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1 Economics 240C Due 5-28-2002 Answers to Problem Set Three 1. This MA(1) process can be expressed as: x(t) = [ 1 - (-0.5Z)] e(t) or equivalently as: [ 1 - (-0.5Z)]-1x(t) = e(t), [1 + (-0.5Z) + (-0.5Z)2 + (-0.5Z)3 + ...] x(t) = e(t), x(t) -0.5 x(t-1) + 0.25 x(t-2) - 0.125 x(t-3) + ... = e(t) 2. This ARMA(1, 1) process can be expressed as: x(t) - 0.9 Z x(t) = e(t) - 0.5 Z e(t), [ 1 - 0.9 Z] x(t) = [ 1 - 0.5 Z] e(t) or equivalently as: x(t) = [ 1 - 0.9 Z]-1 [ 1 - 0.5 Z] e(t) or x(t)=[1 +0.9Z + 0.81Z2 + 0.729Z3 + ...] [ 1 - 0.5 Z] e(t) and multiplying the two polynomials in Z, [1 +0.9Z + 0.81Z2 + 0.729Z3 + ...] [ 1 - 0.5 Z] -------------------------------------------- x(t) = [ 1 + 0.4 Z + 0.36 Z2 + 0.324 Z3 - ... ] e(t) 3. The ARMA(1,1) process can be expressed as [ 1- a Z] x(t) = [ 1- b Z] e(t) As a approaches b the polynomials cancel and x(t) approaches e(t). 4. For the ARIMA(1,1) process: x(t) = 0.5 x(t-1) + e(t) - 0.9 e(t-1) or equivalently, (1) x(t) = [e(t) - 0.9 e(t-1)]/[1-0.5Z] and hence x(t-2) will depend upon e(t-2) and earlier realizations of the orthogonal process e(t). Thus, multiplying Eq. (1) by x(t-2), x(t) x(t-2) = 0.5 x(t-1) x(t-2) + x(t-2) e(t) - 0.9 x(t-2) e(t-1) and taking covariances, xx(2) = 0.5 xx(1) or 2 Economics 240C Due 5-28-2002 Answers to Problem Set Three xx(2) = 0.5 xx(1) . Similarly, xx(j) = 0.5 xx(j-1) for j ≥ 2. Of course xx(0) = 1. That just leaves xx(1). Multiplying Eq. (1) by x(t-1), x(t) x(t-1) = 0.5 x(t-1) x(t-1) + x(t-1) e(t) - 0.9 x(t-1) e(t-1) and taking covariances, xx(1) = 0.5 xx(0) - 0.9 ex(0). Multiplying Eq. (1) by e(t): x(t) e(t) = 0.5 x(t-1) e(t) + e(t) e(t) - 0.9 e(t) e(t-1) and taking covariances, ex(0) = e2, thus xx(1) = 0.5 xx(0) - 0.9 e2. (2) Multiplying Eq. (1) by x(t), x(t) x(t) = 0.5 x(t-1) x(t) + x(t) e(t) - 0.9 x(t) e(t-1) and taking covariances, xx(0) = 0.5 xx(1) + ex(0) - 0.9 xe(-1) or xx(0) = 0.5 xx(1) + e2 - 0.9 xe(-1) Multiplying Eq. (1) by e(t-1): x(t) e(t-1) = 0.5 x(t-1) e(t-1) + e(t) e(t-1) - 0.9 e(t-1) e(t-1) and taking covariances, xe(-1) = 0.5 ex(0) - 0.9e2 or xe(-1) = 0.5 e2 - 0.9e2 = -0.4 e2 and hence xx(0) = 0.5 xx(1) + e2 - 0.9 (-0.4 e2) or xx(0) = 0.5 xx(1) + 1.36 e2 Solving equations (2) and (3), xx(1) = - 0.293 e2 and xx(0) = 1.21 e2 and thus xx(1) = -0.293/1.21 = -0.242 (3) 3 Economics 240C Due 5-28-2002 Answers to Problem Set Three Note the autocorrelation function has a negative spike at lag 1, capturing the MA(1) part of the ARMA(1,1) and then decays out like an AR(1). 5. For the moving average process, x(t) = -0.8 e(t-1) + e(t) where e(t) is white noise, square both sides of the expression for x(t) and take expectations; E[x(t)x(t)]= E[0.64e(t)e(t)-0.8e(t-1)e(t)-0.8e(t-1)e(t)+e(t)e(t)] x,x(0) = 0.64 + 0 + 0 + = 1.64 Lagging the expression for x(t) by one: x(t-1) = -0.8 e(t-2) + e(t-1) and multiplying x(t) by x(t-1) and the corresponding expressions on the right sides of the equality and taking expectations, E[x(t)x(t-1)]= E[0.64e(t-1)e(t-2)-0.8e(t-2)e(t)-0.8e(t-1)e(t-1)+e(t)e(t-1)] x,x(1) = 0 + 0 -0.8 = -0.8 Thus x,x(1) =-0.8/1.64 = -0.488 Lagging the expression for x(t) by two: x(t-2) = -0.8 e(t-3) + e(t-2) and multiplying x(t) by x(t-2) and the corresponding expressions on the right sides of the equality and taking expectations, E[x(t)x(t-2)]= E[0.64e(t-1)e(t-3)-0.8e(t-3)e(t)-0.8e(t-1)e(t-2)+e(t)e(t-2)] x,x(2) = 0 + 0 + 0 = 0 Thus x,x(2) =0/1.64 = 0 and autocorrelations will be zero, as will autocorrelations at higher lags. 4 Economics 240C Due 5-28-2002 Answers to Problem Set Three Autoc orre lation 2 1 0 -1 0 1 2 Lag 3 The MA(1) time series, x(t) = -0.8 e(t-1) + e(t) can be simulated on TSP with: » CREATE U 1000 » Genr wn = nrnd » Smpl 1 1 » Genr maone = wn » Smpl 2 1000 » Genr maone = -0.8*WN(-1) + WN » Smpl 1 50 » PLOT (A) MAONE » Smpl 1 1000 » IDENT (24) MAONE » 4 5 5 Economics 240C Due 5-28-2002 Answers to Problem Set Three Trace of MAONE: = -0.8 WN(-1) + WN 3 2 1 0 -1 -2 -3 10 20 30 40 50 TIME MAONE