Estimation of the pointwise Hölder exponent of hidden multifractional Brownian motion using wavelet coefficients

Abstract

We propose a wavelet-based approach to construct consistent estimators of the pointwise Hölder exponent of a multifractional Brownian motion, in the case where this underlying process is not directly observed. The relative merits of our estimator are discussed, and we introduce an application to the problem of estimating the functional parameter of a nonlinear model.

Appendix

1.1 Proofs of (1.8) and (1.9)

First by using triangle inequality, we get

$$\begin{aligned}&|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|\nonumber \\&\qquad \le 2^{j/2}\sum _{l=0}^{2^{n-j}-1}\int _{\frac{l}{2^n}}^{\frac{l+1}{2^n}}|\psi (2^jt) ||Y(l2^{-n}+k2^{-j})-Y(t+k2^{-j})|{\,\mathrm {d}} t. \end{aligned}$$

(4.1)

Recall that $Y(t)=\Phi (X(t))$ for $t\in [0,1]$. Define the random variable $\Vert X\Vert _{\infty }$ to be

$$\begin{aligned} \Vert X\Vert _{\infty }=\sup _{t\in [0,1]}|X(t)|. \end{aligned}$$

(4.2)

Since $\theta$ is continuous and not equal to 0 almost everywhere, then $\{X(t)\}_{t\in [0,1]}$ is a Gaussian process with continuous trajectories, by applying Dudley’s theorem and Borell’s inequality (more precisely, with the same arguments for the proof of $\mathbb {E}(e^{\widetilde{V}})<+\infty$ on pp. 1445–1446 in Rosenbaum (2008). See also Ledoux and Talagrand (2010)), we can show that $\mathbb {E}(e^{\Vert X\Vert _{\infty }})<+\infty .$ This means all of $\Vert X\Vert _{\infty }$’s moments are finite. Hence, using the mean value theorem, we get

$$\begin{aligned} |Y(l2^{-n}+k2^{-j})-Y(t+k2^{-j})|\le C_1|X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|, \end{aligned}$$

(4.3)

where $C_1=\sup \limits _{s\in [-\Vert X\Vert _{\infty },\Vert X\Vert _{\infty }]}|\Phi '(s)|$ is a random variable. It follows from (4.1) and (4.3) that

$$\begin{aligned}&|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|\nonumber \\&\quad \le C_12^{j/2}\sum _{l=0}^{2^{n-j}-1}\int _{\frac{l}{2^n}}^{\frac{l+1}{2^n}} |\psi (2^jt)||X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|{\,\mathrm {d}} t. \end{aligned}$$

(4.4)

In order to get (1.8), we need (1.7), from which we see there exists a positive random variable $C_2$ with all finite moments such that

$$\begin{aligned}&|X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|\nonumber \\&\quad \le |\theta (l2^{-n}+k2^{-j})||B_{H(l2^{-n}+k2^{-j})}(l2^{-n}+k2^{-j}) -B_{H(t+k2^{-j})}(t+k2^{-j})|\nonumber \\&\qquad +\,|\theta (l2^{-n}+k2^{-j})-\theta (t+k2^{-j})||B_{H(t+k2^{-j})} (t+k2^{-j})|\nonumber \\&\quad \le C_2 \left( |l2^{-n}-t|^{H(t+k2^{-j})}|\log |l2^{-n}-t||^{1/2}+|l2^{-n}-t|\right) . \end{aligned}$$

(4.5)

Observe that for $t\in [l2^{-n},(l+1)2^{-n}]$,

$$\begin{aligned} |l2^{-n}-t|^{H(t+k2^{-j})}|\log |l2^{-n}-t||^{1/2}\ge |l2^{-n}-t| \end{aligned}$$

and

$$\begin{aligned} \sup \limits _{{}^{n\ge 0,j\le n,l\le 2^{n-j}-1,k\le 2^j-1}_{l2^{-n}\le t\le (l+1)2^{-n}}}|l2^{-n}-t|^{H(t+k2^{-j})-H(k2^{-j})}|\log |l2^{-n}-t||^{1/2}<+\infty . \end{aligned}$$

(4.6)

This together with the fact that $|l2^{-n}-t|\le 2^{-n}$ for $t\in [l2^{-n},(l+1)2^{-n}]$ yields there exists a positive random variable $C_3$ with all finite moments such that,

$$\begin{aligned} |X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|\le C_3 2^{-nH(2^{-j}k)}n^{1/2}. \end{aligned}$$

(4.7)

Then (1.8) results from (4.4) and (4.7). $\square$

Now we are going to prove (1.9). For $r\ge 1$, we consider the r-order moment of $|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|$ in (4.4). By applying the following two versions of Jensen’s inequality:

$$\begin{aligned} \left( \sum _{i=1}^n|a_i|\right) ^r\le n^{r-1}\sum _{i=1}^n|a_i|^r \quad \text{ and } \quad \left( \int _a^b|f(s)|{\,\mathrm {d}} s\right) ^r\le |b-a|^{r-1}\int _a^b|f(s)|^r{\,\mathrm {d}} s, \end{aligned}$$

(4.8)

and Cauchy–Schwarz inequality, we obtain

$$\begin{aligned}&\mathbb {E}|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|^r\nonumber \\&\quad \le 2^{jr/2}2^{-j(r-1)}\sum _{l=0}^{2^{n-j}-1} \int _{\frac{l}{2^n}}^{\frac{l+1}{2^n}}|\psi (2^jt)|^r\mathbb {E}\left( C_1|X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|\right) ^r{\,\mathrm {d}} t\nonumber \\&\quad \le \left( \sup _{s\in [0,1]}|\psi (s)|^r\right) 2^{-j(r/2-1)}\nonumber \\&\qquad \times \sum _{l=0}^{2^{n-j}-1}\int _{\frac{l}{2^n}}^{\frac{l+1}{2^n}} \left( \mathbb {E}\left( C_1^{2r}\right) \right) ^{1/2}\left( \mathbb {E}|X(l2^{-n}+k2^{-j})-X(t+k2^{-j}) |^{2r}\right) ^{1/2}{\,\mathrm {d}} t. \end{aligned}$$

(4.9)

Note that by Lemma 2.12 (i) in Ayache et al. (2011), there exists a constant $c_1>0$ which does not depend on n, l, j, k and H such that

$$\begin{aligned}&\mathbb {E}|B_{H(l2^{-n}+k2^{-j})}(l2^{-n}+k2^{-j})-B_{H(t+k2^{-j})}(t+k2^{-j})|^2\\&\quad \le c_1|l2^{-n}+k2^{-j}-(t+k2^{-j}) |^{2\max \{H(l2^{-n}+k2^{-j}),H(t+k2^{-j})\}}\\&\quad \le c_1|l2^{-n}+k2^{-j}-(t+k2^{-j}) |^{2H(l2^{-n}+k2^{-j})}\\&\quad =c_1|l2^{-n}-t|^{2H(k2^{-j})}|l2^{-n}-t |^{2H(l2^{-n}+k2^{-j})-2H(k2^{-j})}. \end{aligned}$$

Therefore by using again (4.6) and the fact that $|l2^{-n}-t|\le 2^{-n}$, there exists some constant $c>0$ such that

$$\begin{aligned} \mathbb {E}|B_{H(l2^{-n}+k2^{-j})}(l2^{-n}+k2^{-j})-B_{H(t+k2^{-j})}(t+k2^{-j})|^2\le c2^{-2nH(k2^{-j})}. \end{aligned}$$

(4.10)

Using (4.10) and similar computations as in (4.5), we obtain there exists $c_2>0$ such that

$$\begin{aligned} \mathbb {E}|X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|^2\le c_22^{-2nH(k2^{-j})}. \end{aligned}$$

(4.11)

By using the fact that all the moments of Gaussian variable are equivalent, we get there exists some constant $c_3>0$ (only depending on r) such that

$$\begin{aligned} \mathbb {E}|X(l2^{-n}+k2^{-j})-X(t+k2^{-j})|^{2r}\le c_32^{-2rnH(k2^{-j})}. \end{aligned}$$

(4.12)

Finally it results from (4.9) and (4.12) that

$$\begin{aligned} \mathbb {E}|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|^r\le c 2^{-r(nH(2^{-j}k)+j/2)}, \end{aligned}$$

where $c=\big (\sup _{s\in [0,1]}|\psi (s)|^r\big )\big (c_3\mathbb {E}(C_1^{2r})\big )^{1/2}$. Therefore (1.9) has been proven. $\square$

1.2 Proof of (1.11)

First notice that, by the definition of $\nu _{t_0,2^j}$, we have

$$\begin{aligned} |card(\nu _{t_0,2^j})- 2^{j+1}\epsilon _j|\le 3 \end{aligned}$$

(4.13)

as $j\rightarrow +\infty$, because $2^{j}\epsilon _j\ge 1$.

It follows from (4.8), Cauchy–Schwarz inequality, (4.13) and the fact that $(a+b)^4\le 2^3(a^4+b^4)$ that

$$\begin{aligned}&\mathbb {E}|\widehat{V_{n,t_0,j}}-V_{Y,t_0,j}|^2\le card(\nu _{t_0,2^j})\sum _{k\in \nu _{t_0,2^j}}\mathbb {E}\big |\widehat{d_{Y,n}}(2^{-j},k)^2-d_Y(2^{-j},k)^2\big |^2\nonumber \\&\quad \le 3\times 2^j\epsilon _j\sum _{k\in \nu _{t_0,2^j}}2^{3/2}\left( \mathbb {E}|\widehat{d_{Y,n}}(2^{-j},k)|^4+\mathbb {E}|d_Y(2^{-j},k)|^4\right) ^{1/2}\nonumber \\&\qquad \times \left( \mathbb {E}|\widehat{d_{Y,n}}(2^{-j},k)-d_Y(2^{-j},k)|^4\right) ^{1/2}. \end{aligned}$$

(4.14)

Roughly speaking (and it can be proven without efforts), since the trajectory $\{\Phi (X(t))\}_{t\ge 0}$ is at least as smooth as $\{X(t)\}_{t\ge 0}$, then for $r\ge 1$, there exists a constant $c_4>0$ (only depending on r) such that,

$$\begin{aligned}&\mathbb {E}|\widehat{d_{Y,n}}(2^{-j},k)|^r\le c_42^{-jr\left( H(k2^{-j})+1/2\right) };\nonumber \\&\mathbb {E}|d_Y(2^{-j},k)|^r\le c_42^{-jr\left( H(k2^{-j})+1/2\right) }. \end{aligned}$$

(4.15)

Then it results from (4.14), (4.15), (1.9) and (4.13) that

$$\begin{aligned} \mathbb {E}|\widehat{V_{n,t_0,j}}-V_{Y,t_0,j}|^2\le 9\times 2^{2j}\epsilon _j^2\left( (8c_4)^{1/2}2^{-j(2H(k2^{-j})+1)}\right) \left( c^{1/2}2^{-2nH(k2^{-j})-j}\right) . \end{aligned}$$

(4.16)

In view of the equivalence relation between $H(k2^{-j})$ and $H(t_0)$ as $k\in \nu _{t_0,2^j}$ and $j\rightarrow +\infty$, (1.11) finally results from (4.16). $\square$

1.3 Proof of Lemma 2

By using Chebyshev’s inequality, (2.9) and the condition that $\epsilon _j=\mathcal {O}(j^{-1})$, we get for any $\eta >0$,

$$\begin{aligned}&\mathbb {P}\left( 2^{j(2H(t_0)+1/2)}\epsilon _j^{-1/2}\big |V_{X,t_0,j}-\mathbb {E}(V_{X,t_0,j})\big |\ge \eta \right) \le \frac{ 2^{j(4H(t_0)+1)}\epsilon _j^{-1} Var \left( V_{X,t_0,j}\right) }{\eta ^2}\nonumber \\&\qquad \le c\frac{2^{j(4H(t_0)+1)}2^{-j(4H(t_0)+1)}}{\eta ^2}= \frac{c}{\eta ^2}, \end{aligned}$$

(4.17)

where $c>0$ is some constant which does not depend on j. This implies

$$\begin{aligned} V_{X,t_0,j}=\mathbb {E}(V_{X,t_0,j})+\mathcal {O}_{\mathbb {P}}\left( 2^{-j(2H(t_0)+1/2)}\epsilon _j^{1/2}\right) . \end{aligned}$$

(4.18)

Then it follows from (2.8), (4.18) and the fact that $\lim _{j\rightarrow +\infty }2^{-j/2}\epsilon _{j}^{-1/2}=0$ that

$$\begin{aligned}&V_{X,t_0,j}=2c_2(t_0)2^{-2jH(t_0)}\epsilon _j+\mathcal {O}_{a.s.}\left( 2^{-2jH(t_0)}\left( 2^{-j}+j\epsilon _j^2+2^{-j}j^4\epsilon _j\right) \right) \nonumber \\&\qquad \qquad \quad +\,\mathcal {O}_{\mathbb {P}}(2^{-j(2H(t_0)+1/2)}\epsilon _j^{1/2})\nonumber \\&\qquad \qquad =2c_2(t_0)2^{-2jH(t_0)}\epsilon _j+\mathcal {O}_{a.s.}\left( 2^{-2jH(t_0)}\left( j\epsilon _j^2+2^{-j}j^4\epsilon _j\right) \right) \nonumber \\&\qquad \qquad \quad +\,\mathcal {O}_{\mathbb {P}}\left( 2^{-j(2H(t_0)+1/2)}\epsilon _j^{1/2}\right) . \end{aligned}$$

(2.11) has been proven. Note that (2.12) follows straightforwardly from (2.11). Now we only need to show (2.13) holds. From (4.18) and Chebyshev’s inequality, we observe that there exists a constant $c>0$ which does not depend on $\eta$ nor on j such that for any $\eta >0$,

$$\begin{aligned} \mathbb {P}\left( (2^j\epsilon _j)^{(1-\delta )/2}\left| \frac{V_{X,t_0,j}}{\mathbb {E}(V_{X,t_0,j})}-1\right| >\eta \right) \le \frac{c(2^j\epsilon _j)^{-\delta }}{\eta ^2}. \end{aligned}$$

(4.19)

Since $\sum \limits _{j=1}^{+\infty }(2^j\epsilon _j)^{-\delta }<+\infty$, then applying Borel–Cantelli’s lemma leads to

$$\begin{aligned} \frac{V_{X,t_0,j}}{\mathbb {E}(V_{X,t_0,j})}-1=\mathcal {O}_{a.s.} \left( (2^j\epsilon _j)^{-1/2+\delta /2}\right) . \end{aligned}$$

Further observe that

$$\begin{aligned} \mathbb {E}(V_{X,t_0,j})=\mathcal {O}_{a.s.}\left( 2^{-2jH(t_0)}\epsilon _j\right) . \end{aligned}$$

Therefore (2.13) follows. Lemma 2 has been proven. $\square$

1.4 Proof of Theorem 2

The proof of Theorem 2 relies heavily on the following Propositions 4 and 5.

Proposition 4

For any $t_0\in (0,1)$ and any integer $j\ge 1$, denote by

$$\begin{aligned} U_{t_0,j}=\sqrt{card(\nu _{t_0,2^j})}\left( \frac{1}{card(\nu _{t_0,2^j})} \sum _{k\in \nu _{t_0,2^j}}\frac{d_X(2^{-j},k)^2}{\mathbb {E}(d_X(2^{-j},k)^2)}-1\right) . \end{aligned}$$

Then

$$\begin{aligned} \big (U_{t_0,j},U_{t_0,j+1}\big )\xrightarrow [j\rightarrow +\infty ]{dist}\mathcal N(0,\Sigma ), \end{aligned}$$

where $\Sigma =(\sigma _{ij})_{i,j\in \{1,2\}}$ with $\sigma _{11}=\sigma _{22}=c_3(t_0)/(c_2(t_0))^2$ and

$$\begin{aligned} \sigma _{12}= & {} \sigma _{21}=c_4(t_0):=\frac{(2c_0)^{1/2}}{c_2(t_0)^2}\left( C_2(t_0,1/2)^22^{2H(t_0)+1}+2c_1(t_0,t_0)^2\right. \nonumber \\&\left. +\,2^{2Q-2H(t_0)+1}C_1(H(t_0),H(t_0),Q,2)^2\theta (t_0)^4\sum _{l\in \mathbb {Z},|l|\ge 2}\frac{1}{|l|^{4Q-4H(t_0)}}\right) . \end{aligned}$$

(4.20)

Proof

Following the similar method as in Bardet (2000), we show that for the empirical average of $d_X( 2^{-j},k)^2/\mathbb {E}(d_X(2^{-j},k)^2)$, a multivariate central limit theorem holds thanks to a Lindeberg’s condition. More precisely, for j big enough and for $k\in \nu _{t_0,2^j}$, $k'\in \nu _{t_0,2^{j+1}}$, denote by

$$\begin{aligned} T_{j,k,k'}=Cov\left( \frac{d_X( 2^{-j},k)^2}{\mathbb {E}(d_X(2^{-j},k)^2)} ,\frac{d_X( 2^{-{(j+1)}},k')^2}{\mathbb {E}(d_X(2^{-{(j+1)}},k')^2)} \right) , \end{aligned}$$

(4.21)

a Lindeberg’s condition thus can be deduced from the following relations:

• For $|2k-k'|=0$,

  $$\begin{aligned} T_{j,k,k'}=2^{2H(2^{-j}k)+2}\left( \frac{C_2(2^{-j}k,1/2)}{c_2(2^{-j}k)}\right) ^2 +\mathcal {O}(2^{-j}j^4). \end{aligned}$$

  (4.22)

• For $|2k-k'|=1$,

  $$\begin{aligned} T_{j,k,k'}=2\left( \frac{c_1(2^{-j}k,2^{-(j+1)}k')^2}{c_2(2^{-j}k)c_2 (2^{-(j+1)}k')}\right) +\mathcal {O}\left( T(2k,k',Q,2^{-j},2^{-j})\right) . \end{aligned}$$

  (4.23)

• For $|2k-k'|\ge 2$,

  $$\begin{aligned} T_{j,k,k'}= & {} 2\left( \frac{(C_1(H(2^{-j}k)+H(2^{-(j+1)}k'),Q,2)\theta (2^{-j}k) \theta (2^{-(j+1)}k'))^2}{c_2(2^{-j}k)c_2(2^{-(j+1)}k')|2k-k'|^{4Q-2H(2^{-j}k) -2H(2^{-(j+1)}k')}}\right) \nonumber \\&+\,\mathcal {O}\left( T(k,k',Q,2^{-j},2^{-(j+1)})\right) . \end{aligned}$$

  (4.24)

To show (4.22)–(4.24) hold, we first recall that if $(Z,Z')$ has a zero-mean joint normal distribution, then (see e.g., Lemma 5.3.4 in Peng (2011b))

$$\begin{aligned} Cov\big (Z^2,{Z'}^2\big )=2\big (Cov(Z,Z')\big )^2. \end{aligned}$$

(4.25)

We thus observe, from (4.21) and (4.25), that

$$\begin{aligned} T_{j,k,k'}=2\left( Cov\left( \frac{d_X( 2^{-j},k)}{\sqrt{\mathbb {E}(d_X(2^{-j},k)^2)}} , \frac{d_X( 2^{-{(j+1)}},k')}{\sqrt{\mathbb {E}(d_X(2^{-{(j+1)}},k')^2)}} \right) \right) ^2. \end{aligned}$$

(4.26)

Therefore (4.22) results from (4.26), (2.3) and (2.6). In order to obtain (4.23), it suffices to take $l=2k$ for $k\in \nu _{t_0,2^j}$, then $l,k'$ belong to $\nu _{t_0,2^{j+1}}$ and satisfy $\sup _{s,t\in [0,1]}|(t-s)/(l-k')|=\sup _{s,t\in [0,1]}|s-t|\le 1$. This entails that (2.2) can be applied on $(a,b,l,k')$ by setting $a=b=2^{-(j+1)}$ and $|l-k'|=1$. As a consequence (4.23) follows from (4.26), (2.4) and (2.6). For proving (4.24), we just plug $a=2^{-j}$, $b=2^{-(j+1)}$ into (2.2) and then use (4.26) and (2.6). Using this identification of $T_{j,k,k'}$’s, we show the Lindeberg’s condition (the same as in Bardet (2000)) is verified and the central limit theorem holds. Now it remains to show

$$\begin{aligned}&\displaystyle Cov\big (U_{t_0,j},U_{t_0,j+1}\big )\xrightarrow [j\rightarrow +\infty ]{} c_4(t_0) \quad (\text{ given } \text{ in } \text{(5.20) });&\end{aligned}$$

(4.27)

$$\begin{aligned}&\displaystyle Var(U_{t_0,j})\xrightarrow [j\rightarrow +\infty ]{}\frac{c_3(t_0)}{c_2(t_0)^2}.&\end{aligned}$$

(4.28)

We only prove (4.27) holds since (4.28) can be followed by quite a similar way. Remark that $\lim _{j\rightarrow +\infty }\sup _{k\in \nu _{t_0,2^j}}2^{-j}k=t_0$; and for j large enough,

$$\begin{aligned}&card\left( \{k\in \nu _{t_0,2^j},k'\in \nu _{t_0,2^{j+1}},|2k-k'|=0\}\right) =card(\nu _{t_0,2^j});\\&card\left( \{k\in \nu _{t_0,2^j},k'\in \nu _{t_0,2^{j+1}},|2k-k'|=l\}\right) =2card(\nu _{t_0,2^j})+\mathcal {O}(1) \quad \text{ for }~l\ge 1; \end{aligned}$$

and for a function f continuous on $t_0$,

$$\begin{aligned} \lim \limits _{j\rightarrow +\infty }\frac{1}{card(\nu _{t_0,2^j})} \sum \limits _{k\in \nu _{t_0,2^j}}f(2^{-j}k)=f(t_0). \end{aligned}$$

Considering all the above facts and (4.22)–(4.24), (5.49)–(5.51) in Jin et al. (2015), $card(\nu _{t_0,2^j})\sim 2^{j}\epsilon _j$ and assumption (A3), we obtain

$$\begin{aligned}&Cov\left( U_{t_0,j},U_{t_0,j+1}\right) =\frac{1}{\sqrt{card(\nu _{t_0,2^j}) card(\nu _{t_0,2^{j+1}})}}\nonumber \\&\qquad \times \left( \sum _{(k\in \nu _{t_0,2^j},k'\in \nu _{t_0,2^{j+1}},2k=k')}T_{j,k,k'} +\sum _{|2k-k'|=1}T_{j,k,k'}\right. \\&\left. \qquad +\sum _{l=2}^{+\infty }\sum _{|2k-k'|=l}T_{j,k,k'}\right) \xrightarrow [j\rightarrow +\infty ]{}c_4(t_0)~~~~\text{(given } \text{ in } \text{(5.20)) }. \end{aligned}$$

Finally, we have proved Proposition 4. $\square$

Proposition 5

(Multivariate delta rule, see e.g., Oehlert (1992)) Let the estimators $\{(X_n,Y_n)\}_{n\in \mathbb N}$ (valued in $(0,+\infty )^2$) of $(\theta _1,\theta _2)$ satisfy the following central limit theorem:

$$\begin{aligned} h(n)\left( (X_n,Y_n)-(\theta _1,\theta _2)\right) \xrightarrow [n\rightarrow +\infty ]{dist}\mathcal N(0,\Sigma ), \end{aligned}$$

where $\Sigma$ denotes the covariance matrix of the limit distribution and $(h(n))_n$ is a sequence of positive numbers tending to infinity. Let $g:~(0,+\infty )^2\rightarrow \mathbb R^p$ ($p=1$ or 2) belong to $C^1((0,+\infty )^2)$, then the following convergence in law holds:

$$\begin{aligned} h(n)\left( g(X_n,Y_n)-g(\theta _1,\theta _2)\right) \xrightarrow [n\rightarrow +\infty ]{dist}\mathcal N\left( 0,\nabla g(\theta _1,\theta _2)^T\Sigma \nabla g(\theta _1,\theta _2)\right) , \end{aligned}$$

where $\nabla g(\theta _1,\theta _2)^T$ denotes the transpose of the gradient of g on $(\theta _1,\theta _2)$.

Note that if $p=2$ in Proposition 5, the gradient of g becomes a Jacobian matrix.

Proof of Theorem 2

For simplifying notation we denote by

$$\begin{aligned} \widehat{U_{t_0,j}}=\frac{1}{2^{j+1}\epsilon _j} \frac{\sum _{k\in \nu _{t_0,2^j}}d_X(2^{-j},k)^2}{c_2(t_0)2^{-j(2H(t_0)+1)}}-1. \end{aligned}$$

Therefore the following decomposition holds:

$$\begin{aligned}&\sqrt{2^{j+1}\epsilon _j}\widehat{U_{t_0,j}}-U_{t_0,j}=\frac{1}{\sqrt{2^{j+1}\epsilon _j}}\sum _{k\in \nu _{t_0,2^j}}d_X(2^{-j},k)^2\nonumber \\&\qquad \times \left( \frac{1}{c_2(t_0) 2^{-j(2H(t_0)+1)}}-\frac{1}{\mathbb {E}(d_X(2^{-j},k)^2)}\sqrt{\frac{2^{j+1}\epsilon _j}{card(\nu _{t_0,2^j})}}\right) \nonumber \\&\qquad +\left( \sqrt{card(\nu _{t_0,2^j})}-\sqrt{2^{j+1}\epsilon _j}\right) \nonumber \\&\quad =\frac{1}{\sqrt{2^{j+1}\epsilon _j}}\sum _{k\in \nu _{t_0,2^j}}d_X(2^{-j},k)^2\nonumber \\&\qquad \times \left( \frac{\mathbb {E}\left( d_X(2^{-j},k)^2\right) \sqrt{card(\nu _{t_0,2^j})}-c_2(t_0) 2^{-j(2H(t_0)+1)}\sqrt{2^{j+1}\epsilon _j}}{c_2(t_0) 2^{-j(2H(t_0)+1)}\mathbb {E}\left( d_X(2^{-j},k)^2\right) \sqrt{card(\nu _{t_0,2^j})}}\right) \nonumber \\&\qquad +\left( \sqrt{card(\nu _{t_0,2^j})}-\sqrt{2^{j+1}\epsilon _j}\right) . \end{aligned}$$

(4.29)

Since the fact that $card(\nu _{t_0,2^j})=2^{j+1}\epsilon _j+\mathcal {O}(1)$ implies

$$\begin{aligned} \sqrt{card(\nu _{t_0,2^j})}=\sqrt{2^{j+1}\epsilon _j}+\mathcal {O}\left( 2^{-j/2}\epsilon _j^{-1/2}\right) . \end{aligned}$$

(4.30)

And also by taking $r=4$ in (4.15), we have

$$\begin{aligned} \mathbb {E}|d_X(2^{-j},k)|^4=\mathcal {O}\left( 2^{-j(4H(t_0)+2)}\right) . \end{aligned}$$

(4.31)

By (2.6) (4.29), (4.8), (4.30) and (4.31), we then obtain

$$\begin{aligned}&\mathbb {E}\left| \sqrt{2^{j+1}\epsilon _j}\widehat{U_{t_0,j}}-U_{t_0,j}\right| ^2 =\frac{card(\nu _{t_0,2^j})^2}{card(\nu _{t_0,2^j})}\mathcal {O}\left( 2^{-j(4H(t_0)+2)}\right) \nonumber \\&\qquad \times \left( \frac{\big (c_2(t_0)2^{-j(2H(t_0)+1\big )}+\mathcal {O}\big (2^{-j(2H(t_0) +2)}j^4)\big )(\sqrt{2^{j+1}\epsilon _j}+\mathcal {O}\big (2^{-(j+1)/2}\epsilon _j^{-1/2})\big )}{\big (c_2(t_0) 2^{-j(2H(t_0)+1)}\big )^2\sqrt{2^{j+1}\epsilon _j}}\right. \nonumber \\&\left. \qquad -\,\frac{1}{c_2(t_0) 2^{-j(2H(t_0)+1)}}\right) ^2+\left( \sqrt{2^{j+1}\epsilon _j}+\mathcal {O}\big (2^{-(j+1)/2} \epsilon _j^{-1/2}\big )-\sqrt{2^{j+1}\epsilon _j}\right) ^2\nonumber \\&\quad =\mathcal {O}\big (2^{-j}\epsilon _jj^8+(2^{j}\epsilon _j)^{-1}\big ). \end{aligned}$$

It follows from Markov’s inequality that there exists $c>0$ such that

$$\begin{aligned} \mathbb {P}\left( \big |\sqrt{2^{j+1}\epsilon _j}\widehat{U_{t_0,j}}-U_{t_0,j}\big |> \eta \right)&\le \eta ^{-2}\mathbb {E}\big |\sqrt{2^{j+1}\epsilon _j}\widehat{U_{t_0,j}}-U_{t_0,j}\big |^2\\&\le c\left( 2^{-j}\epsilon _jj^8+(2^{j}\epsilon _j)^{-1}\right) . \end{aligned}$$

Assumption (A2) then allows us to apply Borel–Cantelli’s lemma to obtain

$$\begin{aligned} \big |\sqrt{2^{j+1}\epsilon _j}\widehat{U_{t_0,j}}-U_{t_0,j}\big |\xrightarrow []{a.s.}0. \end{aligned}$$

Therefore, it follows from Proposition 4 and continuous mapping theorem that

$$\begin{aligned} \sqrt{2^{j+1}\epsilon _j}\left( \widehat{U_{t_0,j}},\sqrt{2\epsilon _{j+1}/\epsilon _j} \widehat{U_{t_0,j+1}}\right) \xrightarrow [j\rightarrow +\infty ]{dist}\mathcal {N}(0,\Sigma ). \end{aligned}$$

Since $\lim _{j\rightarrow +\infty }\sqrt{2\epsilon _{j+1}/{\epsilon _j}}=\sqrt{2c_0}$, using Slutsky’s theorem, we get

$$\begin{aligned} \sqrt{2^{j+1}\epsilon _j}\left( \widehat{U_{t_0,j}},\sqrt{2c_0} \widehat{U_{t_0,j+1}}\right) \xrightarrow [j\rightarrow +\infty ]{dist}\mathcal N(0,\Sigma ). \end{aligned}$$

This is in fact equivalent to

$$\begin{aligned} \sqrt{2^{j+1}\epsilon _j}\left( \widehat{U_{t_0,j}},\widehat{U_{t_0,j+1}}\right) \xrightarrow [j\rightarrow +\infty ]{dist}\mathcal N(0,\widetilde{\Sigma }), \end{aligned}$$

(4.32)

with $\widetilde{\Sigma }=\left( {}_{(2c_0)^{-1/2}\sigma _{21}}^{\sigma _{11}} ~{}_{(2c_0)^{-1}\sigma _{22}}^{(2c_0)^{-1/2}\sigma _{12}}\right)$. Then by applying Proposition 5 to (4.32) with $g_1(x,y)=(\log (x),\log (y))$, $(X_j,Y_j)=\big (\widehat{U_{t_0,j}}+1,\widehat{U_{t_0,j+1}}+1\big )$, $(\theta _1,\theta _2)=(1,1)$ and $h(j)=\sqrt{2^{j+1}\epsilon _j}$, we get

$$\begin{aligned}&\sqrt{2^{j+1}\epsilon _j}\left( \log \left( \sum _{k\in \nu _{t_0,2^i}}d_X(2^{-i},k)^2\right) +2iH(t_0)\log (2)-\log (2c_2(t_0)\epsilon _i)\right) _{i=j,j+1}\nonumber \\&\quad ~~\xrightarrow [j\rightarrow +\infty ]{dist}\mathcal N(0,\widetilde{\Sigma }), \end{aligned}$$

(4.33)

because the Jacobian matrix $\nabla g_1(1,1)=Id$.

Now we apply again Proposition 5 to (4.33) with $g_2(x,y)=\frac{x-y}{2\log 2}$, to get

$$\begin{aligned} \sqrt{2^{j+1}\epsilon _j}\Big (\widehat{H_{X,2^j}}(t_0)-H(t_0)\Big ) \xrightarrow [j\rightarrow +\infty ]{dist}\mathcal N\big (0,\tilde{c}(t_0)\big ), \end{aligned}$$

where

$$\begin{aligned} \tilde{c}(t_0)=\frac{\big ({}_{-1}^{~1}\big )^T\widetilde{\Sigma } \big ({}_{-1}^{~1}\big )}{(2\log 2)^2}=\frac{1}{(2\log 2)^2}\Big (\big ((2c_0)^{-1}+1\big )\frac{c_3(t_0)}{c_2(t_0)^2} -2(2c_0)^{-1/2}c_4(t_0)\Big ), \end{aligned}$$

(4.34)

because $\nabla g_2(0,0)=\frac{1}{2\log 2}\big ({}_{-1}^{~1}\big )$. $\square$

1.5 Proof of Theorem 3

(3.8) and (3.9) are obvious by using Lemma 3 and the fact that convergences in probability and almost surely are preserved under continuous transformations. In order to show (3.10), we only need to verify

$$\begin{aligned} \sqrt{2^{j}\epsilon _j}|\widehat{H_{Y,2^j}}(t_0)-\widehat{H_{X,2^j}} (t_0)|\xrightarrow [j\rightarrow +\infty ]{a.s.}0. \end{aligned}$$

(4.35)

Equivalently, it suffices to show

$$\begin{aligned} \sqrt{2^{j}\epsilon _j}\Big |\log \left( \frac{V_{Y,t_0,j}}{|\Phi '(X(t_0)) |^2V_{X,t_0,j}}\right) -\log (1)\Big |\xrightarrow [j\rightarrow +\infty ]{a.s.}0. \end{aligned}$$

(4.36)

We have, by using the mean value theorem on $\log (\cdot )$,

$$\begin{aligned} \sqrt{2^{j}\epsilon _j}\left| \log \left( \frac{V_{Y,t_0,j}}{|\Phi '(X(t_0)) |^2V_{X,t_0,j}}\right) -\log (1)\right| =\frac{\sqrt{2^{j}\epsilon _j}}{\gamma _j}\left| \frac{V_{Y,t_0,j}}{|\Phi '(X(t_0))|^2V_{X,t_0,j}}-1\right| , \end{aligned}$$

where $\gamma _j$ is some random variable valued in the open interval with ending points $|\Phi '(X(t_0))|^2V_{X,t_0,j}/V_{Y,t_0,j}$ and 1. Since $|\gamma _j|$ tends to 1 a.s. as $j\rightarrow +\infty$, then according to (3.4) and (A1), the right-hand side of the above equation can be bounded by

$$\begin{aligned} c2^{j(H(t_0)+1/2)}\epsilon _j^{2H(t_0)+1/2}j^{1/2}|\log \epsilon _j| \quad \text{ with } \text{ some }~c>0, \end{aligned}$$

which converges to 0 as $j\rightarrow +\infty$, thanks to assumption (A4). $\square$

1.6 Proof of Theorem 4

In order to prove (3.11) and (3.12), we rely on the following relation: under assumptions (A1)–(A2),

$$\begin{aligned} \frac{\widehat{V_{n,t_0,J_n}}}{V_{Y,t_0,J_n}}-1=\mathcal {O}_{\mathbb {P}} \left( 2^{(J_n-n)H(t_0)}\right) . \end{aligned}$$

(4.37)

This is because, by using Markov’s inequality, Cauchy-Schwarz inequality, (1.11), (3.7) and the dominated convergence theorem,

$$\begin{aligned}&\mathbb {P}\left( \left| \frac{\widehat{V_{n,t_0,J_n}}-V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}} \right| >\eta \right) \le \frac{1}{\eta }\mathbb {E}\left| \frac{\widehat{V_{n,t_0,J_n}}-V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}}\right| \nonumber \\&\quad \le \frac{1}{\eta }\left( \mathbb {E}\left| \frac{\widehat{V_{n,t_0,J_n}} -V_{Y,t_0,J_n}}{2^{-2J_nH(t_0)}\epsilon _{J_n}}\right| ^2\right) ^{1/2}\left( \mathbb {E}\left| \frac{2^{-2J_nH(t_0)}\epsilon _{J_n}}{V_{Y,t_0,J_n}}\right| ^2\right) ^{1/2}\nonumber \\&\quad \le c2^{(J_n-n)H(t_0)}/\eta . \end{aligned}$$

(4.38)

Similarly to (4.38), we also obtain for any $\delta '>0$ arbitrarily small,

$$\begin{aligned} \mathbb {P}\left( 2^{(n-J_n-\delta ' n)H(t_0)}\left| \frac{\widehat{V_{n,t_0,J_n}}-V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}}\right| >\eta \right) \le c2^{-\delta ' nH(t_0)}/\eta . \end{aligned}$$

The fact that $\beta <1$ implies $\sum _{n\in \mathbb N}c2^{-\delta ' n H(t_0)}/\eta <+\infty$, then by Borel–Cantelli’s lemma,

$$\begin{aligned} \frac{\widehat{V_{n,t_0,J_n}}}{V_{Y,t_0,J_n}}-1=\mathcal {O}_{a.s.} \left( 2^{(J_n-n+\delta ' n)H(t_0)}\right) =\mathcal {O}_{a.s.}\left( 2^{(\beta -1+\delta ')nH(t_0)}\right) . \end{aligned}$$

(4.39)

Therefore, (3.11) (resp. (3.12)) follows from the following 2 decompositions:

$$\begin{aligned} \widehat{H_{Y,2^{J_n},n}}(t_0)-H(t_0)=\left( \widehat{H_{Y,2^{J_n},n}}(t_0) -\widehat{H_{Y,2^{J_n}}}(t_0)\right) +\left( \widehat{H_{Y,2^{J_n}}}(t_0)-H(t_0)\right) ; \end{aligned}$$

$$\begin{aligned}&\frac{\widehat{V_{n,t_0,J_n}}}{2|\Phi '(X(t_0))|^2c_2(t_0) 2^{-2J_nH(t_0)}\epsilon _{J_n}}-1\nonumber \\&\quad = \left( \frac{V_{Y,t_0,J_n}}{2|\Phi '(X(t_0))|^2c_2(t_0)2^{-2J_nH(t_0)}\epsilon _j} -1\right) \left( \frac{\widehat{V_{n,t_0,J_n}}}{V_{Y,t_0,J_n}}\right) +\left( \frac{\widehat{V_{n,t_0,J_n}}-V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}}\right) ; \end{aligned}$$

and equations (3.8), (4.37) (resp. (3.9), (4.39)).

For showing (3.13), we only need to show

$$\begin{aligned} \sqrt{2^{J_n+1}\epsilon _{J_n}}\big |\widehat{H_{Y,2^{J_n},n}}(t_0) -\widehat{H_{Y,2^{J_n}}}(t_0)\big |\xrightarrow [n\rightarrow +\infty ]{\mathbb {P}}0. \end{aligned}$$

(4.40)

By using the same idea we have taken to prove (4.35), we just need to verify

$$\begin{aligned} \sqrt{2^{J_n+1}\epsilon _{J_n}}\Big |\frac{\widehat{V_{n,t_0,J_n}} -V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}}\Big |\xrightarrow [n\rightarrow +\infty ]{\mathbb {P}}0. \end{aligned}$$

This is true since, according to (4.38) and the fact that (by assumption (A4)) $\epsilon _{J_n}=o\big (2^{-(\frac{2H(t_0)+1}{4H(t_0)+1})J_n}\big )$, the left-hand side of the above term can be bounded in probability by

$$\begin{aligned} c2^{J_n/2}\epsilon _{J_n}^{1/2}2^{(J_n-n)H(t_0)}=o\left( 2^{H(t_0) (J_n\frac{4H(t_0)+2}{4H(t_0)+1}-n)}\right) . \end{aligned}$$

The fact that $0<\beta \le \frac{4H(t_0)+1}{4H(t_0)+2}$ entails $J_n\frac{4H(t_0)+2}{4H(t_0)+1}-n\le 0$. Consequently,

$$\begin{aligned} \sqrt{2^{J_n+1}\epsilon _{J_n}}\Big |\frac{\widehat{V_{n,t_0,J_n}} -V_{Y,t_0,J_n}}{V_{Y,t_0,J_n}}\Big |\xrightarrow [n\rightarrow +\infty ]{\mathbb {P}}0, \end{aligned}$$

hence (3.13) holds. $\square$