Rate of Convergence for Discretization of Integrals with Respect to Fractional Brownian Motion

Abstract

In this article, a uniform discretization of stochastic integrals $\int _{0}^{1} f^{\prime }_-(B_t)\mathrm d B_t$, where $B$ denotes the fractional Brownian motion with Hurst parameter $H \in (\frac{1}{2},1)$, is considered for a large class of convex functions $f$. In Azmoodeh et al. (Stat Decis 27:129–143, 2010), for any convex function $f$, the almost sure convergence of uniform discretization to such stochastic integral is proved. Here, we prove $L^r$-convergence of uniform discretization to stochastic integral. In addition, we obtain a rate of convergence. It turns out that the rate of convergence can be brought arbitrarily close to $H - \frac{1}{2}$.

Appendix: Proofs of Lemmas 3.1 and 3.2

We begin with the following lemma which we use in the proof.

Lemma 4.3

Let $H>\frac{1}{2}$ and fix $0<s\le t\le 1$. Put

$$\begin{aligned} R(t,s) = \frac{1}{2}\left[ t^{2H}+s^{2H} - (t-s)^{2H}\right] . \end{aligned}$$

Then, there exists a constant $C$, such that

$$\begin{aligned} 1-\frac{R(s,s)}{R(t,s)}\le C (t-s)^H s^{-H}. \end{aligned}$$

Proof

Note that since $H>\frac{1}{2}$, we have $R(s,s)\le R(t,s)$. Let now $t>2s$. Then

$$\begin{aligned} \frac{(t-s)^H}{s^H} \ge \frac{(2s-s)^H}{s^H} = 1 \ge 1 -\frac{R(s,s)}{R(t,s)}. \end{aligned}$$

Hence, it is sufficient to consider the case $s\le t \le 2s$. In this case, we have

$$\begin{aligned} \frac{R(t,s)}{R(s,s)}\le \frac{R(2s,s)}{R(s,s)}=2^{2H-1}. \end{aligned}$$

Hence, we only have to prove that

$$\begin{aligned} 1-\frac{R(s,s)}{R(t,s)}\le C \frac{(t-s)^H}{s^{H}}\frac{R(t,s)}{R(s,s)}. \end{aligned}$$

By putting $k=\frac{t}{s}$, this is equivalent to

$$\begin{aligned} \left[ k^{2H}-1-(k-1)^{2H}\right] \le C(k-1)^H\left[ k^{2H} + 1 - (k-1)^{2H}\right] ^2. \end{aligned}$$

Now, we have $k\in [1,2]$. Hence,

$$\begin{aligned}&\frac{\left[ k^{2H}-1-(k-1)^{2H}\right] }{(k-1)^H\left[ k^{2H} + 1 - (k-1)^{2H}\right] ^2}\\&\quad \le \frac{k^{2H}-1}{(k-1)^H}\le \frac{k^{2H}-1}{k^H-1}\\&\quad = k^H+1 \le 2^H+1. \end{aligned}$$

This completes the proof.$\square$

Proof of lemma 3.1

Let $R(t,s)$ denotes the covariance function of fractional Brownian motion given by

$$\begin{aligned} R(t,s) = \frac{1}{2}\left[ t^{2H}+s^{2H} - (t-s)^{2H}\right] . \end{aligned}$$

We make use of decomposition

$$\begin{aligned} B_t = \frac{R(t,s)}{R(s,s)}B_s + \sigma Y, \end{aligned}$$

where $Y$ is $N(0,1)$ random variable independent of $B_s$ and

$$\begin{aligned} \sigma ^2 = \frac{R(t,t)R(s,s)-R(t,s)^2}{R(s,s)}. \end{aligned}$$

Assume that

$$\begin{aligned} \frac{R(s,s)}{R(t,s)}(a-1) < a. \end{aligned}$$

Then, we obtain

$$\begin{aligned} \mathbb{P }(B_t > a > B_s)&= \int \limits _{-\infty }^a \mathbb{P }\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{x^2}{2s^{2H}}}\mathrm d x\\&= \int \limits _{-\infty }^{\frac{R(s,s)}{R(t,s)}(a-1)} \mathbb{P }\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{x^2}{2s^{2H}}}\mathrm d x\\&+ \int \limits _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \mathbb{P }\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{x^2}{2s^{2H}}}\mathrm d x\\&:= I_1 + I_2. \end{aligned}$$

We begin with $I_1$. By Lemma 4.2 we have

$$\begin{aligned} \mathbb{P }\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \le \frac{1}{\sqrt{2\pi }A(x)}e^{-\frac{A(x)^2}{2}}, \end{aligned}$$

where $A(x) = \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }$. Hence

$$\begin{aligned} I_1&\le \int \limits _{-\infty }^{\frac{R(s,s)}{R(t,s)}(a-1)} \frac{1}{\sqrt{2\pi }A(x)}e^{-\frac{A(x)^2}{2}}\frac{1}{\sqrt{2\pi }s^H}e^{-\frac{x^2}{2s^{2H}}}\mathrm d x\\&\le \frac{\sigma }{s^H}e^{-\frac{a^2}{2}}\int \limits _{-\infty }^{\frac{R(s,s)}{R(t,s)}(a-1)} \frac{1}{2\pi }e^{-\frac{A(x)^2}{2}-\frac{x^2}{2s^{2H}} + \frac{a^2}{2}}\mathrm d x\\&= \frac{R(s,s)}{R(t,s)}\frac{\sigma }{s^H}e^{-\frac{a^2}{2}}\int \limits _{1}^{\infty } \frac{1}{2\pi }e^{-\frac{y^2}{2\sigma ^2}-\frac{\left[ \frac{R(s,s)}{R(t,s)}(a-y)\right] ^2}{2s^{2H}} + \frac{a^2}{2}}\mathrm d y \end{aligned}$$

Note that $\sigma \le (t-s)^H$ and $R(s,s)\le R(t,s)$, so it remains to show that the integral is bounded by a constant independent of $s, t$ and $a$. It is easy to see that

$$\begin{aligned}&-\frac{y^2}{2\sigma ^2}-\frac{\left[ \frac{R(s,s)}{R(t,s)}(a-y)\right] ^2}{2s^{2H}} + \frac{a^2}{2} \\&= -\frac{1}{2\sigma ^2}\left[ \left( y-a\frac{R(s,s)}{R(t,s)^2}\bar{\sigma }^2\right) ^2 + a^2\left( \frac{R(s,s)}{R(t,s)^2}\bar{\sigma }^2 - \bar{\sigma }^2 - \frac{R(s,s)^2}{R(t,s)^4}\bar{\sigma }^4\right) \right] , \end{aligned}$$

where

$$\begin{aligned} \frac{1}{\bar{\sigma }^2}=\frac{1}{\sigma ^2} + \frac{R(s,s)}{R(t,s)^2}. \end{aligned}$$

Now

$$\begin{aligned} \frac{1}{\bar{\sigma }^2}\ge 1 \end{aligned}$$

and

$$\begin{aligned} \left( \frac{R(s,s)}{R(t,s)^2}\bar{\sigma }^2 - \bar{\sigma }^2 - \frac{R(s,s)^2}{R(t,s)^4}\bar{\sigma }^4\right) \ge 0. \end{aligned}$$

Hence

$$\begin{aligned}&\int \limits _{1}^{\infty } \frac{1}{2\pi }e^{-\frac{y^2}{2\sigma ^2}-\frac{\left[ \frac{R(s,s)}{R(t,s)}(a-y)\right] ^2}{2s^{2H}} + \frac{a^2}{2}}\mathrm d y\\&\le \int \limits _{1}^{\infty } \frac{1}{2\pi }e^{-\frac{1}{2\sigma ^2}\left( y-a\frac{R(s,s)}{R(t,s)^2}\bar{\sigma }^2\right) ^2}\mathrm d y\\&\le \frac{1}{\sqrt{2\pi }}. \end{aligned}$$

Hence for $I_1$, there exists a constant $C$ such that

$$\begin{aligned} I_1 \le C e^{-\frac{a^2}{2}} \ \frac{(t-s)^H}{s^H}. \end{aligned}$$

We proceed to study the term $I_2$. Note that $\sigma ^2\ge 0$. Hence

$$\begin{aligned} \frac{R(s,s)}{R(t,s)^2}\ge \frac{1}{R(t,t)}\ge 1. \end{aligned}$$

As a consequence, there exists a constant¹ $C$ such that

$$\begin{aligned} e^{-\frac{x^2}{2s^{2H}}} \le Ce^{-\frac{\min \{ a^2,(a-1)^2 \} }{2}} \end{aligned}$$

for every $a$ and every $x\in \left[ \frac{R(s,s)}{R(t,s)}(a-1),a\right]$. Hence

$$\begin{aligned} I_2&= \int \limits _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int \limits _{A(x)}^{\infty }\frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}}\mathrm d y \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{x^2}{2s^{2H}}}\mathrm d x\\&\le \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{\min \{ a^2,(a-1)^2 \} }{2}}\int \limits _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int \limits _{A(x)}^{\infty }\frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}} \frac{1}{\sqrt{2\pi }}\mathrm d y\mathrm d x. \end{aligned}$$

By applying Tonelli’s theorem, the integral can be written as

$$\begin{aligned}&\int \limits _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int \limits _{A(x)}^{\infty }\frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}} \mathrm d y\mathrm d x\\&= \int \limits _{\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}}\left[ a-\frac{R(s,s)}{R(t,s)}(a-\sigma y)\right] \mathrm d y\\&\quad + \int \limits _{\frac{1}{\sigma }}^{\infty }\frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}}\left[ a-\frac{R(s,s)}{R(t,s)}(a-1)\right] \mathrm d y\\&=: I_{2,1}+I_{2,2}. \end{aligned}$$

For $I_{2,2}$, by Lemma 4.2, we obtain

$$\begin{aligned} I_{2,2}&\le C\left[ \left( 1-\frac{R(s,s)}{R(t,s)}\right) |a|+\frac{R(s,s)}{R(t,s)}\right] \sigma \\&\le C\max (1,|a|)(t-s)^H. \end{aligned}$$

For $I_{2,1}$, by applying Lemma 4.3, we obtain

$$\begin{aligned} I_{2,1}&\le C\left( 1-\frac{R(s,s)}{R(t,s)}\right) |a|+C\sigma \\&\le C\max (1,|a|)\frac{(t-s)^H}{s^H}. \end{aligned}$$

Hence, we have the result. To conclude the proof, we note that if

$$\begin{aligned} \frac{R(s,s)}{R(t,s)}(a-1) > a, \end{aligned}$$

then we proceed as for $I_1$ and obtain the result.$\square$

Proof of lemma 3.2

By following the proof of Lemma 3.1, we obtain that for every $H\ge \frac{1}{2}$

$$\begin{aligned} \mathbb{P }(B_t >a>B_s) \le I_1 + \frac{1}{\sqrt{2\pi }s^H}e^{-\frac{\min \{ a^2,(a-1)^2 \} }{2}}(I_{2,1}+I_{2,2}). \end{aligned}$$

Moreover, we have

$$\begin{aligned} I_1 \le C e^{-\frac{a^2}{2}} \ \frac{(t-s)^H}{s^H}. \end{aligned}$$

The claim follows using the fact that when $H=\frac{1}{2}$, we have $R(s,s)=R(t,s)$ for $s \le t$. Hence,

$$\begin{aligned} I_{2,1}&\le C\left( 1-\frac{R(s,s)}{R(t,s)}\right) |a|+C\sigma \\&= C\sigma \le C(t-s)^H, \end{aligned}$$

and

$$\begin{aligned} I_{2,2}&\le C\left[ \left( 1-\frac{R(s,s)}{R(t,s)}\right) |a|+\frac{R(s,s)}{R(t,s)}\right] \sigma \\&= C\sigma \le C(t-s)^H. \end{aligned}$$

$\square$