Growth model with Epstein-Zin preferences and stochastic volatility

Transcription

1 Growth model with Epstein-Zin preferences and stochastic volatility Håkon Tretvoll July 8, Introduction This document goes through a method of solving a growth model with Epstein-Zin preferences and stochastic volatility through a log-linear approximation. 2 Model 2.1 Household preferences Recursive preferences: U t = V[c t, µ t U t+1 ] = [1 βc ρ t + βµ t U t+1 ρ ] 1/ρ 1 where the certainty equivalent µ is specified by µ t U t+1 = [E t U α t+1] 1/α Conventional interpretation: ρ < 1 captures time preference and α < 1 captures risk aversion. Intertemporal elasticity of substitution = 1 1 ρ Coefficient of relative risk aversion = 1 α Note that the household has no disutility of labor and provides a constant labor supply. 1

2 2.2 Firm production and capital evolution Production a function of capital and labor: y t = fk t, a t n where f is homogeneous of degree 1. Since household labor supply is constant we normalize n = 1.For now, we do not consider capital adjustment costs, and we have the following equation for the evolution of capital: and the following overall resource constraint: k t+1 = 1 δk t + i t 2 fk t, a t n = c t + i t 3 [By setting δ = 1 full depreciation, and setting α = ρ = 0 we obtain the Brock-Mirman model which has an analytical solution.] 2.3 Technology Labor augmentinechnology growing at rate g: at log = log = log g + e x t 4 a t 1 x t+1 = Ax t + v 1/2 t Bw 1,t+1 5 v t+1 = 1 φ v v + φ v v t + τw 2,t+1 6 where w 1,t+1 N0, I, w 2,t+1 N0, 1, and these are assumed to be uncorrelated. We call x t the predictable component of consumption growth, and v t+1 is a process for the volatility of this predictable component this is potentially a model with stochastic volatility. In appendix A we discuss the process for technology used by Kaltenbrunner and Lochstoer 2010, RFS and how it compares to the one we use. Even with constant volatility in equation 6 we cannot map our process into the process used by KL

3 2.4 Rescale problem We rescale the problem with a t 1, and define the rescaled variables as: k t = k t a t 1 ; c t = c t a t ; ĩ t = i t a t t = U t a t The rescaled specification of utility is: t = [1 β c ρt + βµ t +1 t+1 ρ ] 1/ρ 7 The rescaled capital evolution equation is: k t+1 = 1 δ k t + ĩ t 8 The rescaled resource constraint is: kt ỹ t = f, 1 = c t + ĩ t First-order conditions The problem we solve is then written [ ρ ] 1/ρ t k t, x t, v t = max 1 β c ρ t + βµ t +1 t+1 k t+1, x t+1, v t+1 c t 10 s.t. kt+1 = 1 δ k t + ĩ t 11 g t kt f, 1 = c t + ĩ t 12 1 This corresponds with the scaling in KL when φ = 1. When φ < 1 so their process for technology is trend-stationary, they rescale with the time trend expµt only which in our case is. 3

4 The first-order condition w.r.t. c t is: 1 β c ρ 1 t The envelope condition is: t = β 1 ρ t µ ρ α t E t k t = βµ ρ α t E t g α t+1 α 1 t+1 t+1 k t+1 g α t+1 α 1 t+1 t+1 k t+1 1 δ kt + f 1, The derivations of these are in appendix B. 3 A log-linear approximation To obtain log-linear approximations we start by makinhe following guess t : k t k,t = t expp 1 k p k 1 t exp p x x t + p v v t k t This implies the following log-linear approximation for the derivative of the value function: log k,t p 1 + p k 1 log k t + p x x t + p v v t A second log-linear approximation The guess for the derivative of the value function also implies an expression for the value function itself: t p 0 + expp 1 log t log p k p 0 + expp 1 exp p k kp k t exp p x x t + p v v t p k log k t exp p x x t + p v v t 16 To get a log-linear expression for the value function we take a log-linear approximation of the above expression around the steady state values of log k t, x t, and v t. 4

5 log t q 1 + q k log k t + q x x t + q v v t 17 The expressions relatinhe q i coefficients to the p i coefficients of the original guess are in appendix C.1. Note however that a log-linear approximation of t about the deterministic steady state has the form: log t log k k log k v v + k k log k t + x x t + v v t Since we know the values of t, k,t, and k t in steady state we can immediately pin down the coefficient q k : q k = k k = p 0 p k 18 where the last equation uses equation 41 to relate this expression to the p i coefficients. As we can see we have an equation relating p 0 and p k. p 0 = k k p k 19 4 Investigatinhe shapes of t and k,t To get an idea of the shapes of the functions we are tryino approximate, we use Dynare to solve for a third-order approximation to the model. We start by considerinhe model without stochastic volatility and with iid shocks to the growth rate. That is, we set v t = v t, e = B = 1, and A = 0. We ll also assume the following functional form for the production function fk t, a t n: fk t, a t n = k θ t a t n 1 θ Note that here we define the rescaled variables: c t = ct a t and k t = kt a t 1. Rescaling in this way k t+1 determined at time t. 5

6 The equations that describe our model are then: ρ ] 1/ρ t = [1 β c ρt + βµ t +1 t+1 α ] 1/α µ t +1 t+1 = E t [+1 t+1 k t+1 = 1 δ k θ t kt + c t 1 β c ρ 1 t log +1 = log g + v 1/2 w 1,t+1 = βµ ρ α t E t t = 1 β c ρ 1 t k t g α t+1 α 1 t+1 1 ρ t t+1 k t+1 1 δ In this version the steady-state relationships are: θ 1 kt 1 + θ k = g βθ g 1 ρ β1 δ 1 1 θ c = 1 δ k g + θ k g k 1/ρ 1 β = c 1 βg ρ k = 1 β β µ = g c 1 ρ 1 In this version of the model, the state variables are k t and w 1,t. 4.1 Euler equation Combininhe FOC w.r.t. c and the envelope condition yields: t = k 1 ρ t 1 β c ρ 1 1 δ θ 1 kt 1 t + θ t Usinhis equation at t+1 to substitute out for t+1 k t+1 gives the Euler equation: 6 g ρ in the FOC w.r.t. c

7 c ρ 1 t = βµ ρ α t E t g α 1 t+1 α ρ t+1 c ρ 1 t+1 1 δ + θ kt+1 +1 θ 1 20 where c t = 1 δ k t + θ kt k t Special case: α = ρ When agents have additive preferences α = ρ the Euler equation becomes: βe t 1 δ k t + g ρ 1 t+1 θ kt t+1 k 1 δ k t kt+1 +1 ρ 1 = θ k t+2 θ 1 kt+1 1 δ + θ +1 ρ Dynare++ Using Dynare++ we calculate third-order approximations to t and k,t. Dynare++ calculates approximations around the deterministic steady state of the form: 7

8 log t log = g 1,U1 log k t log k + g 1,U 2w 1,t + g 2,U 1 log k k 2 t log + g2,u 2 log k t log k w 1,t + g 2,U 3w1,t 2 + g 3,U 1 log k k 3 t log + g3,u 2 log k k 2 t log w1,t + g 3,U 3 log k t log k w1,t 2 + g 3,U 4w1,t 3 log k,t log k = g 1,Uk 1 log k t log k + g 1,Uk 2w 1,t + g 2,Uk 1 log k t log k 2 + g2,uk 2 log k t log k w 1,t + g 2,Uk 3w1,t 2 + g 3,Uk 1 log k t log k 3 + g3,uk 2 log k k 2 t log w1,t + g 3,Uk 3 log k t log k w1,t 2 + g 3,Uk 4w1,t 3 where g i,x j is the coefficient on the j-th term of order i in the equation for variable x. 4.3 Log-linearisation A log-linear approximation of t in this version of the model is of the form: log t log + k k log k t log k + w w t = q 1 + q k log k t + q w w t Hence, in this case we can solve directly for some of the coefficients from steady-state quantities: q k = k k ; and q 1 = log q k log k With this approximation for the level of the value function, the implied approximation for the derivative k,t is log k,t q 1 + log q k + q k 1 log k t + q w w t 8

9 We cannot solve for q w from steady-state values as we don t know w. We can however compare our approximation to the 3 rd order approximation calculated by Dynare Comparison Figures 1-4 makes various comparisons between the output from Dynare++ and a log-linear approximation to the function involved. Figure 1 plots log t vs. log k t while figure 2 plots the same relationship in levels. Similarly, figure 3 plots log kt vs. log k t while figure 4 plots the relationship in levels. Note that with the log-linear approximation above we can only approximate the value funciton and its derivative for the case where w 1,t = 0, since we have not yet solved for q w Shape of log value function log U t log k t Figure 1: Blue: solution from Dynare++. Red: log-linear approximation 4.5 More log-linearisations log c t The first-order conditions in section 4 can be combined into the following equation: 9

10 0.66 Shape of value function U t k t Figure 2: Blue: solution from Dynare++. Red: log-linear approximation 3.8 Shape of log derivative of value function log U kt log k t Figure 3: Blue: solution from Dynare++. Red: log-linear approximation 10

11 0.02 Shape of derivative of value function U kt k t Figure 4: Blue: solution from Dynare++. Red: log-linear approximation 1 β c ρ 1 t Taking logs and rearranging we get log c t = log1 β 1 ρ = 1 ρ t θ θ 1 kt 1 log t log θ 1 ρ + 1 θ 1 ρ log k t + θ 1 ρ log Plugging in the approximation for log t, we have log c t = log1 β log θ + θ log g 1 θ q ρ 1 ρ q k log k θv 1/2 t + 1 ρ q w w t log k t+1 The capital evolution equation implies: log k t+1 = log kt θ c t 11

12 Taking a first-order approximation of the RHS we get: log θ kt c t = log + θ k c g k g θ k θ g c By rearranginhis equation we get that k g c θ log c t log c c θ [ log k t log k ] log log g where log k t+1 = ˆκ 0k + ˆκ ck log c t + ˆκ kk log k t + ˆκ gk log c ˆκ ck = θ k g c ˆκ kk = k g ˆκ gk = θ k θ g c k g θ θ k θ g c θ θ ˆκ 0k = log k c g ˆκ ck log c ˆκ kk log k ˆκ gk log g Insertinhe approximation for log c t and the law of motion for log, we have: 12

13 . 13

14 5 Solving for the approximations TODO: REREAD,UPDATE!!! EVERYTHING BELOW THIS IS OLD!! Given the guesses for log k,t and log t, we have expressions in appendix C.1 that allow us to solve for the q i coefficients as functions of the p i coefficients. Next we derive an approximation for log of the form of 17 and equate coefficients to obtain more equations that will allow us to solve for the p i s. After this step we have 3+dimx t equations from appendix C.1 and 3 + dimx t equations from equating coefficients. We have 7 + dimx t variables however, so we need one more equation to pin down p 0. For this we ll use the steady state values and impose that in steady state our approximation for must be exact. An outline of the steps involved is as follows: Obtain an approximation for the Bellman equation. Obtain an approximation for log c t by approximatinhe first-order and envelope conditions. Obtain an approximation for log µ t by usinhe guess for log t, the assumed law of motion for technology, and an approximation for the capital accumulation equation. By insertinhe approximations for log c t and log µ t into the approximate Bellman equation we can equate coefficients with the log-linearisation 17. To simplify notation in the next couple of sections, we define new coefficients when necessary. These take the form κ term,eqn to indicate the term and equation this coefficient corresponds to. All coefficients are defined in appendix C. 5.1 Approximatinhe Bellman equation First we rewrite equation 7: [ ] ρ t = 1 β exp ρ log c t + β exp ρ log µ t +1 t+1 Then we take a first-order approximation in log c t and log µ t around their non-stochastic steady-state values c and µ: 14

15 ρ log t ρ log + 1 βρ cρ 1 β c ρ + βµ log c ρ t log c 21 βρµ ρ + log µ 1 β c ρ + βµ ρ t +1 t+1 log µ log t κ 0U + κ cu log c t + κ µu log µ t +1 t+1 22 From equation 22 we see that we need log-linear approximations to log c t and log µ t +1 t+1. The κ iu s are defined in appendix C Approximation for log c t Combining equations 13 and 14 we get 1 ρ t 1 β = k,t c t f k,t Taking logs and rearranging: log c t = log1 β 1 ρ + log t 1 1 ρ log k,t ρ log f k,t 23 Now to eliminate log f k,t from this expression, we take a first-order approximation around the steady state value k: log f k,t explog k t log f k + f kk k log f k t log k k log f k,t κ 0f + κ kf log k t 24 We insert 24 and equations 15 and 17 into equation 23: log c t κ 0c + κ kc log k t + κ xcx t + κ vc v t 25 The κ if s and κ ic s are defined in appendix C.3. 15

16 5.3 Approximation for log µ t +1 t+1 To obtain the a log-linear approximation for log µ t +1 t+1 we plug in the law of motion for +1 from equation 4 and the log-linear approximation of the value function 17. log µ t +1 t+1 = 1 [ [ ]] α log E t exp α log +1 + log t+1 26 First we work out the sum log +1 + log t+1: log +1 + log t+1 log +1 + q 1 + q k log k t+1 + qx x t+1 + q v v t+1 = log +1 + q 1 + q k log f k t, 1 c t log +1 + q x x t+1 + q v 1 φ v v + φ v v t + τw 2,t+1 Before we proceed further, we obtain an approximation to the term log f k t, 1 c t : log f k t, 1 c t log f k, 1 c + f k k f k, 1 c log k t log k c f k, 1 c log c t log c log f k t, 1 c t κ 0fc + κ kfc log k t + κ cfc log c t 27 Insertinhis into the above and collecting some terms, we have: log +1 + log t+1 q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log +1 + q k κ kfc log k t + q k κ cfc log c t + qx x t+1 + q v φ v v t + q v τw 2,t+1 = q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log g + 1 q k e + q x Ax t + v 1/2 t Bw 1,t+1 + q k κ kfc log k t + q k κ cfc log c t + q v φ v v t + q v τw 2,t+1 log +1 + log t+1 κ 0gU + κ kgu log k t + κ xgux t + κ vgu v t + v 1/2 t 1 q k e + q x Bw 1,t+1 + q v τw 2,t

17 Now it is useful to calculate the conditional expectation and conditional variance of α log +1 + log t+1 : E t [α V t [α log +1 + log t+1 log +1 + log t+1 ] = α κ 0gU + κ kgu log k t + κ xgux t + κ vgu v t ] = α 2 v t 1 q k e + q x BB 1 q k e + q x + q 2vτ 2 With these expressions we can evaluate the expectations in equation 26: log µ t +1 t+1 κ 0µ + κ kµ log k t + κ xµx t + κ vµ v t 29 The κ ifc s, the κ igu s, and the κ iµ s are defined in appendix C Equating coefficients for log t We insert the approximations for log c t and log µ t +1 t+1, equations 25 and 29, into our approximation for the Bellman equation, equation 22. log t κ 0U + κ cu κ 0c + κ µu κ 0µ + κ cu κ kc + κ µu κ kµ log k t + κ cu κ xc + κ µu κ xµ xt + κ cu κ vc + κ µu κ vµ v t 30 Equating coefficients between equations 17 and 30 gives us the following equations to solve: q 1 = κ 0U + κ cu κ 0c + κ µu κ 0µ 31 q k = κ cu κ kc + κ µu κ kµ 32 q x = κ cu κ xc + κ µu κ xµ 33 q v = κ cu κ vc + κ µu κ vµ 34 With these equations and the equations linkinhe q i s to the p i s in appendix C.1 we have 6+2 dimx t equations, but 7+2 dimx t unknowns. So to solve this system we need one more equation. 17

18 6 Solving for p i s In this section we undertake the somewhat ambitious task of tryino solve the set of equations for the set of coefficients {p 0, p 1, p k, p x, p v }. We break this down and solve for one coefficient at a time. 6.1 Solving for p 1 Assume we have solved for {p 0, p k, p x, p v }. Then we see from equations that we can solve for {q 1, q k, q x, q v }. Hence we can solve for p 1 from equation 31: q 1 = κ 0U + κ cu log1 β 1 ρ p 1 1 ρ κ cu + κ µu q k κ cfc = κ 0U + κ cu + κ µu q k κ cfc p 1 1 ρ + q 1 + κ 0f 1 ρ + κ µu q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log g [ log1 β + q k κ cfc p 1 1 ρ 1 ρ + q 1 + κ ] 0f + α2 1 ρ q2vτ 2 log1 β + q 1 + κ 0f 1 ρ 1 ρ + κ µu q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log g + α 2 q2 vτ 2 We have an equation for p 1 given {p 0, p k, p x, p v }. 6.2 Solving for p v Assume we have solved for {p 0, p k, p x }. Then we see from equations 41 and 42 that we also have {q k, q x }. Hence, we can solve for p v from equation 34: q v = κ cu κ vc + κ µu q k κ cfc κ vc + q v φ v + α 2 1 q ke + q x BB 1 q k e + q x = κ cu + κ µu q k κ cfc p v 1 ρ + q v + κ µu φ v q v + κ µu α 2 1 q ke + q x BB 1 q k e + q x 18

19 Inserting for q v from equation 43 and rearranging gives the following expression for p v : p v = κ µu α 1 q 2 ke + q x BB 1 q k e + q x + κ cu + κ µu q k κ cfc 1 κ µu φ v p0 1 1 ρ p 0 35 We have an equation for p v given {p 0, p k, p x }. 6.3 Solving for p x Assume we have solved for {p 0, p k }. Then we see from equations 41 and 42 that we also have {q k, q x }. Hence, we can solve for p x from equation 33: q x = κ cu κ xc + κ µu 1 qk e + q x A + q k κ cfc κ xc = κ cu + κ µu q k κ cfc 1 1 ρ p x + qx + κ µu 1 q k e A + κ µu qx A Inserting for qx from equation 42 and rearranging gives the following equation that can be solved for p x : [ p0 ] 1 + κ cu + κ µu q k κ cfc 1 ρ p 0 p x p0 κ µu p x A = κ µu 1 q k e A 36 We have an equation we can solve for p x given {p 0, p k } When x t is a scalar... In the special case where x t is a scalar, so is p x, e and A. Equation 36 then becomes: p x = p 0 κ µu 1 q k ea + κ cu + κ µu q k κ cfc 1 1 ρ p 0 κ p0 µu A 19

20 6.4 Solving for p k Assume we have solved for p 0. Then we can solve for p k from equation 32: q k = κ cu κ kc + κ µu q k κ kfc + κ cfc κ kc 1 κ µu κ kfc q k = κ cu + κ µu κ cfc q k κ kc Inserting for q k from equation 41 and for κ kc gives: 1 κ µu κ kfc p 0 p k = This is a quadratic in p k : 0 = ap 2 k + bp k + c p 0 κ cu + κ µu κ cfc p k ] [ p0 1 p k ρ 1 ρ + κ kf 1 ρ p 0 p0 a = κ µu κ cfc 1 1 ρ p 0 1 b = κ µu κ cfc 1 ρ 1 + κ kf p0 + κ cu 1 1 κ µu κ kfc p 0 1 ρ c = κ cu 1 ρ 1 + κ kf Given a value for p 0 we can solve for the roots of this equation 6.5 Solving for p 0 We have shown that given a value for p 0 we can solve for {p k, p x, p v, p 1 }. To solve for p 0 given {p k, p x, p v, p 1 }, we can use equation 16, which must hold in the non-stochastic steady state. Then we have 20

21 p 0 = expp 1 p k kp k exp p v v 37 Hence, we have an iterative procedure to solve for the coefficients {p 0, p 1, p k, p x, p v }. 21

22 7 The Equations The equations we have to solve are: For the q i s: q 1 = log p 0 q k = p 0 p k qx = p 0 p x q v = p 0 p v For p k : 0 = ap 2 k + bp k + c where p k log k + p v v p 0 p0 a = κ µu κ cfc 1 1 ρ p 0 1 b = κ µu κ cfc 1 ρ 1 + κ kf p0 + κ cu 1 1 κ µu κ kfc p 0 1 ρ c = κ cu 1 ρ 1 + κ kf For p x when x is a scalar: p x = p 0 κ µu 1 q k ea + κ cu + κ µu q k κ cfc 1 1 ρ p 0 κ p0 µu A For p v : p v = κ µu α 1 q 2 ke + q x BB 1 q k e + q x + κ cu + κ µu q k κ cfc 1 κ µu φ v p0 1 1 ρ p 0 22

23 For p 1 : p 1 log1 β 1 ρ κ cu + κ µu q k κ cfc = κ 0U + κ cu + κ µu q k κ cfc 1 ρ + q 1 + κ 0f 1 ρ + κ µu q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log g + α 2 q2 vτ 2 Steady-state relationships: p 0 + expp 1 p k kp k exp p v v k expp 1 k p k 1 exp p v v 23

24 A Mappino Kaltenbrunner and Lochstoer 2010, RFS We can map the process for technology specified above to the one used in KL sort of.... Their process for technology is specified as: Z t = expµt + z t 38 z t = φz t 1 + ε t 39 where ε t N0, σ 2. We can obtain a similar process by settinhe parameters in section 2.3 as follows: a t = Z t log g = µ e = 1 x t = z t z t 1 A = φ φ v = τ = 0 v = 1 Bw 1,t+1 = ε t+1 ε t Note the difference in the shocks however. The time t + 1 shocks in KL have conditional expectation of 0 and conditional variance of σ 2 conditioning on time t information, while our shocks have conditional expectation ε t and conditional variance σ 2. In KL the unconditional expectation and variance are equal to the conditional ones, while for our shocks the unconditional expectation is 0 and the unconditional variance is 2σ 2. So in some sense we cannot map between our specification and the one used in KL. Note that to get a stationary problem, KL use a different rescaling of the value function depending on whether φ = 1 or φ < 1 see their technical appendix. 24

25 B Deriving equations 13 and 14 To derive equations 13 and 14 it is useful to work out some derivatives in detail: µ t c t E t +1 t+1 α c t E t +1 t+1 α k t = 1 E t +1 t+1 α α µ1 α t c t µ t = 1 E t +1 t+1 α k t α µ1 α t k t = E t gt+1α α t+1 α 1 t+1 k t+1 k t+1 c t α 1 t+1 = αe t +1 t+1 = E t = αe t gt+1α α t+1 α 1 t+1 k t+1 k t+1 k t+1 k t f α 1 t+1 +1 t+1 k t+1 Combininhese we get expressions for the derivatives of the certainty equivalent: µ t = µ 1 α α 1 t+1 t E t +1 t+1 c t k t+1 µ t = µ 1 α α 1 t+1 f t E t +1 t+1 k t k t+1 k t The first-order condition w.r.t. c t gives: 1 1 βρ c ρ 1 t ρ1 ρ t And the envelope condition gives: t = β 1 ρ t k t 25 + βρµ ρ 1 t µ ρ 1 t µ t k t µ t = 0 c t k t

26 Insertinhe expressions for the derivatives w.r.t. the certainty equivalents into the above, gives equations 13 and 14. C Definition of coefficients This appendix keeps track of definitions of the coefficients defined througout. C.1 Relating q i s to p i s Steady state values are denoted, k, x, and v. The equation we are approximating is log t log p 0 + expp 1 exp p k log p k t exp p x x t + p v v t k Some useful derivatives: log t log k t kt= k = expp 1 k p k exp pv v p 0 + expp 1 kp p k k expp vv = p 0 p k log t log x t xt=x = expp 1 kp k p k exp pv v p 0 + expp 1 kp p k k expp vv p x = p 0 p x log t v t vt=v = p 0 p v The log-linear approximation of log t is then given by 26

27 log t log + p 0 p k log k t log k + p x x t + p v v t v log t = log p 0 p k log k + p v v + p 0 p k log k t + p x x t + p v v t log t q 1 + q k log k t + q x x t + q v v t This gives the following relationship between the q i s and the p i s: q 1 = log p 0 p k log k + p v v 40 q k = p 0 p k 41 q x = p 0 p x 42 q v = p 0 p v 43 C.2 From section 5.1 κ 0U = ρ log κ cu log c κ µu log µ κ cu = 1 βρ c ρ 1 β c ρ + βµ ρ κ µu = βρµ ρ 1 β c ρ + βµ ρ 27

28 C.3 From section 5.2 κ 0f = log f k κ kf log k κ kf = f kk k f k κ 0c = log1 β 1 ρ κ kc = q k p k 1 1 ρ + κ kf 1 ρ p x κ xc = qx 1 ρ κ vc = q v p v 1 ρ + q 1 p 1 1 ρ + κ 0f 1 ρ 28

29 C.4 From section 5.3 κ 0fc = log f k, 1 c κ kfc log k κ cfc log c f k k κ kfc = f k, 1 c c κ cfc = f k, 1 c κ 0gU = q 1 + q k κ 0fc + q v 1 φ v v + 1 q k log g + q k κ cfc κ 0c κ kgu = q k κ kfc + κ cfc κ kc κ xgu = [ ] 1 q k e + q x A + q k κ cfc κ xc κ vgu = q k κ cfc κ vc + q v φ v κ 0µ = κ 0gU + α 2 q2 vτ 2 κ kµ = κ kgu κ xµ = κ xgu κ vµ = κ vgu + α 2 1 q ke + q x BB 1 q k e + q x D Steady-state calculations We need expressions for the steady state values of, k, c, and µ since these show up in the equations for the coefficients. We ll use the non-stochastic steady state, so we set w 1,t+1 = w 2,t+1 = 0 in equations 5 and 6. From the definition of the certainty equivalent we have µ g = g From the captial accumulation equation 2 we have c = f k, 1 kg 29

30 From the rescaled utility equation 7 we have 1/ρ 1 β = c 1 βg ρ From the first-order condition 13 we can solve for k: k = 1 β β 1 ρ g c From the envelope condition 14 we can solve for k: f k = g1 ρ β so if f is Cobb-Douglas with capital share θ, we have θβ k = g 1 ρ 1 1 θ Hence, when solving for the p i s, we have expressions for all the steady state terms involved. By inspectinhe various coefficients we have defined above, we see that some coefficients are simply functions of parameters and steady state values. The following coefficients can be solved independently of the approximation coefficients q i s and p i s: κ 0U, κ cu, κ µu, κ 0f, κ kf, κ 0fc, κ kfc, κ cfc 30