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Econ 240C Lecture 16
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2 Part I. VAR Does the Federal Funds Rate Affect Capacity Utilization?
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3 The Federal Funds Rate is one of the principal monetary instruments of the Federal Reserve Does it affect the economy in “real terms”, as measured by capacity utilization
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4 Preliminary Analysis
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5 The Time Series, Monthly, January 1967through May 2003
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6 Federal Funds Rate: July 1954-April 2006
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7 Capacity Utilization Manufacturing: Jan. 1972- April 2006
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8 Changes in FFR & Capacity Utilization
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9 Contemporaneous Correlation
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10 Dynamics: Cross-correlation
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11 Granger Causality
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12 Granger Causality
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13 Granger Causality
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14 Estimation of VAR
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23 Estimation Results OLS Estimation each series is positively autocorrelated –lags 1 and 24 for dcapu – lags 1, 2, 7, 9, 13, 16 each series depends on the other –dcapu on dffr: negatively at lags 10, 12, 17, 21 –dffr on dcapu: positively at lags 1, 2, 9, 10 and negatively at lag 12
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24 Correlogram of DFFR
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25 Correlogram of DCAPU
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26 We Have Mutual Causality, But We Already Knew That DCAPU DFFR
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27 Interpretation We need help Rely on assumptions
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28 What If What if there were a pure shock to dcapu –as in the primitive VAR, a shock that only affects dcapu immediately
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Primitive VAR
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30 The Logic of What If A shock, e dffr, to dffr affects dffr immediately, but if dcapu depends contemporaneously on dffr, then this shock will affect it immediately too so assume is zero, then dcapu depends only on its own shock, e dcapu, first period But we are not dealing with the primitive, but have substituted out for the contemporaneous terms Consequently, the errors are no longer pure but have to be assumed pure
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31 DCAPU DFFR shock
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32 Standard VAR dcapu(t) = ( /(1- ) +[ ( + )/(1- )] dcapu(t-1) + [ ( + )/(1- )] dffr(t-1) + [( + (1- )] x(t) + (e dcapu (t) + e dffr (t))/(1- ) But if we assume then dcapu(t) = + dcapu(t-1) + dffr(t-1) + x(t) + e dcapu (t) +
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33 Note that dffr still depends on both shocks dffr(t) = ( /(1- ) +[( + )/(1- )] dcapu(t-1) + [ ( + )/(1- )] dffr(t- 1) + [( + (1- )] x(t) + ( e dcapu (t) + e dffr (t))/(1- ) dffr(t) = ( +[( + ) dcapu(t-1) + ( + ) dffr(t-1) + ( + x(t) + ( e dcapu (t) + e dffr (t))
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34 DCAPU DFFR shock e dcapu (t) e dffr (t) Reality
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35 DCAPU DFFR shock e dcapu (t) e dffr (t) What If
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36 EVIEWS
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38 Interpretations Response of dcapu to a shock in dcapu –immediate and positive: autoregressive nature Response of dffr to a shock in dffr –immediate and positive: autoregressive nature Response of dcapu to a shock in dffr –starts at zero by assumption that –interpret as Fed having no impact on CAPU Response of dffr to a shock in dcapu –positive and then damps out –interpret as Fed raising FFR if CAPU rises
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39 Change the Assumption Around
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40 DCAPU DFFR shock e dcapu (t) e dffr (t) What If
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41 Standard VAR dffr(t) = ( /(1- ) +[( + )/(1- )] dcapu(t-1) + [ ( + )/(1- )] dffr(t- 1) + [( + (1- )] x(t) + ( e dcapu (t) + e dffr (t))/(1- ) if then, dffr(t) = dcapu(t-1) + dffr(t-1) + x(t) + e dffr (t)) but, dcapu(t) = ( + ( + ) dcapu(t- 1) + [ ( + ) dffr(t-1) + [( + x(t) + (e dcapu (t) + e dffr (t))
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43 Interpretations Response of dcapu to a shock in dcapu –immediate and positive: autoregressive nature Response of dffr to a shock in dffr –immediate and positive: autoregressive nature Response of dcapu to a shock in dffr –is positive (not - ) initially but then damps to zero –interpret as Fed having no or little control of CAPU Response of dffr to a shock in dcapu –starts at zero by assumption that –interpret as Fed raising FFR if CAPU rises
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44 Conclusions We come to the same model interpretation and policy conclusions no matter what the ordering, i.e. no matter which assumption we use, or So, accept the analysis
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45 Understanding through Simulation We can not get back to the primitive fron the standard VAR, so we might as well simplify notation y(t) = ( /(1- ) +[ ( + )/(1- )] y(t-1) + [ ( + )/(1- )] w(t-1) + [( + (1- )] x(t) + (e dcapu (t) + e dffr (t))/(1- ) becomes y(t) = a 1 + b 11 y(t-1) + c 11 w(t-1) + d 1 x(t) + e 1 (t)
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46 And w(t) = ( /(1- ) +[( + )/(1- )] y(t-1) + [ ( + )/(1- )] w(t-1) + [( + (1- )] x(t) + ( e dcapu (t) + e dffr (t))/(1- ) becomes w(t) = a 2 + b 21 y(t-1) + c 21 w(t-1) + d 2 x(t) + e 2 (t)
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47 Numerical Example y(t) = 0.7 y(t-1) + 0.2 w(t-1)+ e 1 (t) w(t) = 0.2 y(t-1) + 0.7 w(t-1) + e 2 (t) where e 1 (t) = e y (t) + 0.8 e w (t) e 2 (t) = e w (t)
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48 Generate e y (t) and e w (t) as white noise processes using nrnd and where e y (t) and e w (t) are independent. Scale e y (t) so that the variances of e 1 (t) and e 2 (t) are equal –e y (t) = 0.6 *nrnd and –e w (t) = nrnd (different nrnd) Note the correlation of e 1 (t) and e 2 (t) is 0.8
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49 Analytical Solution Is Possible These numerical equations for y(t) and w(t) could be solved for y(t) as a distributed lag of e 1 (t) and a distributed lag of e 2 (t), or, equivalently, as a distributed lag of e y (t) and a distributed lag of e w (t) However, this is an example where simulation is easier
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50 Simulated Errors e 1 (t) and e 2 (t)
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51 Simulated Errors e 1 (t) and e 2 (t)
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52 Estimated Model
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58 Y to shock in w Calculated 0.8 0.76 0.70
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Impact of shock in w on variable y
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