$$\begin{aligned} \big |{\check{R}}_{2}^{n}({\tilde{\theta }}) -\sigma _{1}\sigma _{2}R_{2}^{n}({\tilde{\theta }})\big |&\le \sum _{i: \overline{I_{i}^{1,n}}\le T} \big | A^{1}\big (I_{i}^{1,n}\big ) A^{2}\big ({\mathcal {I}}_{-{\tilde{\theta }}}^{2,n}\big (I_{i}^{1,n}\big )_{{\tilde{\theta }}}\big ) \big | \\&\quad + \sum _{i: \overline{I_{i}^{1,n}}\le T} \big | \sigma _{2}A^{1}\big (I_{i}^{1,n}\big ) B^{2}\big ({\mathcal {I}}_{-{\tilde{\theta }}}^{2,n}\big (I_{i}^{1,n}\big )_{{\tilde{\theta }}}\big ) \big | \\&\quad + \sum _{j} \big | \sigma _{1} B^{1}\big ( {\mathcal {I}}_{\theta }^{1,n,\le T} \big (\big (I_{j}^{2,n}\big )_{\theta -{\tilde{\theta }}}\big ) \big ) A^{2}\big (I_{j}^{2,n}\big ) \big | \\&\lesssim r_{n} + r_{n}^{H_{1}-\epsilon } + r_{2}^{H_{2}-\epsilon }\\&= r_{n}^{H_{1}+H_{2}-1+\mu } \big ( r_{n}^{2-H_{1}-H_{2}-\mu } + r_{n}^{1-H_{2}-\mu -\epsilon } + r_{n}^{1-H_{1}-\mu -\epsilon } \big ). \end{aligned}$$

We can choose $\mu >0$ and $\epsilon >0$ such that $1-(H_{1}\vee H_{2}) -\mu -\epsilon >0$ so that we conclude (A.27) by (B3). $\square$