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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 + 21 + 42 3 = 0 + 31 + 92 4 = 0 + 41 + 162 Polynomial lags - estimation OLS, only need to estimate p parameters: 0,...,p STEP 1: impose restriction yt = +0xt + 0 + 1 + 2xt-1+(0 +21 +42)xt-2+(0+31 +92)xt-3+ (0 +41 + 162)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 i0 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 |

language: | English |

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