Econometric take home APPS_Part_32

# Econometric take home APPS_Part_32 - | GARCH MODEL | |...

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+---------------------------------------------+ | GARCH MODEL | | Maximum Likelihood Estimates | | Model estimated: Jul 31, 2002 at 01:19:14PM.| | Dependent variable PT | | Weighting variable None | | Number of observations 190 | | Iterations completed 22 | | Log likelihood function -135.5043 | | Restricted log likelihood -147.6465 | | Chi squared 24.28447 | | Degrees of freedom 2 | | Prob[ChiSqd > value] = .5328953E-05 | | GARCH Model, P = 1, Q = 1 | | Wald statistic for GARCH = 521.483 | +---------------------------------------------+ +---------+--------------+----------------+--------+---------+----------+ |Variable | Coefficient | Standard Error |b/St.Er.|P[|Z|>z] | Mean of X| +---------+--------------+----------------+--------+---------+----------+ Regression parameters Constant .1308478127 .61887183E-01 2.114 .0345 PT1 .1749239917 .70912277E-01 2.467 .0136 .98810078 PT2 .2532191617 .73228319E-01 3.458 .0005 .98160455 PT3 .1552879436 .68274176E-01 2.274 .0229 .97782066 PT4 .2751467919 .63910272E-01 4.305 .0000 .97277700 Unconditional Variance Alpha(0) .1005125676E-01 .11653271E-01 .863 .3884 Lagged Variance Terms Delta(1) .8556879884 .89322732E-01 9.580 .0000 Lagged Squared Disturbance Terms Alpha(1) .1077364862 .60761132E-01 1.773 .0762 Equilibrium variance, a0/[1-D(1)-A(1)] EquilVar .2748082674 2.0559946 .134 .8937 127

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Chapter 20 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Models with Lagged Variables Exercises 1. For the first, the mean lag is .55(.02)(0) + .55(.15)(1) + ... + .55(.17)(4) = 1.31 periods. The impact multiplier is .55(.02) = .011 while the long run multiplier is the sum of the coefficients, .55. For the second, the coefficient on x t is .6, so this is the impact multiplier. The mean lag is found by applying (18-9) to B ( L ) = [.6 + 2 L ]/[1 - .6 L + .5 L 2 ] = A ( L )/ D ( L ). Then, B (1)/ B (1) = {[ D (1) A (1) - A (1) D (1)]/[ D (1)] 2 } / [ A (1)/ D (1)] = A (1)/ A (1) - D (1)/ D (1) = (2/2.6) / (.4/.9) = 1.731 periods. The long run multiplier is B (1) = 2.6/.9 = 2.888 periods. For the third, since we are interested only in the coefficients on x t , write the model as y t = α + β x t [1 + γ L + γ 2 L 2 + ...] + δ z t * + u t . The lag coefficients on x t are simply β times powers of γ . 2. We would regress y t on a constant, x t , x t-1 , ..., x t-6 . Constrained least squares using 1 -5 10 -10 5 -1 0 0 0 R = 0 1 -5 10 -10 5 -1 0 , q = 0 0 0 1 -5 10 -10 5 -1 0 would produce the PDL estimates. 3. The ratio of polynomials will equal B ( L ) = [.6 + 2 L ]/[1 - .6 L + .5 L 2 ]. This will expand to B ( L ) = β 0 + β 1 L + β 2 L 2 + .... Multiply both sides of the equation by (1 - .6 L + .5 L 2 ) to obtain ( β 0 + β 1 L + β 2 L 2 + .... )(1 - .6 L + .5 L 2 ) = .6 + 2 L . Since the two sides must be equal, it follows that β 0 = .6 (the only term not involving L ) -.6 β 0 + β 1 = 2 (the only term involving only L . Therefore, β 1 = 2.36. All remaining terms, involving L 2 , L 3 , ... must equal zero. Therefore, β j - .6 β j -1 + .5 β j -2 = 0 for all j > 1, or β j = .6 β j- 1 - .5 β j- 2 . This provides a recursion for all remaining coefficients. For the specified coefficients, β 2 = .6(2.36) - .5(.3) = 1.266. β 3 = .6(1.266) - .5(2.36) = -.4204, β 4 = .6(-.4204) - .5(1.266) = -.88524 and so on.
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