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# gmmtimeseriesSlides - Single Equation Linear GMM with...

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Single Equation Linear GMM with Serially Correlated Moment Conditions Eric Zivot November 2, 2009

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Univariate Time Series Let { y t } be an ergodic-stationary time series with E [ y t ]= μ and var ( y t ) < . A fundamental decomposition result is the following: Wold Representation Theorem. y t has the representation y t = μ + X j =0 ψ j ε t j = μ + ε t + ψ 1 ε t 1 + ··· ψ 0 =1 , X j =0 ψ 2 j < ε t MDS(0 2 )
Remarks 1. The Wold representation shows that y t has a linear structure. As a result, theWo ldrep resentat ioniso ftenca l ledthe linear process representation of y t 2. P j =0 ψ 2 j < is called square-summability and controls the memory of the process. It implies that | ψ j | 0 as j →∞ at a su ciently fast rate.

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Variance γ 0 =v a r ( y t ) a r X k =0 ψ k ε t k = X k =0 ψ 2 k var( ε t ) = σ 2 X k =0 ψ 2 k <
Autocovariances γ j = E [( y t μ )( y t j μ )] = E X k =0 ψ k ε t k X h =0 ψ h ε t h j E [( ψ 0 + ψ 1 ε t 1 + ··· + ψ j ε t j + ) × ( ψ 0 ε t j + ψ 1 ε t j 1 + )] = σ 2 X k =0 ψ j + k ψ k ,j =0 , 1 , 2 ,...

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Ergodicity Ergodicity requires X j =0 | γ j | < It can be shown that X j =0 ψ 2 j < implies P j =0 | γ j | < .
Example: MA(1) process (truncated Wold form) Y t = μ + ε t + θε t 1 , | θ | < 1 ε t iid (0 2 ) Then ψ 1 = θ, ψ k =0 for k> 1 E [ Y t ]= μ γ 0 = E [( Y t μ ) 2 σ 2 (1 + θ 2 ) γ 1 = E [( Y t μ )( Y t 1 μ )] = σ 2 θ γ k ,k> 1 Clearly, X j =0 ψ 2 j =1+ θ 2 < , X j =0 | γ j | = σ 2 (1 + θ 2 + | θ | ) < so that { Y t } is both weakly stationary and ergodic.

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Example: AR(1) process Mean adjusted form: Y t μ = φ ( Y t 1 μ )+ ε t t WN(0 2 ) , | φ | < 1 E [ Y t ]= μ Regression form: Y t = c + φY t 1 + ε t ,c = μ (1 φ ) Solution by recursive substitution: Y t μ = φ t +1 ( Y 1 μ φ t ε 0 + ··· + φε t 1 + ε t = φ t +1 ( Y 1 μ t X i =0 φ i ε t i = φ t +1 ( Y 1 μ t X i =0 ψ i ε t i i = φ i
Stability and Stationarity Conditions If | φ | < 1 then lim j →∞ φ j = lim j →∞ ψ j =0 lim j →∞ φ j ( Y 1 μ )=0 and the stationary solution (Wold form) for the AR(1) becomes. Y t = μ + X j =0 φ j ε t j = X j =0 ψ j ε t j ψ j = φ j This is a stable (non-explosive) solution.

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Remarks 1. P j =0 ψ 2 j = P j =0 φ 2 j = 1 1 φ 2 < 2. For the stationary solution, think of the process as having started in the in f nite past
AR(1) in Lag Operator Notation (1 φL )( Y t μ )= ε t If | φ | < 1 then (1 φL ) 1 = X j =0 φ j L j =1+ φL + φ 2 L 2 + ··· such that (1 φL ) 1 (1 φL )=1

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Trick to f nd Wold form: Y t μ =( 1 φL ) 1 (1 φL )( Y t μ )=(1 φL ) 1 ε t = X j =0 φ j L j ε t = X j =0 φ j ε t j = X j =0 ψ j ε t j j = φ j
Trick for calculating moments: use stationarity properties E [ Y t ]= E [ Y t j ] for all j cov( Y t ,Y t j )=c o v ( Y t k t k j ) for all k, j Mean of AR(1) E [ Y t c + φE [ Y t 1 ]+ E [ ε t ] = c + φE [ Y t ] E [ Y t c 1 φ = μ

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Variance of AR(1) γ 0 =v a r ( Y t )= E [( Y t μ ) 2 ] = E [(
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## gmmtimeseriesSlides - Single Equation Linear GMM with...

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