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146 CHAPTER 5. INTERNATIONAL REAL BUSINESS CYCLES+ 1 2 R 22(k t − k) 2 + 1 2 R 33(A t − A) 2 + R 12 (k t+1 − k)(k t − k)+R 13 (k t+1 − k)(A t − A)+R 23 (k t − k)(A t − A).Suppose we let q =(R 1 ,R 2 ,R 3 ) 0 be the 3 × 1 row vector of partialderivatives (the gradient) of R, andQ be the 3 × 3matrixofsecondpartial derivatives (the Hessian) multiplied by 1/2 whereQ ij = R ij /2.Then the approximate period utility function can be compactly writtenin matrix form asR(λ t )=R(λ)+[q +(λ t − λ) 0 Q](λ t − λ). (5.21)The problem is now to maximizeE t ∞ Xj=0The Þrst order conditions are for all tβ j R(λ t+j ). (5.22)0 = (βR 2 + R 1 )+βR 12 (k t+2 − k)+(R 11 + βR 22 )(k t+1 − k)+R 12 (k t − k)+βR 23 (A t+1 − 1) + R 13 (A t − 1). (5.23)If you compare (5.23) to (5.19), you’ll see that a 0 = βR 2 + R 1 ,a 1 = βR 12 ,a 2 = R 11 + βR 22 ,a 3 = R 12 ,a 4 = βR 23 ,a 5 = R 13 . ThisveriÞes that the two approaches are indeed equivalent.Now to solve the linearized Þrst-order conditions, work with (5.19).Since the data that we want to explain are in logarithms, you can convertthe Þrst-order conditions into near logarithmic form. Letã i = ka i for i =1, 2, 3, and let a “hat” denote the approximate logdifference from the steady state so that ˆk t =(k t − k)/k ' ln(k t /k)and Ât = A t − 1 (since the steady state value of A =1). Nowletb 1 = −ã 2 /ã 1 , b 2 = −ã 3 /ã 1 , b 3 = −a 4 /ã 1 , and b 4 = −a 4 /ã 1 .The second—order stochastic difference equation (5.19) can be writtenas(1 − b 1 L − b 2 L 2 )ˆk t+1 = W t , (5.24)whereW t = b 3 Â t+1 + b 4 Â t .