Solving a Dynamic (Stochastic) General Equilibrium Model under the Discrete Time Framework

Solving a Dynamic (Stochastic) General Equilibrium Model under the Discrete Time Framework Dongpeng Liu Nanjing University Sept 2016 D. Liu (NJU) Solving D(S)GE 09/16 1 / 63

Introduction Targets of the Lecture Understanding the basic ideas of D(S)GE models Understanding business cycle analysis Capable of solving dynamic optimization problems (under the discrete-time framework) Capable of solving an easy D(S)GE model D. Liu (NJU) Solving D(S)GE 09/16 2 / 63

Introduction What Does D(S)GE Stand for Dynamic: time is a variable Stochastic: systematic shocks (e.g. productivity, monetary & etc.) General: all markets Equilibrium: equilibrium path only + Microeconomic foundations (also of vital importance!) Why can t an economy achieve the golden rule level of capital and consumption in a Ramsey model? D. Liu (NJU) Solving D(S)GE 09/16 3 / 63

Introduction How Do We Use D(S)GE Models Convergence of growth analysis and cyclical analysis Undergraduate: long-run growth and short-run cycles are two parallel lines Graduate: Growth: saddle path of a DGE model toward its steady state Cycle: effects of shocks on a DGE model around its steady state DSGE models are widely used to analyze business cycles RBC = DGE + productivity shocks New Keynesian = DGE + price stickiness + monetary shocks Animal Spirit = DGE + expectation shocks D. Liu (NJU) Solving D(S)GE 09/16 4 / 63

DGE Basic Settings and Notations Households have a discounting factor β Households own all the capital (K t ) and make investment decisions every period (K 0 is known) Investments in the current period will become capital stock in the next period Labor supply (N) is inelastic (for simplicity, can be easily extended) Firms hire workers and rent capital Households collect wages (w t N) and rental income (r t K t ) Capital stock depreciates at the rate of δ Total factor productivity (TFP) A t For now, Let s assume that there is no uncertainty wrt A t D. Liu (NJU) Solving D(S)GE 09/16 5 / 63

DGE The Representative Household s Problem Flow utility function: ln C t In each period t, the household try to maximize its expected lifetime utility by choosing C t and K t+1, under the constraint of C t + K t+1 (1 δ)k t = w t N + r t K t D. Liu (NJU) Solving D(S)GE 09/16 6 / 63

The Firm s Problem DGE Firms try to maximize their profits in each period t by choosing N t and K t Cobb-Douglas production function Y t = A t K α t N 1 α D. Liu (NJU) Solving D(S)GE 09/16 7 / 63

DGE What to Expect in the Equilibrium The equilibrium of the model is a set of prices {w t, r t } t=0, a set of TFP {A t } t=0, a set of input factors {N t, K t } t=0, and a set of consumptions {C t } t=0, such that Given w t, r t and A t, C t and K t+1 solve the representative household s problem in each period t Given w t, r t and A t, N t and K t solve the firm s problem in each period t Labor market clears in each period t, that is N t N The market for capital stock clears in each period t The goods market clears in each period t: C t + K t+1 (1 δ)k t = Y t D. Liu (NJU) Solving D(S)GE 09/16 8 / 63

How to Solve the Model 1 Solve the household s problem (dynamic) 2 Solve the firm s problem (static) 3 Combine the first order conditions and the TFP process to form a non-linear difference equation system 4 Solve for the steady state 5 Log-linearize the system around its steady state 6 Determine the stability of the steady state 7 Solve the linearized system D. Liu (NJU) Solving D(S)GE 09/16 9 / 63

Dynamic Optimization Solving the Household s Problem with DP Let s temporarily forget the TFP shock for now (no uncertainty) The life-time utility of a household measured in period t can be written as ln C t + β ln C t+1 + β 2 ln C t+2 + β 3 ln C t+3 + = ln C t + β(ln C t+1 + β ln C t+2 + β 2 ln C t+3 + Terms in the parentheses are the life time utility of the same household measured in period t + 1 D. Liu (NJU) Solving D(S)GE 09/16 10 / 63

Dynamic Optimization Solving the Household s Problem with DP (Cont d) Define H(C t ; K t ) as the life-time utility of a household (with optimal level of consumption) measured in period t Bellman equation H(C t ; K t ) = max C t {ln C t + βh(c t+1 ; K t+1 )} Note that C t is the choice variable in period t, whereas K t is a state variable A state variable is a variable showing the current state of the economy. The value of an endogenous state variable was determined in the previous period D. Liu (NJU) Solving D(S)GE 09/16 11 / 63

Dynamic Optimization Solving the Household s Problem with DP (Cont d) Recall the period-by-period budget constraint C t + K t+1 (1 δ)k t = w t N + r t K t The Bellman equation can be rewritten as H(C t ; K t ) = max K t+1 {ln(w t N + (1 + r t δ)k t K t+1 ) +βh(c t+1 ; K t+1 )} D. Liu (NJU) Solving D(S)GE 09/16 12 / 63

Dynamic Optimization Solving the Household s Problem with DP (Cont d) FOC w.r.t. K t+1 1 C t + β H(C t+1; K t+1 ) K t+1 = 0 Note that if we specify disutility from work or utility from leisure in the flow utility function, then H would be affected by N t and there would be one more FOC with respect to N t. The extra FOC is nothing but the labor supply curve. D. Liu (NJU) Solving D(S)GE 09/16 13 / 63

Dynamic Optimization Solving the Household s Problem with DP (Cont d) Envelope condition H(C t ; K t ) K t = 1 C t (1 + r t δ) 1-period-ahead version of the envelope condition H(C t+1 ; K t+1 ) K t+1 = 1 C t+1 (1 + r t+1 δ) Combining the 1-period-ahead version of the envelope condition and the FOC w.r.t. K t+1 yields the Consumption Euler Equation 1 = β 1 (1 + r t+1 δ) C t C t+1 D. Liu (NJU) Solving D(S)GE 09/16 14 / 63

Consumption Euler Equation Dynamic Optimization Intuition: In period t, if I invest 1 more unit of my income, then my consumption in the current period will decrease by 1 unit, leading to a utility loss of 1/C t. (Why?) In the next period, however, I will have one more unit of capital. The gross return of the extra unit of capital is 1 + r t+1 δ. Hence, I will have a discounted utility gain of β(1 + r t+1 δ)/c t+1 in the next period. (why?) If the current utility loss is larger, then making the last unit of investment is not optimal. If the discounted future utility gain is larger, than I can increase my life-time utility by increasing my investment. I am making the optimal amount of investment only if the two terms equate each other. D. Liu (NJU) Solving D(S)GE 09/16 15 / 63

A Second Method: Lagrangian Dynamic Optimization Still, let s temporarily forget the productivity shock We can solve the household s dynamic optimization problem by constructing a Lagrangian L 0 = t=0 + β t (ln C t ) t=0 {λ t [w t N + (1 + r t δ)k t K t+1 C t ]} D. Liu (NJU) Solving D(S)GE 09/16 16 / 63

Dynamic Optimization A Second Method: Lagrangian (Cont d) FOC w.r.t. C t β t /C t λ t = 0 One-period-ahead version of the FOC above β t+1 /C t+1 λ t+1 = 0 FOC w.r.t. K t+1 λ t+1 (1 + r t+1 δ) λ t = 0 Consumption Euler Equation 1 = β 1 (1 + r t+1 δ) C t C t+1 D. Liu (NJU) Solving D(S)GE 09/16 17 / 63

Dynamic Optimization with Uncertainty (DSGE) The Household s Problem with Uncertainty Now, let s consider productivity shocks The process of TFP: A t+1 = ε t A ρ t, where A 0 = 1, ρ (0, 1) and ε t i.i.d. ln N (0, σ) (Why not a simple AR(1) process with a normally distributed innovation term?) With productivity shocks, we do not know our exact level income in the next period As a result, we do not know our exact level of consumption in the next period In each period, we can only maximize our expected discounted life-time utility The prevailing way to model the formation of expectation is called Rational Expectations D. Liu (NJU) Solving D(S)GE 09/16 18 / 63

Dynamic Optimization with Uncertainty (DSGE) Rational Expectations Rational expectations is a hypothesis in economics which states that agents predictions of the future value of economically relevant variables are not systematically wrong in that all errors are random Agents expectations equal true statistical expected values Rational expectations are identical to the best guess of the future (the optimal forecast) that uses all available information Rational expectations are model-consistent expectations D. Liu (NJU) Solving D(S)GE 09/16 19 / 63

Dynamic Optimization with Uncertainty (DSGE) The Household s Problem with Uncertainty (Cont d) The Bellman equation with uncertainty H(C t ; K t, A t ) = max C t {ln C t + βe t H(C t+1 ; K t+1, A t+1 )} Note that the flow utility is not affected by the productivity shock; whereas the exact value of future life-time utility is unknown Period-by-period budget constraint is the same D. Liu (NJU) Solving D(S)GE 09/16 20 / 63

Dynamic Optimization with Uncertainty (DSGE) The Household s Problem with Uncertainty (Cont d) Plugging the budget constraint into the Bellman equation yields H(C t ; K t, A t ) = max K t+1 {ln(w t N + (1 + r t δ)k t K t+1 ) +βe t H(C t+1 ; K t+1, A t+1 )} FOC 1 C t + βe t H(C t+1 ; K t+1, A t+1 ) K t+1 = 0 D. Liu (NJU) Solving D(S)GE 09/16 21 / 63

Dynamic Optimization with Uncertainty (DSGE) The Household s Problem with Uncertainty (Cont d) Envelope condition H(C t ; K t ) K t = 1 C t (1 + r t δ) Expected one-period-ahead version of the envelope condition E t H(C t+1 ; K t+1 ) K t+1 = E t 1 C t+1 (1 + r t+1 δ) Euler equation 1 1 = βe t (1 + r t+1 δ) C t C t+1 D. Liu (NJU) Solving D(S)GE 09/16 22 / 63

Firm Solving the Firm s Problem Firm s optimization problem is static FOCs r t = MPK t = αa t Kt α 1 N 1 α w t = MPN t = (1 α)a t Kt α N α Note: for any HD1 production function, it is always the case that r t K t + w t N = Y t D. Liu (NJU) Solving D(S)GE 09/16 23 / 63

Equilibrium Solving the Model The Equilibrium Path Three key variables C t, K t, A t The equilibrium can be characterized by 3 non-linear first-order difference equations and 2 initial conditions (K 0 and A 0 ) 1 1 = βe t [ (1 + r t+1 δ)] C t C t+1 C t + K t+1 (1 δ)k t = Y t A t+1 = ε t A ρ t To make the system manageable, we need to linearize the system around its steady state (SS) D. Liu (NJU) Solving D(S)GE 09/16 24 / 63

The Equilibrium Path Steady State In a DSGE model, we ignore the uncertainty when solving the system for steady state Define the steady state level of C and K as C and K The steady state level of A is approximately 1 (given that σ is small) It follows that 1 = β(1 + αk α 1 N 1 α δ) C + δk = K α N 1 α Note that the steady state values of all the variables are just functions of parameters; thus they are constant numbers D. Liu (NJU) Solving D(S)GE 09/16 25 / 63

The Equilibrium Path Log-Linearization As economists, we do not want to solve non-linear difference equation systems A form of linearization is needed Define X t = (X t X )/X, where X {C, K, A, Y, w, r} We try to linearize the system around its steady state such that the system becomes a set of first-order difference equations with respect to X t and X t+1 Note that 1 + X t = X t /X Hence, ln(x t /X ) = ln X t ln X = ln(1 + X t ) X t This is why the procedure is called log-linearization Log-linearization is consistent with how we process cyclical data D. Liu (NJU) Solving D(S)GE 09/16 26 / 63

Detrending Solving the Model The Equilibrium Path Many macroeconomic variables, for example, real GDP, can be decomposed into trend and cyclical components The process is called detrending a time series A widely adopted method is called Hodrick-Prescott filter (often referred as the HP filter) Assume that y t = τ }{{} t trend component + c }{{} t cyclical component T { τ t=1 min (y t τ t ) 2 T 1 + λ t=2 [(τ t+1 τ t ) (τ t τ t 1 )] 2 } D. Liu (NJU) Solving D(S)GE 09/16 27 / 63

Detrending Solving the Model The Equilibrium Path HP Filter ( lambda=1600) 20,000 16,000 12,000 400 8,000 200 4,000 0 0 200 400 600 50 55 60 65 70 75 80 85 90 95 00 05 10 GDP Trend Cycle Note that the distribution of the cyclical component is not stationary D. Liu (NJU) Solving D(S)GE 09/16 28 / 63

Detrending Solving the Model The Equilibrium Path Cyclical Component 400 300 200 100 0 100 200 300 400 500 50 55 60 65 70 75 80 85 90 95 00 05 10 D. Liu (NJU) Solving D(S)GE 09/16 29 / 63

The Equilibrium Path Detrending The volatility of the cyclical component of real GDP increases over time This is the result of long-run economic growth In a DSGE model, we focus on the cyclical behaviors of the economy and do not model the economic growth When processing the data, we shall isolate the cyclical behaviors from the effects of economic growth such that the model-generated fluctuations are comparable to the data Specifically, we use the relative size of cyclical component (cycle/trend) to represent the magnitude of the fluctuation D. Liu (NJU) Solving D(S)GE 09/16 30 / 63

Detrending Solving the Model The Equilibrium Path Relative Size of Cyclical Component.04.02.00.02.04.06.08 50 55 60 65 70 75 80 85 90 95 00 05 10 D. Liu (NJU) Solving D(S)GE 09/16 31 / 63

The Equilibrium Path Detrending and Log-Linearization Approximately, the trend component and the volatility of cyclical component grow at the the same rate The relative size of the cyclical component therefore is X t Hence, log-linearization is consistent with the way in which we process cyclical data D. Liu (NJU) Solving D(S)GE 09/16 32 / 63

Log-Linearization: An Example The Equilibrium Path Let us consider the consumption Euler equation as an example 1 1 = βe t [ (1 + αa t+1 K C t C t+1 α 1 N1 α δ)] t+1 We want to linearize the equation around the steady state of the system Step 1: Taylor expansion for the left hand side of the equation 1 C t 1 C 1 C 2 (C t C ) = 1 C 1 C Ĉt Step 2: Taylor expansion for the right hand side of the equation D. Liu (NJU) Solving D(S)GE 09/16 33 / 63

Log-Linearization: An Example The Equilibrium Path Step 2: Taylor expansion for the right hand side of the equation 1 βe t [ (1 + αa t+1 K C t+1 α 1 N1 α δ)] t+1 β 1 C (1 + αk α 1 N 1 α δ) 1 βe t [ C (C 2 t+1 C )(1 + αk α 1 N 1 α δ)] βe t [ 1 C α(1 α)k α 2 N 1 α (K t+1 K )] +βe t [ 1 C αk α 1 N 1 α (A t+1 1)] D. Liu (NJU) Solving D(S)GE 09/16 34 / 63

Log-Linearization: An Example The Equilibrium Path Step 2 (Cont d): Taylor expansion for the right hand side of the equation 1 βe t [ (1 + αa t+1 K C t+1 α 1 N1 α δ)] t+1 1 C β 1 C (1 + αk α 1 N 1 α δ)e t Ĉ t+1 β 1 C α(1 α)k α 1 N 1 α K t+1 +β 1 C αk α 1 N 1 α E t  t+1 D. Liu (NJU) Solving D(S)GE 09/16 35 / 63

Log-Linearization: An Example The Equilibrium Path Step 3 (Cont d): Log-linearized consumption Euler equation Ĉ t = E t Ĉ t+1 + ϕ 12 K t+1 + ϕ 13 E t  t+1 Note that ϕ 12 = βα(1 α)k α 1 N 1 α ϕ 13 = βαk α 1 N 1 α D. Liu (NJU) Solving D(S)GE 09/16 36 / 63

The Equilibrium Path Log-Linearization: An Example Following the same logic, we can log-linearize the resource constraint Note that φ 21 Ĉ t + φ 22 K t + φ 23 Â t = ϕ 22 K t+1 φ 21 = C φ 22 = αk α N 1 α + (1 δ)k φ 23 = K α N 1 α ϕ 22 = K D. Liu (NJU) Solving D(S)GE 09/16 37 / 63

The Equilibrium Path Log-Linearization: An Example Now, let s log-linearize the equation governing the evolution of TFP Note that Therefore, we have ρ ln A t = ln A t+1 ln ε t+1 ln A t = ln(1 + A t 1 ) Â t 1 ρâ t = Â t+1 + ɛ t+1 By construction, ɛ t+1 = ln ε t+1. Hence, ɛ t+1 N (0, σ) D. Liu (NJU) Solving D(S)GE 09/16 38 / 63

Pin Down the Equilibrium Path The Equilibrium Path Now, Let s construct a linear first-order difference equation system Ĉ t Φ K t  t = Θ Ĉ t+1 K t+1  t+1 + Γ ξc t+1 ξ A t+1 ɛ t+1 Note that ξ C t+1 = E tĉt+1 Ĉ t+1, and ξ A t+1 = E tât+1  t+1 D. Liu (NJU) Solving D(S)GE 09/16 39 / 63

Pin Down the Equilibrium Path The Equilibrium Path Note that Φ = φ 21 φ 22 φ 23 1 0 0 0 0 ρ Θ = 1 ϕ 12 ϕ 13 0 ϕ 22 0 0 0 1 Γ = 1 ϕ 13 0 0 0 0 0 0 1 D. Liu (NJU) Solving D(S)GE 09/16 40 / 63

Pin Down the Equilibrium Path The Equilibrium Path The system can be rewritten as Ĉ t K t Note that Ω = Φ 1 Θ Ĉ t+1 K t+1 = Φ 1 Θ + Φ 1 Γ Â t  t+1 Ĉ t+1 = Ω K t+1 + Φ 1 Γ Â t+1 ξc t+1 ξ A t+1 ɛ t+1 ξc t+1 ξ A t+1 ɛ t+1 D. Liu (NJU) Solving D(S)GE 09/16 41 / 63

Eigendecomposition Solving the Model The Equilibrium Path Any square matrix with distinct eigenvectors (not necessarily distinct eigenvalues) can be eigendecomposed Ω = QΛQ 1 Λ is a diagonal matrix formed from the eigenvalues of Ω whereas the columns of Q are the corresponding eigenvectors of Ω Note that by definition, ΩQ = QΛ. Hence, ΩQQ 1 = Ω = QΛQ 1 D. Liu (NJU) Solving D(S)GE 09/16 42 / 63

Pin Down the Equilibrium Path The Equilibrium Path The system can be rewritten as Ĉ t K t  t = QΛQ 1 Ĉ t+1 K t+1  t+1 + Φ 1 Γ Pre-multiply Q 1 on both sides of the equation yields Q 1 Ĉ t K t  t = ΛQ 1 Ĉ t+1 K t+1  t+1 + Q 1 Φ 1 Γ ξc t+1 ξ A t+1 ɛ t+1 ξc t+1 ξ A t+1 ɛ t+1 D. Liu (NJU) Solving D(S)GE 09/16 43 / 63

Pin Down the Equilibrium Path The Equilibrium Path The system can be rewritten as Z 1 t Zt 2 = Λ Zt 3 Z 1 t+1 Z 2 t+1 Z 3 t+1 + e 1 t+1 e 2 t+1 e 3 t+1 Note that e 1 t+1 e 2 t+1 e 3 t+1 Zt 1 Zt 2 Zt 3 = Q 1 = Q 1 Φ 1 Γ Ĉ t K t  t ξc t+1 ξ A t+1 ɛ t+1 D. Liu (NJU) Solving D(S)GE 09/16 44 / 63

Pin Down the Equilibrium Path The Equilibrium Path Now, the difference equation system can be separated into three independent equations in the form of Zt i = λ i Zt+1 i + ei t+1, where i {1, 2, 3} [ Note that every element in the vector ξ C t+1 ξ A t+1 ] ɛ t+1 has zero mean (why?) Consequently, E t e i t+1 = 0 Hence, E t Z i t = λ i E t Z i t+1 + E te i t+1 = Z i t = λ i E t Z i t+1 D. Liu (NJU) Solving D(S)GE 09/16 45 / 63

The Equilibrium Path Pin Down the Equilibrium Path Iterate Z i t forwardly Zt i = λ i E t Zt+1 i = λ 2 i E te t+1 Zt+2 i = λ 2 i E tzt+2 i. = λ m i E t Zt+m i If λ i < 1, then we say the equation Z i t = λ i Z i t+1 + ei t+1 forward stable Obviously, when λ i < 1, lim m λ m i = 0. Therefore Z i t = 0 D. Liu (NJU) Solving D(S)GE 09/16 46 / 63

The Equilibrium Path Pin Down the Equilibrium Path It has been proved that the model presented here has only one λ i < 1 The number of eigenvalues inside and outside the unit circle has strong implications on the stability of the model (to be discussed later) Generally speaking, each state varaible is associated with a initial condition Hence, there is one equation among the three that satisfies Z i t = 0 Note that Z i t is a linear combination of Ĉ t, K t and  t Based on this relation, we can write Ĉ t as a function of K t and  t Ĉ t = g( K t,  t ) Note that we are relating the choice variable to the state variables D. Liu (NJU) Solving D(S)GE 09/16 47 / 63

The Equilibrium Path Iterations Now, let s consider how to pin down the values of Ĉ t, K t and  t 1 Note that we know the values of K 0 and  0. Hence, Ĉ 0 = g( K 0,  0 ) solves the optimization problem for the household in period 0 2 Based on the log-linearized version of the resource constraint, we have K 1 = φ 21Ĉ0 + φ 22 K 0 + φ 23  0 ϕ 22 3 At the beginning of period 1, ɛ 1 is realized and observed. Consequently,  1 = ρâ 0 ɛ 1 is known at the same time 4 With known value of K 1 and  1, we can calculate Ĉ 1 5 With the resource constraint, we can calculate K 2 D. Liu (NJU) Solving D(S)GE 09/16 48 / 63

The Equilibrium Path Iterations In general, at the beginning of any period t, we should know the values of K t and  t K t is calculated based on the resource constraint in the last period  t can be calculated as soon as ɛ t is observed Then, we can calculate Ĉ t as Ĉ t = g( K t,  t ) Then, we can calculate K t+1 based on the period-by-period resource constraint By repeating these steps, we can get the values of Ĉ t, K t and  t for any t Note that a random number generator is needed to create the series of ɛ t if we want to simulate the model economy with software packages (Matlab, R, GAUSS and etc) D. Liu (NJU) Solving D(S)GE 09/16 49 / 63

Other Issues Other Issues We may also be interested in other variables, e.g. Ŷ t, ŵ t and r t What are the parameter values? (This is NOT a trivial question) Calibration: choose the parameter values such that certain moment conditions are satisfied Estimation: use econometric techniques to estimate the parameters What to expect in a numerical analysis (simulation)? What if we have a model with more or fewer eigenvalues outside the unit circle? What if we have a model with complex eigenvalues? What if we want to examine the fluctuations of employment? D. Liu (NJU) Solving D(S)GE 09/16 50 / 63

Other Endougenous Variables Other Issues We can log-linearize Y t, w t, and r t around the steady state to figure out the cyclical behaviors of these variables Ŷ t = Â t + α K t ŵ t = Â t + α K t r t = Â t (1 α) K t Note that if the labor supply is not inelastic, then Ŷ t, ŵ t and r t will also depend on the fluctuations of employment D. Liu (NJU) Solving D(S)GE 09/16 51 / 63

Other Issues Calibration We need to fix the value of β, N, δ, α, ρ and σ Since the production function exhibits CRS, we can normalize N to 1 According to the data, the weight of capital s income is 0.36. Hence, α = 0.36 To pin down the value of remaining parameters, we need to fix the length of one period Usually, 1 period = 1 quarter D. Liu (NJU) Solving D(S)GE 09/16 52 / 63

Other Issues Calibration Long-run average quarterly net interest rate r δ is approximately 1% Based on the consumption Euler equation, we have β = 1/1.01 Long-run average quarterly depreciate rate δ is approximately 2.5% ρ and σ can be calibrated to match the autocorrelation and volatility of output D. Liu (NJU) Solving D(S)GE 09/16 53 / 63

Other Issues Simulation Simulation can be performed with software packages (Matlab, Gauss and etc.) Variance Correlation matrix Autocorrelation Theoretical impulse response functions D. Liu (NJU) Solving D(S)GE 09/16 54 / 63

Other Issues Number of Eigenvalues Outside the Unit Circle Generally speaking, if we have more initial conditions than the eigenvalues outside the unit circle, there is no solution to the system. If we have more eigenvalues outside the unit circle than the initial conditions, there are infinitely many equilibria. If the number of initial conditions equates the number of eigenvalues outside the unit circle, there is a unique equilibrium path D. Liu (NJU) Solving D(S)GE 09/16 55 / 63

Other Issues Number of Eigenvalues Outside the Unit Circle Consider a system with 4 difference equations with 2 initial conditions If there is only 1 eigenvalue inside the unit circle, we have 1 forward stable equation. The 2 choice variables will be related to the 2 state variables by only 1 equation Infinately many solutions If there are 3 eigenvalue inside the unit circle, we have 3 forward stable equation. The 2 choice variables will be related to the 2 state variables by 3 linearly independent equations No solution D. Liu (NJU) Solving D(S)GE 09/16 56 / 63

Other Issues Complex Eigenvalues The reason why we call it "unit circle" is due to the possibility of having complex eigenvalues Consider a complex eigenvalue λ j = a + bi. It can be rewritten as λ j = r(cos θ + i sin θ) Note that r = a 2 + b 2 and cos θ = a/r λ j is inside the unit circle if r < 1. λ j is outside the unit circle if r > 1. Note that λ m j = r m (cos θ + i sin θ) m = r m (cos mθ + i sin mθ) sin(α + β) = sin α cos β + cos α sin β cos(α + β) = cos α cos β sin α sin β Obviously, lim m λ m j = 0 if r < 1. Hence, Z j t = λm j E t Z j t+m = 0 D. Liu (NJU) Solving D(S)GE 09/16 57 / 63

Other Issues Complex Eigenvalues For the baseline model, we do not need to worry about complex eigenvalues Complex eigenvalues can only appear in pairs A pair of complex eigenvalues exhibit the same norm If one complex eigenvalue is in(out)side the unit circle, so is the other one If we have all three eigenvalues outside the unit circle, then we have infinitely many solutions. We cannot pin down the equilibrium path without knowing the initial value of C t anyway If both complex eigenvalues are outside the unit circle and the real eigenvalue is inside the unit circle. No problem If both complex eigenvalues are inside the unit circle, no solution. We do not need to worry about solving the model. D. Liu (NJU) Solving D(S)GE 09/16 58 / 63

Other Issues Complex Eigenvalues For a more complicated model, we may need to consider the case that we have complex eigenvalues Consider a log-linearized system with 4 difference equations 2 initial conditions 2 complex eigenvalues that are both inside the unit circle We have a unique equilibrium path The problem is that the forward stable equations are subject to complex parameters This is because the eigenvectors corresponding to the complex eigenvalues are also complex D. Liu (NJU) Solving D(S)GE 09/16 59 / 63

Complex Eigenvalues Solving the Model Other Issues Without losing generality, assume that λ 1 and λ 2 are both complex and inside the unit circle. Then, it follows that c 11 + d 11 i c 11 d 11 i c 13 c 14 Q = c 21 + d 21 i c 21 d 21 i c 23 c 24 c 31 + d 31 i c 31 d 31 i c 33 c 34 c 41 + d 41 i c 41 d 41 i c 43 c 44 Q 1 = f 11 + g 11 i f 12 + g 12 i f 13 + g 13 i f 14 + g 14 i f 11 g 11 i f 12 g 12 i f 13 g 13 i f 14 g 14 i f 31 f 32 f 33 f 34 f 41 f 42 f 43 f 44 D. Liu (NJU) Solving D(S)GE 09/16 60 / 63

Complex Eigenvalues Solving the Model Other Issues Without losing generality, assume that X 1t and X 2t are choice variables. Meanwhile, X 3t and X 4t are state variables. Then the 2 forward stable equations are 4 j=1 4 j=1 (f 1j + g 1j i)x jt = 0 (f 1j g 1j i)x jt = 0 D. Liu (NJU) Solving D(S)GE 09/16 61 / 63

Other Issues Complex Eigenvalues Obviously, any meaningful economic variables can only take real values Hence, both equations above lead to the following f 11 X 1t + f 12 X 2t = (f 13 X 3t + f 14 X 4t ) g 11 X 1t + g 12 X 2t = (g 13 X 3t + g 14 X 4t ) These 2 equations can be used to relate the choice variables to the state variables We can follow the procedures of iterations accordingly to solve the model Note that all endogenous variables will exhibit an oscillating pattern here D. Liu (NJU) Solving D(S)GE 09/16 62 / 63

Conclusions Concluding Remarks 1 D(S)GE model is the prevailing methodology to conduct business cycle analysis 2 Generally speaking, the dynamic equilibrium of a D(S)GE model is characterized by a set of non-linear (stochastic) difference equations 3 To solve the system, we first need to log-linearize it around its steady state 4 Compare the number of eigenvalues outside the unit circle with the number of initial conditions. Determine the stability of the model 5 After parameterization, a simulation can be performed to examine policy implications and the validity of the model. D. Liu (NJU) Solving D(S)GE 09/16 63 / 63