Parameter estimation for the Langevin equation with stationary-increment Gaussian noise

Abstract

We study the Langevin equation with stationary-increment Gaussian noise. We show the strong consistency and the asymptotic normality with Berry–Esseen bound of the so-called second moment estimator of the mean reversion parameter. The conditions and results are stated in terms of the variance function of the noise. We consider both the case of continuous and discrete observations. As examples we consider fractional and bifractional Ornstein–Uhlenbeck processes. Finally, we discuss the maximum likelihood and the least squares estimators.

Appendix: Proofs of Lemmas

Proof of Lemma 2.1

Let $t>0$ and let $\lfloor t\rfloor$ be the greatest integer not exceeding t. Then

$$\begin{aligned} |G_t| \le |G_t-G_{\lfloor t \rfloor }| + \sum _{k=1}^{\lfloor t \rfloor } | G_k-G_{k-1}|. \end{aligned}$$

By the Minkowski’s inequality and stationary of the increments,

$$\begin{aligned} \sqrt{\mathbf {E}[G_t^2]} \le \sqrt{\mathbf {E}[G_{t-\lfloor t\rfloor }^2]} + \lfloor t \rfloor \sqrt{\mathbf {E}[G_1^2]}. \end{aligned}$$

The claim follows from this.$\square$

Proof of Lemma 3.1

By changing the variable in (2.4) we obtain

$$\begin{aligned} \psi (\theta ) = \frac{1}{2}\int _0^\infty \mathrm {e}^{-t}\,v\left( \frac{t}{\theta }\right) \, \mathrm {d}t. \end{aligned}$$

(6.1)

Since v is strictly increasing, this shows that $\psi$ is also strictly decreasing. Furthermore, $\psi (\theta )\rightarrow 0$ as $\theta \rightarrow \infty$ by the monotone convergence theorem.

By the Lebesgue’s dominated convergence theorem, the function

$$\begin{aligned} \theta \mapsto \int _0^\infty \mathrm {e}^{-\theta t} v(t)\, \mathrm {d}t \end{aligned}$$

is smooth. Consequently, $\psi$ is smooth.

Finally, let us show that $\psi$ is convex. Differentiating $\psi$ we observe

$$\begin{aligned} \psi '(\theta ) = \frac{1}{2}\int _0^\infty \mathrm {e}^{-\theta s}v(s)\, \mathrm {d}s - \frac{\theta }{2} \int _0^\infty s\mathrm {e}^{-\theta s}v(s)\,\mathrm {d}s \end{aligned}$$

from which it follows, by integration by parts, that

$$\begin{aligned} \psi '(\theta ) = -\frac{1}{2} \int _0^\infty s\mathrm {e}^{-\theta s}\, \mathrm {d}v(s), \end{aligned}$$

where $\mathrm {d}v$ is the measure associated to the increasing function v. Similarly, differentiating and using integration by parts again we get

$$\begin{aligned} \psi ''(\theta ) = \frac{1}{2} \int _0^\infty s^2\mathrm {e}^{-\theta s}\, \mathrm {d}v(s) \ge 0. \end{aligned}$$

Consequently, $\psi$ is convex.$\square$

Proof of Lemma 3.2

Let $T\ge 0$ be fixed. Then, by Sottinen and Viitasaari (2016, Theorem 12), there exists a Brownian motion W and a kernel $K_T\in L^2([0,T]^2)$ such that we have the representation

$$\begin{aligned} X_t = \int _0^T K_T(t,s)\, \mathrm {d}W_s, \end{aligned}$$

for all $t\le T$.

Consequently $X_t$ belongs to the 1st Wiener chaos for all $t>0$. Then note that

$$\begin{aligned} X_t^2 - \mathbf {E}\left[ X_t^2\right] = 2H_2(X_t). \end{aligned}$$

Consequently, it belongs to the 2nd Wiener chaos. Finally, note that, because $\gamma$ is continuous, the integral

$$\begin{aligned} \frac{1}{T}\int _0^T 2H_2(X_t)\, \mathrm {d}t \end{aligned}$$

can be defined as a limit in $L^2(\Omega )$ in the 2nd Wiener chaos. The claim follows from this.$\square$

Proof of Lemma 3.3

The claim is an obvious consequence of Isserlis’ theorem.$\square$

Proof of Lemma 3.4

By the symmetry $\gamma (s,t)=\gamma (t,s)$

$$\begin{aligned}&\int _{[0,T]^4} \gamma (t_1,t_2)\gamma (t_2,t_3)\gamma (t_3,t_4)\gamma (t_4,t_1)\, \mathrm {d}t_1\mathrm {d}t_2\mathrm {d}t_3\mathrm {d}t_4 \\&\quad = \int _0^T\!\!\int _0^T \left( \int _0^T \gamma (t_1,t_2)\gamma (t_2,t_3)\, \mathrm {d}t_2\right) ^2 \mathrm {d}t_1 \mathrm {d}t_3 \\&\quad \le \int _0^T\!\!\int _0^T \left( \int _0^T |\gamma (t_1,t_2)|\, \mathrm {d}t_2\right) \left( \int _0^T |\gamma (t_1,t_2)|\gamma (t_2,t_3)^2\, \mathrm {d}t_2\right) \, \mathrm {d}t_1 \mathrm {d}t_3 \\&\quad \le \int _0^T \left( \sup _{t\in [0,T]} \int _0^T |\gamma (t,t_2)|\, \mathrm {d}t_2 \int _0^T\int _0^T |\gamma (t_1,t_2)|\, \mathrm {d}t_1\, \gamma (t_2,t_3)^2\, \mathrm {d}t_2\right) \mathrm {d}t_3 \\&\quad \le \left[ \sup _{t\in [0,T]}\int _0^T |\gamma (t,t_2)|\, \mathrm {d}t_2\right] ^2 \int _0^T\!\!\int _0^T \gamma (t_2,t_3)^2\, \mathrm {d}t_2 \mathrm {d}t_3. \end{aligned}$$

$\square$

Proof of Lemma 3.5

Note that $Q_T/\sqrt{\mathbf {E}[Q_T^2]}$ belongs to the 2nd Wiener chaos and has unit variance. Consequently, by the fourth-moment Proposition 3.2, it suffices to show that

$$\begin{aligned} \frac{\mathbf {E}\left[ Q_T^4\right] }{\mathbf {E}\left[ Q_T^2\right] ^2} - 3 \le C \frac{\left( \sup _{t\in [0,T]} \int _{[0,T]} |\gamma (t,s)|\, \mathrm {d}s\right) ^2}{\int _{[0,T]^2} \gamma (t,s)^2\, \mathrm {d}s\mathrm {d}t}. \end{aligned}$$

Denote

$$\begin{aligned} I_2(T)= & {} \int _{[0,T]^2} \gamma (t_1,t_s)^2\, \mathrm {d}t_1\mathrm {d}t_2, \\ I_4(T)= & {} \int _{[0,T]^4} \gamma (t_1,t_2)\gamma (t_2,t_3)\gamma (t_3,t_4)\gamma (t_4,t_1)\, \mathrm {d}t_1 \mathrm {d}t_2 \mathrm {d}t_3 \mathrm {d}t_4. \end{aligned}$$

Then, by Lemma 3.3,

$$\begin{aligned} \mathbf {E}[Q_T^4]= & {} \frac{12}{T^4} I_2(T)^2 + \frac{24}{T^4} I_4(T), \\ \mathbf {E}[Q_T^2]^2= & {} \frac{4}{T^4} I_2(T)^2, \end{aligned}$$

and, by Lemma 3.4

$$\begin{aligned} I_4(T) \le \left[ \sup _{t\in [0,T]}\int _0^T |\gamma (t,s)|\, \mathrm {d}s\right] ^2 I_2(T). \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{\mathbf {E}\left[ Q_T^4\right] }{\left( \mathbf {E}\left[ Q_T^2\right] \right) ^2} -3= & {} \frac{12 I_2(T)^2+24 I_4(T)-12I_2(T)^2}{4 I_2(T)^2} \\= & {} 6\frac{I_4(T)}{I_2(T)^2} \\\le & {} 6\frac{\left[ \sup _{t\in [0,T]}\int _0^T |\gamma (t,s)|\, \mathrm {d}s\right] ^2}{I_2(T)}. \end{aligned}$$

The claim follows from this.$\square$

Proof of Lemma 3.6

We have

$$\begin{aligned} \frac{\mathbf {E}[(Q^\theta _T)^2}{w_\theta (T)} = 1 + \frac{\frac{1}{T} \int _0^T \int _0^T \left[ \mathbf {E}(X_t X_s)\right] ^2 - r^2(t-s)\mathrm {d}s \mathrm {d}t}{T w_\theta (T)}. \end{aligned}$$

(6.2)

By Proposition 2.2 and ergodicity condition $\frac{1}{T}\int _0^T |r(t)|\mathrm {d}t \rightarrow 0$ it follows that

$$\begin{aligned} \frac{1}{T} \int _0^T \int _0^T \left[ \mathbf {E}(X_t X_s)\right] ^2 - r^2(t-s)\mathrm {d}s \mathrm {d}t \rightarrow 0. \end{aligned}$$

(6.3)

Moreover, by symmetry, change-of-variables and the Fubini theorem

$$\begin{aligned} \int _0^T\!\!\!\!\int _0^T r_\theta (t-s)^2\, \mathrm {d}s \mathrm {d}t= & {} 2\int _0^T\!\!\!\!\int _{t}^T r_\theta (s)^2 \, \mathrm {d}t \mathrm {d}s \\= & {} 2\int _0^T r_\theta (t)^2(T-t)\, \mathrm {d}t. \end{aligned}$$

Consequently, by assuming that $T> 1$,

$$\begin{aligned} \int _0^T r_\theta (t)^2(T-t)\, \mathrm {d}t \ge \int _0^1 r_\theta (t)^2 (T-t)\, \mathrm {d}t \ge C(T-1) \end{aligned}$$

This shows $T w_\theta (T)\ge C$ which, together with (6.2) and (6.3), implies $\mathbf {E}[(Q^\theta _T)^2] \sim w_\theta (T)$. Also, we have shown the equality

$$\begin{aligned} w_\theta (T) = \frac{4}{T^2}\int _0^T r_\theta (t)^2(T-t)\, \mathrm {d}t. \end{aligned}$$

Consider then the case $\int _0^\infty r_\theta (t)^2\, \mathrm {d}t < \infty$. By the equivalence $\mathbf {E}[(Q^\theta _T)^2] \sim \frac{4}{T^2}\int _0^T r_\theta (t)^2(T-t)\, \mathrm {d}t$, we have

$$\begin{aligned} \frac{1}{T}\int _0^T r_\theta (t)^2(T-t)\, \mathrm {d}t = \int _0^T r_\theta (t)^2\, \mathrm {d}t - \frac{1}{T}\int _0^T r_\theta (t)^2t\, \mathrm {d}t. \end{aligned}$$

Here the first term converges to $\int _0^\infty r_\theta (t)^2\, \mathrm {d}t < \infty$. For the second term we have

$$\begin{aligned} \frac{1}{T}\int _0^T r_\theta (t)^2t\, \mathrm {d}t \le \frac{1}{\sqrt{T}}\int _0^{\sqrt{T}}r_\theta (t)^2t\, \mathrm {d}t + \int _{\sqrt{T}}^\infty r_\theta (t)^2t\, \mathrm {d}t \rightarrow 0 \end{aligned}$$

which completes the proof.$\square$

Proof of Lemma 3.7

Let us first split

$$\begin{aligned}&\sup _{x\in \mathbb {R}}\left| \mathbf {P}\left[ \frac{Q^\theta _T}{\sqrt{w_\theta (T)}}\le x\right] -\Phi (x)\right| \\&\quad \le \sup _{x\in \mathbb {R}}\left| \mathbf {P}\left[ \frac{Q^\theta _T}{\sqrt{\mathbf {E}[(Q^\theta _T)^2]}} \le \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} x\right] -\Phi \left( \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} x\right) \right| \\&\qquad + \sup _{x\in \mathbb {R}}\left| \mathbf {P}\left[ \Phi \left( \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} x\right) \right] -\Phi (x) \right| \\&\quad \le \sup _{x\in \mathbb {R}}\left| \mathbf {P}\left[ \frac{Q^\theta _T}{\sqrt{\mathbf {E}[(Q^\theta _T)^2]}} \le x\right] -\Phi \left( x\right) \right| \\&\qquad + \sup _{x\in \mathbb {R}}\left| \Phi \left( \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} x\right) -\Phi (x) \right| \\&\quad = A_1 + A_2. \end{aligned}$$

For the term $A_1$, let us first estimate

$$\begin{aligned}&\int _0^T |\gamma _\theta (t,s)|\, \mathrm {d}s \\&\quad = \int _0^T \left| r_\theta (t-s) + \mathrm {e}^{-\theta (t+s)}r_\theta (0) - \mathrm {e}^{-\theta t}r_\theta (s) - \mathrm {e}^{-\theta s}r_\theta (t) \right| \, \mathrm {d}s \\&\quad \le \int _0^T |r_\theta (t-s)|\, \mathrm {d}s + \frac{\mathrm {e}^{-\theta t}|r_\theta (0)|}{\theta } + \mathrm {e}^{-\theta t}\int _0^T |r_\theta (s)|\, \mathrm {d}s + \frac{|r_\theta (t)|}{\theta }\\&\quad \le \int _0^T |r_\theta (t-s)|\, \mathrm {d}s + \mathrm {e}^{-\theta t}\int _0^T |r_\theta (s)|\, \mathrm {d}s + \frac{\left( \mathrm {e}^{-\theta t}+1\right) |r_\theta (0)|}{\theta }\\&\quad \le \int _0^T |r_\theta (t-s)|\, \mathrm {d}s + \int _0^T|r_\theta (s)| \, \mathrm {d}s + C, \end{aligned}$$

Consequently,

$$\begin{aligned} \sup _{t\in [0,T]} \int _0^T |\gamma _\theta (t,s)|\, \mathrm {d}s\le & {} 2 \sup _{t\in [0,T]} \int _0^T |r_\theta (t-s)|\, \mathrm {d}s +C\\= & {} 2\sup _{t\in [0,T]}\int _{-t}^{T-t} |r_\theta (u)|\, \mathrm {d}u +C \\\le & {} 2\int _{-T}^{T} |r_\theta (u)|\, \mathrm {d}u + C \\= & {} 4\int _0^T |r_\theta (u)|\, \mathrm {d}u + C. \end{aligned}$$

Since we are interested in the case $T\rightarrow \infty$, we can assume that T is bigger than some absolute positive constant. Consequently, since $r_\theta$ continuous with $r_\theta (0)>0$, it follows from the estimate above that

$$\begin{aligned} \sup _{t\in [0,T]} \int _0^T |\gamma _\theta (t,s)|\, \mathrm {d}s \le C\int _0^T |r_\theta (u)| \, \mathrm {d}u. \end{aligned}$$

Therefore, by applying Lemmas 3.5 and 3.6, it follows that $A_1\le C_\theta R_\theta (T)$.

Let us then consider the term $A_2$. Now, by the mean value theorem,

$$\begin{aligned} A_2= & {} \sup _{x\in \mathbb {R}}\left| \Phi \left( \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} x\right) -\Phi (x)\right| \\\le & {} \frac{1}{\sqrt{2\pi }}\sup _{x\in \mathbb {R}}\left( \mathrm {e}^{-1/2 \eta _\theta (T)^2}|x|\right) \, \left| \sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}}-1\right| , \end{aligned}$$

where

$$\begin{aligned} \eta _\theta (T,x) \in \left[ x,x+\sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}}x\right] . \end{aligned}$$

Since $\sqrt{\frac{w_\theta (T)}{\mathbf {E}[(Q^\theta _T)^2]}} \sim 1$, it follows that

$$\begin{aligned} A_2 \le \left| \sqrt{\frac{w_\theta (T)}{\mathbf {E}\left[ (Q^\theta _T)^2\right] }}-1\right| = \frac{\left| \sqrt{w_\theta (T)}-\sqrt{\mathbf {E}\left[ (Q^\theta _T)^2\right] } \right| }{\sqrt{\mathbf {E}\left[ (Q^\theta _T)^2\right] }}. \end{aligned}$$

Consequently, by the asymptotic equivalence of $w_\theta (T)\sim \mathbf {E}[(Q^\theta _T)^2]$, it remains to show that

$$\begin{aligned} T\left| \sqrt{w_\theta (T)}-\sqrt{\mathbf {E}[(Q^\theta _T)^2]}\right| \le C_\theta \int _0^T |r_\theta (t)|\, \mathrm {d}t. \end{aligned}$$

(Actually, we show that the left hand side is bounded.) For this purpose, we estimate, by using the inequality $|\sqrt{a}-\sqrt{b}| \le \sqrt{|a-b|}$ and the identity $a^2-b^2=(a+b)(a-b)$, that

$$\begin{aligned}&T\left| \sqrt{w_\theta (T)}-\sqrt{\mathbf {E}[(Q^\theta _T)^2]}\right| \\&\quad = \sqrt{2} \left| \sqrt{\int _0^T\!\!\!\!\int _0^T r_\theta (t-s)^2\, \mathrm {d}s \mathrm {d}t}- \sqrt{\int _0^T\!\!\!\!\int _0^T \gamma _\theta (t,s)^2\, \mathrm {d}s \mathrm {d}t}\right| \\&\quad \le \sqrt{2} \sqrt{\left| \int _0^T\!\!\!\!\int _0^T \left[ r_\theta (t-s)^2- \gamma _\theta (t,s)^2\, \right] \mathrm {d}s \mathrm {d}t\right| }\\&\quad = \sqrt{2} \sqrt{\left| \int _0^T\!\!\!\!\int _0^T \big (r_\theta (t-s)+\gamma _\theta (t,s)\big )\big (r_\theta (t-s)-\gamma _\theta (t,s)\big ) \mathrm {d}s \mathrm {d}t\right| }. \end{aligned}$$

By applying Proposition 2.2 to the estimate above, we obtain

$$\begin{aligned}&T\left| \sqrt{w_\theta (T)}-\sqrt{\mathbf {E}[(Q^\theta _T)^2]}\right| \\&\quad \le C_\theta \sqrt{\left| \int _0^T\!\!\!\!\int _0^T \big (r_\theta (t-s)+\gamma _\theta (t,s)\big )\mathrm {e}^{-\theta \min (s,t)}\, \mathrm {d}s \mathrm {d}t\right| }. \end{aligned}$$

Now, $|r_\theta (t-s)|\le r_\theta (0)$ and $|\gamma _\theta (t,s)|\le r_\theta (0) +1$, by Proposition 2.2. Consequently, the integral above is bounded, and the proof is finished.$\square$

Proof of Lemma 4.1

Let

$$\begin{aligned} a(t) = a_{H,1}(t) = H \mathrm {e}^{t/H}. \end{aligned}$$

Then

$$\begin{aligned} r_{H,K,\theta }(t) = \frac{1}{2^K}\mathrm {e}^{-\theta t}\left[ \big (a(t)^{2H}+1\big )^K - \big (a(t)-1\big )^{2HK}\right] . \end{aligned}$$

By the Taylor’s theorem

$$\begin{aligned} \big (a(t)^{2H}+1\big )^{K}= & {} a(t)^{2HK} + K \xi (t)^{K-1}, \\ \big (a(t)-1\big )^{2HK}= & {} a(t)^{2HK} - 2HK\eta (t)^{2HK-1}, \end{aligned}$$

for some $\xi (t) \in [a(t), a(t)+1]$ and $\eta (t)\in [a(t)-1,a(t)]$. Consequently,

$$\begin{aligned} r_{H,K,\theta }(t)\sim & {} C_{H,K,\theta } \mathrm {e}^{-\theta t}\left[ a(t)^{K-1} + a(t)^{2HK-1} \right] \\\sim & {} C_{H,K,\theta } \mathrm {e}^{-\theta t \max \left\{ \frac{1}{HK}-1,1+\frac{1}{HK}-\frac{1}{H}\right\} }, \end{aligned}$$

which shows the exponential decay.$\square$