The Gumbel test and jumps in the volatility process

Abstract

In the framework of jump detection in stochastic volatility models the Gumbel test based on extreme value theory has recently been introduced. Compared to other jump tests it possesses the advantages that the direction and location of jumps may also be detected. Furthermore, compared to the Barndorff–Nielsen and Shephard test based on bipower variation the Gumbel test possesses a larger power. However, so far one assumption was that the volatility process is Hölder continuous, though there is empirical evidence for jumps in the volatility as well. In this paper we derive that the Gumbel test still works under the setting of finitely many jumps not exceeding a certain size. This maximal jump size depends on the relative sampling frequencies involved in the definition of the test statistics. Furthermore, we show that the given bound on the jump size is sharp and investigate the details of the phase transition at this critical bound.

Appendix: Auxiliary results

For completeness, we provide a proof that a sequence of stopping times $(S_l)_{l\geqslant 0}$ as stated in the Assumption 2.5 exists, N is measurable and that K can be chosen as a measurable function. Assume for this purpose that the Assumption 2.5 holds and set for $l\geqslant 1$

$$\begin{aligned} S_l(\omega )= {\left\{ \begin{array}{ll} \text {Position of the}\,l{\text {-th jump in}}\,\sigma (\omega ), &{} \sigma (\omega )\,\text {has at least}\,l\,\text {jumps}, \\ \infty , &{} \text {else}, \end{array}\right. } \end{aligned}$$

and set $S_0 = 0.$ To understand that each $S_l$ is a stopping time, an inductive argument is provided. Firstly, define for $r,\,s,\,u,\,v\in \mathbb {Q}$ and $m,\,n\in \mathbb {N}$ the sets

$$\begin{aligned} I_{u,v,n}= & {} \left\{ (r,\,s)\in \mathbb {Q}^2{\text {:}}\,u<r,\,s<v\; \text {and}\,|r-s|<\frac{1}{n}\right\} , \\ A_{r,s,m}= & {} \left\{ \omega \in \Omega {\text {:}}\, \left| \sigma _r(\omega )-\sigma _s(\omega )\right| >\frac{1}{m}\right\} . \end{aligned}$$

$S_0$ is obviously a stopping time. Assume for the induction step that $S_l$ is also a stopping time for some $l\in \mathbb {N}.$ Observe, furthermore, for $t>0$ and

$$\begin{aligned} C_{u,v} =\bigcup _{m\in \mathbb {N}}\bigcap _{n\in \mathbb {N}}\bigcup _{(r,s)\in I_{u,v,n}} A_{r,s,m}\in \mathcal {F}_v,\quad 0<u,\,v<1, \end{aligned}$$

the relation

$$\begin{aligned} \left\{ S_{l+1}< t\right\} = \bigcup _{\mathop {0<s<t,}\limits _{s\in \mathbb {Q}}}\left\{ S_l<s\right\} \cap C_{s,t} \in \mathcal {F}_t, \end{aligned}$$

which proves that $S_{l+1}$ is a stopping time due to the right continuity of the filtration $(\mathcal {F}_t).$ Next

$$\begin{aligned} \{N=n\} = \left\{ S_n <\infty \right\} \cap \left\{ S_{n+1}=\infty \right\} \in \mathcal {F},\quad n\in \mathbb {N}_0, \end{aligned}$$

yields that N is measurable. It remains to establish that K can be chosen as a measurable function. To understand this, set

$$\begin{aligned} \psi _t=\sigma _t-\sum _{l=1}^\infty \Delta \sigma _{S_l} {{\mathrm{\mathbbm {1}}}}_{(S_l,1]}(t),\quad 0\leqslant t\leqslant 1,\quad \Delta \sigma _\infty \buildrel \mathrm{def}\over =0, \end{aligned}$$

and observe that $\psi$ is $(\mathcal {F}_t)$ adapted since $\sigma$ is $(\mathcal {F}_t)$ adapted and $(S_l)$ are stopping times as proven previously. Note that $\psi$ is simply $\sigma$ without jumps. Since $\psi$ is pathwise $\alpha$-Hölder continuous, we can define

$$\begin{aligned} K(\omega )=\left( \sup _{\mathop {0\leqslant s<t\leqslant 1,}\limits _{s,t\in \mathbb {Q}}} \frac{|\psi _t(\omega )-\psi _s(\omega )|}{|t-s|^\alpha }\right) \vee \sup _{\mathop {0\leqslant t\leqslant 1,}\limits _{t\in \mathbb {Q}}} \left| \sigma _t(\omega )\right| <\infty ,\quad \omega \in \Omega . \end{aligned}$$

K is obviously measurable and fulfills the requirements of the Assumption 2.5. Compare for similar results in this context also Chap. I, Proposition 1.32 in Jacod and Shiryaev (2002).

Next, we state the crucial lemma needed for the proof of Proposition 3.3. Note that the upper bound C in Lemma 4.2 is easy to see. However, for the lower bound we need a deep result about Euler’s totient function.

Lemma 4.2

Set $\Lambda _m = \left\{ \frac{k}{l}{\text {:}}\, 1\leqslant k \leqslant l \leqslant m\right\} .$ Then, there are two constants $0<c<C,$ such that

$$\begin{aligned} \frac{|\Lambda _m|}{m^2}\in (c,\,C),\quad m\in \mathbb {N}. \end{aligned}$$

Proof

Set

$$\begin{aligned} \widetilde{B}_l = \left\{ \frac{k}{l}{\text {:}}\,1\leqslant k\leqslant l\right\} ,\quad l\in \mathbb {N}. \end{aligned}$$

Then, it holds $\Lambda _m = \bigcup _{l=1}^m \widetilde{B}_l$ with a non disjunct union. Next, set

$$\begin{aligned} B_l = \left\{ \frac{k}{l}{\text {:}}\, 1\leqslant k\leqslant l,\,\gcd (k,\,l) = 1\right\} \subseteq \widetilde{B}_l,\quad l\in \mathbb {N}. \end{aligned}$$

(47)

Obviously, we have

$$\begin{aligned} \Lambda _m = \bigcup _{l=1}^m \widetilde{B}_l = \biguplus _{l=1}^m B_l,\quad m\in \mathbb {N}, \end{aligned}$$

where the last union is disjunct, so that we have

$$\begin{aligned} \left| \Lambda _m\right| = \sum _{l=1}^m \left| B_l\right| = \sum _{l=1}^m|\{1\leqslant k\leqslant l{\text {:}}\, \gcd (k,\,l) = 1\}| = \sum _{l=1}^m \varphi (l), \end{aligned}$$

(48)

and $\varphi$ denotes Euler’s totient function. Next, consider the asymptotics

$$\begin{aligned} \sum _{l=1}^m \varphi (l) = \frac{3}{\pi ^2}{m^2} + O\left( m^\delta \right) ,\quad m\rightarrow \infty , \end{aligned}$$

(49)

for some $\delta \in (1,\,2),$ cf. Sándor et al. (2006, p. 24). Eventually, (48) and (49) prove the claim of this lemma.$\square$