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.

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 $\widehat{H}_k(a,b)$.$\square$