When one stock share is a biological individual: a stylized simulation of the population dynamics in an order-driven market

Abstract

The demand–supply relationship plays an important role in an order-driven stock market. In this paper, we propose a stylized model by defining demand (supply) over a stock at a certain time as how many shares are on the bid (ask) side, which includes all buy (sell) limit orders and buy (sell) market orders. Also, we will treat the two types of shares as two different species with interaction (a single share corresponds to an individual of one species) and will construct and apply generalized Lotka–Volterra equations (Hofbauer and Sigmund in Evolutionary games and population dynamics, Cambridge University Press, Cambridge, 1998) to simulate how their population evolve based on some properties or assumptions of an order-driven market, and also on the heterogenous beliefs among traders. The model suggests that the population of bid and ask shares moves either to a fixed point or periodically without the impact of external information. Also, our model gives a reason, though it is not perfect, explaining why stock prices can behave chaotically.

Appendix

1.1 The proof of periodical solution

As there are two critical points on the boarder of the target area, Poincare–Bendixson theorem is failed here. This proof is almost the same as the proof of periodical solution for prey–predator model. We largely refer to the method in book of Hirsch et al. (2012). Therefore, the proof below is very similar to the proof that every solution of the prey–predator system is a closed orbit. All information needed to understand the proof below can be found there (page 242, Hirsch et al. 2012).

We let

$$ L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right) = r\ln Y - r\ln \left( {K - Y} \right) - \sigma Y - \left[ {r\ln X - r\ln \left( {K - X} \right) - \sigma X} \right]. $$

The solution with the initial point $\left( {X_{0} ,Y_{0} } \right)$ is different than $\left( {L_{2} ,L_{1} } \right)$. There are two properties about $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right)$: (1) Trajectory of this solution is not a limit circle, as $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right)$ is not constant on any open set. (2) It is easy to check $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {L_{2} ,L_{1} } \right)$ is a strict local maximum point. We proved that the trajectory inside V will be always be trapped in this area already (invariant set). That is to say, the trajectory moves spirally around critical point $\left( {L_{2} ,L_{1} } \right)$. It indicates the trajectory will move across the nullcline X = L₂ countless times. So, there is a doubly infinite sequence $\ldots < t_{ - 1} < t_{0} < t_{1} \ldots$ and $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right)$ will move across X = L₂ countless times. If $\left( {X_{0} ,Y_{0} } \right)$ is not on a close orbit, when $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right)$ intersects with X = L₂, the intersections are monotone on X = L₂. Since there is no limit circle, $L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X,Y} \right)$ should approach to (L₂, L₁), when either $n \to + \infty ,\;{\text{or}}\;n \to - \infty$. Also, we have $L_{{t_{n} }}^{{\left( {x_{0} ,y_{o} } \right)}} \left( {x,y} \right)$ is a constant along a certain solution; this indicates

$$ L_{{t_{n} }}^{{\left( {X_{0} ,Y_{o} } \right)}} \left( {X_{0} ,Y_{0} } \right) = L\left( {L_{2} , L_{1} } \right) $$

However, this contradicts with (L₂, L₁) is a strict local maximum point.

1.2 Proof of system having continuous first-order partial derivative meets the local Lipschitz condition

Here, I concentrate on plane autonomous systems which are applied in the paper. Readers can check Perko (2013) for more general cases. Let the system be

$$ \left\{ {\begin{array}{*{20}c} {\frac{{{\text{d}}x}}{{{\text{d}}t}} = f\left( {x,y} \right)} \\ {\frac{{{\text{d}}y}}{{{\text{d}}t}} = g\left( {x,y} \right)} \\ \end{array} } \right. $$

Both $f\left( {x,y} \right)$ and $g\left( {x,y} \right)$ have continuous first-order partial derivatives for x and y. Then we have

$$ \frac{{||G\left( {t_{2} ;X_{2} } \right) - G\left( {t_{1} ;X_{1} } \right)||}}{{||X_{2} - X_{1} ||}} = \sqrt {\frac{{\left[ {f\left( {x_{2} ,y_{2} } \right) - f\left( {x_{1} ,y_{1} } \right)} \right]^{2} + \left[ {g\left( {x_{2} ,y_{2} } \right) - g\left( {x_{1} ,y_{1} } \right)} \right]^{2} }}{{\left( {x_{2} - x_{1} } \right)^{2} + \left( {y_{2} - y_{1} } \right)^{2} }}} , $$

where $G\left( {t_{i} , X_{i} } \right),\quad i = 1,2$ is the state of the system. Suppose that, when t₁ and t₂ are very close, then we can linearize the part

$$ \left[ {f\left( {x_{2} ,y_{2} } \right) - f\left( {x_{1} ,y_{1} } \right)} \right]^{2} + \left[ {g\left( {x_{2} ,y_{2} } \right) - g\left( {x_{1} ,y_{1} } \right)} \right]^{2} $$

as follow

$$ \left[ {\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {x_{2} - x_{1} } \right) + \frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {y_{2} - y_{1} } \right)} \right]^{2} + \left[ {\frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {x_{2} - x_{1} } \right) + \frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {y_{2} - y_{1} } \right)} \right]^{2} $$

which is less than or equal to

$$ 2\left[ {\left( {\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} } \right]\left( {x_{2} - x_{1} } \right)^{2} + 2\left[ {\left( {\frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} } \right]\left( {y_{2} - y_{1} } \right)^{2} $$

Based on our assumption of continuous first-order partial derivatives for x and y, $\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} ,\quad \frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} ,\quad \frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} ,\quad \frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }}$should be all bounded in the compact set $[0,K] \times [0,K]$. Thus, we have

$$ \left( {\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} $$

and

$$ \left( {\frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} $$

are bounded, let both of them less than L (L ≥ 0). Then

$$ \begin{aligned} & \left[ {\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {x_{2} - x_{1} } \right) + \frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {y_{2} - y_{1} } \right)} \right]^{2} \\ & \quad + \left[ {\frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {x_{2} - x_{1} } \right) + \frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} \left( {y_{2} - y_{1} } \right)} \right]^{2} \\ & \quad \le 2\left[ {\left( {\frac{\partial f}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial x}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} } \right]\left( {x_{2} - x_{1} } \right)^{2} \\ & \quad + 2\left[ {\left( {\frac{\partial f}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} + \left( {\frac{\partial g}{\partial y}{\mid }_{{x = x_{1} ,y = y_{1} }} } \right)^{2} } \right]\left( {y_{2} - y_{1} } \right)^{2} \\ & \quad \le 2L\left[ {\left( {x_{2} - x_{1} } \right)^{2} + \left( {y_{2} - y_{1} } \right)^{2} } \right]. \\ \end{aligned} $$

Therefore, we have

$$ \frac{{||G\left( {t_{2} ;X_{2} } \right) - G\left( {t_{1} ;X_{1} } \right)||}}{{||X_{2} - X_{1} ||}} = \sqrt {\frac{{\left[ {f\left( {x_{2} ,y_{2} } \right) - f\left( {x_{1} ,y_{1} } \right)} \right]^{2} + \left[ {g\left( {x_{2} ,y_{2} } \right) - g\left( {x_{1} ,y_{1} } \right)} \right]^{2} }}{{\left( {x_{2} - x_{1} } \right)^{2} + \left( {y_{2} - y_{1} } \right)^{2} }}} \le \sqrt {2L} , $$

where $\sqrt {2L}$ can be treated as the Lipschitz constant.

1.3 Proof of Proposition 2

The Jacobian matrix of the system is:

$$ \left[ {\begin{array}{*{20}c} {\left( {1 - \frac{2X}{K}} \right)\left( {r - \sigma Y + \frac{{\sigma Y^{2} }}{K}} \right)} & { - \sigma X\left( {1 - \frac{X}{K}} \right)\left( {1 - \frac{2Y}{K}} \right)} \\ { - \sigma Y\left( {1 - \frac{Y}{K}} \right)\left( {1 - \frac{2X}{K}} \right)} & {\left( {1 - \frac{2Y}{K}} \right)\left( {r - \sigma X + \frac{{\sigma X^{2} }}{K}} \right)} \\ \end{array} } \right] $$

The critical point (0, 0): At this point we have the matrix

$$ \left[ {\begin{array}{*{20}c} r & 0 \\ 0 & r \\ \end{array} } \right]. $$

So, we obtain the two eigenvalues $\lambda_{1} = \lambda_{2} = r > 0$. Therefore, this is an unstable node.

The critical point (K, K): At this point we have the matrix

$$ \left[ {\begin{array}{*{20}c} { - r} & 0 \\ 0 & { - r} \\ \end{array} } \right]. $$

So, we obtain the two eigenvalues $\lambda_{1} = \lambda_{2} = - r < 0$. Therefore, this is a stable node.

The critical points (K, 0) and (0, K): At these two points, we have the matrixes

$$ \left[ {\begin{array}{*{20}c} { - r} & 0 \\ 0 & r \\ \end{array} } \right]\;{\text{and}}\;\left[ {\begin{array}{*{20}c} r & 0 \\ 0 & { - r} \\ \end{array} } \right]. $$

Apparently, the eigenvalues of these two matrixes are two real numbers with different signs (r and − r), so (K, 0) and (0, K) are saddle points.

The critical points ($L_{1} ,L_{1}$) and ($L_{2} ,L_{2}$): At these two points, we have the matrixes

$$ \left[ {\begin{array}{*{20}c} 0 & { - \sigma L_{i} \left( {1 - \frac{{L_{i} }}{k}} \right)\left( {1 - \frac{{2L_{i} }}{K}} \right)} \\ { - \sigma L_{i} \left( {1 - \frac{{L_{i} }}{k}} \right)\left( {1 - \frac{{2L_{i} }}{K}} \right)} & 0 \\ \end{array} } \right]. $$

For each point, we get the eigenvalues $\lambda = \pm \sigma L_{i} \left( {1 - \frac{{L_{i} }}{K}} \right)\left( {1 - \frac{{2L_{i} }}{K}} \right)$, i = 1, 2, so ($L_{1} ,L_{1}$) and ($L_{2} ,L_{2}$) are saddle points.

The critical point ($L_{1} ,L_{2}$): At this point we have the matrix

$$ \left[ {\begin{array}{*{20}c} 0 & { - \sigma L_{1} \left( {1 - \frac{{L_{1} }}{K}} \right)\left( {1 - \frac{{2L_{2} }}{K}} \right)} \\ { - \sigma L_{2} \left( {1 - \frac{{L_{2} }}{K}} \right)\left( {1 - \frac{{2L_{1} }}{K}} \right)} & 0 \\ \end{array} } \right]. $$

We have

$$ \lambda^{2} = \sigma^{2} L_{2} L_{1} \left( {1 - \frac{{L_{2} }}{K}} \right)\left( {1 - \frac{{2L_{1} }}{K}} \right)\left( {1 - \frac{{L_{1} }}{K}} \right)\left( {1 - \frac{{2L_{2} }}{K}} \right). $$

Notice that

$$ L_{1} < \frac{K}{2}\;{\text{and}}\;L_{2} > \frac{K}{2}, $$

thus

$$ \sigma^{2} L_{2} L_{1} \left( {1 - \frac{{L_{2} }}{K}} \right)\left( {1 - \frac{{2L_{1} }}{K}} \right)\left( {1 - \frac{{L_{1} }}{K}} \right)\left( {1 - \frac{{2L_{2} }}{K}} \right) < 0. $$

Let

$$ \sigma^{2} L_{2} L_{1} \left( {1 - \frac{{L_{2} }}{K}} \right)\left( {1 - \frac{{2L_{1} }}{K}} \right)\left( {1 - \frac{{L_{1} }}{K}} \right)\left( {1 - \frac{{2L_{2} }}{K}} \right) = - a, $$

and we have

$$ \lambda = \pm \sqrt a i. $$

The case is similar to the case of the point ($L_{2} ,L_{1}$). Thus, they are two spiral points.