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How Many Agents are Rational in China’s Economy? Evidence from a Heterogeneous Agent-Based New Keynesian Model

Abstract

A new Keynesian model built on an agent-based approach is considered and employed to investigate China’s monetary policy and macroeconomic fluctuations. The assumption of perfect rationality used in standard dynamic stochastic general equilibrium (DSGE) models is abandoned. The expectation’s heterogeneity, caused by agents behaving according to individual rules through adaptive learning, is one of the agent-based model (ABM) characteristics inserted into the DSGE model. Differential evolution (DE) algorithm is employed to estimate the parameters of an agent-based new Keynesian (ABNK) model, which combined the ABM and the new Keynesian DSGE models. The primary contribution of this study is that the degree of rationality in the economy has been estimated using a model with heterogeneous bounded rationality and adaptive learning. In addition, the determinacy properties of ABNK models with different degrees of heterogeneity are analyzed, which shows that the models that are determinate under the assumptions of rationality may become indeterminate in the presence of heterogeneous expectations.

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Acknowledgements

We thank Hans Amman (the editor in charge), anonymous referees, Zhiwei Xu and Jun Wen for their helpful comments and suggestions. Wei Zhao acknowledges the financial support from the Doctoral Research Foundation of Northwest A&F University (No. Z109021803). Genfu Feng acknowledges the financial support from the National Social Science Foundation of China (No. 14BJY00).

Author information

Correspondence to Wei Zhao.

Appendices

Appendix

A.1 Derivation of Consumption Rule

The Lagrangian function is given by:

$$\begin{aligned} L_f= & {} E_0 \sum \limits _{t = 0}^\infty {\beta ^t \left\{ {\frac{{C_{i_h,t}^{1 - \sigma } }}{{1 - \sigma }} - \frac{{L_{i_h ,t}^{1 + \varphi } }}{{1 + \varphi }} + \left( {1 + g_m } \right) ^t e^{d_t } \log \left( {\frac{{M_{i_h,t} }}{{P_t }}} \right) } \right\} } \\&+\, E_0 \sum \limits _{t = 0}^\infty \beta ^t \lambda _{i_h,t} \left\{ \frac{{M_{i_h,t - 1} }}{{P_t }} + \frac{{B_{i_h,t - 1} }}{{P_t }} - C_{i_h,t} + w_{i_h,t} L_{i_h,t} + Tr_{i_h,t} - \left( {1 + \gamma } \right) \frac{{M_{i_h,t} }}{{P_t }}\right. \nonumber \\&\left. -\, \left( {1 + \gamma } \right) b_t \frac{{B_{i_h,t} }}{{P_t }} \right\} \end{aligned}$$

Then, we have the first-order conditions of household \(i_h\):

$$\begin{aligned}&\frac{{\partial L_f }}{{\partial C_{i_h,t} }} = 0\mathrm{{:}}\quad C_{i_h,t}^{ - \sigma } = \lambda _{i_h,t}\\&\frac{{\partial L_f }}{{\partial L_{i_h,t} }} = 0\mathrm{{:}}\quad L_{i_h ,t}^\varphi = \lambda _{i_h,t} w_{i_h,t}\\&\frac{{\partial L_f }}{{\partial M_{i_h,t} }} = 0\mathrm{{:}} \quad \left( {1 + g_m } \right) ^t e^{d_t } \left( {\frac{{M_{i_h,t} }}{{P_t }}} \right) ^{ - 1} \frac{1}{{P_t }} = \lambda _{i_h,t} \left( {1 + \gamma } \right) \frac{1}{{P_t }} - E_{i_h,t} \left[ {\beta \lambda _{i_h,t + 1} \frac{1}{{P_{t + 1} }}} \right] \\&\frac{{\partial L_f }}{{\partial B_{i_h ,t} }} = 0\mathrm{{:}}\quad \lambda _{i_h ,t} \left( {1 + \gamma } \right) b_t \frac{1}{{P_t }} = E_{i_h ,t} \left[ {\beta \lambda _{i_h ,t + 1} \frac{1}{{P_{t + 1} }}} \right] \end{aligned}$$

Eliminating the multipliers, we obtain:

$$\begin{aligned} L_{i_h ,t}^\varphi= & {} C_{i_h ,t}^{ - \sigma } w_{i_h ,t}\\ 1 + \gamma= & {} \beta E_{i_h ,t} \left[ {\frac{{C_{i_h ,t + 1}^{ - \sigma } }}{{C_{i_h ,t}^{ - \sigma } }}\frac{{P_t }}{{P_{t + 1} }} \cdot \frac{1}{{b_t }}} \right] \\ \left( {1 + g_m } \right) ^t e^{d_t } \left( {\frac{{M_{i_h ,t} }}{{P_t }}} \right) ^{ - 1}= & {} C_{i_h ,t}^{ - \sigma } \left( {1 + \gamma } \right) \left( {1 - b_t } \right) \end{aligned}$$

which can be log-linearized around the steady state to obtain:

$$\begin{aligned} w_{i_h ,t} - p_t = \sigma c_{i_h ,t} + \varphi l_{i_h ,t} \end{aligned}$$

The Euler equation of household \(i_h\) is as follows:

$$\begin{aligned} 1 + \gamma = \beta E_{i_h ,t} \left[ {\frac{{C_{i_h ,t + 1}^{ - \sigma } }}{{C_{i_h ,t}^{ - \sigma } }}\frac{{P_t }}{{P_{t + 1} }} \cdot \frac{1}{{b_t }}} \right] \end{aligned}$$

where \(b_t = \exp \left\{ { - i_t } \right\} \). Note that \(i_t\) corresponds to the logarithm of the gross yield on the one-period bond; henceforth, this is referred to as the nominal interest rate. (The yield on the one-period bond is defined by \(b_t = \left( {1 + yield_t } \right) ^{ - 1}\). Note that \(i_t = - \log b_t = \log \left( {1 + yield_t } \right) ^{ - 1} \simeq yield_t\), where the latter approximation will be accurate as long as the nominal yield is “small”.) Substituting this into the above equation, we obtain:

$$\begin{aligned} e^{ - i_t } = \frac{\beta }{{1 + \lambda _{i_h ,t} }}E_{i,t} \left[ {\frac{{C_{i_h ,t + 1}^{ - \sigma } }}{{C_{i_h ,t}^{ - \sigma } }}\frac{{P_t }}{{P_{t + 1} }}} \right] \end{aligned}$$

In the zero-inflation steady state, we have:

$$\begin{aligned} e^{ - i_t } = \frac{\beta }{{1 + \lambda _{i_h ,t} }} \end{aligned}$$

The above equation can be rewritten as:

$$\begin{aligned} i = -\, \log \beta + \log \left( {1 + \gamma } \right) \end{aligned}$$

Thus, the consumption Euler equation can be log-linearized as follows:

$$\begin{aligned} c_{i_h ,t} = E_{i_h ,t} c_{i_h ,t + 1} - \frac{1}{\sigma }\left( {i_t - E_{i_h ,t} \pi _{t + 1} - i} \right) \end{aligned}$$

The money demand of household \(i_h\) is:

$$\begin{aligned} \left( {1 + g_m } \right) ^t e^{d_t } \left( {\frac{{M_{i_h ,t} }}{{P_t }}} \right) ^{ - 1} = C_{i_h ,t}^{ - \sigma } \left( {1 + \gamma } \right) \left( {1 - b_t } \right) \end{aligned}$$

Defining the constant money stock as and substituting \(b_t = \exp \left\{ { - i_t } \right\} \) into it, we obtain:

$$\begin{aligned} \frac{{M_{i_h ,t}^g }}{{P_t }} = \frac{1}{{1 + \gamma }} \cdot \frac{{e^{d_t } C_{i_h ,t}^\sigma }}{{1 - e^{ - i_t } }} \end{aligned}$$

In the steady state, we obtain:

$$\begin{aligned} \frac{{M^g }}{P} = \frac{1}{{1 + \gamma }} \cdot \frac{{e^d C^\sigma }}{{1 - e^{ - i} }} \end{aligned}$$

Taking the logarithm of the above equation, we have:

$$\begin{aligned} m_t^g - p_t = d_t + \sigma c_{i,t} - \frac{{e^{ - i} }}{{1 - e^{ - i} }}i_t + const. \end{aligned}$$

By defining , we obtain the linearized money demand function as follows:

$$\begin{aligned} m_t^g - p_t = \sigma c_{i,t} - \eta i_t + d_t \end{aligned}$$

where \(m_t^g = m_t - g_m t\).

A.2 Derivation of Final-Goods Firms Problem

The maximization problem of final-goods firms is:

$$\begin{aligned} \mathop {\max }\limits _{\left\{ {y_{i_f ,t} } \right\} _{i_f } } P_t \left[ {\sum \limits _{i_f = 1}^{100} {y_{i_f ,t} ^{\left( {\varepsilon - 1} \right) /\varepsilon } } } \right] ^{\varepsilon /\left( {\varepsilon - 1} \right) } - \sum \limits _{i_f = 1}^{100} {P_{i_f ,t} y_{i_f ,t} } \end{aligned}$$

The first-order condition for the problem is given by:

Rearranging the terms, we finally obtain:

$$\begin{aligned} y_{i_f ,t} \left( i \right) = \left( {\frac{{P_{i_f ,t} }}{{P_t }}} \right) ^{ - \varepsilon } Y_t \end{aligned}$$

The above equation is the demand curve of the intermediate-goods firm \(i_f\). Plugging it into the profit function of the final-goods firm, we obtain:

$$\begin{aligned} P_t \left( {\sum \limits _{i_f = 1}^{100} {\left( {\frac{{P_{i_f ,t} }}{{P_t }}} \right) ^{ - \varepsilon } } } \right) ^{\frac{\varepsilon }{{\varepsilon - 1}}} - \sum \limits _{i_f = 1}^{100} {P_{i_f ,t} \left( {\frac{{P_{i_f ,t} }}{{P_t }}} \right) ^{ - \varepsilon } } = 0 \end{aligned}$$

Simplifying the above equation, we obtain:

A.3 Derivation of Eq. (14)

Firms that re-optimize their prices will choose the same price \(P_t^*\). By Eq. (13), we obtain:

$$\begin{aligned} P_t= & {} \left[ {\sum \limits _{i_f = 0}^\theta {P_{t - 1,i_f } ^{1 - \varepsilon } } + \left( {1 - \theta } \right) \left( {P_t^* } \right) ^{1 - \varepsilon } } \right] ^{\frac{1}{{\left( {1 - \varepsilon } \right) }}} \\= & {} \left[ {\theta \left( {P_{t - 1} } \right) ^{1 - \varepsilon } + \left( {1 - \theta } \right) \left( {P_t^* } \right) ^{1 - \varepsilon } } \right] ^{\frac{1}{{\left( {1 - \varepsilon } \right) }}} \end{aligned}$$

Dividing both sides of the above equation by \(P_{t - 1}\) gives us:

$$\begin{aligned} \left( {{\overline{\prod }} e^{\pi _t } } \right) ^{1 - \varepsilon } = \theta + \left( {1 - \theta } \right) \left( {\frac{{{\overline{P}} ^* e^{p_t^* } }}{{{\overline{P}} e^{p_t } }}} \right) ^{1 - \varepsilon } \end{aligned}$$

Considering a zero-inflation steady state (\({\overline{\prod }} = 1\) and ) around which we linearize the above equation, we obtain:

$$\begin{aligned} e^{\left( {1 - \varepsilon } \right) \pi _t }= & {} \theta + \left( {1 - \theta } \right) e^{\left( {1 - \varepsilon } \right) \left( {p_t^* - p_t } \right) }\\&1 + \left( {1 - \varepsilon } \right) \pi _t \approx \theta + \left( {1 - \theta } \right) \left( {1 + \left( {1 - \varepsilon } \right) \left( {p_t^* - p_t } \right) } \right) \end{aligned}$$

Thus, we finally obtain:

$$\begin{aligned} \pi _t = \left( {1 - \theta } \right) \left( {p_t^* - p_t } \right) \end{aligned}$$

A.4 Derivation of Eq. (16)

Note that:

$$\begin{aligned} \frac{{\partial Y_{t + k|t} }}{{\partial P_t^* }}= & {} -\, \varepsilon \left( {\frac{{P_t^* }}{{P_{t + k} }}} \right) ^{ - \varepsilon - 1} \cdot C_{t + k} \cdot \frac{1}{{P_{t + k} }} \\= & {} -\, \varepsilon \left( {\frac{{P_t^* }}{{P_{t + k} }}} \right) ^{ - \varepsilon } C_{t + k} \cdot \frac{{P_{t + k} }}{{P_t^* }} \cdot \frac{1}{{P_{t + k} }} = \frac{{ - \varepsilon }}{{P_t^* }}Y_{t + k|t} \end{aligned}$$

Thus:

$$\begin{aligned}&\frac{\partial }{{\partial P_t^* }}\left( {P_t^* Y_{t + k|t} - \varPsi _{t + k} \left( {Y_{t + k|t} } \right) } \right) = Y_{t + k|t} + P_t^* \frac{{ - \varepsilon }}{{P_t^* }}Y_{t + k|t} - \varPsi _{t + k|t} \frac{{ - \varepsilon }}{{P_t^* }}Y_{t + k|t} \\&\quad = Y_{t + k|t} \left( {1 - \varepsilon + \varPsi _{t + k|t} \frac{\varepsilon }{{P_t^* }}} \right) = \frac{{1 - \varepsilon }}{{P_t^* }}Y_{t + k|t} \left( {P_t^* - \frac{\varepsilon }{{1 - \varepsilon }}\varPsi _{t + k|t} } \right) \end{aligned}$$

A.5 Derivation of Eq. (18)

In the zero-inflation steady state, we obtain , \(\prod _{t - 1,t + k} = 1\), \(P_t^* = P_{t + k}\), \(Y_{t + k|t} = Y\) and \(MC_{t + k|t} = MC\). We also obtain \(\Delta _{t,t + k} = \beta ^k\). Thus, in the steady state, we have:

$$\begin{aligned} \sum \limits _{k = 0}^\infty {\left( {\beta \theta } \right) ^k Y\left( {1 - \mu \cdot \overline{MC} } \right) } = 0 \end{aligned}$$

From above equation, it is obtained that . Eq. (17) can be log-linearized to obtain:

We know \(E_t {\hat{p}}_t^* = {\hat{p}}_t^*\), \(E_t {\hat{p}}_{t - 1} = {\hat{p}}_{t - 1}\) and , so:

The above equation can be rewritten as:

A.6 Derivation of Eq. (29)

First, we obtain:

$$\begin{aligned} \frac{{{\bar{M}}_t }}{{P_t }} = \frac{{{\bar{M}}_{t - 1} }}{{P_{t - 1} }} \cdot \frac{{{\bar{M}}_t }}{{{\bar{M}}_{t - 1} }} \cdot \frac{{P_{t - 1} }}{{P_t }} \end{aligned}$$

where \({\overline{M}} _t\) denotes the money stock in the economy. Dividing both sides of the above equation by the total effective labor, we have:

$$\begin{aligned} \left( {1 + \gamma } \right) \frac{{M_t }}{{P_t }} = \frac{{M_{t - 1} }}{{P_{t - 1} }} \cdot \frac{{{\bar{M}}_t }}{{{\bar{M}}_{t - 1} }} \cdot \frac{{P_{t - 1} }}{{P_t }} \end{aligned}$$

Taking the logarithm of both sides and noting that \(\ln \left( {1 + \gamma } \right) \approx \gamma \), we have:

$$\begin{aligned} m_t - p_t = m_{t - 1} - p_{t - 1} + \Delta m_t - \pi _t \end{aligned}$$

where \(\Delta m_t = \Delta {\bar{m}}_t - \gamma \). Substituting \(m_t^g = m_t - g_m t\) into the above equation, we obtain:

$$\begin{aligned} m_t^g - p_t = m_{t - 1}^g - p_{t - 1} + \Delta m_t^g - \pi _t \end{aligned}$$

where \(\Delta m_t^g = \Delta m_t - g_m = \Delta {\bar{m}}_t - \gamma - g_m\). By substituting \(h_t^g \equiv m_t^g - p_t\) into the above equation, we obtain:

$$\begin{aligned} h_t^g = h_{t - 1}^g + \Delta m_t^g - \pi _t \end{aligned}$$

A.7 Index of the Parameters

First, we obtain:

$$\begin{aligned} \frac{{{\bar{M}}_t }}{{P_t }} = \frac{{{\bar{M}}_{t - 1} }}{{P_{t - 1} }} \cdot \frac{{{\bar{M}}_t }}{{{\bar{M}}_{t - 1} }} \cdot \frac{{P_{t - 1} }}{{P_t }} \end{aligned}$$

where \({\overline{M}} _t\) denotes the money stock in the economy. Dividing both sides of the above equation by the total effective labor, we have:

$$\begin{aligned} \left( {1 + \gamma } \right) \frac{{M_t }}{{P_t }} = \frac{{M_{t - 1} }}{{P_{t - 1} }} \cdot \frac{{{\bar{M}}_t }}{{{\bar{M}}_{t - 1} }} \cdot \frac{{P_{t - 1} }}{{P_t }} \end{aligned}$$

Taking the logarithm of both sides and noting that \(\ln \left( {1 + \gamma } \right) \approx \gamma \), we have:

$$\begin{aligned} m_t - p_t = m_{t - 1} - p_{t - 1} + \Delta m_t - \pi _t \end{aligned}$$

where \(\Delta m_t = \Delta {\bar{m}}_t - \gamma \). Substituting \(m_t^g = m_t - g_m t\) into the above equation, we obtain:

$$\begin{aligned} m_t^g - p_t = m_{t - 1}^g - p_{t - 1} + \Delta m_t^g - \pi _t \end{aligned}$$

where \(\Delta m_t^g = \Delta m_t - g_m = \Delta {\bar{m}}_t - \gamma - g_m\). By substituting \(h_t^g \equiv m_t^g - p_t\) into the above equation, we obtain:

$$\begin{aligned} h_t^g = h_{t - 1}^g + \Delta m_t^g - \pi _t \end{aligned}$$

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Zhao, W., Lu, Y. & Feng, G. How Many Agents are Rational in China’s Economy? Evidence from a Heterogeneous Agent-Based New Keynesian Model. Comput Econ 54, 575–611 (2019) doi:10.1007/s10614-018-9844-3

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Keywords

  • New Keynesian model
  • Agent-based model
  • Heterogeneous expectations
  • Adaptive learning