Adjustment Dynamics in Network Games with Stochastic Parameters

Abstract

In this paper we introduce stochastic parameters into the network game model with production and knowledge externalities. This model was proposed by V. Matveenko and A. Korolev as a generalization of the two-period Romer model. Agents differ in their productivities which have deterministic and stochastic (Wiener) components. We study the dynamics of a single agent and the dynamics of a dyad where two agents are aggregated. We derive explicit expressions for the dynamics of a single agent and dyad dynamics in the form of Brownian random processes, and qualitatively analyze the solutions of stochastic equations and systems of stochastic equations.

Appendix

The proof of Proposition 6.4.

Proof

The system of differential equations in the deterministic case has the form

$$\displaystyle \begin{aligned} \left\{ \begin{array}{c} {\dot{k}_1=\left(\frac{A_1}{2a}-1\right)k_1+\frac{A_1}{2a}k_2-\frac{\varepsilon(1-2a)}{2a},} \\ {\dot{k}_2=\frac{A_2}{2a}k_1+\left(\frac{A_2}{2a}-1\right)k_2-\frac{\varepsilon(1-2a)}{2a}.} \end{array} \right. \end{aligned} $$

(6.24)

The characteristic equation for system (6.11) is as follows

$$\displaystyle \begin{aligned} (\lambda+1)^2-\frac{A_1+A_2}{2a}(\lambda+1)=0, \end{aligned}$$

therefore eigenvalues are

$$\displaystyle \begin{aligned} \lambda_1=-1;\quad \lambda_2=-1+\frac{\bar{A}}{a}, \end{aligned}$$

where $\bar {A}=\frac {A_1+A_2}{2}$. Obviously, we can choose as the eigenvectors of the matrix A the vectors

$$\displaystyle \begin{aligned} e_1= \left( \begin{array}{c} 1 \\ -1 \end{array} \right),\quad e_2= \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right). \end{aligned}$$

So the transition matrix is

$$\displaystyle \begin{aligned} S= \left( \begin{array}{rr} 1 &\quad A_1 \\ -1 &\quad A_2 \end{array} \right), \end{aligned}$$

then

$$\displaystyle \begin{aligned} AS=SJ,\quad e^{tA}=Se^{tJ}S^{-1}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} J= \left( \begin{array}{cc} -1 & 0 \\ 0 & \quad -1+\frac{\bar{A}}{a} \end{array} \right),\quad e^{tJ}= \left( \begin{array}{cc} e^{-t} & 0 \\ 0 & \quad \exp\left(\left(\frac{\bar{A}}{a}-1\right)t\right) \end{array} \right), \end{aligned}$$

$$\displaystyle \begin{aligned} S^{-1}=\frac{1}{A_1+A_2} \left( \begin{array}{cc} A_2 & -A_1 \\ 1 & 1 \end{array} \right). \end{aligned}$$

The general solution of system (6.11) is as follows

$$\displaystyle \begin{aligned} \left( \begin{array}{c} k_1 \\ k_2 \end{array} \right)=C_1 \left( \begin{array}{c} 1 \\ -1 \end{array} \right)\exp(-t)+C_2 \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right)\exp\left(\left(\frac{\bar{A}}{a}-1\right)t\right)+ \left( \begin{array}{c} D_1 \\ D_2 \end{array} \right). \end{aligned}$$

We find the constants D ₁ and D ₂ by solving the system of equations

$$\displaystyle \begin{aligned} \left\{ \begin{array}{c} {\left(\frac{A_1}{2a}-1\right)k_1+\frac{A_1}{2a}k_2=\frac{\varepsilon(1-2a)}{2a},} \\ {\frac{A_2}{2a}k_1+\left(\frac{A_2}{2a}-1\right)k_2=\frac{\varepsilon(1-2a)}{2a}.} \end{array} \right. \end{aligned}$$

It is easy to verify that they are determined by the expression (6.18). We find the integration constants C ₁ and C ₂ from the initial conditions:

$$\displaystyle \begin{aligned} \left\{ \begin{array}{l} {k_1^0=C_1+A_1 C_2+D_1,} \\ {k_2^0=-C_1+A_2 C_2+D_2,} \end{array} \right. \end{aligned}$$

so

$$\displaystyle \begin{aligned} C_2=\frac{\bar{k}^0-\bar{D}}{\bar{A}}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} \bar{k}^0=\frac{k_1^0+k_2^0}{2},\quad \bar{D}=\frac{D_1+D_2}{2}=\frac{\varepsilon(1-2a)}{2(\bar{A}-a)},\quad A_2D_1-A_1D_2=\frac{\varepsilon(1-2a)(A_1-A_2)}{2a}. \end{aligned}$$

Then

$$\displaystyle \begin{aligned} C_1=\frac{A_2k_1^0-A_1k_2^0}{4\bar{A}}-\frac{\varepsilon(1-2a)(A_1-A_2)}{8a\bar{A}}. \end{aligned}$$

Thus, the solution is determined by expression (6.17). ■

The proof of Theorem 6.3.

Proof

It is clear that the matrices A and α commute; therefore, for the matrix exponentials, the relation

$$\displaystyle \begin{aligned} e^{At}e^{\alpha W_t}=e^{At+\alpha W_t} \end{aligned}$$

holds and we can solve the matrix equation (6.16) by multiplying from the left by the matrix exponent

$$\displaystyle \begin{aligned} e^{-At-\alpha W_t+\frac{\alpha ^2}{2}t}. \end{aligned}$$

Denote, as in the one-dimensional case, for brevity

$$\displaystyle \begin{aligned} \varPsi=-At-\alpha W_t+\frac{\alpha^2}{2}t. \end{aligned}$$

Then we have

$$\displaystyle \begin{aligned} d\left(e^\varPsi k\right)=e^\varPsi dk+de^\varPsi\dot k+de^\varPsi\dot dk= \end{aligned}$$

$$\displaystyle \begin{aligned} =e^\varPsi\left(Akdt+\alpha kdW_t+\bar{E}dt\right) +e^\varPsi\left(-Adt-\alpha dW_t+\frac{\alpha^2}{2}dt +\frac{\alpha^2}{2}dt\right)k-e^\varPsi k\alpha^2dt= \end{aligned}$$

$$\displaystyle \begin{aligned} =e^\varPsi\bar{E}dt. \end{aligned}$$

Thus, Eq. (6.16) takes the form

$$\displaystyle \begin{aligned} d\left(e^{-At-\alpha W_t+\frac{\alpha^2}{2}t}k\right)= e^{-At-\alpha W_t+\frac{\alpha^2}{2}t}\bar{E}dt, \end{aligned}$$

therefore, the solution of matrix equation (6.8) can be written as

$$\displaystyle \begin{aligned} k(t)=e^{At+\alpha W_t-\frac{\alpha^2}{2}t}k_0 +e^{At+\alpha W_t-\frac{\alpha^2}{2}t} \left(\int_0^t e^{-A\tau-\alpha W_\tau+\frac{\alpha^2}{2}\tau}d\tau\right)\bar{E}. \end{aligned} $$

(6.25)

Notice, that

$$\displaystyle \begin{aligned} \alpha^2=\frac{1}{2a} \left( \begin{array}{cc} \alpha_1 & \alpha_1 \\ \alpha_2 & \alpha_2 \end{array} \right)\cdot\frac{1}{2a} \left( \begin{array}{cc} \alpha_1 & \alpha_1 \\ \alpha_2 & \alpha_2 \end{array}\cdot \right)=\frac{1}{4a^2} \left( \begin{array}{cc} \alpha_1^2+\alpha_1\alpha_2 & \quad \alpha_1^2+\alpha_1\alpha_2 \\ \alpha_1\alpha_2+\alpha_2^2 & \quad \alpha_1\alpha_2+\alpha_2^2 \end{array}\cdot \right). \end{aligned}$$

The eigenvalues of the matrix

$$\displaystyle \begin{aligned} A-\frac{\alpha^2}{2}= \left( \begin{array}{cc} \frac{A_1}{2a}-\frac{\alpha_1^2+\alpha_1\alpha_2}{8a^2}-1 & \frac{A_1}{2a}-\frac{\alpha_1^2+\alpha_1\alpha_2}{8a^2}\\ \frac{A_2}{2a}-\frac{\alpha_1^2+\alpha_1\alpha_2}{8a^2} & \frac{A_2}{2a}-\frac{\alpha_1^2+\alpha_1\alpha_2}{8a^2}-1 \end{array}\cdot \right). \end{aligned}$$

are obviously λ ₁ = −1 and $\lambda _2=\frac {\bar {A}}{a}-\frac {(\alpha _1+\alpha _2)^2}{8a^2}-1$. As eigenvectors we can take

$$\displaystyle \begin{aligned} e_1= \left( \begin{array}{c} 1 \\ -1 \end{array} \right) \end{aligned}$$

and

$$\displaystyle \begin{aligned} e_2= \left( \begin{array}{c} A_1 \\ A_2 \end{array} \right) \end{aligned}$$

or in view of (6.14)

$$\displaystyle \begin{aligned} e_2= \left( \begin{array}{c} \alpha_1 \\ \alpha_2 \end{array} \right). \end{aligned}$$

The eigenvalues of the matrix α are λ ₁ = 0 and λ ₂ = α ₁ + α ₂, and obviously we can choose the same e ₁ and e ₂ as eigenvectors as for the matrix $A-\frac {\alpha ^2}{2}$. Therefore, to reduce to the diagonal form of the matrices $\left (A-\frac {\alpha ^2}{2}\right )t$ and αW _t we can use the same transition matrices

$$\displaystyle \begin{aligned} S= \left( \begin{array}{rr} 1 & \quad A_1 \\ -1 & A_2 \end{array} \right),\quad S^{-1}=\frac{1}{A_1+A_2} \left( \begin{array}{rr} 1 & \quad A_1 \\ -1 & A_2 \end{array} \right), \end{aligned}$$

so we get

$$\displaystyle \begin{aligned} \left(A-\frac{\alpha^2}{2}\right)t+\alpha W_t=S(Jt+\varLambda W_t)S^{-1}, \end{aligned}$$

where

$$\displaystyle \begin{aligned} J= \left( \begin{array}{cc} -1 & 0 \\ \quad 0 & \quad \frac{\bar{A}}{a}-\frac{(\alpha_1+\alpha_2)^2}{8a^2}-1 \end{array} \right),\quad \varLambda= \left( \begin{array}{cc} 0 & 0 \\ 0 & \frac{\alpha_1+\alpha_2}{2a} \end{array} \right), \end{aligned}$$

and correspondingly

$$\displaystyle \begin{aligned} \exp\left(\left(A-\frac{\alpha^2}{2}\right)t+\alpha W_t\right)= \end{aligned}$$

$$\displaystyle \begin{aligned} =S \left( \begin{array}{cc} \exp(-t) & 0 \\ 0 & \quad \exp\left(\left(\frac{\bar{A}}{a}-\frac{(\alpha_1+\alpha_2)^2}{8a^2}-1\right)t +\frac{\alpha_1+\alpha_2}{2a}W_t\right) \end{array} \right) S^{-1}. \end{aligned} $$

(6.26)

Substituting (6.26) into (6.25) we obtain

$$\displaystyle \begin{aligned} \left( \begin{array}{c} k_1(t) \\ k_2(t) \end{array} \right)=\frac{1}{A_1+A_2} \left( \begin{array}{cc} 1 & \quad A_1 \\ -1 & \quad A_2 \end{array} \right)\times \end{aligned}$$

$$\displaystyle \begin{aligned} \times \left( \begin{array}{cc} \exp(-t) & 0 \\ 0 & \quad \exp\left(\left(\frac{\bar{A}}{a}- \frac{(\alpha_1+\alpha_2)^2}{8a^2}-1\right)t +\frac{\alpha_1+\alpha_2}{2a}W_t\right) \end{array} \right) \left( \begin{array}{cc} A_2 & -A_1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{c} k_1^0 \\ k_2^0 \end{array} \right)- \end{aligned}$$

$$\displaystyle \begin{aligned} -\frac{e(1-2a)}{2a}\cdot\frac{1}{A_1+A_2} \left( \begin{array}{cc} 1 & \quad A_1 \\ -1 & \quad A_2 \end{array} \right)\times \end{aligned}$$

$$\displaystyle \begin{aligned} \times \left( \begin{array}{cc} \exp(-t) & 0 \\ 0 & \quad \exp\left(\left(\frac{\bar{A}}{a}- \frac{(\alpha_1+\alpha_2)^2}{8a^2}-1\right)t +\frac{\alpha_1+\alpha_2}{2a}W_t\right) \end{array} \right)\times \end{aligned}$$

$$\displaystyle \begin{aligned} \times \left( \begin{array}{cc} \int_0^t\exp(\tau)d\tau & 0 \\ 0 & \quad \int_0^t \exp\left(\left(-\frac{\bar{A}}{a} +\frac{(\alpha_1+\alpha_2)^2}{8a^2}+1\right)\tau -\frac{\alpha_1+\alpha_2}{2a}W_\tau\right)d\tau \end{array} \right)\times \end{aligned}$$

$$\displaystyle \begin{aligned} \times \left( \begin{array}{cc} A_2 & -A_1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{c} 1 \\ 1 \end{array} \right). \end{aligned} $$

(6.27)

Calculating expression (6.27) we get expressions (6.19)–(6.20). ■