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					MACROECONOMETRICS




        LAB 3 – DYNAMIC MODELS
ROADMAP

   What if we know that the effect lasts in
    time?
   Distributed lags
    –   ALMON
    –   KOYCK
    –   ADAPTIVE EXPECTATIONS
    –   PARTIAL ADJUSTMENT
   STATA not really too complicated here 
How to do lags?

   Infinite?
    –   how many lags do we take?
    –   how to know?
   Unrestricted?
    –   do we impose any structure on the lags?
    –   this structure might be untrue?
    –   but there is also cost to unrestricted approach...
Unrestricted lags (no structure)
  –   It is always finite!
yt =  + 0 xt + 1 xt-1 + 2 xt-2 + . . . +n xt-n + et
          N lags and no structure in parameters
          OLS works
       BUT
          n observations lost
          high multicollinearity

          imprecise, large s.e., low t, lots of d.f. Lost

  STRUCTURE COULD HELP
Arithmetic lag

   The effect of X eventually zero
   Linearly!
   The coefficients not independent of each other
    –   effect of each lag less than previous
    –   exactly like arithmetic series: un=u1+d(n-1)
Arithmetic lag - structure

                 i
0 = (n+1)
   1 = n
                              Linear
2 = (n-1)                      lag
     .                                  structure
     .
     .
   n = 

             0        1   2    .   .      .   .     .   n   n+1 i
Arithmetic lag - maths
   X (log of) money supply and Y (log of) GDP, n=12 and  is
    estimated to be 0.1
   the effect of a change in x on GDP in the current period is
    0=(n+1)=1.3
   the impact of monetary policy one period later has declined to
    1=n=1.2
   n periods later, the impact is n= =0.1
   n+1 periods later the impact is zero

                             E ( yt )
                       i =
                             xt i
Arithmetic lag - estimation
   OLS, only need to estimate one parameter: 
   STEP 1: impose restriction

       yt =  + 0 xt + 1 xt-1 + 2 xt-2 + . . . +n xt-n + et
   STEP 2: factor out the parameter
       yt =  +  [(n+1)xt + nxt-1 + (n-1)xt-2 + . . . + xt-n] + et
   STEP 3: define z
       zt = [(n+1)xt + nxt-1 + (n-1)xt-2 + . . . + xt-n]
   STEP 4: decide n (no. of lags)          ???
        For n = 4:      zt = [ 5xt + 4xt-1 + 3xt-2 + 2xt-3 + xt-4]
Arithmetic lag – pros & cons

   Advantages:
    –   Only one parameter to be estimated!
            t-statistics ok., better s.e., results more reliable
    –   Straightforward interpretation
   Disadvantages:
    –   If restriction untrue, estimators biased and inconsistent


   Solution? F-test! (see: end of the notes)
Polynomial lag (ALMON)

   If we want a different shape of IRF...
    –   It’s just a different shape
    –   Still finite: the effect eventually goes to zero
        (by DEFINITION and not by nature!)
    –   The coefficients still related to each other BUT not a uniform
        pattern (decline)
Polynomial lag - structure


                              i = 0 + 1i + 2i
                                                        2
  i             2
                                       E ( yt )
       0
            . . .
            1           3

                                        xt i
                                                 = i
       .                          4
                              .

       0 1       2    3       4               i
Polynomial lag - maths
   n – the lenght of the lag
   p – degree of the polynomial

    i = 0 + 1i + 2i2 +...+ pip, where i=1, . . . , n
   For example a quadratic polynomial

    i = 0 + 1i + 2i2 , where p=2       and   n=4


    0 =  0                        1 =  0 +  1 +  2
    2 = 0 + 21 + 42            3 = 0 + 31 + 92
    4 = 0 + 41 + 162
Polynomial lags - estimation
   OLS, only need to estimate p parameters: 0,...,p
   STEP 1: impose restriction

    yt =  +0xt + 0 + 1 + 2xt-1+(0 +21 +42)xt-2+(0+31 +92)xt-3+
         (0 +41 + 162)xt-4 + et
   STEP 2: factor out the unknown coefficients
    yt =  +0 [xt +xt-1+xt-2+xt-3 +xt-4]+1[xt+xt-1+2xt-2+3xt-3 +4xt-4] +
         2 [xt + xt-1 + 4xt-2 + 9xt-3 + 16xt-4] + et

   STEP 3: define z
    z t0 = [xt + xt-1 + xt-2 + xt-3 + xt-4]      z t1 = [xt + xt-1 + 2xt-2 + 3xt-3 + 4xt- 4 ]
    z t2 = [xt + xt-1 + 4xt-2 + 9xt-3 + 16xt- 4]
   STEP 4: do OLS on yt =  + 0 z t0 + 1 z t1 + 2 z t2 + et
Polynomial lag – pros & cons

   Advantages:
    –   Fewer parameters to be estimated than in the unrestricted lag
        structure
            More precise than unrestricted
    –   If the polynomial restriction likely to be true:
            More flexible than arithmetic DL


   Disadvantages
    –  If the restriction untrue, biased and inconsistent
    (see F-test in the end of the notes)
Arithmetic vs. Polynomial vs. ???

   Conclusion no. 1
    –   Data should decide about the assumed pattern of impulse-
        response function
   Conclusion no. 2
    –   We still do not know, how many lags!
   Conclusion no. 3
    –   We still have a finite no. of lags.
Geometric lag (KOYCK)
   Distributed lag is infinite  infinite lag length (no time limits)
   BUT cannot estimate an infinite number of parameters!

      Restrict the lag coefficients to follow a pattern


    For the geometric lag the pattern is one of continuous decline at
     decreasing rate
    (we are still stuck with the problem of imposing fading out instead
     of observing it – gladly, it is not really painful, as most processes
     behave like that anyway )
Geometric lag - structure

           
     0 = 
           i   .
                       Geometrically
    1 =       .     declining
                            weights
    2 = 2          .
    3 = 3
    4 = 4
                           . .
                0 1    2   3   4       i
Geometric lag - maths
   Infinite distributed lag model
     yt =  + 0 xt + 1 xt-1 + 2 xt-2 + . . . + et
     yt =  + i xt-i + et

   Geometric lag structure
   i = i  where || < 1 and i0
   
   Infinite unstructured geometric lag model
     yt =  + 0 xt + 1 xt-1 + 2 xt-2 + 3 xt-3 + . . . + et
    AND:0=,1=,2=2,3=3 ...

   Substitute i = i => infinite geometric lag
    yt =  + xt +  xt-1 + 2 xt-2 + 3 xt-3 + . . .) + et
Geometric lag - estimation

   Cannot estimate using OLS
     yt-1 is dependent on et-1  cannot alow that (need to instrument)
   Apply Koyck transformation
   Then use 2SLS
   Only need to estimate two parameters: ,
   Have to do some algebra to rewrite the model in form
    that can be estimated.
Geometric lag – Koyck transformation

   Original equation:
   yt =  + xt +  xt-1 + 2 xt-2 + 3 xt-3 + . . .) + et
  Koyck rule: lag everything once, multiply by and
    substract from the original
    yt =  + xt +  xt-1 + 2 xt-2 + 3 xt-3 + . . .) + et
   yt-1 =  +  xt-1 + 2 xt-2 + 3 xt-3 + . . .) +  et-1
     yt   yt-1 = 1 + xt + (et  et-1)
     yt = 1 +  yt-1 + xt + (et  et-1) so yt = 1+ 2 yt-1 + 3xt + t
    yt-1 is dependent on et-1 so yt-1 is correlated with vt-1
    OLS will be consistent (it cannot distinguish between change in yt
     caused by yt-1 that caused by vt)
Geometric lag - estimation

   Regress yt-1 on xt-1 and calculate the fitted value
   Use the fitted value in place of yt-1 in the Koyck
    regression and that is it!
   Why does this work?
    –   from the first stage fitted value is not correlated with et-1 but yt-1
        is so fitted value is uncorrelated with vt =(et -et-1 )
   2SLS will produce consistent estimates of the
    Geometric Lag Model
Geometric lag – pros & cons

   Advantages
    –   You only estimate two parameters!
   Disadvantages
    –   We allow neither for heterogenous nor for unsmooth declining
   It has many well specified versions, among which two
    have particular importance:
    –  ADAPTIVE EXPECTATIONS
    – PARTIAL ADJUSTMENT MODEL
    (for both: see next student presentation)
F-tests of restrictions
1.   Estimate the unrestricted model
2.   Estimate the restricted (any lag) model
3.   Calculate the test statistic
              ( SSER  SSEU ) / df1
           F=
                   SSEU / df 2
4.   Compare with critical value F(df1,df2)
          df1 = number of restrictions
          df2 = number of observations-number of variables in the
           unrestricted model (incl. constant)
5.   H0: residuals are ‘the same’, restricted model OK

				
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posted:12/24/2011
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