Lecture: Solving RBC model - jaceksuda.com · Advanced Macroeconomics University of Warsaw Jacek...

Linearization Solution

Lecture: Solving RBC model

Advanced MacroeconomicsUniversity of Warsaw

Jacek Suda

March 27, 2018

RBC


Plan of the Presentation

1 Linearization

2 SolutionBlanchard and Kahn’s (1980) MethodSims’ (2001) Method

RBC


What we did

We built the model

We have a non-linear expectational difference model

Want to solve it. We can either1 use non-linear solution method,2 use (linear) approximation method.

Today:approximate (linearly) using Taylor-series expansions andsolve the resulting linear system.

If we log-linearize the system, variables are expressed as deviationsfrom steady state values.

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Taylor series approximation

Consider an arbitrary univariate function f : R→ R.Taylor’s theorem says that if f is k-times differentiable then f (x) can beapproximated by a power series around a particular point x∗:

f (x) = f (x∗) +f ′(x∗)

1!(x− x∗) +

f ′′(x∗)2!

(x− x∗)2 +f (3)(x∗)

3!(x− x∗)3 + . . .

+f (k)(x∗)

k!(x− x∗)k + hk(x)(x− x∗)k

f ′(x∗) denotes the first derivative of f with respect to x, evaluated at x∗;f ′′(x∗) is the second derivative at x∗;f (n)(x∗) is n-th derivative at x∗

limx→x∗ hk(x) = 0.The order of approximation tells how many powers are used toapproximate

1st order (linear) approximationf (x) ' f (x∗) + f ′(x∗)(x− x∗)

2nd order (quadratic) approximation

f (x) ' f (x∗) + f ′(x∗)(x− x∗) +f ′′(x∗)

2(x− x∗)2

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Taylor series approximation

Taylor’s theorem holds for multivariate functions.

Consider a function f (x) = f (x1, x2), x = [x1 x2]′ k−times

differentiable at x∗.

Taylor series around x∗ = (x∗1 , x∗2) can be written as

f (x1, x2) = f (x∗) +∂f∂x1

(x∗)(x1 − x∗1 ) +∂f∂x2

(x∗)(x2 − x∗2 )+

+12∂2f∂x2

1(x∗)(x1 − x∗1 )

2 +12∂2f∂x2

2(x∗)(x2 − x∗2 )

2

+∂2f

∂x2∂x1(x∗)(x1 − x∗1 )(x2 − x∗2 ) + . . .

∂f∂xi

denotes the partial derivative of f with respect to xi, evaluated at x∗;

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Log-linearization

Sometimes it is more convenient to first take log of these equations andlinearize them.Consider an equation f (xt) = g(xt). Taking logs of both sides yields

ln f (xt) = ln g(xt)

Now, if we take first-order Taylor expansion around xt = x∗ we get

ln f (xt) ≈ ln f (x∗) +f ′(x∗)f (x∗)

(xt − x∗)

ln g(xt) ≈ ln g(x∗) +g′(x∗)g(x∗)

(xt − x∗)

as [ln f (x)]′ =f ′(x)f (x)

Substituting in

ln f (x∗) +f ′(x∗)f (x∗)

(xt − x∗) = ln g(x∗) +g′(x∗)g(x∗)

(xt − x∗)

and using ln f (x∗) = ln g(x∗) and rearranging terms we getf ′(x∗)x∗

f (x∗)xt − x∗

x∗=

g′(x∗)x∗

g(x∗)xt − x∗

x∗

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Log-linearization

Define xt =xt−x∗

x∗ , or the percentage deviation of xt from x∗ to get

f ′(x∗)x∗

f (x∗)xt =

g′(x∗)x∗

g(x∗)xt

Note that f ′(x∗)x∗

f (x∗) is the elasticity of f with respect to xt at x∗.

Since ln(1 + ε) ≈ ε for small values of ε, we can write xt as

xt ≈ ln (xt + 1) = ln(

xt − x∗

x∗+ 1

)= ln

(xt

x∗

)= ln xt − ln x∗

and, as x = eln x = ln ex,

xt = x∗xt

x∗= x∗eln( xt

x∗ ) ≈ x∗ext .

Lastly, note that the Taylor expansions of exponential function and theproduct of two variables around xt = yt = 0 are

exp(αxt) ≈ exp(αxt)|xt=0 + α exp(αxt)|xt=0xt = 1 + αxt

xt yt ≈ 0 + yt |yt=0(xt − 0) + xt |xt=0(yt − 0) = 0

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Log-linearization

Using these observations we obtain several useful expressions

xt = x∗ext ≈ x∗ · (1 + xt)

xtyt = x∗ext y∗eyt ≈ x∗y∗ · (1 + xt + yt)

xat ≈ (x∗)a · (1 + axt)

xat yb

t ≈ (x∗)a(y∗)b · (1 + axt + byt)

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Back to model - summary

Recall we have a system of 8 equations with 8 endogenous variables (c,r, l, w, k, i, y, z):

c−σt = βEt[c−σt+1(1 + rt+1 − δ)] (1)

wt =lϕt

c−σt(2)

kt = (1− δ)kt−1 + it (3)yt = ct + it (4)

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Back to model - summary

yt = ztkαt−1l1−αt (5)

wt = (1− α)yt

lt(6)

rt = αyt

kt−1(7)

zt = exp(εt)zρt−1 (8)

We computed steady-states last time.Now we will linearize each equation.

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Log-linearize labor - consumption choice

Equation (2)lϕt

c−σt= wt

In the steady state:

(lss)ϕ

(csst )−σ = wss

Let’s log-linearize:

(lss)ϕ

(css)−σ (1 + ϕlt + σct) = wss(1 + wt)

Divide by steady state:

ϕlt + σct = wt

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Log-linearized Euler equation

Equation (1):c−σt = βEt[c−σt+1(1 + rt+1 − δ)]

In the steady state:

1 = β(1 + rss − δ)

When linearized:

(css)−σ

(1− σct) = β (css)−σ Et [(1− σct+1)(1− δ + rss(1 + rt+1))]

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Log-linearized Euler equation

Substitute for rss:

1− σct = βEt

[(1− σct+1)(1− δ + (

1β− 1 + δ)(1 + rt+1))

]

1− σct = Et [(1− σct+1)(β − βδ + (1− β + βδ)(1 + rt+1))]

1− σct = Et [(1− σct+1)(1 + (1− β(1− δ))rt+1)]

Multiply and drop higher order terms:

1− σct = 1− σEtct+1 + (1− β(1− δ))Et rt+1

Rearrange terms:

σ(Etct+1 − ct) = (1− β(1− δ))Et rt+1

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Log-linearized market clearing condition

Equation (4):ct + it = yt

In the steady state:css + iss = yss

Linearize:css(1 + ct) + iss(1 + it) = yss(1 + yt)

Subtract steady state equation:

(1− iss

yss )ct +iss

yss it = yt

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Log-linearized firm’s equilibrium conditions

For labor (equation 6):(1− α)yt

lt= wt

Linearized:yt − lt = wt

For capital (equation 7):α

yt

kt−1= rt

Linearized:yt − kt−1 = rt

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Log-linearized shock process

Equation (8):zt = exp(εt)z

ρt−1

After linearization:zt = ρzt−1 + εt

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Steady state values

Our equations contain steady state ratio iss

yss .These are determined by our parameters

From the Euler equation:

rss = β−1 − (1− δ)

and from the equilibrium condition for capital:

rsskss = αyss

kss

yss =α

rss =α

β−1 − (1− δ)

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Steady state values cont’d

From the capital accumulation equation:

δkss = iss

thusiss

yss = δkss

yss =αδ

β−1 − (1− δ)

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Log-linearized system

We now have a system of 8 linear (difference) equations and 8 variablesHave to solve it

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Linearization Solution Blanchard and Kahn’s (1980) Method Sims’ (2001) Method


1 Linearization

2 Solution

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Solving linear DSGE models

Solving a DSGE model means changing the system of forward-lookingdifference equations that we have ...... into a VAR , a system of backward-looking difference equationsThere are several techniques for solving such systems:e.g. Blanchard and Kahn (1980), Sims (2001), Christiano (2002), Uhlig(2002), Evans and Honkapohja (2001),...(Dynare will solve it for you)

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Blanchard & Kahn condition

One very important issue is the existence and the stability conditionWrite the system in state space form:

A1

[Xt+1

EtPt+1

]= A0

[Xt

Pt

]+ γZt+1

where:Xt: vector n× 1 of state variables (backward-looking)Pt: vector m× 1 of jumpers (forward-looking)Zt: vector k × 1 of shocks (with mean equal to 0 every period)A1, A0: (n + m)× (n + m) matricesγ: (n + m)× k matrix

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Blanchard & Kahn condition cont’d

Assume that A1 is invertible. Then:[Xt+1

EtPt+1

]= A

[Xt

Pt

]+ A−1

1 γZt+1

where A = A−11 A0

The Blanchard-Kahn condition says that for the model to have a uniquesolution, the number of eigenvalues of A lying outside the unit circle(unstable roots) must equal the number of forward looking variables(jumpers)

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1 Linearization

2 SolutionBlanchard and Kahn’s (1980) MethodSims’ (2001) Method

RBC

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