Pathwise Integrals and Itô–Tanaka Formula for Gaussian Processes

Abstract

We prove an Itô–Tanaka formula and existence of pathwise stochastic integrals for a wide class of Gaussian processes. Motivated by financial applications, we define the stochastic integrals as forward-type pathwise integrals introduced by Föllmer and as pathwise generalized Lebesgue–Stieltjes integrals introduced by Zähle. As an application, we illustrate the importance of the Itô–Tanaka formula for pricing and hedging of financial derivatives.

Appendix: Level-Crossing Lemma

The key lemma in our analysis is the following estimate for the probability that a Gaussian process $X$ crosses a fixed level. Actually, in [3] the authors proved the lemma in the particular case of fractional Brownian motion. We extend the result here for more general Gaussian process. We consider only probability ${\mathbb {P}}(X_s < a < X_t)$ and a case $\sup _{s\le T} V(s) \le 1$. However, by considering processes $Y = -X$ and $Y = \frac{X}{\sqrt{V^*}}$ we obtain same bound for ${\mathbb {P}}(X_s > a > X_t)$ and for the general case $\sup _{s\le T} V(s) < \infty$. Also, note that continuous Gaussian processes on compact time intervals satisfy $V^*<\infty$.

Lemma 4.1

Let $X$ be a Gaussian process with positive covariance function $R$. Denote

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

and fix $0 < s < t \le T$ and $a \in {\mathbb {R}}$. Assume that the variance function satisfies

$$\begin{aligned} V^* := \sup _{s_\le T} V(s) \le 1. \end{aligned}$$

• (i) If

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

  then there exists a constant $C$, independent of $s$, $t$, $T$ and $a$, such that

  $$\begin{aligned} {\mathbb {P}}\big ( X_s < a < X_t\big ) \le I_1 + I_2 + I_3 +I_4, \end{aligned}$$

  where

  $$\begin{aligned} I_1&\le C \min [\sqrt{V(s)}\sigma ,\sigma ^2] e^{-\frac{a^2}{2}}, \\ I_2&\le Ce^{-\frac{\min [a^2,(a-1)^2]}{2V^*}}\frac{\sigma }{\sqrt{V(s)}}\left[ {\mathbf {1}}_{|a|>2}+\left( a-\frac{R(s,s)}{R(t,s)}(a-1)\right) {\mathbf {1}}_{|a|\le 2}\right] , \\ I_3&\le C \frac{R(s,s)}{R(t,s)}\frac{\sigma }{\sqrt{V(s)}}e^{-\frac{\min [a^2,(a-1)^2]}{2V^*}}, \\ I_4&\le e^{-\frac{a^2}{2V^*}}\frac{1}{\sqrt{V(s)}}\left| a\left( 1-\frac{R(s,s)}{R(t,s)}\right) \right| , \end{aligned}$$

• (ii) If

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

  then there exists a constant $C$, independent of $s$, $t$, $T$ and $a$, such that

  $$\begin{aligned} {\mathbb {P}}\big ( X_s < a <X_t \big ) \le C \min [\sqrt{V(s)}\sigma ,\sigma ^2] e^{-\frac{a^2}{2}}. \end{aligned}$$

In the proof we use the following well-known estimate.

Lemma 4.2

Let $Z$ be a standard normal random variable and fix $a>0$. Then

$$\begin{aligned} {\mathbb {P}}\big (Z>a\big ) \le \frac{1}{\sqrt{2\pi }a}e^{-\frac{a^2}{2}}. \end{aligned}$$

Proof of Lemma 4.1

We make use of decomposition

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

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

$$\begin{aligned} {\mathbb {P}}(X_s<a<X_t)&= \int _{-\infty }^a {\mathbb {P}}\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x\\&= \int _{-\infty }^{\frac{R(s,s)}{R(t,s)}(a-1)} {\mathbb {P}}\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x\\&\quad + \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} {\mathbb {P}}\left( Y\ge \frac{a-\frac{R(t,s)}{R(s,s)}x}{\sigma }\right) \frac{1}{\sqrt{2\pi }\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x\\&= I_1 + A_1. \end{aligned}$$

Moreover, if

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

then

$$\begin{aligned} {\mathbb {P}}(X_s<a<X_t) \le I_1. \end{aligned}$$

Note that here $I_1$ corresponds the one given in the Lemma and $A_1$ contains $I_2$, $I_3$ and $I_4$. We shall use similar technique for the rest of the proof.

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

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

We proceed to study the integral. Now,

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

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 since $V^*\le 1$, we also have

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

Thus,

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

Hence, we obtain for $I_1$ that

$$\begin{aligned} I_1 \le C\frac{R(s,s)}{R(t,s)}\frac{\sigma }{\sqrt{V(s)}}e^{-\frac{a^2}{2}}\bar{\sigma }. \end{aligned}$$

Now we have

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

and hence for $I_1$ there exists a constant $C$ such that

$$\begin{aligned} I_1 \le C\min [\sqrt{V(s)}\sigma ,\sigma ^2]e^{-\frac{a^2}{2}}. \end{aligned}$$

For the term $A_1$ we have

$$\begin{aligned} A_1&= \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int _{A(x)}^\infty \frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}}\,{\mathrm {d}}y\frac{1}{\sqrt{2\pi }\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x\\&= \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int _{A(x)}^{\frac{1}{\sigma }}\ldots {\mathrm {d}}y{\mathrm {d}}x + \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int _{\frac{1}{\sigma }}^\infty \ldots {\mathrm {d}}y{\mathrm {d}}x\\&= A_{2}+I_{2}. \end{aligned}$$

Consider then $I_2$. Applying Lemma 4.2 we obtain

$$\begin{aligned} I_2 \le C \frac{\sigma }{\sqrt{V(s)}}e^{-\frac{1}{2\sigma ^2}}\int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x. \end{aligned}$$

Note that $\sigma ^2\ge 0$. Therefore,

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

Now if $|a|> 2$, we can apply Lemma 4.2 to obtain

$$\begin{aligned} \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x \le Ce^{-\frac{\min [a^2,(a-1)^2]}{2V^*}}. \end{aligned}$$

As a consequence, we obtain the required upper bound for $I_2$. Now, if $|a| \le 2$ we obtain

$$\begin{aligned} \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x \le Ce^{-\frac{\min [a^2,(a-1)^2]}{2V^*}}\left[ a-\frac{R(s,s)}{R(t,s)}(a-1)\right] . \end{aligned}$$

To conclude we study the term $A_2$. If we have

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

then by applying the Tonelli theorem we obtain

$$\begin{aligned} A_2&= \int _{\frac{R(s,s)}{R(t,s)}(a-1)}^{a} \int _{A(x)}^{\frac{1}{\sigma }} \frac{1}{\sqrt{2\pi }}e^{-\frac{y^2}{2}}\,{\mathrm {d}}y\frac{1}{\sqrt{2\pi }\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}}\,{\mathrm {d}}x\\&= \int _{(\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\int _{(a-\sigma y)\frac{R(s,s)}{R(t,s)}}^a \ldots {\mathrm {d}}x{\mathrm {d}}y\\&= \int _{(\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\int _{(a-\sigma y)\frac{R(s,s)}{R(t,s)}}^{\frac{R(s,s)}{R(t,s)}a}\ldots {\mathrm {d}}x {\mathrm {d}}y + \int _{(\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\int _{\frac{R(s,s)}{R(t,s)}a}^a\ldots {\mathrm {d}}x{\mathrm {d}}y\\&= I_3 + I_4. \end{aligned}$$

Moreover, if

$$\begin{aligned} \frac{R(s,s)}{R(t,s)}a \ge a, \end{aligned}$$

then

$$\begin{aligned} A_2 \le I_3. \end{aligned}$$

Now, for $I_3$ we have

$$\begin{aligned}&\int _{(\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\int _{(a-\sigma y)\frac{R(s,s)}{R(t,s)}}^{\frac{R(s,s)}{R(t,s)}a}e^{-\frac{y^2}{2}}\frac{1}{\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}} \,{\mathrm {d}}x {\mathrm {d}}y \\&\quad \quad \le C\frac{\sigma }{\sqrt{V(s)}}e^{-\frac{\min [a^2,(a-1)^2]}{2V^*}} \frac{R(s,s)}{R(t,s)}\int _{-\infty }^{\infty }|y|e^{-\frac{y^2}{2}}\,{\mathrm {d}}y. \end{aligned}$$

Hence, we have the required upper bound for $I_3$. To conclude, note that for $I_4$ we have

$$\begin{aligned} I_4&= \int _{(\left[ 1-\frac{R(t,s)}{R(s,s)}\right] \frac{a}{\sigma }}^{\frac{1}{\sigma }}\int _{\frac{R(s,s)}{R(t,s)}a}^ae^{-\frac{y^2}{2}}\frac{1}{\sqrt{V(s)}}e^{-\frac{x^2}{2V(s)}} \,{\mathrm {d}}x {\mathrm {d}}y\\&\le C\frac{1}{\sqrt{V(s)}}e^{-\frac{a^2}{2V^*}}|a|\left| 1-\frac{R(s,s)}{R(t,s)}\right| . \end{aligned}$$

This finishes the proof of Lemma 4.1. $\square$