Drawdown and Drawup for Fractional Brownian Motion with Trend

Abstract

We consider the drawdown and drawup of a fractional Brownian motion with trend, which corresponds to the logarithm of geometric fractional Brownian motion representing the stock price in a financial market. We derive the asymptotics of tail probabilities of the maximum drawdown and maximum drawup, respectively, as the threshold goes to infinity. It turns out that the extremes of drawdown lead to new scenarios of asymptotics depending on the Hurst index of fractional Brownian motion.

Appendix

1.1 Appendix A

This subsection is devoted to proofs of Lemmas 3.1–3.2.

Proof of Lemma 3.1

Note that for any $\delta >0$ and u sufficiently large, the maximum of $\sigma _u^-(s,t)$ over $0\le s\le t\le T$ is only attained at $[0,\delta ]\times [T-\delta , T]$. Next we consider the variance function $\sigma _u^-(s,t)$ over $[0,\delta ]\times [T-\delta , T]$. It follows that

$$\begin{aligned} 1-\frac{\sigma _u^-(s,t)}{\sigma _u^-(0,T)}&=1-\frac{|t-s|^H}{u+\mu (t-s)-\frac{1}{2}(t^{2H}-s^{2H})}\frac{u+\mu T-\frac{1}{2}T^{2H}}{T^H}\\&=1-\frac{\frac{|t-s|^H}{T^H}}{\frac{u+\mu (t-s)-\frac{1}{2}(t^{2H}-s^{2H})}{u+\mu T-\frac{1}{2}T^{2H}}}\\&= \left( 1-\frac{|t-s|^H}{T^H}\right) (1+o(1))\\&\quad +\left( \frac{u+\mu (t-s)-\frac{1}{2}(t^{2H}-s^{2H})}{u+\mu T-\frac{1}{2}T^{2H}}-1\right) (1+o(1))\\&= \frac{H}{T}(T-t+s)(1+o(1))\\&\quad +\frac{-\mu (T-t+s)+\frac{1}{2}(2HT^{2H-1}(T-t)+s^{2H})}{u+\mu T-\frac{1}{2}T^{2H}}(1+o(1))\\&= \left( \frac{H}{T}(T-t)+\frac{H}{T}s+\frac{1}{2u}s^{2H}\right) \\&\qquad (1+a(\delta ,u)), (s,t)\in [0,\delta ]\times [T-\delta , T], \end{aligned}$$

as $\delta$ sufficiently small and u sufficiently large, where $\lim _{\delta \rightarrow 0, u\rightarrow \infty }a(\delta ,u)=0$. The fact that

$$\begin{aligned} \frac{H}{T}(T-t)+\frac{H}{T}s+\frac{1}{2u}s^{2H}>0 \end{aligned}$$

for $(s,t)\in ([0,\delta ]\times [T-\delta , T]){\setminus } \{(0,T)\}$ implies that the maximizer of $\sigma _u^-(s,t)$ over $0\le s\le t\le T$ is unique and equals (0, T). This completes the proof. $\square$

Proof of Lemma 3.2

For any $\delta >0$ and u sufficiently large, the maximum of $\sigma _u^+(s,t)$ over $0\le s\le t\le T$ is attained in $[0,\delta ]\times [T-\delta , T]$. Next we focus on $\sigma _u^+(s,t)$ over $[0,\delta ]\times [T-\delta ,T]$. For $\delta >0$ sufficiently small and u sufficiently large,

$$\begin{aligned} 1-\frac{\sigma _u^+(s,t)}{\sigma _u^+(0,T)}&=1-\frac{|t-s|^H}{u-\mu (t-s)+\frac{1}{2}(t^{2H}-s^{2H})}\frac{u-\mu T+\frac{1}{2}T^{2H}}{T^H}\\&=1-\frac{\frac{|t-s|^H}{T^H}}{\frac{u-\mu (t-s)+\frac{1}{2}(t^{2H}-s^{2H})}{u-\mu T+\frac{1}{2}T^{2H}}}\\&= \left( 1-\frac{|t-s|^H}{T^H}\right) (1+o(1))\\&\quad +\left( \frac{u-\mu (t-s)+\frac{1}{2}(t^{2H}-s^{2H})}{u-\mu T+\frac{1}{2}T^{2H}}-1\right) (1+o(1))\\&= \frac{H}{T}(T-t+s)(1+o(1))\\&\quad +\frac{\mu (T-t+s)-\frac{1}{2}(2HT^{2H-1}(T-t)+s^{2H})}{u-\mu T+\frac{1}{2}T^{2H}}(1+o(1))\\&= \left( \frac{H}{T}(T-t)+\frac{H}{T}s\right) (1+a_1(\delta ,u))\\&\quad -\frac{1}{2u}s^{2H}(1+a_2(\delta ,u)), \quad (s,t)\in [0,\delta ]\times [T-\delta , T], \end{aligned}$$

where $\lim _{\delta \rightarrow 0, u\rightarrow \infty }a_i(\delta ,u)=0, i=1,2.$ If $H\ge \frac{1}{2}$, then

$$\begin{aligned} 1-\frac{\sigma _u^+(s,t)}{\sigma _u^+(0,T)} = \left( \frac{H}{T}(T-t)+\frac{H}{T}s\right) (1+a_1(\delta ,u)), \quad (s,t)\in [0,\delta ]\times [T-\delta , T], \end{aligned}$$

which implies that the maximizer of $\sigma _u^+(s,t)$ is unique and equals (0, T). For $0<H<\frac{1}{2}$,

$$\begin{aligned} 1-\frac{\sigma _u^+(s,T)}{\sigma _u^+(0,T)}&=\frac{H}{T}s(1+a_1(\delta ,u))-\frac{1}{2u}s^{2H}(1+a_2(\delta ,u))\\&=\frac{H}{T}s^{2H}\left( s^{1-2H}(1+a_1(\delta ,u))-\frac{1}{2u}(1+a_2(\delta ,u))\right) <0, \end{aligned}$$

as $s<\left( \frac{(1+a_2(\delta ,u))}{2u(1+a_1(\delta ,u))}\right) ^{\frac{1}{1-2H}}\sim (2u)^{-\frac{1}{1-2H}}.$ This implies that the maximum of $\sigma _u^+(s,T)$ over [0, T] is attained over $(0,\delta )$ for $\delta >0$ sufficiently small and u sufficiently large. We denote this point by $s_u$. Using the fact that

$$\begin{aligned}&\frac{\partial \sigma _u^+(s_u,T)}{\partial s}\\&\quad =\frac{-H(T-s_u)^{H-1}\left( u-\mu (T-s_u)+\frac{1}{2}(T^{2H}-s_u^{2H})\right) -(T-s_u)^H(\mu -Hs_u^{2H-1})}{(u-\mu (T-s_u)+\frac{1}{2}(T^{2H}-s_u^{2H}))^2}=0, \end{aligned}$$

we have that

$$\begin{aligned} s_u=\left( \frac{u}{T}+\frac{1}{2}T^{2H-1}+\frac{\mu (1-H)}{H}+\frac{1}{2T}s_u^{2H}-\frac{\mu (1-H)}{TH}s_u\right) ^{\frac{1}{2H-1}}\sim T^{\frac{1}{1-2H}}u^{-\frac{1}{1-2H}}. \end{aligned}$$

Next we show that the maximizer of $\sigma _u^+(s,t)$ is $(s_u,T)$ for $0<H<\frac{1}{2}$ and u sufficiently large. Observe that

$$\begin{aligned} 1-\frac{\sigma _u^+(s,t)}{\sigma _u^+(s_u,T)}=-\frac{\sigma _u^+(s,T)-\sigma _u^+(s_u,T)}{\sigma _u^+(s_u,T)} +\frac{\sigma _u^+(s,T)-\sigma _u^+(s,t)}{\sigma _u^+(s_u,T)}. \end{aligned}$$

Direct calculation gives that, as $u\rightarrow \infty$,

$$\begin{aligned} \sigma _u^+(s_u,T)&\sim \frac{T^H}{u},\\ \sigma _u^+(s,T)-\sigma _u^+(s_u,T)&=\frac{1}{2}\frac{\partial ^2\sigma _u^+(s_u,T)}{\partial ^2 s }(s-s_u)^2(1+o(1))\sim \frac{H(H-1)T^{H-2}}{2u}(s-s_u)^2,\\ \sigma _u^+(s,T)-\sigma _u^+(s,t)&=\frac{\partial \sigma _u^+(s,T)}{\partial t}(T-t)(1+o(1))\sim \frac{HT^{H-1}}{u}(T-t), \quad t\rightarrow T. \end{aligned}$$

Thus we have

$$\begin{aligned} 1-\frac{\sigma _u^+(s,t)}{\sigma _u^+(s_u,T)}= & {} \frac{H(1-H)}{2T^2}(s-s_u)^2(1+o(1))+\frac{H}{T}(T-t)(1+o(1)), \quad \\&u\rightarrow \infty , |s-s_u|, T-t\rightarrow 0, \end{aligned}$$

which implies that the maximizer of $\sigma _u^+(s,t)$ is unique and equals $(s_u,T)$ for u large. This completes the proof. $\square$

Proof of Lemma 3.3

Recalling that $\sigma _H(s,t)=|t-s|^H$, we have, for $(s,t)\ne (s',t')$,

$$\begin{aligned}&1-Corr(B_H(t)-B_H(s),B_H(t')-B_H(s'))\\&\quad = \frac{{\mathbb {E}}\left\{ ((B_H(t)-B_H(s))-(B_H(t')-B_H(s')))^2\right\} -(\sigma _H(s,t)-\sigma _H(s',t'))^2}{2\sigma _H(s,t)\sigma _H(s',t')}\\&\quad = \frac{{\mathbb {E}}\left\{ ((B_H(t)-B_H(t'))-(B_H(s)-B_H(s')))^2\right\} -(\left|t-s \right|^{H}-\left|t'-s' \right|^{H})^2}{2\left|t-s \right|^{H}\left|t'-s' \right|^{H}}\\&\quad = \frac{\left|t-t' \right|^{2H}+\left|s-s' \right|^{2H}+(|t-s|^{2H}+|t'-s'|^{2H}-|t-s'|^{2H}-|t'-s|^{2H}) -(\left|t-s \right|^{H}-\left|t'-s' \right|^{H})^2}{2\left|t-s \right|^{H}\left|t'-s' \right|^{H}}. \end{aligned}$$

Using Taylor formula, we have that for $(s,t)\in [0,\delta _u]\,\times \, [T-\delta _u,T]$, with $\lim _{u\rightarrow \infty }\delta _u=0$ and u sufficiently large

$$\begin{aligned}&|t-s|^{2H}-|t-s'|^{2H}-(|t'-s|^{2H}-|t'-s'|^{2H})\\&\quad =2H(|\theta _1-s|^{2H-1}-|\theta _1-s'|^{2H-1})(t-t')\\&\quad =2H(2H-1)(\theta _1-\theta _2)^{2H-2}(s-s')(t-t'),\\&\qquad (\left|t-s \right|^{H}-\left|t'-s' \right|^{H})^2 =(H\theta _3(t-t'-s+s'))^2, \end{aligned}$$

where $\theta _1\in (t,t')$, $\theta _2\in (s,s')$ and $\theta _3\in (t-s, t'-s')$. Moreover,

$$\begin{aligned} \lim _{u\rightarrow \infty }\lim _{s,t\in [0,\delta _u]\times [T-\delta _u,T]}\left| |t-s|^{H}- T^H\right| =0. \end{aligned}$$

Consequently, for $\lim _{u\rightarrow \infty }\delta _u=0$,

$$\begin{aligned}&\lim _{u\rightarrow \infty }\sup _{(s,t)\ne (s',t'), (s,t), (s',t')\in [0,\delta _u]\times [T-\delta _u, T]}\\&\quad \times \left| \frac{1-Corr\left( B_H(t)-B_H(s), B_H(t')-B_H(s')\right) }{\frac{|s-s'|^{2H}+|t-t'|^{2H}}{2T^{2H}}}-1\right| =0. \end{aligned}$$

$\square$

1.2 Appendix B

In this subsection, we present some useful results derived in [17]. First, we give an adaptation of Theorem 3.2 in [17] to our setting. Let $X_u(s,t), (s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]$ with ${(0,0)}\in \prod _{i=1,2}[a_i(u), b_i(u)]$, be a family of centered continuous Gaussian random fields with variance function $\sigma _u(s,t)$ satisfying

$$\begin{aligned} \sigma _u(0,0)=1, \quad \lim _{u\rightarrow \infty }\sup _{(s,t)\ne (0,0), (s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]}\left| \frac{1-\sigma _u(s,t)}{ \frac{|s|^{\beta _1}}{g_1(u)}+\frac{|t|^{\beta _2}}{g_2(u)}}-1\right| =0,\qquad \end{aligned}$$

(27)

where $\beta _i>0, i=1,2$, $\lim _{u\rightarrow \infty }g_i(u)=\infty , i=1,2$, $\lim _{u\rightarrow \infty }\frac{|a_i(u)|^{\beta _1}}{g_1(u)}+\frac{+|b_i(u)|^{\beta _2}}{g_2(u)}=0, i=1,2,$ and correlation function satisfying

$$\begin{aligned}&\lim _{u\rightarrow \infty }\sup _{(s,t)\ne (s',t'), (s,t), (s',t')\in \prod _{i=1,2}[a_i(u), b_i(u)]}\nonumber \\&\quad \times \left| n^2(u)\frac{1-Corr(X_u(s,t), X_u(s',t'))}{|s-s'|^{\alpha }+|t-t'|^{\alpha }}-1\right| =0, \end{aligned}$$

(28)

with $\alpha \in (0,2]$ and $\lim _{u\rightarrow \infty }n(u)=\infty$.

We suppose that $\lim _{u\rightarrow \infty }\frac{n^2(u)}{g_i(u)}=\nu _i\in [0,\infty ], i=1,2$.

Lemma 3.4

Let $X_u(s,t), (s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]$ with ${(0,0)}\in \prod _{i=1,2}[a_i(u), b_i(u)]$ be a family of centered continuous Gaussian random fields satisfying (27) and (28).

• i) If $\nu _i=0, i=1,2$ and for $i=1,2$,

  $$\begin{aligned}&\lim _{u\rightarrow \infty }\frac{(n(u))^{2/\beta _i}a_i(u)}{(g_i(u))^{1/\beta _i}}=y_{i,1}, \quad \lim _{u\rightarrow \infty }\frac{(n(u))^{2/\beta _i}b_i(u)}{(g_i(u))^{1/\beta _i}}=y_{i,2},\\&\quad \lim _{u\rightarrow \infty }\frac{(n(u))^{2/\beta _i}(a_i^2(u)+b_i^2(u))}{(g_i(u))^{2/\beta _i}}=0, \end{aligned}$$

  with $-\infty \le y_{i,1}<y_{i,2}\le \infty$, then

  $$\begin{aligned}&\mathbb {P}\left\{ \sup _{(s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]}X_u(s,t)>n(u)\right\} \\&\quad \sim \left( {\mathcal {H}}_{\alpha /2}\right) ^2\prod _{i=1}^{2}\int _{y_{i,1}}^{y_{i,2}}e^{-|s|^{\beta _i}}ds \prod _{i=1}^{2}\left( \frac{g_i(u)}{n^2(u)}\right) ^{1/\beta _i}\Psi (n(u)). \end{aligned}$$

• ii) If $\nu _i\in (0,\infty )$ and further $\lim _{u\rightarrow \infty }a_i(u)=a_i\in [-\infty ,0], \lim _{u\rightarrow \infty }b_i(u)=b_i\in [0,\infty ]$, then

  $$\begin{aligned} \mathbb {P}\left\{ \sup _{(s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]}X_u(s,t)>n(u)\right\} \sim \prod _{i=1}^{2} {\mathcal {P}}_{\alpha /2}^{v_{i},\beta _i}([a_i,b_i])\Psi (n(u)), \end{aligned}$$

  where

  $$\begin{aligned} {\mathcal {P}}_{\alpha /2}^{v_{i},\beta _i} ([a_i,b_i])={\mathbb {E}}\left\{ \sup _{t\in [a_i,b_i]}e^{\sqrt{2}B_{\alpha /2}(t)-|t|^{\alpha }-\nu _i|t|^{\beta _i}}\right\} \in (0,\infty ),\ i=1,2. \end{aligned}$$

• iii) If $\nu _i=\infty , i=1,2$, then

  $$\begin{aligned} \mathbb {P}\left\{ \sup _{(s,t)\in \prod _{i=1,2}[a_i(u), b_i(u)]}X_u(s,t)>n(u)\right\} \sim \Psi (n(u)). \end{aligned}$$

Next we give a simpler version of Proposition 2.2 in [17]. Denote by $\Lambda (u)$ a series of index sets depending on u and by $[a_1,a_2]\times [b_1,b_2]$ a rectangle with $a_1<a_2$ and $b_1<b_2$. Let $X_{u,k,l}(s,t), (s,t)\in [a_1,a_2]\times [b_1,b_2], (k,l)\in \Lambda (u)$ be a family of two-dimensional continuous Gaussian random fields with mean 0 and variance function 1. There exist $n_{k,l}(u), (k,l)\in \Lambda (u)$ satisfying

$$\begin{aligned} \lim _{u\rightarrow \infty }\sup _{(k,l), (k',l')\in \Lambda (u)}\left| \frac{n_{k,l}(u)}{n_{k'l'}(u)}-1\right| =0,\quad \lim _{u\rightarrow \infty }\inf _{(k,l)\in \Lambda (u)}n_{k,l}=\infty , \end{aligned}$$

(29)

such that the correlation function satisfies

$$\begin{aligned}&\lim _{u\rightarrow \infty }\sup _{(k,l)\in \Lambda (u)}\sup _{(s,t)\ne (s',t'), (s,t), (s',t')\in [a_1,a_2]\times [b_1,b_2]}\nonumber \\&\quad \times \left| \left( n_{k,l}(u)\right) ^2\frac{1-Corr\left( X_{u,k,l}(s,t), X_{u,k,l}(s',t')\right) }{|s-s'|^{\alpha _1}+|t-t'|^{\alpha _2}}-1\right| =0, \end{aligned}$$

(30)

where $\alpha _i\in (0,2], i=1,2$.

Then Proposition 2.2 in [17] leads to the following result.

Lemma 3.5

Let $X_{u,k,l}(s,t), (s,t)\in E, (k,l)\in \Lambda (u)$ be a family of centered two-dimensional continuous Gaussian random fields with unit variance. Assume further that (29)–(30) hold. Then

$$\begin{aligned}&\lim _{u\rightarrow \infty }\sup _{(k,l)\in \Lambda (u)}\left| \frac{\mathbb {P}\left\{ \sup _{(s,t)\in [a_1,a_2]\times [b_1,b_2]}X_{u,k,l}(s,t)>n_{k,l}(u)\right\} }{\Psi \left( n_{k,l}(u)\right) }\right. \\&\quad \left. -{\mathcal {H}}_{\frac{\alpha _1}{2}}([a_1,a_2]){\mathcal {H}}_{\frac{\alpha _1}{2}}([b_1,b_2])\right| =0. \end{aligned}$$

Finally, we display a lemma concerning the uniform double maximum, a simpler version of Corollary 3.2 in [17]. Let $E_u$ be a family of non-empty compact subset of $\mathbb {R}^2$ and $A_i\subset [0,S]^2, i=1,2$ be two non-empty compact subsets of $\mathbb {R}^2$. Denote by $\Lambda _0(u)=\{(k_1,l_1,k_2, l_2): (k_i,l_i)+A_i\subset E_u, i=1,2\}$. Let n(u) and $n_{k_i,l_i}(u), (k_i,l_i)+A_i\subset E_u$ be a family of positive functions such that

$$\begin{aligned} \lim _{u\rightarrow \infty }\sup _{(k_i,l_i)+A_i\in E_u}\left| \frac{n_{k_i,l_i}(u)}{n(u)}-1\right| =0, \quad i=1,2, \quad \lim _{u\rightarrow \infty } n(u)=\infty . \end{aligned}$$

(31)

Lemma 3.6

Let $X_u(s,t), (s,t)\in E_u$ be a family of centered Gaussian random fields with variance 1 and correlation function satisfying

$$\begin{aligned} \lim _{u\rightarrow \infty }\sup _{(s,t)\ne (s',t'), (s,t),(s',t')\in E_u}\left| (n(u))^2\frac{1-Corr(X_u(s,t), X_u(s',t'))}{|s-s'|^{\alpha _1}+|t-t'|^{\alpha _2}}-1\right| =0. \end{aligned}$$

Moreover, there exists $\delta >0$ such that for u large enough

$$\begin{aligned} Corr(X_u(s,t), X_u(s',t'))>\delta -1, \ (s,t),\ (s',t')\in E_u. \end{aligned}$$

If further (31) is satisfied, then there exit ${\mathcal {C}}, {\mathcal {C}}_1>0$ such that for all u large

$$\begin{aligned}&\sup _{(k_1,l_1,k_2,l_2)\in \Lambda _0(u), A_i\subset [0,S]^2, A_i\ne \emptyset , i=1,2 }\\&\quad \frac{\mathbb {P}\left\{ \sup _{(s,t)\in (k_1,l_1)+A_1}X_u(s,t)>n_{k_1,l_1}(u), \sup _{(s,t)\in (k_2,l_2)+A_2}X_u(s,t)>n_{k_2,l_2}(u)\right\} }{e^{-{\mathcal {C}}_1 (F((k_1,l_1)+A_1, (k_2,l_2)+A_2))^{\frac{1}{2}\min (\alpha _1,\alpha _2)}}S^4\Psi (n_{k_1,l_1,k_2,l_2}(u))}\le {\mathcal {C}}, \end{aligned}$$

where

$$\begin{aligned} F(A,B)=\inf _{s\in A, t\in B}||s-t||, \quad n_{k_1,l_1,k_2,l_2}(u)=\min (n_{k_1,l_1}(u), n_{k_2,l_2}(u)), \end{aligned}$$

and ${\mathcal {C}}$ and ${\mathcal {C}}_1$ are independent of u and S.