Stochastic complex integrals in a two-dimensional flow

1. Introduction

We consider two-dimensional flow of a liquid in the $xy$ plane. Suppose every fluid particle moves with the constant speed $U$ parallel to the $x$-axis. Place a flat plate on the $xy$ plane with its leading edge coinciding with $x=0$, as shown in Fig. 1. For example, see the example on p. 80 in Chorin and Marsden’s book¹ or Sec. 23.20 in Milne-Thomson’s book.²

The fluid velocity on the plate surface is the same as that of the plate, that is, it is equal to zero. On the other hand, the fluid velocity in the regions far from the plate is equal to $U$. The fluid velocity drastically changes in the neighborhood of the plate surface. This region is called a boundary layer. In Fig. 1, $d$ denotes the thickness of the boundary layer. When the flow rate is small, the flow is laminar. Then we denote by $u(x,y)=(u(x,y),v(x,y))$ the velocity of the fluid particle that is moving through $(x,y)$. Here, the flow is steady. When the flow rate increases, we suppose that a random force acts on the flow in the $y$-axis direction. Under this assumption, we investigate the circulation in the boundary layer. We adopt the following stochastic process $\{Z_y\}$ as a random force :

Here, $\{X_y:y\ge 0\}$ and $\{{\tilde{X}}_y:y\ge 0\}$ are independent Lévy processes on $ℝ$ such that where

$$\mathrm{\Psi }(\xi )=-\frac{a_0}{2}{\xi }^2+i\gamma \xi +{\int }_{ℝ}(e^{i\xi s}-1-i\xi s{1}_{\{|s|\le 1\}}(s))\mathrm{\Pi }(ds),$$

$a_0\ge 0$, $\gamma \in ℝ$, ${\int }_{ℝ}min\{1,|s{|}^2\}\mathrm{\Pi }(ds)<\infty$, and $\mathrm{\Pi }(\{0\})=0$. Then we have the decomposition where each $\{Z_y^k:y\in ℝ\}$ is a second-order stochastic process with mean zero and Here, $\{X_y^k:y\ge 0\}$ and $\{{\tilde{X}}_y^k:y\ge 0\}$ are independent Lévy processes on $ℝ$ such that where and, for $k\ge 1$, and In addition, we have ${\gamma }_0=\gamma$, ${\gamma }_1=0$ and, for $k\ge 2$ Here, we note that $\{Z_y^k\}$, $k=0,1,2,\dots$, are independent. Let $0<A<\infty$ and let $C$ be a closed curve in the boundary layer such that The circulation $\mathrm{\Gamma }$ around $C$ is defined to be the line integral where $dt=(dx,d{Z}_y)$. Similarly, we define the flux $\mathrm{\Theta }$ across $C$ by the line integral where $dn=(d{Z}_y,-dx)$. Then the circulation is represented as the real part of the stochastic complex integral where $dZ=dx+id{Z}_y$ and $w=u-iv$. In particular, $w$ is called a complex velocity. Let $\beta >0$. In addition, we consider a stochastic process $\{W_y^k:y\in ℝ\}$ such that

$$W^k(y)-W^k(y_0)=-\beta {\int }_{y_0}^y{W}^k(s)ds+Z^k(y)-Z^k(y_0),$$(1)

for $-\infty <y_0<y<\infty$. This process is called an Ornstein–Uhlenbeck type process (OU type process). The solution of (1) is represented as Each $\{W_y^k:y\in ℝ\}$ is stationary in the sense that, for every $s\in ℝ$ See Refs. 3 and 4 for more details. We define random variables for $n\in \mathbb{Z}$ and $k\in \mathbb{N}\cup \{0\}$, where the symbol ${\varphi }_n(y)$ is found in Proposition 3.3. By virtue of Theorem 6.3 in Sec. 6, we can obtain the circulation in the boundary layer.

Theorem 1.1. Let the length of  $C$  be finite. If  $u$  is of class  $C^1$, the circulation  $\mathrm{\Gamma }$  is given by

Here, l.i.m. stands for limit in quadratic mean.

Remark 1.1. Let $\nu >0$. Suppose the pressure gradient $dp∕dx$ is equal to zero, and consider the boundary layer equation

and the continuity equation where $u(x,0)=v(x,0)=0$ for $x\ge 0$ and $u=U$ as $y\to \infty$. We see from (4) that there is a stream function $\psi$ such that With the change of variables the stream function can be expressed as Hence, (3) implies the Blasius equation The boundary conditions are that $f=f^{\prime }=0$ for $\eta =0$ and $f^{\prime }=1$ as $\eta \to \infty$. Then it follows that

$$\begin{gathered} u(x,y)=U{f}^{\prime }(\eta ), \\ v(x,y)=\frac{1}{2}\sqrt{\frac{\nu U}{x}}(\eta f^{\prime }(\eta )-f(\eta )). \end{gathered}$$

See Ref. 5 for the Blasius problem.

Stochastic integrals of nonrandom functions with respect to additive processes on $ℝ$ were studied in Refs. 6,7, and 8. See also Refs. 9–11. Sato developed this theory in Refs. 12 and 13. These days, a lot of papers related to this theory exist, for example, Refs. 14–18. Stochastic complex integrals are constructed on basis of this theory in Ref. 19. Our study extends to the application to Blasius’ formula of fluid mechanics in Ref. 20. The paper is written to be as self-contained as possible. The terminologies follow Ref. 21.

2. Covariance Functions

We find the covariance functions, before we define the stochastic integrals with respect to $\{W_y^k\}$.

Lemma 2.1. The law of  $W_y^k$  is represented as 

Lemma 2.1 is derived from Proposition 2.6 of Ref. 7. Lemma 2.1 tells us that the mean of $W_y^k$ is equal to 0, that is

Define the covariance function of $W_y^k$ by

$$R^k(s,y)=E[(W_s^k-E[W_s^k])(W_y^k-E[W_y^k])]=E[W_s^k{W}_y^k].$$

Lemma 2.2. The second moment of  $W_y^k$  is represented as 

Proof. From Lemma 2.1, we obtain that

Substituting 0 for $\xi$, we get the lemma. □

Lemma 2.3. For  $h>0$, it holds that 

  and 

$$E\left[{\left({\int }_0^h{e}^{-\beta (h-s)}d{Z}_s^k\right)}^2\right]=-\frac{1-e^{-2\beta h}}{2\beta }{\mathrm{\Psi }}_k^{\prime \prime }(0).$$

Proof. Since we have

the lemma is obtained. □

Lemma 2.4. For  $h>0$, it holds that 

Proof. We have

From Lemmas 2.2 and 2.3, we obtain that

$$\begin{array}{l}E[{(W_0^k+W_h^k)}^2]\\ ={(1+e^{-\beta h})}^{2}E\left[{\left({\int }_{-\infty }^0{e}^{\beta s}d{Z}_s^k\right)}^2\right]\\ +2(1+e^{-\beta h})E\left[{\int }_{-\infty }^0{e}^{\beta s}d{Z}_s^k\right]\\ \times E\left[{\int }_0^h{e}^{-\beta (h-s)}d{Z}_s^k\right]\\ +E\left[{\left({\int }_0^h{e}^{-\beta (h-s)}d{Z}_s^k\right)}^2\right]\\ =(1+e^{-\beta h}{)}^2\left(-\frac{1}{2\beta }{\mathrm{\Psi }}_k^{\prime \prime }(0)\right)-\frac{1-e^{-2\beta h}}{2\beta }{\mathrm{\Psi }}_k^{\prime \prime }(0)\\ =-\frac{1+e^{-\beta h}}{\beta }{\mathrm{\Psi }}_k^{\prime \prime }(0).\end{array}$$

The lemma has been proved. □

We give the covariance function of $W_t^k$.

Theorem 2.1. The covariance function of  $W_y^k$  is represented as 

  where 

Proof. Let $h=|s-y|$. From Lemmas 2.2 and 2.4, we obtain that

Hence, we have The theorem has been proved. □

3. Karhunen–Loève Expansions

Let $-\infty <-A\le a<b\le A<\infty$. We first discuss the existence of the integral

Subdivide $[a,b]$ as and denote this partition by where we take ${\mathrm{\Delta }}_n$ such that We define the size of ${\mathrm{\Delta }}_n$ by Then for the partition ${\mathrm{\Delta }}_n$, we define

$$Q(g,{\mathrm{\Delta }}_n)=\sum _{i=1}^{k_n}g({ỹ}_i^n)W^k({ỹ}_i^n)(y_i^n-y_{i-1}^n),$$

where ${ỹ}_i^n\in [y_{i-1}^n,y_i^n]$ for each $i$.

Definition 3.1. If the quadratic mean limit of $Q(g,{\mathrm{\Delta }}_n)$ exists as $|{\mathrm{\Delta }}_n|\to 0$ and is independent of the choice of $\{{\mathrm{\Delta }}_n\}$ and, for each ${\mathrm{\Delta }}_n$, is independent of the choice of $\{{ỹ}_i^n\}$, then the limit is denoted by

Theorem 3.1. If  $g(y)$  is continuous on  $[a,b]$, then the integral  ${\int }_a^{b}g(y)W^k(y)dy$  exists as a quadratic mean limit.

Proof. We have

Since $R^k(s,y)$ is continuous, this limit exists as $n,m\to \infty$ and is independent of the choice of $\{{\mathrm{\Delta }}_n\}$. Hence, we obtain that

$$E[{(Q(g,{\mathrm{\Delta }}_n)-Q(g,{\mathrm{\Delta }}_m))}^2]=E[Q{(g,{\mathrm{\Delta }}_n)}^2]+E[Q{(g,{\mathrm{\Delta }}_m)}^2]-2E[Q(g,{\mathrm{\Delta }}_n)Q(g,{\mathrm{\Delta }}_m)]\to 0,$$

as $n,m\to \infty$. The theorem has been proved. □

Assume that ${\mathrm{{\rm Y}}}_k\ne 0$. We next consider the integral equation

where $\eta (y)$ satisfies Then $\lambda$ and $\eta (y)$ are called an eigenvalue and the corresponding eigenfunction. We see from (5) that $\eta (y)$ is continuous. Hence, we have

$$\lambda {\int }_{-A}^A\eta {(y)}^{2}dy={\int }_{-A}^A{\int }_{-A}^A{R}^k(s,y)\eta (s)\eta (y)dsdy=E{\left({\int }_{-A}^A{W}^k(y)\eta (y)dy\right)}^2\ge 0.$$

This implies $\lambda \ge 0$. Theorem 2.1 tells us that This integral equation is equal to where $T=\beta A$, $u=\beta s$, $v=\beta y$, $\mu ={\mathrm{{\rm Y}}}_k^{-1}\beta \lambda$, and $f(u)=\eta (u{\beta }^{-1})$. Equation (6) can be written Differentiate both sides with respect to $v$ twice

$$\begin{gathered} \mu f^{\prime }(v)=-{\int }_{-T}^v{e}^{u-v}f(u)du+{\int }_v^T{e}^{v-u}f(u)du, \\ \mu f^{\prime \prime }(v)={\int }_{-T}^T{e}^{-|u-v|}f(u)du-2f(v)=\mu f(v)-2f(v). \end{gathered}$$

Hence where $\mu \ge 0$. If $\mu =0$ or if $\mu =2$, then $f(v)=0$.

Lemma 3.1. The integral equation (6)  is not satisfied for any  $\mu >2$.

Proof. Equation (7) has the general solution

where $c_1,c_2\in ℝ$ and Substituting this expression in (6), we obtain that

$$\begin{gathered} e^{\alpha v}\left[\frac{c_1}{\alpha +1}+\frac{c_1}{1-\alpha }\right]+e^{-\alpha v}\left[\frac{c_2}{\alpha +1}+\frac{c_2}{1-\alpha }\right]+e^{-v}\left[-\frac{c_1{e}^{-(\alpha +1)T}}{\alpha +1}-\frac{c_2{e}^{(\alpha -1)T}}{1-\alpha }\right]+e^v\left[-\frac{c_2{e}^{-(\alpha +1)T}}{\alpha +1}-\frac{c_1{e}^{(\alpha -1)T}}{1-\alpha }\right] \\ =\mu (c_1{e}^{\alpha v}+c_2{e}^{-\alpha v}). \end{gathered}$$

As the coefficients of $e^{-v}$ and $e^v$ are equal to zero, it follows that Adding these equations gives Hence, $c_1+c_2=0$. It follows from (8) that if $c_1\ne 0$ and $c_2\ne 0$ The left-hand side is smaller than 1, but the right-hand side is bigger than 1. This is a contradict. Hence, in the case $\mu >2$, (6) does not have a nontrivial solution. □

Hence, we see from Lemma 3.1 that if (6) has a nontrivial solution, then $0<\mu <2$. We consider the equation

where

Proposition 3.1. Eigenvalues for (5)  are 

  and the corresponding eigenfunctions are    if  $n\ge 0$, and    if  $n\le -1$. Here, if  $n\ge 0$, then  ${\alpha }_n$  satisfies    and if  $n\le -1$, then  ${\alpha }_n$  satisfies 

Proof. Now we have

Hence, the boundary condition is that

$$f(T)+f^{\prime }(T)=0,$$(10)

$$f(-T)-f^{\prime }(-T)=0.$$(11)

Equation (9) has the general solution where $c_3,c_4\in ℝ$. The boundary condition yields that

$$\left(\begin{array}{c}\hfill f(T)+f^{\prime }(T)\hfill \\ \hfill f(-T)-f^{\prime }(-T)\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill c_3-c_4\alpha \hfill & \hfill c_3\alpha +c_4\hfill \\ \hfill -c_3-c_4\alpha \hfill & \hfill -c_3\alpha +c_4\hfill \end{array}\right)\left(\begin{array}{c}\hfill sin(\alpha T)\hfill \\ \hfill cos(\alpha T)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \end{array}\right).$$

Then we have

$$\left|\begin{array}{cc}\hfill c_3-c_4\alpha \hfill & \hfill c_3\alpha +c_4\hfill \\ \hfill -c_3-c_4\alpha \hfill & \hfill -c_3\alpha +c_4\hfill \end{array}\right|=2c_3{c}_4(1+{\alpha }^2).$$

If $c_3{c}_4\ne 0$, then This is a contradiction. If $c_3=0$ and $c_4\ne 0$, then that is If $c_3\ne 0$ and $c_4=0$, then that is Recall that Hence, in the case where $\alpha tan(\alpha T)=1$, we have In the case where $tan(\alpha T)=-\alpha$, we have The proposition has been proved. □

Here, we show the range of each eigenvalue.

Proposition 3.2. If  $n\ge 0$, then 

  If  $n\le -1$, then 

Proof. If $n\ge 0$, then

This implies If $n\le -1$, then This implies It follows from Proposition 3.1 that the proposition holds. □

Now we find the orthonormal eigenfunctions required for the Karhunen–Loève expansion.

Proposition 3.3. Let 

  where  ${ϵ}_n=1$  if  $n\ge 0$  and  ${ϵ}_n=-1$  if  $n<0$. The eigenfunctions  $\{{\varphi }_n(y):n\in \mathbb{Z}\}$  are orthonormal.

Proof. Let $f_n(u)={\eta }_n({\beta }^{-1}u)$. Let $n\ne m$. We see from (9) that

These equations yield Hence, we obtain that

$$\begin{array}{l}{\int }_{-A}^A{\eta }_n(y){\eta }_m(y)dy\\ ={\beta }^{-1}{\int }_{-T}^T{f}_n(u)f_m(u)du\\ =\frac{1}{({\alpha }_m^2-{\alpha }_n^2)\beta }{\int }_{-T}^T(f_m(u)f_n^{\prime \prime }(u)-f_n(u)f_m^{\prime \prime }(u))du\\ =\frac{1}{({\alpha }_m^2-{\alpha }_n^2)\beta }(f_m(T)f_n^{\prime }(T)-f_n(T)f_m^{\prime }(T)\\ -f_m(-T)f_n^{\prime }(-T)+f_n(-T)f_m^{\prime }(-T)).\end{array}$$

Using (10) and (11), we see that

$$\begin{gathered} (f_m(u)+f_m^{\prime }(u))(f_n(u)-f_n^{\prime }(u))=0, \\ (f_n(u)+f_n^{\prime }(u))(f_m(u)-f_m^{\prime }(u))=0 \end{gathered}$$

for $u=T,-T$. Subtracting both sides, we see that for $u=T,-T$. Hence and thereby $\{{\eta }_n(y):n\in \mathbb{Z}\}$ are orthogonal. Furthermore, we have The proposition has been proved. □

Let ${\delta }_{n,m}=1$ if $n=m$ and ${\delta }_{n,m}=0$ if $n\ne m$.

Proposition 3.4. The random variables  $\{{\mathrm{\Phi }}_n^k\}$  are orthogonal, that is

Proof. We have

If $k\ne l$, then $W^k(y)$ and $W^l(s)$ are independent. Hence This implies $E[{\mathrm{\Phi }}_n^k{\mathrm{\Phi }}_m^l]=0$.

Let $k=l$. We have

The proposition has been proved. □

The Karhunen–Loève expansion is represented as follows: see Refs. 22 and 23 for the Karhunen–Loève expansion.

Theorem 3.2. We have 

  uniformly for  $-A\le y\le A$, and 

Theorem 3.2 tells us a proposition.

Proposition 3.5. For any  $m$, we have 

Proof. As $N\to \infty$, we see that

Theorem 3.2 yields that

$$E[W_y^k{\mathrm{\Phi }}_m^k]=\underset{N\to \infty }{lim}E\left[\sum _{n=-N}^N{\varphi }_n(y){\mathrm{\Phi }}_n^k{\mathrm{\Phi }}_m^k\right]={\varphi }_m(y)E[{\mathrm{\Phi }}_m^k{\mathrm{\Phi }}_m^k]={\varphi }_m(y){\lambda }_m^k.$$

The proposition has been proved. □

4. Stochastic Integrals with Respect to Momentum

Let $-\infty <-A\le a<b\le A<\infty$. Subdivide $[a,b]$ as

and denote this partition by where we take ${\mathrm{\Delta }}_n$ such that For the partition ${\mathrm{\Delta }}_n$, we define where ${ỹ}_i^n\in [y_{i-1}^n,y_i^n]$ for each $i$.

Definition 4.1. If the quadratic mean limit of $S(g,{\mathrm{\Delta }}_n)$ exists as $|{\mathrm{\Delta }}_n|\to 0$ and is independent of the choice of $\{{\mathrm{\Delta }}_n\}$ and, for each ${\mathrm{\Delta }}_n$, is independent of the choice of $\{{ỹ}_i^n\}$, then the limit is denoted by

and $g$ is said to be $\{W_y^k\}$-integrable.

We introduce two half planes

Define

$$D^k(I)=R^k(t,v)-R^k(s,v)-R^k(t,u)+R^k(s,u)={\mathrm{{\rm Y}}}_k(e^{-\beta |t-v|}-e^{-\beta |s-v|}-e^{-\beta |t-u|}+e^{-\beta |s-u|}),$$

for any rectangle $I=[s,t]\times [u,v]$. Let for $1\le i,j\le k_n$. Denote by $|I_{i,j}^n|$ the length of the diagonal of $I_{i,j}^n$.

Lemma 4.1. (i) If  $i\ne j$, then we have 

  Here, $l(I_{i,j}^n)$uniformly converges to 0 as$|I_{i,j}^n|\to 0$.

(ii) If  $i=j$, then we have 

  Here, $L(I_{i,i}^n)$is uniformly bounded as$|I_{i,i}^n|\to 0$.

Proof. We first prove (i). We consider the case $I_{i,j}^n\subset H^{+}$. Then

Let $y_i^n-y_{i-1}^n\le y_j^n-y_{j-1}^n$. There exists such that

$$e^{-\beta (y_i^n-y_j^n)}-e^{-\beta (y_{i-1}^n-y_j^n)}=-\beta (y_i^n-y_{i-1}^n)e^{-\beta (y_{i-1}^n-y_j^n)}+\frac{{\beta }^2{(y_i^n-y_{i-1}^n)}^2}{2}{e}^{-\beta {\xi }_1},$$

and exists such that

$$e^{-\beta (y_i^n-y_{j-1}^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)}=-\beta (y_i^n-y_{i-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}+\frac{{\beta }^2{(y_i^n-y_{i-1}^n)}^2}{2}{e}^{-\beta {\xi }_2}.$$

In addition, we have

$$e^{-\beta (y_{i-1}^n-y_j^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)}=\beta (y_j^n-y_{j-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}+\frac{{\beta }^2{(y_j^n-y_{j-1}^n)}^2}{2}{e}^{-\beta {\xi }_3},$$

for some ${\xi }_3\in (y_{i-1}^n-y_j^n,y_{i-1}^n-y_{j-1}^n)$. Hence, we obtain that

$$\begin{array}{l}\frac{D^k(I_{i,j}^n)}{{\mathrm{{\rm Y}}}_k}=-\beta (y_i^n-y_{i-1}^n)(e^{-\beta (y_{i-1}^n-y_j^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)})\\ +\frac{{\beta }^2(y_i^n-y_{i-1}^n{)}^2}{2}(e^{-\beta {\xi }_1}-e^{-\beta {\xi }_2})\\ =-{\beta }^2(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}\\ +\frac{{\beta }^2(y_i^n-y_{i-1}^n{)}^2}{2}(e^{-\beta {\xi }_1}-e^{-\beta {\xi }_2})\\ -\frac{{\beta }^3}{2}(y_i^n-y_{i-1}^n){(y_j^n-y_{j-1}^n)}^2{e}^{-\beta {\xi }_3}.\end{array}$$

Next, let $y_i^n-y_{i-1}^n>y_j^n-y_{j-1}^n$. There exists such that

$$e^{-\beta (y_i^n-y_j^n)}-e^{-\beta (y_i^n-y_{j-1}^n)}=\beta (y_j^n-y_{j-1}^n)e^{-\beta (y_i^n-y_{j-1}^n)}+\frac{{\beta }^2{(y_j^n-y_{j-1}^n)}^2}{2}{e}^{-\beta {\xi }_4},$$

and exists such that

$$e^{-\beta (y_{i-1}^n-y_j^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)}=\beta (y_j^n-y_{j-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}+\frac{{\beta }^2{(y_j^n-y_{j-1}^n)}^2}{2}{e}^{-\beta {\xi }_5}.$$

In addition, we have

$$e^{-\beta (y_i^n-y_{j-1}^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)}=-\beta (y_i^n-y_{i-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}+\frac{{\beta }^2{(y_i^n-y_{i-1}^n)}^2}{2}{e}^{-\beta {\xi }_6},$$

for some ${\xi }_6\in (y_{i-1}^n-y_{j-1}^n,y_i^n-y_{j-1}^n)$. Hence, we obtain that

$$\begin{array}{l}\frac{D^k(I_{i,j}^n)}{{\mathrm{{\rm Y}}}_k}=\beta (y_j^n-y_{j-1}^n)(e^{-\beta (y_i^n-y_{j-1}^n)}-e^{-\beta (y_{i-1}^n-y_{j-1}^n)})\\ +\frac{{\beta }^2(y_j^n-y_{j-1}^n{)}^2}{2}(e^{-\beta {\xi }_4}-e^{-\beta {\xi }_5})\\ =-{\beta }^2(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)e^{-\beta (y_{i-1}^n-y_{j-1}^n)}\\ +\frac{{\beta }^2(y_j^n-y_{j-1}^n{)}^2}{2}(e^{-\beta {\xi }_4}-e^{-\beta {\xi }_5})\\ +\frac{{\beta }^3}{2}(y_j^n-y_{j-1}^n){(y_i^n-y_{i-1}^n)}^2{e}^{-\beta {\xi }_6}.\end{array}$$

This implies that (i) holds.

We next consider the case $I_{i,j}^n\subset H^{-}$. Then we have

Hence, exchanging $i$ for $j$ in the case $I_{i,j}^n\subset H^{+}$, we have

$$\frac{D^k(I_{i,j}^n)}{{\mathrm{{\rm Y}}}_k}=-{\beta }^2(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)e^{-\beta (y_{j-1}^n-y_{i-1}^n)}+(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)l(I_{i,j}^n).$$

Assertion (i) remains true.

Finally, we prove (ii). We consider the case $i=j$. Then we have

for some ${\xi }_7\in (0,y_i^n-y_{i-1}^n)$. The lemma has been proved. □

Lemma 4.2. Then  ${\sum }_{1\le i,j\le k_n}|D^k(I_{i,j}^n)|$  is uniformly bounded regardless of how  $[a,b]\times [a,b]$  is divided.

Proof. Let $i\ne j$. From Lemma 4.1(i), we see that there is $M_1>0$ such that

Let $i=j$. From Lemma 4.1(ii), we see that there is $M_2>0$ such that

$$|D^k(I_{i,i}^n)|\le |{\mathrm{\Psi }}_k^{\prime \prime }(0)|(y_i^n-y_{i-1}^n)+M_2{(y_i^n-y_{i-1}^n)}^2\le (|{\mathrm{\Psi }}_k^{\prime \prime }(0)|+M_2(b-a))(y_i^n-y_{i-1}^n).$$

Consequently, we have

$$\begin{array}{l}\sum _{1\le i,j\le k_n}|D^k(I_{i,j}^n)|\\ \le \left(\frac{|{\mathrm{\Psi }}_k^{\prime \prime }(0)|\beta }{2}+M_1\right)\sum _{i\ne j}(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)\\ +(|{\mathrm{\Psi }}_k^{\prime \prime }(0)|+M_2(b-a))\sum _{i=1}^{k_n}(y_i^n-y_{i-1}^n)\\ \le \left(\frac{|{\mathrm{\Psi }}_k^{\prime \prime }(0)|\beta }{2}+M_1\right){(b-a)}^2\\ +(|{\mathrm{\Psi }}_k^{\prime \prime }(0)|+M_2(b-a))(b-a).\end{array}$$

Hence, the lemma is true. □

Through this paper, we use “piecewise continuous” as the following meaning:

Definition 4.2. Let $a_1,\dots ,a_n$ be discontinuous points of $g(y)$ on $(a,b)$. For each interval $[a_{i-1},a_i]$, we define a function ${ğ}_i(y)$ by

Here, $a_0=a$ and $a_{n+1}=b$. If ${ğ}_i(y)$ is continuous on $[a_{i-1},a_i]$ for $i=1,2,\dots ,n+1$, $g(y)$ is said to be piecewise continuous.

Theorem 4.1. If  $g(y)$  is piecewise continuous on  $[a,b]$, then the integral  ${\int }_a^{b}g(y)d{W}_y^k$  exists as a quadratic mean limit.

Proof. Let $n>m$. Since ${\mathrm{\Delta }}_n\supset {\mathrm{\Delta }}_m$, we have

where $g_{n,m}(i)=g({ỹ}_i^n)-g({ỹ}_j^m)$ if $[y_{i-1}^n,y_i^n]\subset [y_{j-1}^m,y_j^m]$. Let $a_1,\dots ,a_n$ be discontinuous points of $g(y)$ on $(a,b)$ and in addition, $a_0=a$ and $a_{n+1}=b$. Since ${ğ}_i(y)$ is uniformly continuous on $[a_{i-1},a_i]$, it follows from Lemma 4.2 that

$$\begin{array}{l}\underset{n,m\to \infty }{lim}E[{(S(g,{\mathrm{\Delta }}_n)-S(g,{\mathrm{\Delta }}_m))}^2]\\ =\underset{n,m\to \infty }{lim}\sum _{i=1}^{k_n}\sum _{j=1}^{k_n}{g}_{n,m}(i)g_{n,m}(j)(R^k(y_i^n,y_j^n)\\ -R^k(y_i^n,y_{j-1}^n)-R^k(y_{i-1}^n,y_j^n)+R^k(y_{i-1}^n,y_{j-1}^n))\\ =\underset{n,m\to \infty }{lim}\sum _{i=1}^{k_n}\sum _{j=1}^{k_n}{g}_{n,m}(i)g_{n,m}(j)D^k(I_{i,j}^n)=0.\end{array}$$

Here, we used the fact that the number of discontinuous points of $g$ is finite. The theorem has been proved. □

The following theorem is obvious. The proof is omitted.

Theorem 4.2. If  $f$  and  $g$  are  $\{W_y^k\}$-integrable, then  $c_{1}f+c_{2}g$  is  $\{W_y^k\}$-integrable and 

  for any  $c_1,c_2\in ℝ$.

We show one lemma to prove Proposition 4.1. The proof is omitted.

Lemma 4.3. Let  $X_n$  and  $X$  be random variables and suppose  ${sup}_{n}E{X}_n^2<\infty$. If  $X_n$  converges to  $X$  in quadratic mean, then 

Proposition 4.1. Let  $[a,b]\supset [a_j,b_j]$  for  $j=1,2$. If  $f(y)$  and  $g(y)$  are piecewise continuous on  $[a_1,b_1]$  and on  $[a_2,b_2]$, respectively, then we have 

  and 

$$E\left[{\int }_{a_1}^{b_1}f(y)d{W}_y^k{\int }_{a_2}^{b_2}g(y)d{W}_y^k\right]=\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}{\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}f(s)g(y)e^{-\beta |s-y|}dyds-{\mathrm{\Psi }}_k^{\prime \prime }(0){\int }_{[a_1,b_1]\cap [a_2,b_2]}f(y)g(y)dy.$$

Proof. We have

Hence, it follows that Let $h(y)$ be piecewise continuous on $[a,b]$. From Lemma 4.1, we obtain that

$$\begin{array}{l}E[S{(h,{\mathrm{\Delta }}_n)}^2]\\ =\sum _{i=1}^{k_n}\sum _{j=1}^{k_n}h({ỹ}_i^n)h({ỹ}_j^n)D^k(I_{i,j}^n)\\ =\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}\sum _{i=1}^{k_n}\sum _{j=1}^{k_n}h({ỹ}_i^n)h({ỹ}_j^n)e^{-\beta |y_{i-1}^n-y_{j-1}^n|}\\ \times (y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)\\ -\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}\sum _{i=1}^{k_n}h{({ỹ}_i^n)}^2{(y_i^n-y_{i-1}^n)}^2\\ +\sum _{i\ne j}h({ỹ}_i^n)h({ỹ}_j^n)l(I_{i,j}^n)(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)\\ -{\mathrm{\Psi }}_k^{\prime \prime }(0)\sum _{i=1}^{k_n}h{({ỹ}_i^n)}^2(y_i^n-y_{i-1}^n)e^{-\beta (y_i^n-y_{i-1}^n)}\\ +\sum _{i=1}^{k_n}h{({ỹ}_i^n)}^{2}L(I_{i,i}^n){(y_i^n-y_{i-1}^n)}^2.\end{array}$$

Hence, it follows from Lemmas 4.2 and 4.3 that

$$E\left[{\left({\int }_a^{b}h(y)d{W}_y^k\right)}^2\right]=\underset{|{\mathrm{\Delta }}_n|\to 0}{lim}E[S{(h,{\mathrm{\Delta }}_n)}^2]=\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}{\int }_a^b{\int }_a^{b}h(s)h(y)e^{-\beta |s-y|}dyds-{\mathrm{\Psi }}_k^{\prime \prime }(0){\int }_a^{b}h{(y)}^{2}dy.$$

Here, we define $f_0$ and $g_0$ by and respectively. Theorem 4.2 tells us that

$$\begin{array}{l}E\left[{\left({\int }_a^b{f}_0(y)d{W}_y^k+{\int }_a^b{g}_0(y)d{W}_y^k\right)}^2\right]\\ =E\left[{\left({\int }_a^b(f_0(y)+g_0(y))d{W}_y^k\right)}^2\right]\\ =\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}{\int }_a^b{\int }_a^b(f_0(s)+g_0(s))(f_0(y)+g_0(y))\\ \times e^{-\beta |s-y|}dyds-{\mathrm{\Psi }}_k^{\prime \prime }(0){\int }_a^b{(f_0(y)+g_0(y))}^{2}dy.\end{array}$$

Hence, we obtain that

$$\begin{array}{l}E\left[{\int }_a^b{f}_0(y)d{W}_y^k{\int }_a^b{g}_0(y)d{W}_y^k\right]\\ =\frac{1}{2}\left(E\left[{\left({\int }_a^b{f}_0(y)d{W}_y^k+{\int }_a^b{g}_0(y)d{W}_y^k\right)}^2\right]\right.\\ -E\left[{\left({\int }_a^b{f}_0(y)d{W}_y^k\right)}^2\right]\\ \left.-E\left[{\left({\int }_a^b{g}_0(y)d{W}_y^k\right)}^2\right]\right)\\ =\frac{{\mathrm{\Psi }}_k^{\prime \prime }(0)\beta }{2}{\int }_a^b{\int }_a^b{f}_0(s)g_0(y)e^{-\beta |s-y|}dyds\\ -{\mathrm{\Psi }}_k^{\prime \prime }(0){\int }_a^b{f}_0(y)g_0(y)dy.\end{array}$$

The proposition has been proved. □

Theorem 4.3. If  $g(y)$  is of class  $C^1$, then 

  almost surely.

Proof. From Theorem 4.1, we see that $g$ is $\{W_y^k\}$-integrable. For the partition ${\mathrm{\Delta }}_n$, we take

Then we see that

$$\begin{array}{l}S(g,{\mathrm{\Delta }}_n)=g(b)W^k(b)-g(y_1^n)W^k(a)\\ -\sum _{i=2}^{k_n}(g(y_i^n)-g(y_{i-1}^n))W^k(y_{i-1}^n)\\ =g(b)W^k(b)-g(y_1^n)W^k(a)\\ -\sum _{i=2}^{k_n}(g^{\prime }(y_{i-1}^n)+{ϵ}_i)(y_i^n-y_{i-1}^n)W^k(y_{i-1}^n).\end{array}$$

Here, ${ϵ}_i\to 0$ uniformly in $i$ as $|{\mathrm{\Delta }}_n|\to 0$. Hence, the theorem holds. □

5. Representation Through OU Type Processes

Stochastic integrals based on $\{Z_y\}$ are represented in terms of $W_y^k$.

Theorem 5.1. Let  $-\infty <a<b<\infty$. If  $g(y)$  is a function of class  $C^1$, then 

In order to prove Theorem 5.1, we prepare a lemma.

Lemma 5.1. For any function  $g(y)$  of class  $C^1$, we have 

Proof. We have

Let $ϵ>0$. As we take $|I_{i,j}^n|$ small enough, we have in $(y,s)\in I_{i,j}^n=[y_{i-1}^n,y_i^n]\times [y_{j-1}^n,y_j^n]$. Then it follows that

$$K\le ϵ\sum _{i,j}|g(y_{i-1}^n)g(y_{j-1}^n)|(y_i^n-y_{i-1}^n)(y_j^n-y_{j-1}^n)\to ϵ{\left({\int }_a^b|g(y)|dy\right)}^2\mathrm{\text{as}}|{\mathrm{\Delta }}_n|\to 0.$$

Since $ϵ$ can be taken arbitrarily, $K$ goes to zero as $n\to \infty$. □

Now we prove Theorem 5.1.

Proof. Proof of Theorem 5.1

Since $\{Z_y^k\}$, $k=0,1,2,\dots$, are independent, the integrals

are independent. By virtue of Lévy’s theorem, we have almost surely. Now we have Here, p-lim stands for limit in probability. On the other hand, we see from (1) that

$$\begin{array}{l}\sum _{i=1}^{k_n}g(y_{i-1}^n)(Z_{y_i^n}^k-Z_{y_{i-1}^n}^k)\\ =\sum _{i=1}^{k_n}g(y_{i-1}^n)\left(W^k(y_i^n)-W^k(y_{i-1}^n)+\beta {\int }_{y_{i-1}^n}^{y_i^n}{W}^k(y)dy\right)\\ =\sum _{i=1}^{k_n}g(y_{i-1}^n)(W^k(y_i^n)-W^k(y_{i-1}^n))\\ +\beta \sum _{i=1}^{k_n}g(y_{i-1}^n)W^k(y_{i-1}^n)(y_i^n-y_{i-1}^n)\\ +\beta \sum _{i=1}^{k_n}g(y_{i-1}^n){\int }_{y_{i-1}^n}^{y_i^n}(W^k(y)-W^k(y_{i-1}^n))dy.\end{array}$$

Theorems 3.1 and 4.1 and Lemma 5.1 tell us that almost surely The theorem has been proved. □

Theorem 5.1 is also represented as follows:

Theorem 5.2. Let  $-\infty <-A\le a<b\le A<\infty$. If  $g(y)$  is a function of class  $C^1$, then 

  almost surely.

Proof. From Theorems 3.2 and 4.3, we obtain that

almost surely. The theorem has been proved. □

6. Stochastic Complex Integrals

In this section, $z$ is any complex number and represented as $z=x+iy$, where $x$ and $y$ are real numbers. Furthermore, $z=x+iy\in ℂ$ is identified with $(x,y)\in {ℝ}^2$. Hence, we often use the representation $g(x,y)$ instead of any complex function $g(z)$. Only in this section, $g$ is a complex-valued function. A curve $\mathrm{\Lambda }$ is represented as a function

which is of class $C^1$. Let $\mathrm{\Lambda }$ be a curve with an initial point $z_0=(a,c)$ and a terminal point $z_1=(b,d)$ such that where $\tilde{D}$ is some open set. From now on, we write so we have $z_0=z(0)=(a,c)$ and $z_1=z(1)=(b,d)$. We subdivide $[0,1]$ as and denote this partition by where we suppose ${\mathrm{\Delta }}_{m+1}\supset {\mathrm{\Delta }}_m$. We define and

Definition 6.1. If $Q(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })$ converges in quadratic mean as $|{\mathrm{\Delta }}_m|\to 0$, and if the limit does not depend on the choice of the sequence $\{{\mathrm{\Delta }}_m\}$, then the limit is denoted by

To show quadratic mean convergence of stochastic integrals, the following criterion is useful. See Ref. 22.

Lemma 6.1. The Loève criterion

The random variable sequence  $X_n$  converges in quadratic mean if and only if  $E[X_m{\bar{X}}_n]$  has a finite limit, when  $m$and  $n$  tend to infinity independently of each other.

Remark 6.1. For any $z\in ℂ$, we denote by $\bar{z}$ the complex-conjugate of $z$. Hence, ${\bar{X}}_n$ means the complex-conjugate of $X_n$.

The following lemma is known as the Mercer’s theorem. See Refs. 24 and 25.

Lemma 6.2. In our setting, we have 

For any random variable $X$, the norm of $X$ is defined by

Lemma 6.3. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g$  is continuous on  $\tilde{D}$, then 

  where  $ϵ(N)$  converges to 0 as  $N\to \infty$.

Proof. We obtain from Proposition 3.5 that

Here

$$ϵ({\mathrm{\Delta }}_m,N)=\sum _{i=1}^{k_m}\sum _{j=1}^{k_m}g(z(t_{i-1}^m))\overline{g(z(t_{j-1}^m))}\times \left(R^k(y(t_{i-1}^m),y(t_{j-1}^m))-\sum _{n=-N}^N{\lambda }_n^k{\varphi }_n(y(t_{i-1}^m)){\varphi }_n(y(t_{j-1}^m))\right)\times \mathrm{\Delta }y(t_i^m)\mathrm{\Delta }y(t_j^m).$$

Since the length of $\mathrm{\Lambda }$ is finite, Lemma 6.2 tells us that $ϵ({\mathrm{\Delta }}_m,N)$ goes to 0 uniformly in $m$. The lemma has been proved. □

Lemma 6.4. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g$  is continuous on  $\tilde{D}$, then 

Proof. We have

Hence, we obtain that

$$\sum _{n=-\infty }^{\infty }{\lambda }_n^k{\left|{\int }_{\mathrm{\Lambda }}g(z){\varphi }_n(y)dy\right|}^2\le {\int }_0^1|g(z(t)){|}^2\left|\frac{dy}{dt}\right|dt{\int }_0^1\sum _{n=-\infty }^{\infty }{\lambda }_n^k{\varphi }_n{(y)}^2\left|\frac{dy}{dt}\right|dt\le {\int }_0^1|g(z(t)){|}^2\left|\frac{dy}{dt}\right|dt{\int }_0^1{R}^k(y,y)\left|\frac{dy}{dt}\right|dt.$$

The lemma is true. □

Theorem 6.1. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g$  is continuous on  $\tilde{D}$, then the integral  ${\int }_{\mathrm{\Lambda }}g(z)W^k(y)dy$  exists as a quadratic mean limit, which is represented as 

Proof. From Lemma 6.2, we see that

By using uniform convergence in $m$ and $\tilde{m}$, it follows from Lemma 6.4 that the limit is equal to as $|{\mathrm{\Delta }}_m|\to 0$ and $|{\mathrm{\Delta }}_{\tilde{m}}|\to 0$. We obtain from Lemma 6.1 that Now we have

$$∥{\int }_{\mathrm{\Lambda }}g(z)W^k(y)dy-\sum _{n=-N}^N{\mathrm{\Phi }}_n^k{\int }_{\mathrm{\Lambda }}g(z){\varphi }_n(y)dy∥\le ∥{\int }_{\mathrm{\Lambda }}g(z)W^k(y)dy-Q(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })∥+∥Q(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })-\sum _{n=-N}^N{\mathrm{\Phi }}_n^k{\int }_{\mathrm{\Lambda }}g(z){\varphi }_n(y)dy∥.$$

Hence, Lemma 6.3 tells us that the theorem holds. □

Definition 6.2. We define

If $S(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })$ converges in quadratic mean as $|{\mathrm{\Delta }}_m|\to 0$, and if the limit does not depend on the choice of the sequence $\{{\mathrm{\Delta }}_m\}$, then the limit is denoted by

We prepare two lemmas.

Lemma 6.5. Suppose the length of  $\mathrm{\Lambda }$  is finite. Let  $H_m(t)$  be a complex-valued random variable sequence. If 

  then 

$$\begin{gathered} \underset{m\to \infty }{lim}∥\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}{H}_m(t)\frac{dx}{dt}dt∥=0, \\ \underset{m\to \infty }{lim}∥\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}{H}_m(t)\frac{dy}{dt}dt∥=0. \end{gathered}$$

Proof. There is $\delta >0$ such that

for sufficiently large $m$. If $m$ is sufficiently large, then we have

$${∥\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}{H}_m(t)\frac{dx}{dt}dt∥}^2\le \sum _{i=1}^{k_m}\sum _{j=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}{\int }_{t_{j-1}^m}^{t_j^m}\sqrt{E|H_m(t){|}^{2}E|H_m(s){|}^2}\times \left|\frac{dx(s)}{ds}\right|\cdot \left|\frac{dx(t)}{dt}\right|dsdt\le \delta {\left(\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}\left|\frac{dx(t)}{dt}\right|dt\right)}^2.$$

Since the length of $\mathrm{\Lambda }$ is finite, the last expression is bounded. As $\delta \to 0$, we get the first limit of the lemma. The second limit can be obtained as well. □

Lemma 6.6. Suppose that a random function  $H_m(t)$  on  $[0,1]$  is defined as follows:

if$t\in (t_{i-1}^m,t_i^m]$. If$g$is of class$C^1$on$\tilde{D}$, then

Remark 6.2. In the case where

the assertion of Lemma 6.6 remains true.

Proof. Let $t\in (t_{i-1}^m,t_i^m]$. We have

Hence, the lemma is true. □

Theorem 6.2. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g$  is of class  $C^1$  on  $\tilde{D}$, then the integral  ${\int }_{\mathrm{\Lambda }}g(z)d{W}_y^k$  exists as a quadratic mean limit, which is represented as 

Remark 6.3. If $g$ is continuous on $\tilde{D}$, the integral ${\int }_{\mathrm{\Lambda }}g(z)W^k(y)dx$ exists as a quadratic mean limit, which is represented as

Proof. Let $\mathrm{\Delta }g(z(t_i^m))=g(z(t_i^m))-g(z(t_{i-1}^m))$. We have

Here, ${ϵ}_i^j\to 0$ uniformly in $i$ as $|{\mathrm{\Delta }}_m|\to 0$ for $j=1,2$. This implies that

$$\begin{array}{l}∥S(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })-g(z_1)W^k(d)+g(z_0)W^k(c)\\ +{\int }_{\mathrm{\Lambda }}{W}^k(y)\left(\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy\right)∥\\ \le |g(z_0)-g(z(t_1^m))|\parallel W^k(c)\parallel \\ +\left|\frac{\partial g}{\partial x}(z_0)\mathrm{\Delta }x(t_1^m)\right|\parallel W^k(c)\parallel \\ +∥\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}\left(\frac{\partial g}{\partial x}(z(t))W^k(y(t))\right.\\ \left.-\frac{\partial g}{\partial x}(z(t_{i-1}^m))W^k(y(t_{i-1}^m))\right)\frac{dx}{dt}dt∥\\ +\left|\frac{\partial g}{\partial y}(z_0)\mathrm{\Delta }y(t_1^m)\right|\parallel W^k(c)\parallel \\ +∥\sum _{i=1}^{k_m}{\int }_{t_{i-1}^m}^{t_i^m}\left(\frac{\partial g}{\partial y}(z(t))W^k(y(t))\right.\\ \left.-\frac{\partial g}{\partial y}(z(t_{i-1}^m))W^k(y(t_{i-1}^m))\right)\frac{dy}{dt}dt∥\\ +∥\sum _{i=2}^{k_m}({ϵ}_i^1\mathrm{\Delta }x(t_i^m)+{ϵ}_i^2\mathrm{\Delta }y(t_i^m))W^k(y(t_{i-1}^m))∥.\end{array}$$

We obtain from Lemmas 6.5 and 6.6 that the above expression goes to 0 as $m\to \infty$. The theorem has been proved. □

Finally, we introduce line integrals with respect to $Z_y^k$, which is not assumed that $\mathrm{\Lambda }$ is closed.

Definition 6.3. We define

If $S_0(g,{\mathrm{\Delta }}_m,\mathrm{\Lambda })$ converges in quadratic mean as $|{\mathrm{\Delta }}_m|\to 0$, and if the limit does not depend on the choice of the sequence $\{{\mathrm{\Delta }}_m\}$, then the limit is denoted by

Proposition 6.1. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g(z)$  is of class  $C^1$  on  $\tilde{D}$, then the integral  ${\int }_{\mathrm{\Lambda }}g(z)d{Z}_y^k$  exists as a quadratic mean limit and 

  almost surely.

Proof. We see from (1) that

Since a lemma similar to Lemma 5.1 holds, we can get the proposition. □

We do not assume that $\mathrm{\Lambda }$ is closed. Let $g$ be a function of class $C^1$ on $\tilde{D}$ and let $g_1$ and $g_2$ be the real part and the imaginary part of $g$, respectively. By virtue of Proposition 6.1, we have

for $j=1,2$. This implies that As a result, we obtain that

$$\underset{n\to \infty }{lim}E\left[e^{i\xi \sum _{k=0}^n({\gamma }_k{\int }_{\mathrm{\Lambda }}{g}_j(z)dy+{\int }_{\mathrm{\Lambda }}{g}_j(z)d{Z}_y^k)}\right]=exp\left[{\int }_{\mathrm{\Lambda }}\mathrm{\Psi }(g_j(z)\xi )dy\right].$$

We here note that ${\gamma }_k{\int }_{\mathrm{\Lambda }}{g}_j(z)dy+{\int }_{\mathrm{\Lambda }}{g}_j(z)d{Z}_y^k,k=0,1,2,\dots ,$ are independent. Since ${\sum }_{k=0}^n({\gamma }_k{\int }_{\mathrm{\Lambda }}{g}_j(z)dy+{\int }_{\mathrm{\Lambda }}{g}_j(z)d{Z}_y^k)$ converges in probability, we can define line integrals along $\mathrm{\Lambda }$ with respect to $Z$ as follows:

$${\int }_{\mathrm{\Lambda }}g(z)dZ={\int }_{\mathrm{\Lambda }}g(z)dx+i\sum _{k=0}^{\infty }\left({\gamma }_k{\int }_{\mathrm{\Lambda }}g(z)dy+{\int }_{\mathrm{\Lambda }}g(z)d{Z}_y^k\right).$$

Theorem 6.3. Suppose the length of  $\mathrm{\Lambda }$  is finite. If  $g$  is of class  $C^1$  on  $\tilde{D}$, then 

Proof. Proposition 6.1 tells us that almost surely

Hence, it follows from Theorems 6.1 and 6.2 that the theorem holds. □

Acknowledgments

The author appreciates the referee’s comments for improving the paper.

Competing Interests

The author declares that he has no competing interests.

Contribution

This work has been done alone. The author read and approved the final paper.