advmacro Takeki Sunakawa RBC

Real Business Cycle model

Takeki Sunakawa

Advanced Macroeconomics at Tohoku University

Stochastic neoclassical growth model

A stochastic version of the RCK (neoclassical growth) model is originally developed by Brock and Mirman (1972).

The model is adapted to the analysis of economic fluctuations by Kydland and Prescott (1982).

There are only two differences from the basic RCK model: Stochastic TFP process and Euler equation.

Recap: Stochastic TFP

We assume that the TFP follows a stochastic process: log At+1= (1 − ρ) log ¯A + ρ log At+ εt+1, where εt∼ N (0, σ_ε²).

Note that

At+1= ¯A^1−ρA^ρ_te^εt+1, holds.

The stochastic process of At is estimated by OLS using a time series of TFP. A typical quarterly estimate for the U.S. economy is ρ = 0.95 and σε= 0.008.

Stochastic Euler equation

The Euler equation becomes

c⁻_t¹= βEtc⁻_t+1¹ (1 + rt+1− δ) .

Recall: The LHS is the benefit from one unit of consumption today. The RHS is the benefit from one unit of saving today and consuming the return on saving tomorrow.

We, modelers, are uncertain about the return on saving and hence consumption tomorrow. How do we estimate them?

We assume rational expectations: Agents in the model has a perfect

knowledge of the model economy, and correctly estimate the expected values of future variables.

Equilibrium

Now we have the following equilibrium conditions: 1 = βEt

n ct

ct+1^{(1 + rt+1− δ)}

o, rt= αyt/kt−1,

yt= Atk^α_t−1, ct+ kt− (1 − δ)kt−1= yt,

log At= ρ log At−1+ εt.

There are five unknowns and five equations, so we can solve the model.

The steady state

Recall: In the steady state,

1 = β 1 + αAk^α−1− δ
, 0 = Ak^α− δk − c, hold.

The steady-state conditions can be solved for

k^∗=

 _αβA

1 − β(1 − δ)

_1−α¹ , c^∗= A(k^∗)^α− δk^∗.

Log-linearization

To solve the rational expectation equilibrium of the model, we approximate the model around the steady state.

Use the formula of approximation

xt≡ x exp ˆxt≈ x(1 + ˆxt),

where x is the steady state of xtand ˆxt is percent deviation from the steady state.

Log-linearization, cont’d

Other useful formulae:

x_ty_t = xy exp(ˆx_t+ ˆy_t)

≈ xy(1 + ˆxt+ ˆyt), xt/yt = (x/y) exp(ˆxt− ˆyt)

≈ (x/y)(1 + ˆxt− ˆyt), y_t^a = y^aexp(aˆyt)

≈ y^a(1 + aˆyt).

ˆ

xⁿ_t = 0 for n > 1, xˆtyˆt= 0,

Ety^a_t+1 = Ety^aexp(aˆyt+1).

≈ y^a(1 + aEtyˆt+1).

Log-linearization: Production function

Production function:

yt= Atk_t−1^α . It can be written as

y exp(ˆyt) = Ak exp(ˆat+ αˆkt−1). In the steady state, y = Ak holds. Then,

ˆ

yt= ˆat+ αˆkt−1. [Note: This is not approximation.]

Log-linearization: Euler equation

Euler equation:

1 = βEt

 _c

t

ct+1

Rt+1

 , where Rt+1≡ 1 + rt+1− δ. It can be written as

1 = βREt

nexp(ˆct− ˆct+1+ ˆRt+1)^o.

Use the formula of approximation 1 = βREt

n1 + ˆct− ˆct+1+ ˆRt+1

o.

In the steady state, 1 = βR holds. Then, ˆ

ct− Etˆct+1+ EtR^ˆt+1= 0.

Log-linearization: Gross interest rate

Gross interest rate (rate of return from capital): Rt+1= 1 + αAt+1k^α−1_t − δ. It can be written as

R exp( ˆRt+1) = 1 + α(y/k) exp(ˆat+1+ (α − 1)ˆkt) − δ. Use the formula of approximation

R(1 + ˆRt+1) = 1 + α(y/k)(1 + ˆat+1+ (α − 1)ˆkt) − δ. In the steady state, R = β⁻¹= 1 + α(y/k) − δ holds. Then we have

R ˆRt+1= α(y/k)(ˆat+1+ (α − 1)ˆkt), or, _Rˆ_t+1= βα(y/k)(ˆat+1+ (α − 1)ˆkt).

Log-linearization: Resource constraint

Resource constraint:

ct+ kt− (1 − δ)kt−1= yt. It can be written as

c exp(ˆct) + k exp(ˆkt) − (1 − δ)k exp(ˆkt−1) = y exp(ˆyt). Use the formula of approximation

c(1 + ˆct) + k(1 + ˆkt) − (1 − δ)k(1 + ˆkt−1) = y(1 + ˆyt). In the steady state, c + k − (1 − δ)k = y holds. Then we have

cˆct+ kˆkt− (1 − δ)kˆkt−1= y ˆyt.

Log-linearization: Summary

After all, the log-linealized equlibrium conditions are: ˆ

ct− Etˆct+1+ EtR^ˆt+1= 0,

Rˆt+1= βα(y/k)(ˆat+1+ (α − 1)ˆk_t), ˆ

yt= ˆat+ αˆkt−1,

cˆct+ kˆkt− (1 − δ)kˆkt−1= y ˆyt. Or,

ˆ

ct− Etcˆt+1+ βα(y/k)(Etˆat+1+ (α − 1)ˆkt) = 0, (c/k)ˆct+ ˆkt= (α(y/k) + 1 − δ)ˆkt−1+ (y/k)ˆat.

Summary

We have

ˆ

ct− Etˆct+1+ cckk^ˆt+ ccaEtaˆt+1= 0, ckccˆt+ ˆkt= ckk^ˆkt−1+ ckaˆat. where

c_ck= −βα(1 − α)(y/k), c_kc= c/k, cca= βα(y/k), ckk= 1/β,

cka= y/k.

A brute force method

We use a brute force method, which is a variant of undetermined coefficient methods.

Conjecture the solution ˆ

ct = γckk^ˆt−1+ γcaˆat, kˆt = γkkk^ˆt−1+ γkaˆat. Substitute them into the equilibrium conditions,

γck^ˆkt−1+γcaaˆt− (γckk^ˆt+γcaEtˆat+1) + cck^ˆkt+ ccaEtaˆt+1= 0, ckc(γck^ˆkt−1+γcaˆat) + ˆkt= ckkk^ˆt−1+ ckaˆat.

A brute force method, cont’d

Terms related to ˆkt−1and ˆateach are put together [γck+ (cck− γck)γkk]ˆkt−1

+[γca+ (cck− γck)γka+ (cca−γca)ρ]ˆat= 0,

[ckcγck+γkk]ˆkt−1+ [ckcγca+γka]ˆat= ckk^ˆkt−1+ ckaˆat. These equations must hold for any ˆkt−1 and ˆat, which implies

γck+ cckγkk−γckγkk= 0, ckcγck+γkk= ckk,

γca+ (cck−γck)γka+ (cca−γca)ρ = 0, ckcγca+γka= cka.

There are four equations and four unknowns(γck, γca, γkk, γka).

Solving for the dynamics

The first two equations become

(1 − γkk)(−γkk+ ckk)/ckc+ cckγkk= 0,

⇔ −(1 − γkk)γkk+ (1 − γkk)ckk+ ckccckγkk= 0,

∴ _γkk² _{− (1 + ckk− ckccck)γkk+ ckk= 0.} Applying the quadratic formula,

γkk=^{1 + ckk− ckccck±p(1 + ckk− ckccck)}

2_{− 4c} kk

2

This equation has two roots. We pick the stable one γkk∈ (−1, 1).

Solving for the dynamics, cont’d

Once we solve for γkk, we also obtain

γck= −cckγkk/(1 − γkk), γ_ca=^{−ρcca− cckcka+ ckaγck}

1 − ρ − cckckc+ ckcγck

, γka= cka− ckcγca,

for the conjectured solution ˆct= γck^ˆkt−1+ γcaˆat and ˆkt= γkk^ˆkt−1+ γkaˆat.

Calibration

We set β = 0.99, α = 0.36 and δ = 0.025. For the stochastic process, we use ρ = 0.95 and σ = 0.008.

With these parameter values, we obtain

ˆkt = 0.9653ˆkt−1+ 0.0754ˆat, ˆ

ct = 0.6182ˆkt−1+ 0.3052ˆat.

Impulse responses

Stochastic simulation

Two methods to solve linearized models

Blanchard and Khan’s method

Uhlig’s method of undetermined coefficients

A general form

A linear model can be written as

((n+m)×(n+m))B





 xt+1 (n×1)

Etyt+1 (m×1)





⁼((n+m)×(n+m))^A



 xt (n×1)

y_t

(m×1)



+ G

((n+m)×k)_(k×1)^{εt ,}

where

xt: vector of predetermined variables,

Etyt₊₁: vector of expectations for non-predetermined (jump) variables εt: vector of stochastic shocks.

Predetermined variables do not depend on shocks in t + 1, while jump variables do. That is why we have an expectational operator on yt+1.

Blanchard and Khan

If B is invertible,

 xt+1

Etyt+1



= B⁻¹A

 xt

yt



+ B⁻¹Gε_t,

Z = B⁻¹A can be decomposed into Z = M ΛM⁻¹, where diagonal elements of Λ are theeigenvaluesof Z and M is a matrix of the right eigenvectors (eigenvalue decomposition).

In Matlab, the command [M Lambda] = eig(Z) computes the eigenvalues and eigenvectors.

Notes on Blanchard and Khan

Reorder the eigenvalues from smallest to largest as ¯Λ and the corresponding M .¯

The Blanchard and Khan condition: (# of eigenvalues that have an absolute value greater than one) = (# expectational variables), m.

When the BK condition is satisfied, one can impose conditions so that there exists a stable solution.

Deterministic case

Consider a deterministic case with Etyt+1= yt+1,

 xt+1

yt+1



= ¯M ¯Λ ¯M⁻¹

 x_t yt

 , M¯⁻¹

 xt+1

yt+1



= ¯Λ ¯M⁻¹

 xt

yt

 , and partition matrices as

M¯⁻¹=





 Mˆ11 (n×n)

Mˆ12 (n×m)

Mˆ21 (m×n)

Mˆ22 (m×m)





^{, Λ =¯}



 Λ¯11 (n×n)

012 (n×m)

021 (m×n)

Λ¯22 (m×m)



.

Note that the BK condition is satisfied, i.e., m =(# of the explosive roots).

Deterministic case, cont’d

Using this partition, there are two matrix equations Mˆ11

(n×n)_(n×1)^{xt+1 + ˆ}(n×m)^M12 _(m×1)^{yt+1 =}

Λ¯11 (n×n)

" Mˆ11

(n×n)(n×1)^{xt + ˆ}(n×m)^M12 (m×1)^yt

# ,

Mˆ21 (m×n)

xt+1 (n×1)

+ M^ˆ22 (m×m)

yt+1 (m×1)

= Λ^¯22 (m×m)

" Mˆ21 (m×n)

xt (n×1)

+ M^ˆ22 (m×m)

yt (m×1)

# .

Given that the elements of ¯Λ22are greater than one, iff ˆM21xt+ ˆM22yt= 0, the solution is stable. Then we have

y^{t =} −^{ ˆ}M22

⁻1

Mˆ21x^t,

xt₊₁ =



Mˆ11−^Mˆ12^{ ˆ}M22

⁻1

Mˆ21

⁻1

Λ¯₁₁



Mˆ11−^Mˆ12^{ ˆ}M22

⁻1

Mˆ21

 xt.

Stochastic case

Consider a stochastic case with E_tyt+1:

 xt+1

Etyt+1



= ¯M ¯Λ ¯M⁻¹

 xt

yt



+ B⁻¹Gεt,

⇔ M^¯−1

 xt+1

Etyt+1



= ¯Λ ¯M⁻¹

 xt

yt



+ ¯M⁻¹B⁻¹Gεt, or,

 _ˆ

M11 M^ˆ12

Mˆ21 M^ˆ22

  xt₊₁

Etyt₊₁



=

 _¯ Λ₁₁ 0

0 Λ^¯₂₂

  _ˆ

M11 M^ˆ12

Mˆ21 M^ˆ22

  xt

yt

 +

 _ˆ G1

Gˆ2

 εt,

where





 Gˆ1 (n×k)

Gˆ2 (m×k)





^{= ¯M}

−₁

B⁻¹G.

Stochastic case, cont’d

The lower partition is written as Mˆ21

(m×n)_(n×1)^xt+1+

Mˆ22 (m×m)_(m×1)^yt+1

= ¯Λ22 (m×m)

" Mˆ21 (m×n)

xt (n×1)

+ M^ˆ22 (m×m)

yt (m×1)

# + G^ˆ2

(m×k)

εt (k×1)

,

or, letting λt= ˆM21xt+ ˆM22yt,

Etλt+1= ¯Λ22λt+ ˆG2εt. It can be solved forward and obtain λt= −¯Λ⁻₂₂¹G^ˆ2εt, or

yt= − ˆM₂₂⁻¹M^ˆ21xt− ˆM₂₂⁻¹Λ^¯−₂₂¹G^ˆ2εt.

Stochastic case, cont’d

The upper partition is written as Mˆ11

(n×n)

xt+1 (n×1)

+ ˆM12 (n×m)

yt+1 (m×1)

= ¯Λ11 (n×n)

" Mˆ11 (n×n)

xt (n×1)

+ ˆM12 (n×m)

yt (m×1)

# + ˆG1

(n×k)

εt (k×1)

,

Substitute ytand Etyt+1= − ˆM₂₂⁻¹M^ˆ21xt+1, we have xt₊₁ =



Mˆ11−^Mˆ12^{ ˆ}M22

⁻1

Mˆ21

⁻1

Λ¯₁₁



Mˆ11−^Mˆ12^{ ˆ}M22

⁻1

Mˆ21

 xt

−



Mˆ11−^Mˆ12^{ ˆ}M22

⁻1

Mˆ21

⁻1

Λ¯₁₁M^ˆ12^{ ˆ}M22

⁻1

Λ¯⁻₂₂¹G^ˆ2−^Gˆ1

 εt.

Example

Recall the stochastic neoclassical growth model ˆ

c_t− E_tcˆt+1+ c_ck^ˆk_t+ c_caρˆa_t= 0, ckccˆt+ ˆkt= ckk^ˆkt−1+ ckaˆat, ˆ

at= ρˆat−1+ εt.

Consider the following matrix difference equation





1 0 0

−c^caρ −c^{ck 1}

−c^{ka 1 0}







 at

kt

Etct₊₁



 =





ρ 0 0

0 0 1

0 ckk _−ckc







 at−1

kt−₁

ct



 +



 1 0 0



ε^t.

It can be solved for

kˆt = 0.9653ˆkt−1+ 0.0716ˆat−1+ 0.0754εt, ˆ

ct = 0.6182ˆkt−1+ 0.2900ˆat−1+ 0.3052εt.

Undetermined coefficient methods

The model is written as

0 = Ax_t+ Bxt−1+ Cy_t+ Dz_t,

0 = Et[F xt+1+ Gxt+ Hxt−1+ Jyt+1+ Kyt+ Lzt+1+ M zt] , and stochastic process zt+1= N zt+ εt+1.

Conjecture the solution

xt = P xt−1+ Qzt, y_t = Rxt−1+ Sz_t.

The problem is to find the values for the matrices P, Q, R and S.

Undetermined coefficient methods, cont’d

Substitute the conjectured solution,

0 = [AP + B + CR] xt−1+ [AQ + CS + D] zt, 0 = [F P P + GP + H + JRP + KR] xt−1

+ [F P Q + F QN + GQ + JRQ + JSN + KS + LN + M ] zt, Both equation must equal zero for any xt−1and zt. Then we have four equations to find the four matrices:

0 = AP + B + CR, 0 = AQ + CS + D,

0 = F P²+ GP + H + JRP + KR,

0 = F P Q + F QN + GQ + JRQ + JSN + KS + LN + M.

Undetermined coefficient methods, cont’d

The solution is given by

R = −C⁻¹[AP + B] , S = −C⁻¹[AQ + D] , P is the solution ofthe matrix quadratic equation

0 =F − JC⁻¹A P²−JC⁻¹B − G + KC⁻¹A P − KC⁻¹B + H, and Q is constructed from vec(Q), where

vec(Q) = N^′⊗ (F − JC⁻¹A) + Ik⊗ (F P + G + JR − KC⁻¹A)
⁻¹

×vec (JC⁻¹D − L)N + KC⁻¹D − M
, where ⊗ is a kronecker product. [See the text for derivation.]

Example

Again, recall the stochastic neoclassical growth model ˆ

ct− Etcˆt+1+ cck^ˆkt+ ccaρˆat= 0, ckccˆt+ ˆkt= ckk^ˆkt−1+ ckaˆat, ˆ

at= ρˆat−1+ εt. The model is written as

0 = Axt+ Bxt−1+ Cyt+ Dzt,

0 = Et[F xt+1+ Gxt+ Hxt−1+ Jyt+1+ Kyt+ Lzt+1+ M zt] , where A = 1, B = −ckk, C = ckc, D = −cka, F = 0, G = cck, H = 0, J = −1, K = 1, L = cca, M = 0 and N = ρ.

Example, cont’d

The solution is

ˆkt = 0.9653ˆkt−1+ 0.0754ˆat, ˆ

ct = 0.6182ˆkt−1+ 0.3052ˆat. Exactly the same as in the other methods.

Hansen (1985) “Indivisible labor and the business cycle”

The original studies of the real business cycle theory (c.f., Kydland and Prescott, 1982) fail to account for some important labor market phenomena. In the existing papers,

There is no unemployment (only the “intensive margin”)

Fluctuations in hours worked are small relative to productivity fluctuations In micro studies using panel data, low intertemporal substitution of leisure is detected.

Business cycle statistics

Output is more volatile than consumption and less volatile than investment. Correlation between output, consumption and investment is high.

Hours are as much volatile as output is.

Hours and labor productivity are less correlated with output.

Fluctuations in hours worked

Fluctuations in total hours worked is largely explained by the “extensive margin”, not the intensive margin

var(log Ht) = var(log ht) + var(log Nt) + 2cov(log ht, log Nt),

= 20% + 55% + 25%. Most people either work full time or not at all.

This paper: Large fluctuations of total hours worked are explained by the number of employed and low elasticity of substitution by individuals.

Hours worked (from Shimer, 2010)

An economy with indivisible labor

The social planner has the problem of:

∞

X

t=0

β^tlog ct+ Aαtlog(1 − h0) subject to

kt= (1 − δ)kt−1+ it, yt= λtk^θ_t−1h^1−θ_t ≥ ct+ it, ht= αth0

where h_tis total hours worked and h0 is hours worked by individuals. α_tis the probablity of being employed.

The technology shock follows an AR(1) process:

λt= (1 − γ) + γλt−1+ εt, ε ∼ N (0, σ_ε²).

Linear disutility of labor

The problem can be rewritten as

∞

X

t=0

β^tlog ct+ B(1 − ht) subject to

kt= (1 − δ)kt−1+ it, yt= λtk_t−1^θ h^1−θ_t ≥ ct+ it, where B = −A log(1 − h0)/h0> 0.

Lagrangean

We set up the Lagrangean as

L0 ≡ E0

∞

X

t=0

β^t{log ct+ B(1 − ht)

−φt ct+ kt− (1 − δ)kt−1− λtk_t−1^θ h^1−θ_t
. φtis the Lagrange multiplier.

Taking the derivatives of the Lagrangean and set them to zero,

∂ct: φt= 1/ct,

∂h_t: φ_t(1 − θ)λ_tk^θ_t−1h⁻_t^θ= B,

∂kt: φt= βφt+1 1 + θλt+1k^θ−1_t h^1−θ_t − δ
,

∂φt: ct+ kt− (1 − δ)kt−1− Atk^θ_t−1h^1−θ_t = 0.

Equilibrium

The equilibrium conditions are 1 = βEt

 _c

t

ct+1



1 + θ^yt+1 kt

− δ

 , (1 − θ)^yt

ht

= Bct, yt= λtk_t−1^θ h^1−θ_t ,

ct+ kt− (1 − δ)kt−1− yt= 0, λt= (1 − γ) + γλt−1+ εt,

There are five unknowns and five equations, so we can solve the model.

The steady state

The equilibrium conditions are

1 = β^1 + θ^y k^{− δ}

, (1 − θ)^y

h ^{= Bc,} y = k^θh^1−θ, ct+ δk = y.

These equations can be solved for (k, h, y, c). Note h is not normailzed in Hansen (1985).

Assignment #4 (due: December 2)

Read Hansen (1985) and use the model with indivisible labor. Also use his calibration: θ = 0.36, δ = 0.025, β = 0.99, A = 2, h0= 0.53, γ = 0.95 and σε= 0.00712.

1 Derive the steady state values of (k, h, y, c) and the log-linearized equlibrium conditions.

2 Replicate Table 1 in the paper (column of “the economy with indivisible labor”).

3 Compare the result with the one in the basic stochastic growth model without labor. How different are business cycle statistics?