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Asymptotic behavior of mixed power variations and statistical estimation in mixed models

Abstract

We obtain results on both weak and almost sure asymptotic behaviour of power variations of a linear combination of independent Wiener process and fractional Brownian motion. These results are used to construct strongly consistent parameter estimators in mixed models.

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Acknowledgments

This work has been partially supported by the Commission of the European Committees Grant PIRSES-GA-2008-230804 within the program “Marie Curie Actions”. The authors are also grateful to Rim Touibi and anonymous referees for their careful reading of the manuscript and helpful suggestions, which helped to improve the article.

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Correspondence to Georgiy Shevchenko.

Appendix: Proofs

Appendix: Proofs

Proof of Theorem 3.1

Write

$$\begin{aligned} \log _2 V^{H,2}_{2^k} = \log _2 \left( a^2 T^{2H} 2^{(1-2H)k}\right) + \log _2 \left( 1+ \frac{b^2}{a^2}T^{1-2H}2^{-(1-2H)k} + \zeta _k\right) , \end{aligned}$$

where

$$\begin{aligned} \zeta _k = \frac{a^2 \left( V_{2^k}^{H,0,2}-T^{2H}2^{(1-2H)k}\right) + b^2\left( V_{2^k}^{H,2,0}-T\right) + 2a b V_{2^k}^{H,1,1}}{a^2 T^{2H} 2^{(1-2H)k}}. \end{aligned}$$

From Proposition 2.1 it follows that for any \(\varepsilon >0\) \(\zeta _k = o(2^{(-1/2+\varepsilon )k}) + o(2^{(2H-3/2+\varepsilon )k})+ o(2^{(H-1+\varepsilon )k})=o(2^{(-1/2+\varepsilon )k})\), \(k\rightarrow \infty \). Hence we have

$$\begin{aligned} \log _2 V^{H,2}_{2^k} = 2 \log _2 a + 2H \log _2 T + (1-2H) k + O(2^{(2H-1)k})+ o(2^{(-1/2+\varepsilon )k}), k\rightarrow \infty .\nonumber \\ \end{aligned}$$
(18)

In particular,

$$\begin{aligned} \log _2 V^{H,2}_{2^k} \sim (1-2H)k,\quad k\rightarrow \infty , \end{aligned}$$

whence the result immediately follows.\(\square \)

Proof of Proposition 3.1

Write

$$\begin{aligned} \widetilde{H}_k-H&= \frac{1}{2}\left(\log _2 \frac{V_{2^{k-1}}^{H,2}}{V_{2^{k}}^{H,2}}-(2H-1)\right) = \frac{1}{2}\log _2 \frac{V_{2^{k-1}}^{H,2}}{2^{2H-1}V_{2^{k}}^{H,2}} \\&= \frac{1}{2}\log _2 \left( \frac{V_{2^{k-1}}^{H,2}-2^{2H-1}V_{2^{k}}^{H,2}}{2^{2H-1}V_{2^{k}}^{H,2}}+1\right) . \end{aligned}$$

Since by (12)

$$\begin{aligned} \zeta _k:= \frac{V_{2^{k-1}}^{H,2}-2^{2H-1}V_{2^{k}}^{H,2}}{2^{2H-1}V_{2^{k}}^{H,2}}\rightarrow 0,\quad k\rightarrow \infty , \end{aligned}$$

we obtain

$$\begin{aligned} \widetilde{H}_k-H = \zeta _k \left( \frac{1}{2\log 2}+o(1)\right) ,\quad k\rightarrow \infty . \end{aligned}$$

Now write

$$\begin{aligned} V_{2^{k-1}}^{H,2}-2^{2H-1}V_{2^{k}}^{H,2} = a^2 R_k^{H,0,2} + 2ab R_k^{H,1,1} + b^2 R_k^{H,2,0}, \end{aligned}$$

where

$$\begin{aligned} R_k^{H,i,j} = V_{2^{k-1}}^{H,i,j} - 2^{2H-1}V_{2^{k}}^{H,i,j},\quad i,j\in \left\{ 0,1,2\right\} . \end{aligned}$$

By Proposition 2.1 we have for any \(\varepsilon \in (0,H)\) \(V_{n}^{H,1,1} = o(n^{-H+\varepsilon }),\, n\rightarrow \infty \), whence \(R_k^{H,1,1} = o(2^{(-H+\varepsilon )k}) = o(1)\), \(k\rightarrow \infty \). Therefore,

$$\begin{aligned} \frac{2ab R_k^{H,1,1}}{2^{2H-1}V^{H,2}_{2^{k}}}\sim \frac{2b R_k^{H,1,1}}{a T^{2H} 2^{(1-2H)(k-1)}} = o(2^{(2H-1)k}),\quad k\rightarrow \infty . \end{aligned}$$

Further, by Proposition 2.1, \(V_n^{H,2,0} \rightarrow T\), \(n\rightarrow \infty \), so

$$\begin{aligned} \frac{b^2 R_k^{H,2,0}}{2^{2H-1}V^{H,2}_{2^{k}}}\sim \frac{b^2 R_k^{H,2,0}}{a^2 T^{2H} 2^{(k-1)(1-2H)}} = O(2^{(2H-1)k}),\quad k\rightarrow \infty . \end{aligned}$$

Thus, we get

$$\begin{aligned}&2^{k/2}\left( \widetilde{H}_k-H\right) =\left( \frac{a^2 2^{k/2} R_k^{H,0,2}}{2^{2H-1}V_{2^{k}}^{H,2}} + O(2^{(2H-1/2)k}) \right) \left( \frac{1}{2\log 2 }+o(1)\right) \nonumber \\&\quad \quad = \frac{2^{(2H-1/2)(k-1)} R_k^{H,0,2}}{\sqrt{2} T^{2H}\log 2 } + o(1),\quad k\rightarrow \infty . \end{aligned}$$
(19)

Now write

$$\begin{aligned} R_k^{H,0,2} = \sum _{m=0}^{2^{k-1}-1} \left( \left( \Delta ^{2^{k-1}}_m B^H\right) ^2 - 2^{2H-1}\left( \Delta ^{2^{k}}_{2m} B^H\right) ^2 - 2^{2H-1}\left( \Delta ^{2^{k}}_{2m+1} B^H\right) ^2\right) . \end{aligned}$$

In view of the self-similarity of \(B^H\),

$$\begin{aligned} R_k^{H,0,2} \overset{d}{=} 2^{2H(1-k)}T^{2H}\sum _{m=0}^{2^{k-1}-1} \eta _m, \end{aligned}$$

where

$$\begin{aligned} \eta _m = \left(B_{m+1}^H-B_{m}^H\right)^2 - 2^{2H-1}\left(B_{m+1/2}^H-B_{m}^H\right)^2 - 2^{2H-1}\left(B_{m+1}^H-B_{m+1/2}^H\right)^2. \end{aligned}$$

So we can apply CLT for stationary Gaussian sequence (see Breuer and Major 1983) and deduce that

$$\begin{aligned} \frac{2^{(2H-1/2)(k-1)} R_k^{H,0,2}}{ T^{2H} } \overset{d}{=} 2^{(1-k)/2}\sum _{m=0}^{2^{k-1}-1}\eta _m \Rightarrow N(0,\sigma ^2),\quad k\rightarrow \infty , \end{aligned}$$

where

$$\begin{aligned} \sigma ^2 = \mathsf{E}\left[ \,\eta _0^2\,\right] + 2\sum _{m=0}^\infty \mathsf{E}\left[ \,\eta _0\eta _m\,\right] = \rho '_{H,0} + 2\sum _{m=0}^\infty \rho '_{H,m}. \end{aligned}$$

Using this convergence and (19), we get the required statement with the help of Slutsky’s theorem.\(\square \)

Proof of Theorem 3.2

The proof is similar to that of Proposition 3.1, so we will omit some details. Using the same transformations as there, we get

$$\begin{aligned} \widetilde{H}^{(2)}_k-H = \frac{U_{k-1}^{H,2}-2^{2H-1}U_{k}^{H,2}}{2^{2H-1}U_{k}^{H,2}} \left( \frac{1}{2\log 2}+o(1)\right) ,\quad k\rightarrow \infty . \end{aligned}$$

Expand

$$\begin{aligned} U_{k-1}^{H,2}-2^{2H-1}U_{k}^{H,2} = a^2 P_k^{H,0,2} + 2ab P_k^{H,1,1} + b^2 P_k^{H,2,0}, \end{aligned}$$

where

$$\begin{aligned} P_k^{H,i,j} = V_{2^{k-2}}^{H,i,j} - (c_H+1)V_{2^{k-1}}^{H,i,j} + c_H V_{2^{k}}^{H,i,j},\quad i,j\in \left\{ 0,1,2\right\} . \end{aligned}$$

Similarly to \(R_k^{H,1,1}\) in Proposition 3.1, for any \(\varepsilon >0\) \(P_k^{H,1,1} = o(2^{(-H+\varepsilon )k})\), \(k\rightarrow \infty \). Further, \(P_{k}^{H,2,0}\) has a generalized chi-square distribution with \(\mathsf{E}\left[ \,P_{k}^{H,2,0}\,\right] =0\) and \(\mathsf{E}\left[ \,\left( P_{k}^{H,2,0}\right) ^2\,\right] = O(2^{-k})\), \(k\rightarrow \infty \). As in Proposition 2.1, we deduce that for any \(\varepsilon >0\) \(P_k^{H,0,2} = o(2^{(-1/2+\varepsilon )k})\), \(k\rightarrow \infty \).

Further, from (12) \(U_k^{H,2}\sim a^2 T^{2H} (2^{2H-1}-1)2^{(1-2H)k}\), \(k\rightarrow \infty \). Combining the obtained asymptotics, we can write

$$\begin{aligned} 2^{k/2}\left( \widetilde{H}^{(2)}_k-H\right) = \frac{2^{(2H-1/2)(k-2)} P_k^{H,0,2}}{T^{2H}(1-2^{1-2H})\log 2 } + o(1),\quad k\rightarrow \infty , \end{aligned}$$

whence we deduce the asymptotic normality exactly as in Proposition 3.1.

The estimate (16) is obtained as in Proposition 2.1.\(\square \)

Proof of Proposition 3.2

First, observe that

$$\begin{aligned} \frac{\widetilde{a}^2_k}{a^2} \sim 2^{2(\widetilde{H}_k- H)k} T^{2(H-\widetilde{H}_k)} \rightarrow 1,\quad k\rightarrow \infty , \end{aligned}$$

since \((\widetilde{H}_k-H)k\rightarrow 0\), \(k\rightarrow \infty \), by (15). Hence we get the strong consistency of \(\widetilde{a}^2_k\).

Concerning \(\widetilde{b}^2_k\), define

$$\begin{aligned} \widehat{b}^2_k = \frac{2^{1-2H}V^{H,2}_{2^{k-1}}-V^{H,2}_{2^{k}}}{\big (2^{1-2H}-1\big )T}. \end{aligned}$$

It easily follows from (12) that \(\widehat{b}^2_k\rightarrow b^2\), \(k\rightarrow \infty \). So it is enough to show that \(\widetilde{b}^2_k - \widehat{b}^2_k\rightarrow 0\), \(k\rightarrow \infty \). To this end, write

$$\begin{aligned} \widetilde{b}^2_k - b^2 =\frac{\left(2^{1-2\widetilde{H}^{(2)}_k}\!-\! 2^{1\!-\!2H}\right) V^{H,2}_{2^{k-1}}}{\big (2^{1-2\widetilde{H}^{(2)}_k}-1\big )T} \quad +T^{-1}\left(2^{1-2H}V^{H,2}_{2^{k-1}}\!-\!V^{H,2}_{2^{k}}\right) \left(\big (2^{1-2\widetilde{H}^{(2)}_k}-1\big )^{-1}\!-\!\big (2^{1-2H}\!-\!1\big )^{-1} \right). \end{aligned}$$

Obviously, the second term converges to zero. Due to (12) and (16), for any \(\varepsilon >0\)

$$\begin{aligned}&\big (2^{1-2\widetilde{H}^{(2)}_k}-2^{1-2H}\big )V^{H,2}_{2^k} \sim -2^{2-2H} (\widetilde{H}^{(2)}_k-H) a^2 T^{2H} 2^{(1-2H)(k-1)}\log 2 \\&\quad \quad =2^{(1-2H)k}o(2^{(-1/2+\varepsilon )k}),\quad k\rightarrow \infty , \end{aligned}$$

whence we deduce the strong consistency of \(\widetilde{b}_k^2\) for \(H\in (1/4,1/2)\), since \(1-2H <1/2\).

\(\square \)

Proof of Theorem 3.3

Write

$$\begin{aligned} U_k^{H,2} = a^2 Q_k^{H,0,2} + 2ab Q_k^{H,1,1} + b^2 Q_k^{H,2,0}, \end{aligned}$$

where \(Q_k^{H,i,j} = V_{2^{k-1}}^{H,i,j} - V_{2^{k}}^{H,i,j}\), \(i,j\in {0,1,2}\). By Proposition 2.1, \(Q_k^{H,0,2}\sim T^{2H} (2^{2H-1}-1)2^{(1-2H)k}\), \(k\rightarrow \infty \) and for any \(\varepsilon >0\) \(Q_{k}^{H,1,1} = o(2^{(-H+\varepsilon )k})\), \(k\rightarrow \infty \), and

$$\begin{aligned} Q_{k}^{H,2,0} = \left( V_{2^{k-1}}^{H,2,0}-T\right) - \left( V_{2^{k}}^{H,2,0}-T\right) = o(2^{(-1/2+\varepsilon )k}),\quad k\rightarrow \infty . \end{aligned}$$

Thus, we have

$$\begin{aligned} U_k^{H,2} \sim a^2 T^{2H} (2^{2H-1}-1)2^{(1-2H)k},\quad k\rightarrow \infty , \end{aligned}$$
(20)

which yields the proof.\(\square \)

Proof of Theorem 3.4

As in the proof of Theorem 3.2, write

$$\begin{aligned} \widetilde{H}^{(2)}_k-H = \frac{U_{k-1}^{H,2}-2^{2H-1}U_{k}^{H,2}}{2^{2H-1}U_{k}^{H,2}} \left( \frac{1}{2\log 2}+o(1)\right) ,\quad k\rightarrow \infty , \end{aligned}$$

and expand

$$\begin{aligned} U_{k-1}^{H,2}-2^{2H-1}U_{k}^{H,2} = a^2 P_k^{H,0,2} + 2ab P_k^{H,1,1} + b^2 P_k^{H,2,0}, \end{aligned}$$

where

$$\begin{aligned} P_k^{H,i,j} = V_{2^{k-2}}^{H,i,j} - (c_H+1)V_{2^{k-1}}^{H,i,j} + c_H V_{2^{k}}^{H,i,j},\quad i,j\in \left\{ 0,1,2\right\} , \end{aligned}$$

and \(c_H = 2^{2H-1}\).

We have as in the proof of Theorem 3.2 that for any \(\varepsilon >0\) \(P_k^{H,1,1} = o(2^{(-H+\varepsilon )k})\), \(P_{k}^{H,0,2} = o(2^{(1/2-2H+\varepsilon )k})\), \(k\rightarrow \infty \). Therefore, using (20), we get

$$\begin{aligned} 2^{(3/2-2H)k}\left( \widetilde{H}_k^{(2)}-H\right) = \frac{b^2 2^{(k-2)/2} P_k^{H,2,0}}{ a^2 2^{2H-1} T^{2H}(2^{2H-1}-1)\log 2 } + o(1),\quad k\rightarrow \infty . \end{aligned}$$

We can write \(P_k^{H,2,0}=\sum _{m=0}^{2^{k-2}-1}\kappa _{k,m}\), where

$$\begin{aligned}&\kappa _{k,m} = \left( \Delta ^{2^{k-2}}_m W\right) ^2 - (c_H+1) \left( \left( \Delta ^{2^{k-1}}_{2m} W\right) ^2+\left( \Delta ^{2^{k-1}}_{2m+1} W\right) ^2\right) \\&\quad \quad + c_H \left( \left( \Delta ^{2^{k}}_{4m} W\right) ^2+\left( \Delta ^{2^{k}}_{4m+1} W\right) ^2+\left( \Delta ^{2^{k}}_{4m+2} W\right) ^2+\left( \Delta ^{2^{k}}_{4m+3} W\right) ^2\right) . \end{aligned}$$

The random variables \(\left\{ \kappa _{k,m}, m=0,\dots ,2^k-1\right\} \) are iid with \(\mathsf{E}\left[ \,\kappa _{k,m}\,\right] =0\) and \(\mathsf{E}\left[ \,\kappa _{k,m}^2\,\right] = T^2 2^{-2(k-2)}(2^{4H-3}+1)\). Therefore, by the classical CLT,

$$\begin{aligned} 2^{(k-2)/2} P_k^{H,2,0} \Rightarrow N(0,T^2 (2^{4H-3}+1)),\quad k\rightarrow \infty , \end{aligned}$$

whence we get by Slutsky’s theorem,

$$\begin{aligned} 2^{(3/2-2H)k}\left( \widetilde{H}_k^{(2)}-H\right) \Rightarrow N(0,(\sigma _H'')^2),\quad k\rightarrow \infty . \end{aligned}$$

Again, the estimate (17) is obtained as in Proposition 2.1.\(\square \)

Proof of Proposition 3.3

In view of (20),

$$\begin{aligned} \frac{\hat{a}^2_k}{a^2} \sim 2^{2(\widetilde{H}_k^{(2)}- H)k}T^{2(H-\widetilde{H}^{(2)}_k)}\frac{2^{2H-1}-1}{2^{2\widetilde{H}_k^{(2)}-1}-1} \rightarrow 1,\quad k\rightarrow \infty , \end{aligned}$$

since \((\widetilde{H}^{(2)}_k-H)k\rightarrow 0\),    \(k\rightarrow \infty \), by (17). Hence we get the strong consistency of \(\hat{a}^2_k\). The strong consistency of \(\hat{b}^2_k\) is obvious from (13).\(\square \)

Proof of Proposition 3.4

Define

$$\begin{aligned} \xi _k = \frac{2^{k/2}}{\sqrt{2} T}\left(V_{2^k}^{H,2,0} - T\right) = \frac{2^{-k/2}}{\sqrt{2} T}\sum _{i=0}^{2^k-1}\left(2^{k}\left(\Delta _i^{2^k} W\right)^2- T \right). \end{aligned}$$

By the classical CLT, \(\xi _k\Rightarrow N(0,1)\), \(k\rightarrow \infty \), so we need to study the collective behaviour. To this end, observe that the vector \((\xi _k,\xi _{k+1},\dots ,\xi _{k+m})\) can be represented as a sum of independent vectors

$$\begin{aligned} (\xi _k,\xi _{k+1},\dots ,\xi _{k+m}) = \sum _{i=0}^{2^{k}-1} \zeta _{k,i}, \end{aligned}$$

where the \(j\)th coordinate of \(\zeta _{k,i}\), \(j=0,1,2,\dots ,m\), is

$$\begin{aligned} \zeta _{k,i,j} = \frac{2^{-k/2}}{\sqrt{2} T}\sum _{l=0}^{2^j-1} \left(2^{2(k+j)}\left(\Delta _{l + i 2^{j}}^{2^{k+j}} W\right)^2- T \right). \end{aligned}$$

(We simply group terms on the intervals of the partition \(\left\{ i T2^{-k}, i=0,\dots ,2^{k} \right\} \)). Therefore, we can apply a vector CLT and deduce that for every \(m\ge 0\) the vector \((\xi _k,\xi _{k+1},\dots ,\xi _{k+m})\) converges in distribution to an \((m+1)\)-dimensional centered Gaussian vector as \(k\rightarrow \infty \). Consequently, the sequence \((\xi _k,\xi _{k+1},\xi _{k+2},\dots )\) converges to a centered stationary Gaussian sequence as \(k\rightarrow \infty \).

We have seen above that \(V^{H,2}_n = b^2 V^{H,2,0}_n + o(n^{-1/2})\), \(n\rightarrow \infty \). Therefore, \(Z_k = b^2 \left( \sqrt{2}\xi _{k-1}-\xi _{k}\right) + o(1)\), \(k\rightarrow \infty \), so by Slutsky’s theorem the sequence \((Z_k,Z_{k+1},Z_{k+2},\dots )\) also converges to a centered stationary Gaussian sequence. It is straightforward to check that the limit covariance is that of the i.i.d. standard Gaussian sequence, whence the result follows.\(\square \)

Proof of Theorem 3.5

By Proposition 2.1, \(V_n^{H,2,2} \sim T^{2H+1} n^{-2H}\), \(V_n^{H,0,4} \sim 3 T^{4H}n^{1-4H}\), \(n\rightarrow \infty \) and for any \(\varepsilon >0\) \(V_n^{H,4,0}- 3T^2 n^{-1} = o(n^{-3/2+\varepsilon })\), \(V_n^{H,3,1} = o(n^{-1-H+\varepsilon })\), \(V_n^{H,1,3} = o(n^{-3H+\varepsilon })\), \(n\rightarrow \infty \).

Now write

$$\begin{aligned} U_k^{H,4}= \sum _{i=0}^{4} \left( {\begin{array}{c}4\\ i\end{array}}\right) a^i b^{4-i} U^{H,4-i,i}_k, \end{aligned}$$

where \(U^{H,4-i,i}_k = V^{H,4-i,i}_{2^{k-1}} -2 V^{H,4-i,i}_{2^{k}}\), \(i=0,\dots ,4\). We have \(U_k^{H,2,2} \sim T^{2H+1} (2^{2H}-2)2^{-2Hk}\), \(U_k^{H,0,4} = O(2^{(1-4H)k})= o(2^{-2Hk})\), \(k\rightarrow \infty \), and for any \(\varepsilon >0\)

$$\begin{aligned}&\displaystyle U_k^{H,4,0} = \left( V_{2^{k-1}}^{H,4,0}- T^2 2^{-(k-1)}\right) - 2\left( V_{2^{k}}^{H,4,0}- T^2 2^{-k}\right) = o(2^{(-3/2+\varepsilon )k}),\\&\displaystyle U_k^{H,3,1} = o(2^{(-1-H+\varepsilon )k}),\ U_k^{H,1,3} = o(2^{(-3H+\varepsilon )k}),\quad k\rightarrow \infty . \end{aligned}$$

Collecting all the terms, we get

$$\begin{aligned} U_k^{H,4} \sim 4T^{2H+1} (2^{2H}-2)2^{-2Hk}, \quad k\rightarrow \infty . \end{aligned}$$

Hence, the assertion follows.\(\square \)

Proof of Theorem 3.6

The statement for \(\widehat{H}_k(a)\) follows immediately from (18). To prove the statement for \(\widetilde{H}_k(b)\) and \(\widehat{H}_k(a,b)\), note that, in view of (12), \(V^{H,2}_{2^k}>b^2T\) for sufficiently large \(k\). Therefore, we can write, as in the proof of Theorem 3.1,

$$\begin{aligned} \log _{2}(V^{H,2}_{2^k}-b^2T) = \log _2 \left( a^2 T^{2H} 2^{k(1-2H)}\right) + \log _2 \left( 1 + \zeta _k\right) , \end{aligned}$$

with the same \(\zeta _k\); in particular, for \(H\in (0,1/2]\) and any \(\varepsilon >0\), \(\zeta _k = o(2^{k(-1/2+\varepsilon )})\), \(k\rightarrow \infty \). For \(H\in (1/2,3/4)\), \(\zeta _k = o(2^{k(-H+\varepsilon )})+ o(2^{k(2H-3/2+\varepsilon )})+ o(2^{k(H-1+\varepsilon )}) = o(2^{k(2H-3/2+\varepsilon )})\), \(k\rightarrow \infty \). This implies the statement for both \(\widetilde{H}_k(b)\) and \(\widehat{H}_k(a,b)\).\(\square \)

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Dozzi, M., Mishura, Y. & Shevchenko, G. Asymptotic behavior of mixed power variations and statistical estimation in mixed models. Stat Inference Stoch Process 18, 151–175 (2015). https://doi.org/10.1007/s11203-014-9106-5

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Keywords

  • Power variation
  • Fractional Brownian motion
  • Hurst parameter
  • Wiener process
  • Consistent estimator

Mathematics Subject Classification

  • 60G22
  • 62M09
  • 60G15
  • 62F25