Here the sequence of coefficients β β 1 β k

This preview shows page 14 - 19 out of 28 pages.

Here, the sequence of coefficients { β 0 , β 1 , . . . , β k } constitutes the impulse- response function of the mapping from x ( t ) to y ( t ). That is to say, if we imagine that, on the input side, the signal x ( t ) is a unit impulse of the form (35) x ( t ) = . . . , 0 , 1 , 0 , . . . , 0 , 0 . . . } which has zero values at all but one instant, then the output of the transfer function would be (36) r ( t ) = . . . , 0 , β 0 , β 1 , . . . , β k , 0 , . . . . 14
Image of page 14

Subscribe to view the full document.

EC3062 ECONOMETRICS Another concept that helps in understanding the dynamic response is the step-response of the transfer function. Imagine that the input sequence is zero-valued up to a point in time when it assumes a constant unit value: (37) x ( t ) = . . . , 0 , 1 , 1 , . . . , 1 , 1 . . . } . The output of the transfer function would be the sequence (38) s ( t ) = . . . , 0 , s 0 , s 1 , . . . , s k , s k , . . . , where (39) s 0 = β 0 , s 1 = β 0 + β 1 , . . . s k = β 0 + β 1 + · · · + β k . Here, the value s k , which is attained by the sequence when the full adjust- ment has been accomplished after k periods, is called the (steady-state) gain of the transfer function; and it is denoted by γ = s k . 15
Image of page 15
EC3062 ECONOMETRICS The distributed-lag formulation of equation (34) is profligate in its use of parameters; and, given that the sequence x ( t ) is likely to show strong serial correlation, we may expect to encounter problems of multicollinearity. The Geometric Lag Structure Another early approach to the problem of defining a lag structure, which depends on two parameters β and φ , is the geometric lag scheme of Koyk: (40) y ( t ) = β x ( t ) + φx ( t 1) + φ 2 x ( t 2) + · · · + ε ( t ) . The impulse-response function of the Koyk model is a geometrically de- clining sequence { β, φβ, φ 2 β, . . . } The gain of the transfer function, which is obtained by summing the geometric series, has the value of (41) γ = β 1 φ . 16
Image of page 16

Subscribe to view the full document.

EC3062 ECONOMETRICS By lagging the equation by one period and multiplying by φ , we get (42) φy ( t 1) = β φx ( t 1)+ φ 2 x ( t 2)+ φ 3 x ( t 3)+ · · · + φε ( t 1) . Taking the latter from (40) gives y ( t ) φy ( t 1) = βx ( t ) + ε ( t ) φε ( t 1) or (1 φL ) y ( t ) = βx ( t ) + (1 φL ) ε ( t ) of which the rational form is (45) y ( t ) = β 1 φL x ( t ) + ε ( t ) . The following expansion can be used to recover equation (40): (46) β 1 φL x ( t ) = β 1 + φL + φ 2 L 2 + · · · x ( t ) = β x ( t ) + φx ( t 1) + φ 2 x ( t 2) + · · · 17
Image of page 17
EC3062 ECONOMETRICS Equation (43) cannot be estimated consistently by OLS regression because the composite disturbance term { ε ( t ) φε ( t 1) } is correlated with the lagged dependent variable y ( t 1). A simple consistent estimation procedure is based on the equation under (40). The elements of y ( t ) within the sample may be expressed as (47) y t = β i =0 φ i x t i + ε t = θφ t + β t 1 i =0 φ i x t i + ε t = θφ t + βz t + ε t . Here (48) θ = β x 0 + φx 1 + φ 2 x 2 + · · · is a nuisance parameter, which embodies the presample elements of the sequence x ( t ), whereas (49) z t = x t + φx t 1 + · · · + φ t 1 x 1 is a synthetic variable based on the observations x t , x t 1 , . . . , x 1 and on the value attributed to φ .
Image of page 18

Subscribe to view the full document.

Image of page 19
You've reached the end of this preview.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern