state space models and filteringstate space models and filtering jes´us fern´andez-villaverde...
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State Space Models and Filtering
Jesus Fernandez-VillaverdeUniversity of Pennsylvania
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State Space Form
• What is a state space representation?
• States versus observables.
• Why is it useful?
• Relation with filtering.
• Relation with optimal control.
• Linear versus nonlinear, Gaussian versus nongaussian.2
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State Space Representation
• Let the following system:
— Transition equation
xt+1 = Fxt +Gωt+1, ωt+1 ∼ N (0, Q)
— Measurement equation
zt = H0xt + υt, υt ∼ N (0, R)
— where xt are the states and zt are the observables.
• Assume we want to write the likelihood function of zT = ztTt=1.3
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The State Space Representation is Not Unique
• Take the previous state space representation.
• Let B be a non-singular squared matrix conforming with F .
• Then, if x∗t = Bxt, F ∗ = BFB−1, G∗ = BG, and H∗ =¡H0B
¢0, wecan write a new, equivalent, representation:
— Transition equation
x∗t+1 = F ∗x∗t +G∗ωt+1, ωt+1 ∼ N (0, Q)
— Measurement equation
zt = H∗0x∗t + υt, υt ∼ N (0, R)
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Example I
• Assume the following AR(2) process:zt = ρ1zt−1 + ρ2zt−2 + υt, υt ∼ N
³0,σ2υ
´
• Model is not apparently not Markovian.
• Can we write this model in different state space forms?
• Yes!
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State Space Representation I
• Transition equation:
xt =
"ρ1 1ρ2 0
#xt−1 +
"10
#υt
where xt =hyt ρ2yt−1
i0
• Measurement equation:zt =
h1 0
ixt
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State Space Representation II
• Transition equation:
xt =
"ρ1 ρ21 0
#xt−1 +
"10
#υt
where xt =hyt yt−1
i0
• Measurement equation:zt =
h1 0
ixt
• Try B =
"1 00 ρ2
#on the second system to get the first system.
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Example II
• Assume the following MA(1) process:zt = υt + θυt−1, υt ∼ N
³0,σ2υ
´, and Eυtυs = 0 for s 6= t.
• Again, we have a more conmplicated structure than a simple Markov-ian process.
• However, it will again be straightforward to write a state space repre-sentation.
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State Space Representation I
• Transition equation:
xt =
"0 10 0
#xt−1 +
"1θ
#υt
where xt =hyt θυt
i0
• Measurement equation:zt =
h1 0
ixt
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State Space Representation II
• Transition equation:
xt = υt−1
• where xt = [υt−1]0
• Measurement equation:zt = θxt + υt
• Again both representations are equivalent!10
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Example III
• Assume the following random walk plus drift process:
zt = zt−1 + β + υt, υt ∼ N³0,σ2υ
´
• This is even more interesting.
• We have a unit root.
• We have a constant parameter (the drift).
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State Space Representation
• Transition equation:
xt =
"1 10 1
#xt−1 +
"10
#υt
where xt =hyt β
i0
• Measurement equation:zt =
h1 0
ixt
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Some Conditions on the State Space Representation
• We only consider Stable Systems.
• A system is stable if for any initial state x0, the vector of states, xt,
converges to some unique x∗.
• A necessary and sufficient condition for the system to be stable is that:|λi (F )| < 1
for all i, where λi (F ) stands for eigenvalue of F .
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Introducing the Kalman Filter
• Developed by Kalman and Bucy.
• Wide application in science.
• Basic idea.
• Prediction, smoothing, and control.
• Why the name “filter”?
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Some Definitions
• Let xt|t−1 = E³xt|zt−1
´be the best linear predictor of xt given the
history of observables until t− 1, i.e. zt−1.
• Let zt|t−1 = E³zt|zt−1
´= H0xt|t−1 be the best linear predictor of zt
given the history of observables until t− 1, i.e. zt−1.
• Let xt|t = E³xt|zt
´be the best linear predictor of xt given the history
of observables until t, i.e. zt.
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What is the Kalman Filter trying to do?
• Let assume we have xt|t−1 and zt|t−1.
• We observe a new zt.
• We need to obtain xt|t.
• Note that xt+1|t = Fxt|t and zt+1|t = H0xt+1|t, so we can go backto the first step and wait for zt+1.
• Therefore, the key question is how to obtain xt|t from xt|t−1 and zt.16
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A Minimization Approach to the Kalman Filter I
• Assume we use the following equation to get xt|t from zt and xt|t−1:
xt|t = xt|t−1 +Kt³zt − zt|t−1
´= xt|t−1 +Kt
³zt −H0xt|t−1
´
• This formula will have some probabilistic justification (to follow).
• What is Kt?
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A Minimization Approach to the Kalman Filter II
• Kt is called the Kalman filter gain and it measures how much we
update xt|t−1 as a function in our error in predicting zt.
• The question is how to find the optimal Kt.
• The Kalman filter is about how to build Kt such that optimally updatext|t from xt|t−1 and zt.
• How do we find the optimal Kt?
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Some Additional Definitions
• Let Σt|t−1 ≡ Eµ³xt − xt|t−1
´ ³xt − xt|t−1
´0 |zt−1¶ be the predictingerror variance covariance matrix of xt given the history of observables
until t− 1, i.e. zt−1.
• Let Ωt|t−1 ≡ E
µ³zt − zt|t−1
´ ³zt − zt|t−1
´0 |zt−1¶ be the predictingerror variance covariance matrix of zt given the history of observables
until t− 1, i.e. zt−1.
• Let Σt|t ≡ Eµ³xt − xt|t
´ ³xt − xt|t
´0 |zt¶ be the predicting error vari-ance covariance matrix of xt given the history of observables until t−1,i.e. zt.
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Finding the optimal Kt
• We want Kt such that minΣt|t.
• It can be shown that, if that is the case:Kt = Σt|t−1H
³H0Σt|t−1H +R
´−1
• with the optimal update of xt|t given zt and xt|t−1 being:
xt|t = xt|t−1 +Kt³zt −H0xt|t−1
´
• We will provide some intuition later.20
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Example I
Assume the following model in State Space form:
• Transition equationxt = µ+ υt, υt ∼ N
³0,σ2υ
´
• Measurement equationzt = xt + ξt, ξt ∼ N
³0,σ2ξ
´
• Let σ2ξ = qσ2υ.
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Example II
• Then, if Σ1|0 = σ2υ, what it means that x1 was drawn from the ergodic
distribution of xt.
• We have:K1 = σ2υ
1
1 + q∝ 1
1 + q.
• Therefore, the bigger σ2ξ relative to σ2υ (the bigger q) the lower K1and the less we trust z1.
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The Kalman Filter Algorithm I
Given Σt|t−1, zt, and xt|t−1, we can now set the Kalman filter algorithm.
Let Σt|t−1, then we compute:
Ωt|t−1 ≡ Eµ³zt − zt|t−1
´ ³zt − zt|t−1
´0 |zt−1¶
= E
H0
³xt − xt|t−1
´ ³xt − xt|t−1
´0H+
υt³xt − xt|t−1
´0H +H0
³xt − xt|t−1
´υ0t+
υtυ0t|zt−1
= H0Σt|t−1H +R
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The Kalman Filter Algorithm II
Let Σt|t−1, then we compute:
E
µ³zt − zt|t−1
´ ³xt − xt|t−1
´0 |zt−1¶ =
EµH0
³xt − xt|t−1
´ ³xt − xt|t−1
´0+ υt
³xt − xt|t−1
´0 |zt−1¶ = H0Σt|t−1
Let Σt|t−1, then we compute:
Kt = Σt|t−1H³H0Σt|t−1H +R
´−1Let Σt|t−1, xt|t−1, Kt, and zt then we compute:
xt|t = xt|t−1 +Kt³zt −H0xt|t−1
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The Kalman Filter Algorithm III
Let Σt|t−1, xt|t−1, Kt, and zt, then we compute:
Σt|t ≡ Eµ³xt − xt|t
´ ³xt − xt|t
´0 |zt¶ =
E
³xt − xt|t−1
´ ³xt − xt|t−1
´0−³xt − xt|t−1
´ ³zt −H0xt|t−1
´0K0t−
Kt³zt −H0xt|t−1
´ ³xt − xt|t−1
´0+
Kt³zt −H0xt|t−1
´ ³zt −H0xt|t−1
´0K0t|zt
= Σt|t−1 −KtH0Σt|t−1
where, you have to notice that xt−xt|t = xt−xt|t−1−Kt³zt −H0xt|t−1
´.
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The Kalman Filter Algorithm IV
Let Σt|t−1, xt|t−1, Kt, and zt, then we compute:
Σt+1|t = FΣt|tF 0 +GQG0
Let xt|t, then we compute:
1. xt+1|t = Fxt|t
2. zt+1|t = H0xt+1|t
Therefore, from xt|t−1, Σt|t−1, and zt we compute xt|t and Σt|t.26
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The Kalman Filter Algorithm V
We also compute zt|t−1 and Ωt|t−1.
Why?
To calculate the likelihood function of zT = ztTt=1 (to follow).
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The Kalman Filter Algorithm: A Review
We start with xt|t−1 and Σt|t−1.
The, we observe zt and:
• Ωt|t−1 = H0Σt|t−1H +R
• zt|t−1 = H0xt|t−1
• Kt = Σt|t−1H³H0Σt|t−1H +R
´−1• Σt|t = Σt|t−1 −KtH0Σt|t−1
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• xt|t = xt|t−1 +Kt³zt −H0xt|t−1
´
• Σt+1|t = FΣt|tF 0 +GQG0
• xt+1|t = Fxt|t
We finish with xt+1|t and Σt+1|t.
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Some Intuition about the optimal Kt
• Remember: Kt = Σt|t−1H³H0Σt|t−1H +R
´−1
• Notice that we can rewrite Kt in the following way:Kt = Σt|t−1HΩ−1
t|t−1
• If we did a big mistake forecasting xt|t−1 using past information (Σt|t−1large) we give a lot of weight to the new information (Kt large).
• If the new information is noise (R large) we give a lot of weight to theold prediction (Kt small).
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A Probabilistic Approach to the Kalman Filter
• Assume:
Z|w = [X 0|w Y 0|w]0 ∼ NÃ"
x∗y∗
#,
"Σxx ΣxyΣyx Σyy
#!
• then:X|y,w ∼ N
³x∗ +ΣxyΣ
−1yy (y − y∗) ,Σxx −ΣxyΣ
−1yy Σyx
´
• Also xt|t−1 ≡ E³xt|zt−1
´and:
Σt|t−1 ≡ Eµ³xt − xt|t−1
´ ³xt − xt|t−1
´0 |zt−1¶31
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Some Derivations I
If zt|zt−1 is the random variable zt (observable) conditional on zt−1, then:
• Let zt|t−1 ≡ E³zt|zt−1
´= E
³H0xt + υt|zt−1
´= H0xt|t−1
• LetΩt|t−1 ≡ E
µ³zt − zt|t−1
´ ³zt − zt|t−1
´0 |zt−1¶ =
E
H0
³xt − xt|t−1
´ ³xt − xt|t−1
´0H+
υt³xt − xt|t−1
´0H+
H0³xt − xt|t−1
´υ0t+
υtυ0t|zt−1
= H0Σt|t−1H +R
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Some Derivations II
Finally, let
Eµ³zt − zt|t−1
´ ³xt − xt|t−1
´0 |zt−1¶ =E
µH0
³xt − xt|t−1
´ ³xt − xt|t−1
´0+ υt
³xt − xt|t−1
´0 |zt−1¶ == H0Σt|t−1
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The Kalman Filter First Iteration I
• Assume we know x1|0 and Σ1|0, thenÃx1z1|z0!N
Ã"x1|0H0x1|0
#,
"Σ1|0 Σ1|0HH0Σ1|0 H0Σ1|0H +R
#!
• Remember that:X|y,w ∼ N
³x∗ +ΣxyΣ
−1yy (y − y∗) ,Σxx −ΣxyΣ
−1yy Σyx
´
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The Kalman Filter First Iteration II
Then, we can write:
x1|z1, z0 = x1|z1 ∼ N³x1|1,Σ1|1
´where
x1|1 = x1|0 +Σ1|0H³H0Σ1|0H +R
´−1 ³z1 −H0x1|0
´and
Σ1|1 = Σ1|0 −Σ1|0H³H0Σ1|0H +R
´−1H0Σ1|0
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• Therefore, we have that:
— z1|0 = H0x1|0
— Ω1|0 = H0Σ1|0H +R
— x1|1 = x1|0 +Σ1|0H³H0Σ1|0H +R
´−1 ³z1 −H0x1|0
´— Σ1|1 = Σ1|0 −Σ1|0H
³H0Σ1|0H +R
´−1H0Σ1|0
• Also, since x2|1 = Fx1|1 +Gω2|1 and z2|1 = H0x2|1 + υ2|1:
— x2|1 = Fx1|1
— Σ2|1 = FΣ1|1F 0 +GQG0
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The Kalman Filter th Iteration I
• Assume we know xt|t−1 and Σt|t−1, thenÃxtzt|zt−1
!N
Ã"xt|t−1H0xt|t−1
#,
"Σt|t−1 Σt|t−1HH0Σt|t−1 H0Σt|t−1H +R
#!
• Remember that:X|y,w ∼ N
³x∗ +ΣxyΣ
−1yy (y − y∗) ,Σxx −ΣxyΣ
−1yy Σyx
´
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The Kalman Filter th Iteration II
Then, we can write:
xt|zt, zt−1 = xt|zt ∼ N³xt|t,Σt|t
´where
xt|t = xt|t−1 +Σt|t−1H³H0Σt|t−1H +R
´−1 ³zt −H0xt|t−1
´and
Σt|t = Σt|t−1 −Σt|t−1H³H0Σt|t−1H +R
´−1H0Σt|t−1
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The Kalman Filter Algorithm
Given xt|t−1, Σt|t−1 and observation zt
• Ωt|t−1 = H0Σt|t−1H +R
• zt|t−1 = H0xt|t−1
• Σt|t = Σt|t−1 −Σt|t−1H³H0Σt|t−1H +R
´−1H0Σ
• xt|t = xt|t−1 + Σt|t−1H³H0Σt|t−1H +R
´−1 ³zt −H0xt|t−1
´039
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• Σt+1|t = FΣt|tF 0 +GQGt|t−1
• xt+1|t = Fxt|t−1
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Putting the Minimization and the Probabilistic Approaches Together
• From the Minimization Approach we know that:
xt|t = xt|t−1 +Kt³zt −H0xt|t−1
´
• From the Probability Approach we know that:
xt|t = xt|t−1 + Σt|t−1H³H0Σt|t−1H +R
´−1 ³zt −H0xt|t−1
´
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• But since:Kt = Σt|t−1H
³H0Σt|t−1H +R
´−1
• We can also write in the probabilistic approach:
xt|t = xt|t−1 +Σt|t−1H³H0Σt|t−1H +R
´−1 ³zt −H0xt|t−1
´=
= xt|t−1 +Kt³zt −H0xt|t−1
´
• Therefore, both approaches are equivalent.
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Writing the Likelihood Function
We want to write the likelihood function of zT = ztTt=1:log `
³zT |F,G,H,Q,R
´=
TXt=1
log `³zt|zt−1F,G,H,Q,R
´=
−TXt=1
N2log 2π +
1
2log
¯Ωt|t−1
¯+1
2
TXt=1
v0tΩ−1t|t−1vt
vt = zt − zt|t−1 = zt −H0xt|t−1
Ωt|t−1 = H0tΣt|t−1Ht +R
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Initial conditions for the Kalman Filter
• An important step in the Kalman Fitler is to set the initial conditions.
• Initial conditions:
1. x1|0
2. Σ1|0
• Where do they come from?
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Since we only consider stable system, the standard approach is to set:
• x1|0 = x∗
• Σ1|0 = Σ∗
where x solves
x∗ = Fx∗
Σ∗ = FΣ∗F 0 +GQG0
How do we find Σ∗?
Σ∗ = [I − F ⊗ F ]−1 vec(GQG0)45
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Initial conditions for the Kalman Filter II
Under the following conditions:
1. The system is stable, i.e. all eigenvalues of F are strictly less than one
in absolute value.
2. GQG0 and R are p.s.d. symmetric
3. Σ1|0 is p.s.d. symmetric
Then Σt+1|t→ Σ∗.
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Remarks
1. There are more general theorems than the one just described.
2. Those theorems are based on non-stable systems.
3. Since we are going to work with stable system the former theorem is
enough.
4. Last theorem gives us a way to find Σ as Σt+1|t→ Σ for any Σ1|0 westart with.
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The Kalman Filter and DSGE models
• Basic Real Business Cycle model
maxE0
∞Xt=0
βt ξ log ct + (1− ξ) log (1− lt)
ct + kt+1 = kαt (eztlt)
1−α + (1− δ) k
zt = ρzt−1 + εt, εt ∼ N (0,σ)
• Parameters: γ = α,β, ρ, ξ, η,σ
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Equilibrium Conditions
1
ct= βEt
(1
ct+1
³1 + αezt+1kα−1t+1 l
1−αt+1 − η
´)
1− ξ
1− lt=
ξ
ct(1− α) eztkαt l
−αt
ct + kt+1 = eztkαt l
1−αt + (1− η) kt
zt = ρzt−1 + εt
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A Special Case
• We set, unrealistically but rather useful for our point, η = 1.
• In this case, the model has two important and useful features:
1. First, the income and the substitution effect from a productivity
shock to labor supply exactly cancel each other. Consequently, ltis constant and equal to:
lt = l =(1− α) ξ
(1− α) ξ + (1− ξ) (1− αβ)
2. Second, the policy function for capital is kt+1 = αβeztkαt l1−α.
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A Special Case II
• The definition of kt+1 implies that ct = (1− αβ) eztkαt l1−α.
• Let us try if the Euler Equation holds:1
ct= βEt
(1
ct+1
³αezt+1kα−1t+1 l
1−αt+1
´)1
(1− αβ) eztkαt l1−α = βEt
(1
(1− αβ) ezt+1kαt+1l1−α
³αezt+1kα−1t+1 l
1−αt+1
´)1
(1− αβ) eztkαt l1−α = βEt
(α
(1− αβ) kt+1
)αβ
(1− αβ)=
βα
(1− αβ)
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• Let us try if the Intratemporal condition holds1− ξ
1− l =ξ
(1− αβ) eztkαt l1−α (1− α) eztkαt l
−α
1− ξ
1− l =ξ
(1− αβ)
(1− α)
l
(1− αβ) (1− ξ) l = ξ (1− α) (1− l)
((1− αβ) (1− ξ) + (1− α) ξ) l = (1− α) ξ
• Finally, the budget constraint holds because of the definition of ct.
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Transition Equation
• Since this policy function is linear in logs, we have the transition equa-tion for the model: 1
log kt+1zt
= 1 0 0logαβλl1−α α ρ
0 0 ρ
1log ktzt−1
+ 011
²t.
• Note constant.
• Alternative formulations.
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Measurement Equation
• As observables, we assume log yt and log it subject to a linearly additivemeasurement error Vt =
³v1,t v2,t
´0.
• Let Vt ∼ N (0,Λ), where Λ is a diagonal matrix with σ21 and σ22, asdiagonal elements.
• Why measurement error? Stochastic singularity.
• Then:Ãlog ytlog it
!=
Ã− logαβλl1−α 1 0
0 1 0
! 1log kt+1zt
+ Ãv1,tv2,t
!.
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The Solution to the Model in State Space Form
xt =
1log ktzt−1
, zt =Ãlog ytlog it
!
F =
1 0 0logαβλl1−α α ρ
0 0 ρ
, G = 011
, Q = σ2
H0 =Ã− logαβλl1−α 1 0
0 1 0
!, R = Λ
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The Solution to the Model in State Space Form III
• Now, using zT , F,G,H,Q, and R as defined in the last slide...
• ...we can use the Ricatti equations to compute the likelihood functionof the model:
log `³zT |F,G,H,Q,R
´
• Croos-equations restrictions implied by equilibrium solution.
• With the likelihood, we can do inference!
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What do we Do if η 6= 1?
We have two options:
• First, we could linearize or log-linearize the model and apply theKalman filter.
• Second, we could compute the likelihood function of the model usinga non-linear filter (particle filter).
• Advantages and disadvantages.
• Fernandez-Villaverde, Rubio-Ramırez, and Santos (2005).57
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The Kalman Filter and linearized DSGE Models
• We linearize (or loglinerize) around the steady state.
• We assume that we have data on log output (log yt), log hours (log lt),and log investment (log ct) subject to a linearly additive measurement
error Vt =³v1,t v2,t v3,t
´0.
• We need to write the model in state space form. Remember thatbkt+1 = P bkt +Qzt
and blt = Rbkt + Szt58
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Writing the Likelihood Function I
• The transitions Equation:
1bkt+1zt+1
= 1 0 00 P Q0 0 ρ
1bktzt
+ 001
²t.
• The Measurement Equation requires some care.
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Writing the Likelihood Function II
• Notice that byt = zt + αbkt + (1− α)blt• Therefore, using blt = Rbkt + Szt
byt = zt + αbkt + (1− α)(Rbkt + Szt) =(α+ (1− α)R) bkt + (1 + (1− α)S) zt
• Also since bct = −α5blt + zt + αbkt and using again blt = Rbkt + Sztbct = zt + αbkt − α5(R
bkt + Szt) =(α− α5R)
bkt + (1− α5S) zt
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Writing the Likelihood Function III
Therefore the measurement equation is: log ytlog ltlog ct
= log y α+ (1− α)R 1 + (1− α)Slog l R Slog c α− α5R 1− α5S
1bktzt
+
v1,tv2,tv3,t
.
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The Likelihood Function of a General Dynamic Equilibrium Economy
• Transition equation:St = f (St−1,Wt; γ)
• Measurement equation:Yt = g (St, Vt; γ)
• Interpretation.
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Some Assumptions
1. We can partition Wt into two independent sequencesnW1,t
oandn
W2,t
o, s.t. Wt =
³W1,t,W2,t
´and dim
³W2,t
´+dim (Vt) ≥ dim (Yt).
2. We can always evaluate the conditional densities p³yt|Wt
1, yt−1, S0; γ
´.
Lubick and Schorfheide (2003).
3. The model assigns positive probability to the data.
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Our Goal: Likelihood Function
• Evaluate the likelihood function of the a sequence of realizations ofthe observable yT at a particular parameter value γ:
p³yT ; γ
´
• We factorize it as:
p³yT ; γ
´=
TYt=1
p³yt|yt−1; γ
´
=TYt=1
Z Zp³yt|Wt
1, yt−1, S0; γ
´p³Wt1, S0|yt−1; γ
´dWt
1dS0
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A Law of Large Numbers
If
(½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
)Tt=1
N i.i.d. draws fromnp³Wt1, S0|yt−1; γ
´oTt=1,
then:
p³yT ; γ
´'
TYt=1
1
N
NXi=1
pµyt|wt|t−1,i1 , yt−1, st|t−1,i0 ; γ
¶
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...thus
The problem of evaluating the likelihood is equivalent to the problem of
drawing from np³Wt1, S0|yt−1; γ
´oTt=1
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Introducing Particles
•nst−1,i0 , w
t−1,i1
oNi=1
N i.i.d. draws from p³Wt−11 , S0|yt−1; γ
´.
• Each st−1,i0 , wt−1,i1 is a particle and
nst−1,i0 , w
t−1,i1
oNi=1
a swarm of
particles.
•½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
N i.i.d. draws from p³Wt1, S0|yt−1; γ
´.
• Each st|t−1,i0 , wt|t−1,i1 is a proposed particle and
½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
a swarm of proposed particles.
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... and Weights
qit =pµyt|wt|t−1,i1 , yt−1, st|t−1,i0 ; γ
¶PNi=1 p
µyt|wt|t−1,i1 , yt−1, st|t−1,i0 ; γ
¶
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A Proposition
Letnesi0, ewi1oNi=1 be a draw with replacement from
½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
and probabilities qit. Thennesi0, ewi1oNi=1 is a draw from p
³Wt1, S0|yt; γ
´.
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Importance of the Proposition
1. It shows how a draw½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
from p³Wt1, S0|yt−1; γ
´can be used to draw
nst,i0 , w
t,i1
oNi=1
from p³Wt1, S0|yt; γ
´.
2. With a drawnst,i0 , w
t,i1
oNi=1
from p³Wt1, S0|yt; γ
´we can use p
³W1,t+1; γ
´to get a draw
½st+1|t,i0 , w
t+1|t,i1
¾Ni=1
and iterate the procedure.
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Sequential Monte Carlo I: Filtering
Step 0, Initialization: Set tà 1 and initialize p³Wt−11 , S0|yt−1; γ
´=
p (S0; γ).
Step 1, Prediction: Sample N values
½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
from the
density p³Wt1, S0|yt−1; γ
´= p
³W1,t; γ
´p³Wt−11 , S0|yt−1; γ
´.
Step 2, Weighting: Assign to each draw st|t−1,i0 , w
t|t−1,i1 the weight
qit.
Step 3, Sampling: Drawnst,i0 , w
t,i1
oNi=1
with rep. from
½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
with probabilitiesnqit
oNi=1. If t < T set t à t + 1 and go to
step 1. Otherwise stop.
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Sequential Monte Carlo II: Likelihood
Use
(½st|t−1,i0 , w
t|t−1,i1
¾Ni=1
)Tt=1
to compute:
p³yT ; γ
´'
TYt=1
1
N
NXi=1
p
µyt|wt|t−1,i1 , yt−1, st|t−1,i0 ; γ
¶
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A “Trivial” Application
How do we evaluate the likelihood function p³yT |α,β,σ
´of the nonlinear,
nonnormal process:
st = α+ βst−1
1 + st−1+ wt
yt = st + vt
where wt ∼ N (0,σ) and vt ∼ t (2) given some observables yT = ytTt=1and s0.
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1. Let s0,i0 = s0 for all i.
2. Generate N i.i.d. draws½s1|0,i0 , w1|0,i
¾Ni=1
from N (0,σ).
3. Evaluate pµy1|w1|0,i1 , y0, s
1|0,i0
¶= pt(2)
Ãy1 −
Ãα+ β
s1|0,i0
1+s1|0,i0
+ w1|0,i!!.
4. Evaluate the relative weights qi1 =
pt(2)
Ãy1−
Ãα+β
s1|0,i0
1+s1|0,i0
+w1|0,i!!
PNi=1 pt(2)
Ãy1−
Ãα+β
s1|0,i0
1+s1|0,i0
+w1|0,i!!.
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5. Resample with replacement N values of½s1|0,i0 , w1|0,i
¾Ni=1
with rela-
tive weights qi1. Call those sampled valuesns1,i0 , w
1,ioNi=1.
6. Go to step 1, and iterate 1-4 until the end of the sample.
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A Law of Large Numbers
A law of the large numbers delivers:
p³y1| y0,α,β,σ
´' 1
N
NXi=1
pµy1|w1|0,i1 , y0, s
1|0,i0
¶and consequently:
p³yT¯α,β,σ
´'
TYt=1
1
N
NXi=1
p
µyt|wt|t−1,i1 , yt−1, st|t−1,i0
¶
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Comparison with Alternative Schemes
• Deterministic algorithms: Extended Kalman Filter and derivations
(Jazwinski, 1973), Gaussian Sum approximations (Alspach and Soren-
son, 1972), grid-based filters (Bucy and Senne, 1974), Jacobian of the
transform (Miranda and Rui, 1997).
Tanizaki (1996).
• Simulation algorithms: Kitagawa (1987), Gordon, Salmond and Smith(1993), Mariano and Tanizaki (1995) and Geweke and Tanizaki (1999).
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A “Real” Application: the Stochastic Neoclassical Growth Model
• Standard model.
• Isn’t the model nearly linear?
• Yes, but:
1. Better to begin with something easy.
2. We will learn something nevertheless.
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The Model
• Representative agent with utility function U = E0P∞t=0 β
t
³cθt (1−lt)1−θ
´1−τ1−τ .
• One good produced according to yt = eztAkαt l1−αt with α ∈ (0, 1) .
• Productivity evolves zt = ρzt−1 + ²t, |ρ| < 1 and ²t ∼ N (0,σ²).
• Law of motion for capital kt+1 = it + (1− δ)kt.
• Resource constrain ct + it = yt.79
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• Solve for c (·, ·) and l (·, ·) given initial conditions.
• Characterized by:Uc(t) = βEt
hUc(t+ 1)
³1 + αAezt+1kα−1t+1 l(kt+1, zt+1)
α − δ´i
1− θ
θ
c(kt, zt)
1− l(kt, zt)= (1− α) eztAkαt l(kt, zt)
−α
• A system of functional equations with no known analytical solution.
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Solving the Model
• We need to use a numerical method to solve it.
• Different nonlinear approximations: value function iteration, perturba-tion, projection methods.
• We use a Finite Element Method. Why? Aruoba, Fernandez-Villaverdeand Rubio-Ramırez (2003):
1. Speed: sparse system.
2. Accuracy: flexible grid generation.
3. Scalable.81
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Building the Likelihood Function
• Time series:
1. Quarterly real output, hours worked and investment.
2. Main series from the model and keep dimensionality low.
• Measurement error. Why?
• γ = (θ, ρ, τ ,α, δ,β,σ²,σ1,σ2,σ3)
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State Space Representation
kt = f1(St−1,Wt; γ) = etanh−1(λt−1)kαt−1l
³kt−1, tanh−1(λt−1); γ
´1−α ∗1− θ
1− θ(1− α)
³1− l
³kt−1, tanh−1(λt−1); γ
´´l³kt−1, tanh−1(λt−1); γ
´+ (1− δ) kt−1
λt = f2(St−1,Wt; γ) = tanh(ρ tanh−1(λt−1) + ²t)
gdpt = g1(St, Vt; γ) = etanh−1(λt)kαt l
³kt, tanh
−1(λt); γ´1−α
+ V1,t
hourst = g2(St, Vt; γ) = l³kt, tanh
−1(λt); γ´+ V2,t
invt = g3(St, Vt; γ) = etanh−1(λt)kαt l
³kt, tanh
−1(λt); γ´1−α ∗1− θ
1− θ(1− α)
³1− l
³kt, tanh
−1(λt); γ´´
l³kt, tanh
−1(λt); γ´
+ V3,tLikelihood Function
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Since our measurement equation implies that
p (yt|St; γ) = (2π)−32 |Σ|−12 e−
ω(St;γ)2
where ω(St; γ) = (yt − x(St; γ)))0Σ−1 (yt − x(St; γ)) ∀t, we have
p³yT ; γ
´=
(2π)−3T2 |Σ|−T2
Z TYt=1
Ze−
ω(St;γ)2 p
³St|yt−1, S0; γ
´dSt
p (S0; γ) dS1' (2π)−3T2 |Σ|−T2
TYt=1
1
N
NXi=1
e−ω(sit;γ)
2
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Priors for the Parameters
Priors for the Parameters of the ModelParameters Distribution Hyperparameters
θρταδβσ²σ1σ2σ3
UniformUniformUniformUniformUniformUniformUniformUniformUniformUniform
0,10,10,1000,10,0.050.75,10,0.10,0.10,0.10,0.1
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Likelihood-Based Inference I: a Bayesian Perspective
• Define priors over parameters: truncated uniforms.
• Use a Random-walk Metropolis-Hastings to draw from the posterior.
• Find the Marginal Likelihood.
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Likelihood-Based Inference II: a Maximum Likelihood Perspective
• We only need to maximize the likelihood.
• Difficulties to maximize with Newton type schemes.
• Common problem in dynamic equilibrium economies.
• We use a simulated annealing scheme.
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An Exercise with Artificial Data
• First simulate data with our model and use that data as sample.
• Pick “true” parameter values. Benchmark calibration values for thestochastic neoclassical growth model (Cooley and Prescott, 1995).
Calibrated ParametersParameter θ ρ τ α δValue 0.357 0.95 2.0 0.4 0.02Parameter β σ² σ1 σ2 σ3Value 0.99 0.007 1.58*10−4 0.0011 8.66*10−4
• Sensitivity: τ = 50 and σ² = 0.035.
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0.9 0.92 0.94 0.96 0.98
Likelihood cut at ρ
1.5 2 2.5 3 3.5
Likelihood cut at τ
0.38 0.4 0.42
Likelihood cut at α
0.018 0.02 0.022
Likelihood cut at δ
7 8 9 10 11
x 10-3
Likelihood cut at σ
0.98 0.985 0.99
Likelihood cut at β
0.25 0.3 0.35 0.4
Likelihood cut at θ
Nonlinear
Linear
Pseudotrue
Figure 5.1: Likelihood Function Benchmark Calibration
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0.94850.9490.9495 0.95 0.95050.9510.95150
1000
2000
3000
4000
5000ρ
1.996 1.998 2 2.002 2.0040
1000
2000
3000
4000
5000τ
0.3995 0.4 0.40050
2000
4000
α
0.01957 0.0196 0.019630
2000
4000
δ
6.99 6.995 7 7.005 7.01
x 10-3
0
5000σ
0.988 0.989 0.99 0.9910
5000β
0.3564 0.3568 0.3572 0.35760
5000θ
1.578 1.579 1.58 1.581 1.582 1.583 1.584
x 10-4
0
5000
σ1
1.116 1.117 1.118 1.119 1.12
x 10-3
0
2000
4000
σ2
8.645 8.65 8.655 8.66 8.665 8.67 8.675
x 10-4
0
2000
4000
σ3
Figure 5.2: Posterior Distribution Benchmark Calibration
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0.9 0.92 0.94 0.96 0.98
Likelihood cut ρ
40 45 50 55
Likelihood cut τ
0.36 0.38 0.4 0.42 0.44
Likelihood cut α
0.016 0.018 0.02 0.022
Likelihood cut δ
0.03 0.035 0.04
Likelihood cut σ
0.95 0.96 0.97 0.98 0.99
Likelihood cut β
0.3 0.35 0.4
Likelihood cut θ
Nonlinear
Linear
Pseudotrue
Figure 5.3: Likelihood Function Extreme Calibration
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0.9495 0.95 0.95050
2000
4000
6000
ρ
49.95 50 50.050
2000
4000
6000
τ
0.3996 0.3998 0.4 0.40020
2000
4000
6000
α
0.019555 0.019565 0.019575 0.0195850
2000
4000
6000
δ
0.035 0.035 0.035 0.035 0.035 0.035 0.0350
2000
4000
6000
σ
0.989 0.9895 0.99 0.99050
2000
4000
6000
β
0.3567 0.3569 0.3571 0.35730
5000
θ
1.58 1.5805 1.581 1.5815 1.582 1.5825
x 10-4
0
5000
σ1
1.117 1.1175 1.118 1.1185 1.119
x 10-3
0
2000
4000
6000
σ2
8.655 8.66 8.665
x 10-4
0
2000
4000
6000
σ3
Figure 5.4: Posterior Distribution Extreme Calibration
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0 1 2 3 4
x 105
0.4
0.6
0.8
1ρ
0 1 2 3 4
x 105
40
60
80τ
0 1 2 3 4
x 105
0.350.4
0.45
α
0 1 2 3 4
x 105
0.01
0.02
0.03δ
0 1 2 3 4
x 105
0.02
0.03
0.04σ
0 1 2 3 4
x 105
0.9
0.95
1β
0 1 2 3 4
x 105
0.35
0.4θ
0 1 2 3 4
x 105
0
0.05
σ1
0 1 2 3 4
x 105
0
0.005
0.01
0.015
σ2
0 1 2 3 4
x 105
0
0.02
0.04
σ3
Figure 5.5: Converge of Posteriors Extreme Calibration
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0.96 0.97 0.98 0.99 10
5000
10000ρ
1.68 1.7 1.72 1.74 1.760
5000
10000
15000τ
0.32 0.322 0.324 0.326 0.328 0.33 0.3320
1
2x 10
4 α
6.2 6.25 6.3 6.35 6.4
x 10-3
0
5000
10000δ
0.0198 0.02 0.0202 0.02040
5000
10000
15000σ
0.9964 0.9966 0.9968 0.997 0.9972 0.9974 0.99760
5000
10000
15000β
0.385 0.39 0.395 0.40
5000
10000θ
0.0435 0.044 0.0445 0.045 0.0455 0.046 0.04650
5000
10000
σ1
0.014 0.0145 0.015 0.0155 0.0160
5000
10000
σ2
0.037 0.0375 0.038 0.0385 0.039 0.0395 0.040
5000
10000
15000
σ3
Figure 5.6: Posterior Distribution Real Data
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0.39 0.395 0.4 0.405 0.41
-7000
-6000
-5000
-4000
-3000
-2000
-1000
Transversal cut at α
Exact100 Particles1000 Particles10000 Particles
0.97 0.98 0.99 1 1.01
-7000
-6000
-5000
-4000
-3000
-2000
-1000
Transversal cut at β
Exact100 Particles1000 Particles10000 Particles
0.93 0.94 0.95 0.96 0.97
-220
-200
-180
-160
-140
-120
-100
-80
-60
-40
Transversal cut at ρ
Exact100 Particles1000 Particles10000 Particles
6.85 6.9 6.95 7 7.05 7.1 7.15
x 10-3
-220
-200
-180
-160
-140
-120
-100
-80
-60
-40
Transversal cut at σε
Exact100 Particles1000 Particles10000 Particles
Figure 6.1: Likelihood Function
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0 2000 4000 6000 8000 100000
0.2
0.4
0.6
0.8
110000 particles
0 0.5 1 1.5 2
x 104
0
0.2
0.4
0.6
0.8
120000 particles
0 1 2 3
x 104
0
0.2
0.4
0.6
0.8
130000 particles
0 1 2 3 4
x 104
0
0.2
0.4
0.6
0.8
140000 particles
0 1 2 3 4 5
x 104
0
0.2
0.4
0.6
0.8
150000 particles
0 2 4 6
x 104
0
0.2
0.4
0.6
0.8
160000 particles
Figure 6.2: C.D.F. Benchmark Calibration
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0 2000 4000 6000 8000 100000
0.2
0.4
0.6
0.8
110000 particles
0 0.5 1 1.5 2
x 104
0
0.2
0.4
0.6
0.8
120000 particles
0 1 2 3
x 104
0
0.2
0.4
0.6
0.8
130000 particles
0 1 2 3 4
x 104
0
0.2
0.4
0.6
0.8
140000 particles
0 1 2 3 4 5
x 104
0
0.2
0.4
0.6
0.8
150000 particles
0 2 4 6
x 104
0
0.2
0.4
0.6
0.8
160000 particles
Figure 6.3: C.D.F. Extreme Calibration
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0 2000 4000 6000 8000 100000
0.2
0.4
0.6
0.8
110000 particles
0 0.5 1 1.5 2
x 104
0
0.2
0.4
0.6
0.8
120000 particles
0 1 2 3
x 104
0
0.2
0.4
0.6
0.8
130000 particles
0 1 2 3 4
x 104
0
0.2
0.4
0.6
0.8
140000 particles
0 1 2 3 4 5
x 104
0
0.2
0.4
0.6
0.8
150000 particles
0 2 4 6
x 104
0
0.2
0.4
0.6
0.8
160000 particles
Figure 6.4: C.D.F. Real Data
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Posterior Distributions Benchmark CalibrationParameters Mean s.d.
θρταδβσ²σ1σ2σ3
0.3570.9502.0000.4000.0200.9890.007
1.58×10−41.12×10−28.64×10−4
6.72×10−53.40×10−46.78×10−48.60×10−51.34×10−51.54×10−59.29×10−65.75×10−86.44×10−76.49×10−7
89
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Maximum Likelihood Estimates Benchmark CalibrationParameters MLE s.d.
θρταδβσ²σ1σ2σ3
0.3570.9502.0000.4000.0020.9900.007
1.58×10−41.12×10−38.63×10−4
8.19×10−60.0010.020
2.02×10−62.07×10−51.00×10−60.0040.0070.0070.005
90
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Posterior Distributions Extreme CalibrationParameters Mean s.d.
θρταδβσ²σ1σ2σ3
0.3570.95050.000.4000.0200.9890.035
1.58×10−41.12×10−28.65×10−4
7.19×10−41.88×10−47.12×10−34.80×10−53.52×10−68.69×10−64.47×10−61.87×10−82.14×10−72.33×10−7
91
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Maximum Likelihood Estimates Extreme CalibrationParameters MLE s.d.
θρταδβσ²σ1σ2σ3
0.3570.95050.0000.4000.0190.9900.035
1.58×10−41.12×10−38.66×10−4
2.42×10−66.12×10−30.022
3.62×10−77.43×10−61.00×10−50.0150.0170.0140.023
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Convergence on Number of Particles
Convergence Real DataN Mean s.d.
100002000030000400005000060000
1014.5581014.6001014.6531014.6661014.6881014.664
0.32960.25950.18290.16040.14650.1347
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Posterior Distributions Real DataParameters Mean s.d.
θρταδβσ²σ1σ2σ3
0.3230.9691.8250.3880.0060.9970.0230.0390.0180.034
7.976× 10−40.0080.0110.001
3.557× 10−59.221× 10−52.702× 10−45.346× 10−44.723× 10−46.300× 10−4
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Maximum Likelihood Estimates Real DataParameters MLE s.d.
θρταδβσ²σ1σ2σ3
0.3900.9871.7810.3240.0060.9970.0230.0380.0160.035
0.0440.7081.3980.0190.160
8.67×10−30.2240.0600.0610.076
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Logmarginal Likelihood Difference: Nonlinear-Linear
p Benchmark Calibration Extreme Calibration Real Data0.1 73.631 117.608 93.650.5 73.627 117.592 93.550.9 73.603 117.564 93.55
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Nonlinear versus Linear Moments Real DataReal Data Nonlinear (SMC filter) Linear (Kalman filter)
outputhoursinv
Mean s.d1.950.360.42
0.0730.0140.066
Mean s.d1.910.360.44
0.1290.0230.073
Mean s.d1.610.340.28
0.0680.0040.044
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A “Future” Application: Good Luck or Good Policy?
• U.S. economy has become less volatile over the last 20 years (Stockand Watson, 2002).
• Why?
1. Good luck: Sims (1999), Bernanke and Mihov (1998a and 1998b)
and Stock and Watson (2002).
2. Good policy: Clarida, Gertler and Galı (2000), Cogley and Sargent
(2001 and 2003), De Long (1997) and Romer and Romer (2002).
3. Long run trend: Blanchard and Simon (2001).
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How Has the Literature Addressed this Question?
• So far: mostly with reduced form models (usually VARs).
• But:
1. Results difficult to interpret.
2. How to run counterfactuals?
3. Welfare analysis.
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Why Not a Dynamic Equilibrium Model?
• New generation equilibrium models: Christiano, Eichebaum and Evans(2003) and Smets and Wouters (2003).
• Linear and Normal.
• But we can do it!!!
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Environment
• Discrete time t = 0, 1, ...
• Stochastic process s ∈ S with history st = (s0, ..., st) and probability
µ³st´.
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The Final Good Producer
• Perfectly Competitive Final Good Producer that solves
maxyi(st)
µZyi³st´θdi¶1θ −
Zpi³st´yi³st´di.
• Demand function for each input of the form
yi³st´=
pi³st´
p (st)
1
θ−1y³st´,
with price aggregator:
p³st´=
ÃZpi³st´ θθ−1 di
!θ−1θ
.
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The Intermediate Good Producer
• Continuum of intermediate good producers, each of one behaving as
monopolistic competitor.
• The producer of good i has access to the technology:
yi³st´= max
½ez(s
t)kαi³st−1
´l1−αi
³st´− φ, 0
¾.
• Productivity z³st´= ρz
³st−1
´+ εz
³st´.
• Calvo pricing with indexing. Probability of changing prices (before
observing current period shocks) 1− ζ.
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Consumers Problem
Est
∞Xt=0
βt
εc³st´ ³c ³st´− dc ³st−1´´σc
σc− εl
³st´ l ³st´σl
σl+ εm
³st´m ³
st´σm
σm
p³st´ ³c³st´+ x
³st´´+M
³st´+Zst+1
q³st+1
¯st´B³st+1
´dst+1 =
p³st´ ³w³st´l³st´+ r
³st´k³st−1
´´+M
³st−1
´+B
³st´+ Π
³st´+ T
³st´
B³st+1
´≥ B
k³st´= (1− δ) k
³st−1
´− φ
x³st´
k³st−1
´+ x ³st´ .
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Government Policy
• Monetary Policy: Taylor rulei³st´= rgπg
³st´
+a³st´ ³
π³st´− πg
³st´´
+b³st´ ³y³st´− yg
³st´´+ εi
³st´
πg³st´= πg
³st−1
´+ επ
³st´
a³st´= a
³st−1
´+ εa
³st´
b³st´= b
³st−1
´+ εb
³st´
• Fiscal Policy.105
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Stochastic Volatility I
• We can stack all shocks in one vector:ε³st´=³εz³st´, εc
³st´, εl
³st´, εm
³st´, εi
³st´, επ
³st´, εa
³st´, εb
³st´´0
• Stochastic volatility:
ε³st´= R
³st´0.5
ϑ³st´.
• The matrix R³st´can be decomposed as:
R³st´= G
³st´−1
H³st´G³st´.
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Stochastic Volatility II
• H³st´(instantaneous shocks variances) is diagonal with nonzero ele-
ments hi³st´that evolve:
log hi³st´= log hi
³st−1
´+ ςiηi
³st´.
• G³st´(loading matrix) is lower triangular, with unit entries in the
diagonal and entries γij³st´that evolve:
γij³st´= γij
³st−1
´+ ωijνij
³st´.
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Where Are We Now?
• Solving the model: problem with 45 state variables: physical capital,
the aggregate price level, 7 shocks, 8 elements of matrix H³st´, and
the 28 elements of the matrix G³st´.
• Perturbation.
• We are making good progress.
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