Wyklad z Ekonometrii, WNE UW, III rok · †The to ﬁlter ct and gt we have to solve following...

Hodrick-Prescott filter

• Assume that the series yt is the sum of the growt component gt and acyclical component ct

yt = ct + gt for t = 1, . . . , T

• The growth component gt varies smoothly over time

• The cyclical component ct is in average equal to zero

• Assume taht measure of smoothness of {gt} path is the sum of squares ofits second differances

Macroeconometrics, WNE UW, Copyright c©2007 by Jerzy Mycielski 1

• The to filter ct and gt we have to solve following maximization problem

min{gt}T

t=−1

{T∑

t=1

c2t + λ

T∑t=1

∆2gt

}

where ct = yt − gt

• λ is a positive number which panelizes the variability in growth componentseries

• The larger is λ the smoother is the solution, for λ → ∞ the solution of theproblem is the OLS fit of linear trend

• Hodrick and Prescott suggested to use λ = 1600 for quarterly data.Rawn and Uhlig (2002) shown that λ should be proprtional to 4 powerof frequency observation rations.


• This implies λ = 6.25 for annual data and 129600 for monthly data

• The Hodrick-Prescott filter can be derived as special case of Kalman filterif we assume:

– vector of state zt = [ct, gt]– transition equation is of the form:

ct = ε1t

gt = gt−1 + ε2t

– Random components ε1t and ε2t are NID and independ of each other

εt ∼ N

(0,

[σ2

1 00 σ2

2

])


– Smoothing parameter is equal to ratio of variances of ε1 and ε2

λ =σ2

1

σ22

• Typical application of Hodrick-Prescott filter: filtering the output gap


Markov chain definition

• Values of random variable Xn are in finite set {s1, s2, . . . , sr}

• Finite Markov chain

Pr (Xn = j|X0 = i0, X1 = i1, . . . , Xn−1 = in−1) = P (Xn = j|Xn−1 = i)

• This process has a short memory, transition probability only depends onthe state of the process in previous period

• Finite homogenous Markov chain has a following property:

P (Xn = j|Xn−1 = i) = P (Xn+s = j|Xn+s−1 = i) = pij


• For homogenous Markov chain, transition probabilities do not change overtime


Matrix of transition probabilities

• Matrix of transition probabilities has a following form:

P =

p11 p12 · · · p1r

p21 p22 · · · p2r... ... ...

pr1 pr2 · · · prr

• pij has to satisfy following properties:

r∑

j=1

pij = 1

pij ≥ 0


Marginal distribution of Markov chain

• Marginal probability distribution of the Markov chain state:

dnj = Pr (Xn = j)

which can be denoted as a vector:

dn =[

dn1 dn2 · · · dnr

]

where:r∑

i=j

dnj = 1

dnj ≥ 0


• Marginal probabilities for the Markov chain can be calculated as follows

dnj = Pr (Xn = j)

= Pr (Xn = j|Xn−1 = 1) Pr (Xn−1 = 1) + . . . + Pr (Xn = j|Xn−1 = r) Pr (Xn−1 = r)

=r∑

i=1

Pr (Xn = i|Xn−1 = i) Pr (Xn−1 = i) =r∑

i=1

dn−1,ipij

• This formula can be written using matrix notation:

dn = dn−1 P

• Recursively repeating this transformation we get:

dn = d0 P n


P n =

p(n)11 p

(n)12 · · · p

(n)1r

p(n)21 p

(n)22 · · · p

(n)2r... ... ...

p(n)r1 p

(n)r2 · · · p

(n)rr


Stationary distribution of Markov chain

• Stationary distribution of Markow chain is such vector d, for which:

d = dP

• This is the charcteristic equation for matrix P . Vector d is left eigenvectorof matrix P related to eigenvalue λ1 = 1.

• It is possible to prove that each matrix P has at least one d satifyng thisequation.


Ergodic Markov chain

• We say that stochastic process is ergodic if for n → ∞ the process statedoes not depend from process state at n = 0.

• As for n →∞lim

n→∞dn = d0 lim

n→∞P n = d0E = e

where

limn→∞

P n = E


• The Markov chain is ergotic only if for every d0

d0E = e

r∑

j=1

ej = 1

• This is only possible if all rows of E is equal

• Vector e is called the steady state of Markow chain

• As e is the only stationary distribution of ergodic Markov chain

• Therfore steady state for ergodic Markov chain can be found from followingsystem of equations: {

eP = ee1 = 1


Estimation of transition probabilities, individual level data

• Estimation with ML

• Likelihood function

Pr [n (1) , . . . , n (T )] =T∏

t=1

∏

i∈S

ni (t− 1)!∏

j∈S

nij (t)!

∏

j∈S

pnij(t)

ijt

where n (t) = [n1 (t) , . . . , nr (t)] is the vector of empirical counts for states

• The way we estimate depend on whether we assume homogeneity.


• Under homogeneity the ML estimator of transition probabilities hasfollowing intuitive form:

p̂ij =nij

ni

where

nij =T−1∑t=1

nijt

ni =T−1∑t=1

∑

j∈S

nijt

• So the ML estimator for pij is the proportion of transition between state ito j for all times 1, . . . , T , in total number os observations observed in statei for times 1, . . . , T − 1

• ML estimator of transiotion probabilities for nonhomogenous Markow


chain isp̂ijt =

nijt

nit

wherenit =

∑

j∈S

nijt

• The hipotesis of homogeneity of Markov chain can be easly tesed with LRtest

• It is however impossible to test in such a way the hypotesis that pij = 0 orpij = 1.

• In particular it is impossible to test with standard LR, W or LM tests thenumber of states in Markov chain


Estimation of transition probabilities, aggregated data

• For aggregated data we only observe the numbers of units (or fractions)observed in given states in time t

• Conditional expected value of the number of observations in state j in timet from ni observation, which were obseved to be in state i in time t − 1 isequal to:

E [nij (t)|ni (t− 1)] = ni (t− 1) pij

• From independence of transition we obtain

E [nj (t)|N (t− 1)] =∑

i∈S

ni (t− 1) pij = N (t− 1) pj


where pj is j-th column of the matrix of transition probabilities and rowvector N (t− 1) = [n1 (t) , . . . , nr (t)] is a vector of numbers of units beingin each state

• Denote as εit = ni (t)−∑

ni (t− 1) pij, and the we obtain

nj (t) =∑

i∈S

ni (t− 1) pij + εit = N (t− 1)pj + εit for j ∈ S

where εt satisfyE [εt|N (t− 1)] = 0

• This model can be written as

N (t) = N (t− 1) P + εt

andN (t)1 = N (t− 1)1


• As the number of unit should be constant over time the following addtionalrestriction should be valid

N (t)1 = N (t− 1) 1 = N

where N is the total number of units obeserved

• This model can be estimated be OLS applied to each equation or moreefficiently with GMM or GLS apllied to all equation

• The major problem in estimation of this model is related the fact that wehave to impose restriction that all the coefficients pij > 0


Marcov switching models

• Assume that we there are two regimes (e.g. recession and expansion)

• Each of the regimes is decribed with the different model

• We can not directly identify the regime in place in time t

• Denote regime by St ∈ {0, 1}

• Assume thatE (yt|St) = α0 + α1St + βxt

• The regimes differ by constant term α1


• The general case is if the transition probabilities can be influenced by someexogenous variables

• Transition probabilities are then given by:

P t =[

qt 1− qt

pt 1− pt

]

where

qt = Pr (St = 0|St−1 = 0) = Φ (δxt)

pt = Pr (St = 1|St−1 = 1) = Φ (γxt)

• With this kind of model we can:

– filter the probabilities states St (e.g. recession, expansion)


– distinguish two channels of influence: the influence on the probabilityof states and the direct influence on expected value of the dependentvariable


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Wyklad z Ekonometrii, WNE UW, III rok · †The to ﬁlter ct and gt we have to solve following...

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