2 Model set-up

Consider two security price processes $(P_t)_{t\ge 0}$ and $(Q_t)_{t\ge 0}$ in a continuous-time economy. We may regard P and Q as discounted price processes for simplicity. Suppose that fir some constants $\beta ,\lambda \in {\mathbb R}$ ,

(2.1) $$ \begin{align} X_t = \log P_t - \beta \log Q_t-\lambda, \quad t\ge 0, \end{align} $$

follows a one-dimensional diffusion process. Concretely, $X=(X_t)_{t\ge 0}$ satisfies the stochastic differential equation (SDE)

(2.2) $$ \begin{align} dX_t =\mu(X_t)\,dt + \sigma(X_t)\,dW_t, \quad t\ge 0, \quad X_0\in {E}, \end{align} $$

on a filtered probability space $(\Omega ,{\mathcal F},{\mathbb P}, ({\mathcal F}_t)_{t\ge 0})$ endowed with the ${\mathcal F}_t$ -Brownian motion $(W_t)_{t\ge 0}$ , where $E=(l,r)$ ( $-\infty \le l<0<r\le \infty $ ) is the state space. Here, $\mu ,\sigma :E\to {\mathbb R}$ are continuous functions and $\sigma (x)^2>0$ for all $x\in E$ . Associated with X, we define the scale function

$$ \begin{align*} s(x)= \int_0^x \exp\bigg\{ -2\int_0^y \bigg( \frac{\mu}{\sigma^2}\bigg)(z)\,dz\bigg\} \,dy, \quad x\in E, \end{align*} $$

and the speed measure on E by

$$ \begin{align*} m(dx)= \frac{2}{\sigma(x)^2} \exp\bigg\{ 2\int_0^x \bigg( \frac{\mu}{\sigma^2}\bigg)(y)\,dy\bigg\} \,dx. \end{align*} $$

For the basic roles of the scale function and the speed measure in one-dimensional diffusion theory, we can refer to the works of Durret [Reference Durret4, Chapter 6] and Karatzas and Shreve [Reference Karatzas and Shreve9, Section 5.5], for example. We make the following assumptions.

Assumption 2.1 The boundary values of the scale function are

$$ \begin{align*}\lim_{x \downarrow l}s(x) = -\infty, \quad \lim_{x \uparrow r}s(x)=+\infty,\end{align*} $$

and $m(E)<\infty $ .

Remark 2.3 Assumption 2.1 implies that X is recurrent and has the invariant probability measure, given by

(2.3) $$ \begin{align} \bar{m}(dx)=\frac{1}{m(E)} m(dx). \end{align} $$

Hence, X is ergodic, and its stationary distribution is given by (2.3). Moreover, we see that

(2.4) $$ \begin{align} \lim_{t\to\infty}{\mathbb P}_x(X_t < y) =\bar{m}((l,y)) \quad \text{for all } x,y\in E, \end{align} $$

where ${\mathbb P}_x(\cdot )={\mathbb P}(\cdot \mid X_0=x)$ . This relation is seen in Karatzas and Shreve [Reference Karatzas and Shreve9], which is an extension of a result obtained by Pollack and Siegmund [Reference Pollak and Siegmund12]. On the other hand, Assumption 2.2 implies the symmetry of the stationary distribution $\bar {m}$ on E. Further, it implies that s and $s''$ are odd functions and $s'$ is an even function.

Here are some examples which satisfy Assumptions 2.1 and 2.2.

Example 2.4 (Ornstein–Uhlenbeck process)

Let $\sigma \in {\mathbb R}_{++}$ be constant, let $\mu (x)=-\kappa \sigma ^2 x$ with $\kappa \in {\mathbb R}_{++}:=(0,\infty )$ , and let $E={\mathbb R}$ . The associated process is written as

(2.5) $$ \begin{align} dX_t =-\kappa \sigma^2 X_t \,dt +\sigma \,dW_t. \end{align} $$

In this case, we see that

$$ \begin{align*} s'(x)=e^{\kappa x^2}, \quad m(dx)= \frac{2}{\sigma^2} e^{-\kappa x^2} \,dx \end{align*} $$

and the stationary distribution is a centred Gaussian distribution.

Example 2.5 (Pearson diffusion)

Let $\mu (x)=-\kappa \gamma ^2 x$ and $\sigma (x)= \gamma \sqrt {\delta +x^2}$ with $\kappa ,\gamma ,\delta \in {\mathbb R}_{++}$ , and let $E={\mathbb R}$ . The associated process is written as

(2.6) $$ \begin{align} dX_t =-\kappa \gamma^2 X_t \,dt +\gamma \sqrt{\delta+X_t^2} \,dW_t. \end{align} $$

In this case, we see that

$$ \begin{align*} s'(x)=( \delta+x^2)^{\kappa}, \quad m(dx)= \frac{2}{\gamma^2}( \delta+x^2)^{-(1+\kappa)} \,dx, \end{align*} $$

and the stationary distribution is a scaled t-distribution.

Example 2.6 (Jacobi diffusion)

Let $\mu (x)=-\kappa \gamma ^2x$ and $\sigma (x)= \gamma \sqrt {\delta ^2-x^2}$ with $\kappa , \gamma ,\delta \in {\mathbb R}_{++}$ , and let $E=(-\delta ,\delta )$ . The associated process is written as

$$ \begin{align*} dX_t =-\kappa \gamma^2 X_t \,dt +\gamma \sqrt{\delta^2-X_t^2} \,dW_t. \end{align*} $$

In this case, we see that

$$ \begin{align*} s'(x)=(\delta^2-x^2)^{-\kappa}, \quad m(dx)= \frac{2}{\gamma^2} ( \delta+x)^{\kappa-1} ( \delta-x)^{\kappa-1} \,dx, \end{align*} $$

and the stationary distribution is a centred and scaled beta distribution.

Remark 2.7 Forman and Sørensen [Reference Forman and Sørensen7] studied the parametrized one-dimensional diffusion process

$$ \begin{align*} dX_t =-\theta(X_t-\mu)\,dt +\sqrt{2\theta(aX_t^2+b X_t +c)}\, dW_t, \end{align*} $$

with $\theta>0$ , $\mu ,a,b,c\in {\mathbb R}$ , which contains the above three examples, and shows the feasibility of explicit statistical inference of parameters.

3 Expected value formula for first passage time

For $y\in E$ , denote the first hitting time by

$$ \begin{align*} \tau_{y}=\inf\{ t\ge 0 \mid \ X_t =y\}, \end{align*} $$

where we make the interpretation $\inf \emptyset =+\infty $ , with $\emptyset $ being the empty set. Using the simplified notation ${\mathbb E}_x[(\cdot )] ={\mathbb E}[ (\cdot ) \mid X_0=x]$ , we obtain the following theorem.

Theorem 3.1. Under Assumptions 2.1 and 2.2 , and for $l<-\alpha <\beta <\alpha <r$ ,

$$ \begin{align*} {\mathbb E}_\alpha[\tau_{\beta}]+ {\mathbb E}_\beta[\tau_{-\alpha} \wedge \tau_{\alpha}] =\frac{m(E)}{2}\{ s(\alpha)-s(\beta)\}. \end{align*} $$

Remark 3.2 For the OU process given in Example 2.4, Zeng and Lee [Reference Zeng and Lee14] derived the series expansion formula

$$ \begin{align*} {\mathbb E}_\alpha[\tau_{\beta}]+ {\mathbb E}_\beta[\tau_{-\alpha} \wedge \tau_{\alpha}] &=\frac{1}{2\kappa\sigma^2}\sum_{n=0}^\infty \frac{1}{(2n+1)!}\\ &\quad \times \{ ( 2 \sqrt{\kappa}\alpha)^{2n+1} -( 2 \sqrt{\kappa}\beta )^{2n+1}\} \Gamma\bigg( \frac{2n+1}{2}\bigg), \end{align*} $$

where we use the notation

$$ \begin{align*} \Gamma(z)= \int_0^\infty x^{z-1} e^{-x} \,dx \quad ( z\in {\mathbb C}, \ \Re(z)>0 ) \end{align*} $$

for the gamma function. Using Theorem 3.1 for Example 2.4, we have the following different analytic representation:

$$ \begin{align*} {\mathbb E}_\alpha[\tau_{\beta}]+ {\mathbb E}_\beta[\tau_{-\alpha} \wedge \tau_{\alpha}] =\frac{\sqrt{2\pi}}{2\kappa\sigma^2} \int_\beta^\alpha e^{\kappa x^2}\,dx. \end{align*} $$

Proof. Let $l<a<x<b<r$ . Recall that

$$ \begin{align*} {\mathbb E}_x[\tau_{a} \wedge \tau_{b} ]=\int_E G_{a,b}(x,y) m(dy), \end{align*} $$

where we define the Green function by

(3.1) $$ \begin{align} G_{a,b}(x,y)= \begin{cases} \displaystyle \dfrac{(s(x)-s(a))( s(b)-s(y))} {s(b)-s(a)}& \text{if } a\le x\le y\le b, \\[.3cm] \displaystyle \dfrac{(s(y)-s(a))( s(b)-s(x))} {s(b)-s(a)}& \text{if } a\le y\le x\le b, \\ 0& \text{otherwise}, \end{cases} \end{align} $$

(see, for example, Durret [Reference Durret4, Chapter 6] or Karatzas and Shreve [Reference Karatzas and Shreve9, Section 5.5]). Note that

(3.2) $$ \begin{align} {\mathbb E}_x[\tau_{a} \wedge \tau_{b}] &=\frac{s(x)-s(a)}{s(b)-s(a)} \int_x^b ( s(b)-s(y)) m(dy) \notag \\ &\quad +\frac{s(b)-s(x)}{s(b)-s(a)} \int_a^x (s(y)-s(a)) m(dy). \end{align} $$

Letting $b\uparrow r$ and using Assumption 2.1 yields

(3.3) $$ \begin{align} {\mathbb E}_x[\tau_a] =( s(x)-s(a)) \int_x^r m(dy) +\int_a^x (s(y)-s(a)) m(dy). \end{align} $$

Then, combining (3.2) with $x=\beta $ , $a=-\alpha $ , $b=\alpha $ and (3.3) with $x=\alpha $ , $a=\beta $ , we have

$$ \begin{align*} {\mathbb E}_\beta [\tau_{-\alpha} &\wedge \tau_{\alpha}] +{\mathbb E}_\alpha[\tau_{\beta}] \\ =&\frac{s(\beta)-s(-\alpha)}{s(\alpha)-s(-\alpha)} \int_\beta^\alpha ( s(\alpha)-s(y)) m(dy) +\frac{s(\alpha)-s(\beta)}{s(\alpha)-s(-\alpha)}\int_{-\alpha}^\beta (s(y)-s(-\alpha)) m(dy) \\ &+( s(\alpha)-s(\beta)) \int_\alpha^r m(dy) +\int_\beta^\alpha (s(y)-s(\beta)) m(dy) \\ =&\frac{s(\beta)+s(\alpha)}{2s(\alpha)} \int_\beta^\alpha ( s(\alpha)-s(y)) m(dy) +\frac{s(\alpha)-s(\beta)}{2s(\alpha)} \int_{-\alpha}^\beta (s(y)+s(\alpha)) m(dy) \\ &+( s(\alpha)-s(\beta)) \int_\alpha^r m(dy) +\int_\beta^\alpha (s(y)-s(\beta)) m(dy) \\ =&\frac{s(\alpha)-s(\beta)}{2} \bigg( \int_\beta^\alpha + 2 \int_\alpha^r + \int_{-\alpha}^\beta \bigg) m(dy) +\frac{s(\alpha)-s(\beta)}{2s(\alpha)} \bigg( \int_{-\alpha}^\beta + \int_\beta^\alpha\bigg) s(y)m(dy) \\ =&\frac{s(\alpha)-s(\beta)}{2}m(E), \end{align*} $$

where we use the property that s is an odd function, that is, $s(x)=-s(-x)$ for $x\in E$ , and the symmetry of m, that is, $m(A)=m(-A)$ for $A\in {\mathcal B}(E)$ , which follow from Assumption 2.2. This completes the proof.

Remark 3.3 Under Assumption 2.1 (without imposing Assumption 2.2), we have for $l<\beta <\alpha <r$ ,

(3.4) $$ \begin{align} {\mathbb E}_\alpha[\tau_{\beta}] +{\mathbb E}_\beta[ \tau_{\alpha}] = m(E)\{ s(\alpha)-s(\beta)\}. \end{align} $$

This assertion follows by combining equation (3.2) with $x=\alpha $ , $a=\beta $ , and

(3.5) $$ \begin{align} {\mathbb E}_x[\tau_b] =\int_x^b ( s(b)-s(y)) m(dy) +( s(b)-s(x)) \int_l^x m(dy) \end{align} $$

with $x=\beta $ , $b=\alpha $ . Equation (3.5) follows from (3.2), by letting $a\downarrow l$ and using Assumption 2.1. Note that formula (3.4) appeared in Karatzas and Shreve [Reference Karatzas and Shreve9]. Also, Bertram [Reference Bertram1] has derived this formula for the OU process given in Example 2.4.

4 Optimal thresholds for pairs trading

In this section, we apply Theorem 3.1 to compute optimal thresholds for pairs trading. Consider the situation where the stochastic spread process $X=(X_t)_{t\ge 0}$ (2.1) is governed by the one-dimensional SDE (2.2), and that Assumptions 2.1 and 2.2 are satisfied. First, we note that the SDE (2.2) has a weak solution, which is unique in the sense of probability law (for the definition of the uniqueness of the solution of SDE in the sense of probability law, see [Reference Karatzas and Shreve9, Chapter 5]). For the proof, we refer to Karatzas and Shreve [Reference Karatzas and Shreve9], where the case $E={\mathbb R}$ was treated, which can be straightforwardly modified to apply to our situation with $E\subset {\mathbb R}$ . Next, we note that

$$ \begin{align*} X^{(x)}\underset{\text{law}}{\equiv}-X^{(-x)}, \end{align*} $$

where the superscript $(x)$ denotes the starting position of the process X, that is, $x=X_0^{(x)} \in E$ , which follows from the uniqueness of the solution of SDE (2.2) and Assumption 2.2.

Now, inspired by Zeng and Lee [Reference Zeng and Lee14], we consider the following trading strategy. Let $a>0$ and $b \in [-a,a]$ be two preset thresholds, which determine the trading signals. We set $T_0=0$ , and, for $n\in {\mathbb N}$ , determine the starting time $S_n$ and the completion time $T_n$ of the nth trading cycle by

$$ \begin{align*} S_n=&\inf\{ t> T_{n-1} | |X_t|=a\}, \\ T_n=&\inf\{ t>S_{n} | X_t =\mathrm{sgn}(X_{S_n})b\}, \end{align*} $$

respectively. The nth trading cycle consists of the following two steps.

1. (A) At time $S_n$ , if $X_{S_n}=a$ (respectively, $X_{S_n}=-a$ ), set a one dollar short (respectively, long) position of security P and a $\beta $ dollar long (respectively, short) position of security Q. Keep these positions until time $T_n$ .

2. (B) At time $T_n$ , clear the positions and make a profit. Wait until the next starting time $S_{n+1}$ .

The profit of the nth trading cycle is (approximately) given by

$$ \begin{align*} -\text{sgn}(X_{S_n}) \bigg( \frac{P_{T_n}-P_{S_n}}{P_{S_n}} -\beta \frac{Q_{T_n}-Q_{S_n}}{Q_{S_n}} \bigg)-c &\approx -\text{sgn}(X_{S_n}) \bigg( \log \frac{P_{T_n}}{P_{S_n}} -\beta \log \frac{Q_{T_n}}{Q_{S_n}} \bigg)-c \\ &= -\text{sgn}(X_{S_n})( X_{T_n}-X_{S_n})-c=a-b-c, \end{align*} $$

where $\text {sgn}(x)$ denotes the sign of $x\in {\mathbb R}$ , and $c>0$ is the total transaction cost in the trading cycle. Here, recall that

$$ \begin{align*} U_n= T_{n}-T_{n-1}, \quad n\in {\mathbb N}, \end{align*} $$

are independent, identically distributed random variables, which are deduced from the time-homogeneity of X and the strong Markov property (the strong Markov property of the solution of the Markovian SDE can be found in Karatzas and Shreve [Reference Karatzas and Shreve9, Chapter 5]). We then define

$$ \begin{align*} N_t=\mathrm{sup}\{ n\in {\mathbb N} \mid T_n\le t\}, \quad t\ge 0, \end{align*} $$

which is a renewal process. The cumulative profit obtained from the completed trading until time t is

$$ \begin{align*} (a-b-c) N_t, \end{align*} $$

and the long-time averaged profit is

(4.1) $$ \begin{align} \lim_{t\to\infty} \frac{1}{t} (a-b-c)N_t =\frac{a-b-c}{{\mathbb E}[U_1]}=L(a,b), \quad\text{almost surely,} \end{align} $$

where we use the strong law of large numbers result in renewal theory [Reference Borovkov2, Chapter 10]. To determine the optimal threshold levels, we are interested in the maximization problem

(4.2) $$ \begin{align} \max_{(a,b)\,\in\, {\mathcal A}} L(a,b), \end{align} $$

where

(4.3) $$ \begin{align} {\mathcal A}=\{ (x,y)\in E^2\, | -x \le y\le x-c\}, \end{align} $$

and the objective function can be expressed as

(4.4) $$ \begin{align} L(a,b) =\frac{a-b-c}{{\mathbb E}_b[\tau_{a}\wedge \tau_{-a}]+{\mathbb E}_a[\tau_b]} =\frac{2(a-b-c)}{m(E)\{ s(a)-s(b)\}} \end{align} $$

by Theorem 3.1. To solve the problem expressed by (4.2)–(4.4), we impose the following conditions.

Remark 4.3 Assumption 4.1 implies that X is mean-reverting: the drift $\mu $ of X is always directed “inward”. It also implies that for all $x\in (0,r)$ ,

$$ \begin{align*} s''(x)=-2\bigg( \frac{\mu}{\sigma^2}\bigg)(x) s'(x) =-s''(-x)>0, \end{align*} $$

where we recall that for all $x\in E$ ,

$$ \begin{align*} s'(x)=\exp\bigg\{ -2\int_0^x \bigg( \frac{\mu}{\sigma^2}\bigg)(y)\,dy \bigg\} =s'(-x)>0, \end{align*} $$

which follow from Assumption 2.2 and Remark 2.3. Assumption 4.2 ensures that $L(a,b) \to 0$ as $a \uparrow r(=\infty )$ ; the long-time averaged profit becomes small if the threshold a is too large. The details can be found in the proof of Theorem 4.5.

We obtain the following theorem concerning the solution of the maximization problem (4.2).

Theorem 4.5. Suppose that Assumptions 2.1 – 4.2 are satisfied, and let

$$ \begin{align*} h(a)=s(a)-\bigg( a-\frac{c}{2}\bigg)s'(a), \quad a\ge 0. \end{align*} $$

Then there exists a unique constant $a^*>0$ such that

$$ \begin{align*} h(a^*)=0 \quad\text{and}\quad a^* \in \bigg( \frac{c}{2}, r\bigg), \end{align*} $$

and it defines the maximizer for (4.2) with (4.3) and (4.4) as follows:

$$ \begin{align*} \displaystyle \max_{(a,b)\,\in, {\mathcal A}}L(a,b)=L(a^*,-a^*) =\frac{\displaystyle 2( a^*-c/2)}{m(E)s(a^*)} =\frac{2}{m(E)s'(a^*)}. \end{align*} $$

Proof. Let

$$ \begin{align*} f(a,b)=\frac{a-b-c}{s(a)-s(b)}. \end{align*} $$

We see that

$$ \begin{align*} \partial_a f(a,b)=& \frac{\{ s(a)-s(b)\}-(a-b-c)s'(a)}{\{ s(a)-s(b)\}^2}, \\ \partial_b f(a,b)=& \frac{-\{ s(a)-s(b)\}+(a-b-c)s'(b)}{\{ s(a)-s(b)\}^2}, \end{align*} $$

and $\partial _b f(a,b)\le 0$ for any $(a,b)\in {\mathcal A}_{+}=\{ (a,b)\in {\mathcal A} \mid b\ge 0 \}$ . Indeed, the inequality

$$ \begin{align*} -\{ s(a)-s(b)\}+(a-b-c)s'(b) \le -c s'(b)\le 0 \end{align*} $$

holds for $(a,b)\in {\mathcal A}_{+}$ as s is convex on $(0,r)$ . Hence, we deduce that

(4.5) $$ \begin{align} \max_{(a,b)\,\in\, {\mathcal A}} f(a,b) = \max_{(a,b)\,\in\, {\mathcal A}_{-}}f(a,b), \end{align} $$

where $ {\mathcal A}_{-}=\{ (a,b)\in {\mathcal A} \,|\, b\le 0 \}.$ Then we check the following.

1. (i) For $(a,b)\in {\mathcal A}_{-}$ , we see that

   $$ \begin{align*} f(a,b)\le \frac{2a-c}{s(a)}=\bar{f}(a), \end{align*} $$
   as s is increasing. Further, we see that $\lim \nolimits _{a\uparrow r}\bar {f}(a)=0$ by Assumption 2.1 (if $r<\infty $ ) and l’Hôpital’s rule combined with Assumption 4.2 (if $r=\infty $ ).

2. (ii) We compute the boundary values of f on ${\mathcal A}_-$ as

   1. (a) $f(a,b)=f(a,a-c)=0$ on $\displaystyle \{(a,b)\in {\mathcal A} \,|\, b=a-c, a\ge {c}/{2}\}$ .

   2. (b) $\displaystyle f(a,b)=f(a,0)=({a-c})/{s(a)}$ on $\displaystyle \{ (a,b)\in {\mathcal A} \,|\, b=0, a\ge c \}$ .

   3. (c) $\displaystyle f(a,b)=f(a,-a)=({a-c/2})/{s(a)}$ on $\displaystyle \{ (a,b)\in E^2 \,|\, b=-a, a\ge {c}/{2}\}$ .

Hence, we deduce that the maximum of $f(a,b)$ on ${\mathcal A}_{-}$ exists, and that the maximizer(s) exist(s) in ${\mathcal A}_0\cup \partial _1 A$ , where

$$ \begin{align*} {\mathcal A}_0=& \{ (a,b)\in {\mathcal A}_{-} \,|\, \partial_a f(a,b)=\partial_b f(a,b)=0 \}, \\ \partial_1 {\mathcal A} =&\bigg\{ (a,b)\in E^2 \bigm| b=-a, a\ge \frac{c}{2}\bigg\}. \end{align*} $$

Here, we see that ${\mathcal A}_0\subset \partial _1 A$ . Indeed, $\partial _a f(a,b)=\partial _b f(a,b)=0$ for $(a,b)\in {\mathcal A}_-$ implies

$$ \begin{align*} 0=\{ s(a)-s(b)\}-(a-b-c)s'(a) =-\{ s(a)-s(b)\}+(a-b-c)s'(b), \end{align*} $$

from which we deduce the relations $s'(a)-s'(b)=0$ and $b=-a$ . Now, to compute the maximizer of

$$ \begin{align*} g(a)=f(a,-a) =\frac{1}{s(a)}\bigg( a-\frac{c}{2}\bigg) \end{align*} $$

on $\partial _1{\mathcal A}(={\mathcal A}_0\cup \partial _1 A)$ , we first note the results

(4.6) $$ \begin{align} g\bigg( \frac{c}{2}\bigg)=0 \quad\text{and}\quad \lim_{a\uparrow r}g(a)=0, \end{align} $$

where the latter follows from $g(a)\le \bar {f}(a) \to 0$ as $a\uparrow r$ as we have seen. We next compute the derivative as

$$ \begin{align*} g'(a) =\frac{h(a)}{s(a)^2}. \end{align*} $$

Here, note that $g'(c/2)=1/s( c/2)>0$ for $c>0$ and that $h'(a)=-(a-c/2)s''(a)< 0$ for $a> c/2$ . So, combining these observations with (4.6), we deduce that there exists a unique $a^*\in (c/2,r)$ such that $h(a^*)=0=g'(a^*)$ , and we have that $g'(a)>0$ for $a\in ( c/2,a^*)$ and $g'(a)<0$ for $a\in ( a^*,r)$ . Therefore, recalling (4.5), we conclude that

$$ \begin{align*} \max_{a\ge c/2} g(a)=g(a^*) =f(a^*,-a^*) =\max_{(a,b)\,\in\, {\mathcal A}} f(a,b).\\[-42pt] \end{align*} $$

5 Asymptotic arbitrage

Pairs trading is sometimes explored in the context of so-called “statistical arbitrage”, as it is sometimes executed algorithmically at high frequency. In this section, we consider the asymptotic arbitrage property of the threshold strategy, which is a weak form of arbitrage in the long run. Let us consider the threshold strategy described in Section 4. The cumulative gain of the (self-financing) threshold strategy until time t is given by

$$ \begin{align*} & \sum_{j=1}^{N_t} -\text{sgn}( X_{S_j}) \bigg( \frac{P_{T_j}-P_{S_j}}{P_{S_j}} -\beta \frac{Q_{T_j}-Q_{S_j}}{Q_{S_j}} \bigg) -cN_t \\ &\quad +\bigg[ -\!\text{sgn}( X_{S_{N_t+1}}) \bigg( \frac{P_t-P_{S_{N_t+1}}}{P_{S_{N_t+1}}} -\beta \frac{Q_t-Q_{S_{N_{t}+1}}}{Q_{S_{N_t+1}}} \bigg) -c \bigg] 1_{\{S_{N_t+1}\le t<T_{N_t+1}\}}, \end{align*} $$

where we regard P and Q as discounted price processes to cancel the interest rate effect. Here, we make the interpretation that

$$ \begin{align*} S_{N_t+1}=\sum_{j=0}^\infty S_{j+1} 1_{\{ N_t =j\}} \quad\text{and}\quad X_{S_{N_t+1}}=\sum_{j=0}^\infty X_{S_{j+1}} 1_{\{ N_t =j\}}, \end{align*} $$

for example. As we have seen in Section 4, we can approximate the cumulative gain

(5.1) $$ \begin{align} G_t&=\sum_{j=1}^{N_t} -\text{sgn}( X_{S_j}) \bigg( \log \frac{P_{T_j}}{P_{S_j}} -\beta \log \frac{Q_{T_j}}{Q_{S_j}} \bigg) -c N_t \nonumber \\ &\quad +\bigg[ -\!\text{sgn}( X_{S_{N_t+1}}) \bigg( \log \frac{P_t}{P_{S_{N_t+1}}} -\beta \log\frac{Q_t}{Q_{S_{N_t+1}}} \bigg) -c \bigg] 1_{\{S_{N_t+1}\le t<T_{N_t+1}\}} \nonumber \\ &=(a-b-c)N_t +[ - \text{sgn}( X_{S_{N_t+1}})(X_t-a) -c ] 1_{\{S_{N_t+1}\le t<T_{N_t+1}\}}. \end{align} $$

Remark 5.1 On $\{ T_{N_t}\le t<S_{N_t+1}\}$ ,

$$ \begin{align*} G_t =(a-b-c)N_t\ge 0, \end{align*} $$

which implies that the cumulative gain is always nonnegative and constant, until the new $(N_t+1)$ th trading cycle starts at time $S_{N_t+1}$ after the previous $N_t$ th trading cycle completes. On the other hand, note that, on $\{S_{N_t+1}\le t<T_{N_t+1}\}$ ,

$$ \begin{align*} G_t =(a-b-c)N_t - \text{sgn}( X_{S_{N_t+1}})(X_t-a) -c, \end{align*} $$

which implies the cumulative gain can be negative, and hence the threshold strategy becomes risky until the new $(N_t+1)$ th trading cycle completes after starting at time $S_{N_t+1}$ .

We recall the notion of asymptotic arbitrage in the long run, which was introduced by Föllmer and Schachermayer [Reference Föllmer and Schachermayer6].

Let us impose the following condition.

We then obtain the following result.

Proposition 5.4 Suppose Assumptions 2.1 , 2.2 and 5.3 hold and let $a-b-c>0$ . Then the cumulative gain process $(G_t)_{t\ge 0}$ given by (5.1) satisfies

$$ \begin{align*} \lim_{t\to\infty}\frac{G_t}{t} =L(a,b) \quad\text{in } L^1({\mathbb P}), \end{align*} $$

where $L(a,b)$ is given by (4.1), and the existence of an SAA follows.

Proof. First, recall the relation

(5.2) $$ \begin{align} | G_t - (a-b-c)N_t |\le |X_t|+a+c \quad\text{for almost every } (t,\omega)\in {\mathbb R}_+ \times\Omega. \end{align} $$

Further, we see that

(5.3) $$ \begin{align} \lim_{t\to\infty} {\mathbb E}_x[|X_t|]= \int_{E} |y| \bar{m}\,(dy)<\infty \end{align} $$

for any $x\in E$ , where we use Assumption 5.3 and relation (2.4). Combining (5.2), (5.3) and the law of large numbers for the renewal process $(N_t)_{t\ge 0}$ , we deduce that

$$ \begin{align*} \lim_{t\to\infty} {\mathbb E}\bigg[ \bigg| \frac{G_t}{t}- L(a,b)\bigg|\bigg]=0. \end{align*} $$

So, for any $\delta>0$ ,

$$ \begin{align*} \lim_{t\to\infty} {\mathbb P}\bigg( \bigg| \frac{G_t}{t}-L(a,b) \bigg|\ge \delta \bigg)=0 \end{align*} $$

follows from Markov’s inequality. Further, for any $\delta ,\epsilon>0$ sufficiently small, there exists ${T}_*>0$ such that

$$ \begin{align*} {\mathbb P}( G_T\le \{L(a,b) -\delta\}T )<\epsilon \quad\text{for all } T\ge T_*. \end{align*} $$

Hence, for $T\ge \max ( T_*, (\{L(a,b)-\delta \}\epsilon )^{-1})$ ,

$$ \begin{align*} {\mathbb P}(G_{T}>\epsilon^{-1}) =1-{\mathbb P}( G_{T}\le \epsilon^{-1}) \ge 1-{\mathbb P}( G_{T}\le \{ L(a,b)-\delta\} T ) \ge 1- \epsilon, \end{align*} $$

and an SAA is realized.

6 Numerical experiment

In this section, using two examples of the stochastic spread processes, that is, Examples 2.4 and 2.5, we show some numerical experiments. Recall that the OU process (2.5) has the stationary distribution

$$ \begin{align*} \bar{m}(dx)=\frac{m(dx)}{m({\mathbb R})} =\sqrt{\frac{\kappa}{\pi}} e^{-\kappa x^2}\,dx \ \sim N\bigg( 0, \frac{1}{2\kappa}\bigg), \end{align*} $$

the centered normal distribution with variance $1/2\kappa $ , and that Pearson diffusion process (2.6) has the stationary distribution,

$$ \begin{align*} \bar{m}(dx) =\frac{m(dx)}{m({\mathbb R})} =\frac{\Gamma( \kappa+1)} {\displaystyle\sqrt{\delta\pi}\Gamma( \kappa+1/2)} \bigg( 1+ \frac{x^2}{\delta}\bigg)^{-(\kappa+1)} \,dx, \end{align*} $$

which is a scaled t-distribution; concretely, we have

$$ \begin{align*} \lim_{t\to\infty}{\mathbb P}\bigg( \sqrt{\frac{2\kappa+1}{\delta}} X_t\in \,dx \bigg)\sim f_{2\kappa+1}(x) \,dx \end{align*} $$

with

$$ \begin{align*} f_{\nu}(x)= \frac{\displaystyle \Gamma( (\nu+1)/2)} {\displaystyle \sqrt{\nu\pi}\Gamma( \nu/2)} \bigg( 1+\frac{x^2}{\nu}\bigg)^{-(\nu+1)/2}. \end{align*} $$

Hence, the limiting variance and kurtosis are given by

(6.1) $$ \begin{align} \begin{array}{@{}l} V^\infty= \displaystyle \lim_{t\to\infty} {\mathbb V}[X_t]=\frac{\delta}{2\kappa-1}, \\[8pt] K^\infty= \displaystyle \lim_{t\to\infty} {\mathbb K}[X_t]=\frac{6}{2\kappa-3}, \end{array} \end{align} $$

respectively, where ${\mathbb V}[ (\cdot )]$ denotes the variance and ${\mathbb K}[ (\cdot )]$ denotes the kurtosis.

For the optimization problem studied in Section 4, we consider the following numerical experiment for Pearson diffusion model (2.6).

1. (i) Set $\delta =(0.2)^2$ , $\gamma =1$ , and choose several values of $\kappa $ , which control the limiting variance and kurtosis (6.1) of the t-stationary distribution.

2. (ii) For each parameter set, numerically compute the optimal threshold value $a^*_{\mathrm {P}}$ and the optimal long-time averaged profit

   $$ \begin{align*} L_{\mathrm{P}}^*=L_{\mathrm{P}}(a^*_{\mathrm{P}},-a^*_{\mathrm{P}}), \end{align*} $$
   given in Theorem 4.5, where $L_{\mathrm {P}}(a,b)$ is given by (4.4) for the Pearson diffusion model with $c=0.01$ .

Further, as a comparison, we consider the following numerical experiment for the OU process model (2.5).

1. (iii) Set the limiting variance $1/2\kappa_{\mathrm {OU}}$ of the OU process equal to the limiting variance $V^{\infty} =\delta/(2\kappa-1)$ of the Pearson diffusion process, to solve for the parameter value $\kappa_{\mathrm {OU}}$ :

   $$ \begin{align*} \frac{1}{2\kappa_{\mathrm{OU}}}=\frac{\delta}{2\kappa-1} \quad \Leftrightarrow \quad \kappa_{\mathrm{OU}}= \frac{2\kappa-1}{2\delta}. \end{align*} $$
   Set $\sigma =0.2$ .

2. (iv) For each corresponding parameter set, numerically compute the optimal threshold value $a^*_{\mathrm {OU}}$ and the optimal long-time averaged profit

   $$ \begin{align*} L_{\mathrm{OU}}^*=L_{\mathrm{OU}}(a^*_{\mathrm{OU}},-a^*_{\mathrm{OU}}), \end{align*} $$
   given in Theorem 4.5, where $L_{\mathrm {OU}}(a,b)$ is given by (4.4) for the OU model with $c=0.01$ .

3. (v) Compute

   $$ \begin{align*} L_{\mathrm{mis}}=L_{\mathrm{P}}( a^*_{\mathrm{OU}}, -a^*_{\mathrm{OU}}), \end{align*} $$
   which is interpreted as the long-time expected profit for a “misspecified” agent who observes the limiting variance and “misapplies” OU model (2.5) instead of Pearson model (2.6). Also, compute the loss rate,
   $$ \begin{align*} \mathrm{Loss}= \frac{L_{\mathrm{P}}^*-L_{\mathrm{mis}}}{L_{\mathrm{P}}^*}. \end{align*} $$

From the result (Table 1) we see that $L^*_{\mathrm {P}}>L^*_{\mathrm {OU}}$ always holds though a model misspecification in the profit is rather small (as we see in the loss rate, Loss). Also, we see that higher mean-reversion (with larger $\kappa $ and $\kappa _{\mathrm {OU}}$ ) yields higher optimized profits.

Table 1 Pearson model result and comparison with OU model.

7 Concluding remarks

The long-time maximization of profit for pairs trading with thresholds, discussed in Section 4, is an idealized problem, based on the law of large numbers in the long run:

$$ \begin{align*} \lim_{t\to\infty} \frac{N_t}{t} = \lim_{t\to\infty} \frac{{\mathbb E}[N_t]}{t} = \frac{1}{{\mathbb E}[U_1]}. \end{align*} $$

To consider a suitable “mean–variance” optimization of profit is an interesting and important challenge as the fluctuation from the mean value seems to have a considerable effect in realistic situations with a finite time horizon. One such mean–variance optimization is to consider the criterion function

$$ \begin{align*} \mathrm{MV}(a,b)= L(a,b)- \alpha V(a,b), \end{align*} $$

where

$$ \begin{align*} L(a,b)=\lim_{t\to\infty}\frac{{\mathbb E}[(a-b-c)N_t]}{t} =\frac{(a-b-c)}{{\mathbb E}[U_1]} =\frac{2(a-b-c)}{m(E)\{s(a)-s(b)\}} \end{align*} $$

is the long-time limit of the mean-value of profit, which was analysed in Section 4. Note that

(7.1) $$ \begin{align} V(a,b)= \lim_{t\to\infty}\frac{{\mathbb V}[(a-b-c)N_t]}{t} =\frac{(a-b-c)^2{\mathbb V}[U_1]}{{\mathbb E}[U_1]^3} \end{align} $$

is the long-time limit of the variance of profit, and $\alpha>0$ is the risk-aversion parameter. The limiting variance (7.1) is obtained from the central limit theorem

$$ \begin{align*} \frac{\displaystyle \sqrt{t}\bigg( \frac{N_t}{t}- \frac{1}{{\mathbb E}[U_1]}\bigg)} {\displaystyle \sqrt{\frac{{\mathbb V}[U_1]}{{\mathbb E}[U_1]^3}} } \ \Rightarrow \ N(0,1) \end{align*} $$

for the scaled and centred renewal process (see Borovkov [Reference Borovkov2, Chapter 10]). Defining ${\mathbb V}_x[(\cdot )]={\mathbb E}_x[(\cdot )^2]- \{ {\mathbb E}_x[(\cdot )]\}^2$ , the variance ${\mathbb V}[U_1]$ in (7.1), rewritten as

$$ \begin{align*} {\mathbb V}[U_1]={\mathbb V}_a[\tau_b] +{\mathbb V}_b[\tau_a \wedge \tau_{-a}], \end{align*} $$

can be computed by using Kac’s moment formula [Reference Fitzsimmons and Pitman5],

$$ \begin{align*} {\mathbb E}_x [ (\tau_{\alpha}\wedge \tau_{\beta})^2] =2\int_E G_{\alpha,\beta}(x,y) {\mathbb E}_y [ \tau_{\alpha}\wedge \tau_{\beta}] m(dy), \end{align*} $$

where we use the Green function (3.1), combined with the expected value formula (3.2). Although we have an analytic representation of $V(a,b)$ in closed form, the maximization of $\mathrm {MV}(a,b)$ does not seem to be straightforward, so this is left as a future research topic. For a different but related optimization for determining thresholds to trigger trading signals, we refer the reader to the Sharpe ratio maximization problem, discussed by Bertram [Reference Bertram1].