2012-2013 – Master 2 – Macro I

Lecture notes #12 : Solving Dynamic RationalExpectations Models

Franck Portier(based on Gilles Saint-Paul lecture notes)

franck.portier@TSE-fr.eu

Toulouse School of Economics

Version 1.125/11/2012

Changes from version 1.0 are in red

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Disclaimer

These are the slides I am using in class. They are notself-contained, do not always constitute original material and docontain some “cut and paste” pieces from various sources that Iam not always explicitly referring to (not on purpose but because ittakes time). Therefore, they are not intended to be used outside ofthe course or to be distributed. Thank you for signalling me typosor mistakes at franck.portier@TSE-fr.eu.

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Introduction

I Economic agents have to make a decision based on theirforecast of some future variable.

I In the most general case, the entire probability distribution ofthat variable will affect their optimal choice.

I The rational expectations hypothesis states that people takeinto account the actual distribution of the relevant variablethey want to forecast.

I If expectations are rational, we have to solve simultaneouslyfor the equilibrium distribution and for the rest of the model

I We restrict to linear models in which the distribution ofendogenous variables of interest only matters through its firstmoment, i.e. its mathematical expectation.

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1. A simple exampleModel

I Agricultural market where producers have to decide how muchto produce prior to knowing the equilibrium price

ydt = a− bpt + εdt

y st = c + dpat + εst ,

y st = ydt .

I Here, p =price,εd=demand shock, εs=supplyshock,y s =supply, yd =demand,pa = anticipated price.

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1. A simple exampleRational expectations

I The Rational Expectations Hypothesis assumes that

pat = E (pt | It),

where E = mathematical expectation operator and It =information set of the price-setters when they set y st .

I Typically, It ={past values of all the variables, the model, how to solve it}

I If farmers set y st at date t − 1, :

pat = Et−1pt .

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1. A simple exampleSolving the model

I General principle :I take expectations in all the equations and then solve for the

expected variables,I then substitute this in all the equations where expectations

enter,I this allows to solve for the actual equilibrium variables.

I Let us try this here.

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1. A simple exampleSolving the model (continued)

ydt = a− bpt + εdt

y st = c + dpat + εst ,

y st = ydt .

I The model can be collapsed to

a− bpt + εdt = c + dEt−1pt + εst . (1)

I Take expectations :

a− bEt−1pt + Et−1εdt = c + dEt−1pt + Et−1εst .

I Thus

Et−1pt =a− c + Et−1(εdt − εst)

b + d.

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1. A simple exampleSolving the model (continued)

I substitute this into (1) to solve for pt :

pt =1

b(εd − εs)− d

b(b + d)Et−1(εd − εs) +

a− c

b + d. (2)

I This is the solution.

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1. A simple exampleThe method of undetermined coefficients

I If we know the distribution of those shocks, we can useanother method called the method of undeterminedcoefficients

I This is akin to the notion of Markov perfect equilibrium ingame theory.

I Method :

I Identify the relevant state spaceI Look for a solution as a (linear) function of this state vectorI Given such a function, I can compute all the expectations I

needI Subsitute this function and the appropriate formula for

expectations in all places where it is neededI Identify the coefficients on each state variableI This gives us equations in the coefficients that allow us to

compute them

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1. A simple exampleThe method of undetermined coefficients (continued)

I Assume Et−1εdt = Et−1εst = 0.

I The relevant state space is (εd , εs).

I Look for a solution of the form

pt = αεdt + βεst + γ.

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1. A simple exampleThe method of undetermined coefficients (continued)

I Under the assumed solution, we have

Et−1pt = γ.

I Substituting into (1) we get

a− b (αεdt + βεst + γ) + εdt = c + dγ + εst .

I This must hold for all realizations of εd and εs and thereforewe must have

a− bγ = c + dγ

−αb + 1 = 0

−βb = 1.

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1. A simple exampleThe method of undetermined coefficients (continued)

I Therefore

γ =a− c

b + dα = 1/b

β = −1/b.

I Hence the solution

pt =1

b(εd − εs) +

a− c

b + d,

I which is equivalent to (2) if Et−1εdt = Et−1εst = 0.

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2. A dynamic modelModel

I Suppose now the supply curve is

y st = c + dEtpt+1 + εst

I Eliminating y between supply and demand :

pt =a− c

b+−db

Etpt+1 +εdt − εst

b

I and assuming for simplicity that the constant terms are zero,we end up with

pt = ρEtpt+1 + ηt , (3)

where ηt is a shock.

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2. A dynamic modelSolution by forward integration

I (3) implies pt+i = ρEtpt+i+1 + ηt ∀iI Taking expectations

Etpt+i = ρEtpt+i+1 + Etηt+i

I Substitute forward (forward integration) :

pt = ρkEtpt+k +k−1∑j=0

ρjEtηt+j (4)

I Assume stationary shocks and impose Etpt+k is bounded ∀k.I (stochastic equivalent of selecting the non-explosive

saddle-path as the unique solution.)I Then if | ρ |< 1, then limk→∞ ρ

kEtpt+k = 0, hence

pt =∞∑j=0

ρjEtηt+j .

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2. A dynamic modelSolution by forward integration (continued)

pt =∞∑j=0

ρjEtηt+j .

I This is the unique stationary solution to the rationalexpectations equation (3).

I Suppose | ρ |> 1, then the solution cannot be unique.I For any solution pt , pt + xt is also a solution under some

conditions on xI xt is a random variable (sunspot) satisfying

xt+1 =1

ρxt + ut+1, (5)

with u any shock such that Etut+1 = 0.I Furthermore, we can construct at least one non explosive

solution by picking pt+1 = 1ρ(pt − ηt).

I Hence in this case we have a continuum of equilibria.15 / 29

2. A dynamic modelUndetermined coefficients

I We can apply the method of undetermined coefficients to thisproblem if we know the distribution of η.

I Assume η follows an AR1 :

ηt = αηt−1 + ωt ,

with ωt iid and with zero mean.I The only relevant state variables are ηt−1 and ωt so we look

for a solution of the form

pt = ληt−1 + βωt .

I Then

Etpt+1 = Et(ληt + βωt+1)

= ληt

= λ(αηt−1 + ωt).

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2. A dynamic modelUndetermined coefficients (continued)

I We can then rewrite (3) as

ληt−1 + βωt = ρλ(αηt−1 + ωt) + αηt−1 + ωt .

I Identifying, we get

λ = ρλα + α

β = ρλ+ 1

I Hence

λ =α

1− ρα;

β =1

1− ρα.

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2. A dynamic modelUndetermined coefficients (continued)

I We can check that if | ρ |< 1, this is indeed correct :

∞∑j=0

ρjEtηt+j . =∞∑j=0

ρjαjηt

=1

1− ρα(αηt−1 + ωt)

= ληt−1 + βωt

I If | ρ |> 1, this is still a solution, and it is the unique ”linearMarkov” solution, i.e. the only one which is a linearcombination of the relevant state space.

I But many other solutions can be constructed by adding asunspot x satisfying (5).

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3. A general modelModel

I General linear model (Blanchard-Kahn)(Kt

Xt

)= A

(Kt−1EtXt+1

)+ Ut , (6)

I Kt is an m × 1 vector of state (=predetermined) variables,meaning that Kt−1 is fixed at the beginning of period t,

I Xt is an n × 1 vector of non-predetermined variables, i.e.expected future values of those variables, rather than laggedones, affect current outcomes.

I A is an (m + n)× (m + n) matrix

I U an (m + n)× 1 vector of shocks.

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3. A general modelModel transformation

II Let

A =

(A00 A01

A10 A11

), Ut =

(U0t

U1t

),

I then

Kt = A00Kt−1 + A01EtXt+1 + U0t

Xt = A10Kt−1 + A11EtXt+1 + U1t .

I We can solve for EtXt+1 and write the whole system withbackward-looking dynamics :(Kt

EtXt+1

)=

(A00 − A01A

−111 A10 A01A

−111

−A−111 A10 A−111

)(Kt−1Xt

)+

(U0t − A01A

−111 U1t

−A−111 U1t

)= M

(Kt−1Xt

)+ Vt . (7)

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3. A general modelThe diagonal case

(Kt

EtXt+1

)= M

(Kt−1Xt

)+ Vt . (7)

I Let us start with the simple case where M is diagonal, say

M =

(D0 00 D1

), where D0 =

d1 0 ... 00 d2 ... ...... ... ... 00 ... 0 dm

and

D1 =

d ′1 0 ... 00 d ′2 ... ...... ... ... 00 ... 0 d ′n

and Vt =

(V0t

V1t

)

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3. A general modelThe diagonal case (continued)

II Then, iterating (7),

Kt+i = D i+10 Kt−1 +

i∑j=0

D i−j0 V0t+j .

I For Kt+i to remain bounded, we need

| di |< 1, i = 1, ...,m. (8)

I Similarly

EtXt+i = D i1Xt +

i−1∑j=0

D i−1−j1 EtV1t+j

= D i1(Xt +

i−1∑j=0

D−1−j1 EtV1t+j).

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3. A general modelThe diagonal case (continued)

I Consider first a k such that | d ′k |< 1. Let xkt (resp. vkt) bethe k-th term in Xt (resp. V1t).

I Then

Etxt+i = (d ′k)ixt +i−1∑j=0

(d ′k)i−1−jEtvk,t+j

⇐⇒ (d ′−1k )iEtxt+i −i−1∑j=0

(d ′−1k )1+jEtvk,t+j = xt .

I This is similar to (4) and since | d ′−1k |> 1 there is no uniqueinitial value of xt yielding a non-explosive path for subsequentexpectations.

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3. A general modelThe diagonal case (continued)

I Second, assume conversely that

| d ′k |> 1 for k = 1, ..., n. (9)

I Then all the terms in D i1 go to infinity as i →∞.

I Thus it must be that limi→+∞ Xt +∑i−1

j=0D−1−j1 EtV1t+j = 0.

Therefore,

Xt = −∞∑j=0

(D−11

)1+jEtV1t+j .

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3. A general modelThe diagonal case (continued)

| di |< 1, i = 1, ...,m. (8)

| d ′k |> 1 for k = 1, ..., n. (9)

I To summarizeI To have a unique non-explosive solution, (8) and (9) must holdI If (8) is violated, all solutions are explosiveI If (9) is violated, there is a continuum of non-explosive

solutions.

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3. A general modelThe non diagonal case

I Assume now that the matrix M is not diagonal.

I The idea is to transform the model into a diagonal one.

I Generically, M is diagonalizable in the complex plane C.

I Thus there exists P invertible such that

M = P−1DP,

with D diagonal :

D =

d0 0 ... ... ... ... 00 ... ... ... ... ... ...... ... dm 0 ... ... ...... ... 0 dm+1 0 ... ...... ... ... 0 ... ... ...... ... ... ... ... ... 00 ... ... ... ... 0 dm+n

=

(D0 00 D1

)

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3. A general modelThe non diagonal case (continued)

I The diagonal coefficients are the eigenvalues of M.

I Eigenvalues are ranked by ascending module :| d0 |≤| d1 |≤ ... ≤| dm |≤| d ′1 |≤| d ′2 |≤ ... ≤| d ′n | .

I (7) rewrite

P

(Kt

EtXt+1

)= D

(P

(Kt

Xt

))+ PVt .

I Let Zt = P

(Kt−1Xt

);

I We are back to the diagonal case with this change of variable.

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4. How general is the general model ?More lags for K

I Suppose that we have terms in Kt−k , k > 1.

I The solution is to add the variables (Kt−1, ...,Kt−k+1) to thestate vector.

I Suppose we have this equation Kt = aKt−1 + bKt−kI Let us assume to save on notation that K is a scalar.

I Then the vector

Kt

Kt−1...

Kt−k+1

satisfies

Kt

Kt−1...

Kt−k+1

=

a 0 ... b1 0 ... ...0 ... ... ...0 0 1 0

Kt−1Kt−2...

Kt−k

,

which only involves the first lag of this vector.

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4. How general is the general model ?More lags for X

I Suppose we have terms in EtXt+k , k > 1.I Let’s add EtXt+1, ...,EtXt+k−1 to the jump variablesI Suppose that we have

Xt = aEtXt+1 + bEtXt+k .

I Then the vector Ψt = (Xt ,EtXt+1, ...,EtXt+k−1)′ satisfies

Ψt =

Xt

EtXt+1

...EtXt+k−1

=

a 0 ... b1 0 ... ...0 ... ... ...0 0 1 0

EtXt+1

EtXt+2

...EtXt+k

=

a 0 ... b1 0 ... ...0 ... ... ...0 0 1 0

EtΨt+1.

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