tailieunhanh - International Macroeconomics and Finance: Theory and Empirical Methods Phần 5

Các dẫn xuất được đánh giá giá trị trạng thái ổn định. Một lựa chọn thứ hai, nhưng tương đương để có một thứ hai để xấp xỉ Taylor chức năng khách quan xung quanh các trạng thái ổn định và giải quyết vấn đề tối ưu hóa kết quả bậc hai. | . CALIBRATING THE ONE-SECTOR GROWTH MODEL 145 where a0 Ucgi pucg2 0 1 ftUccgigỉ aỉ Uccgỉ U ỉ rng22 3 Uccgigỉ 4 ft Ucg3ỉ Ucc iỉ i3. a5 Uccỹiỹ3- The derivatives are evaluated at steady state values. A second but equivalent option is to take a second-order Taylor approximation to the objective function around the steady state and to solve the resulting quadratic optimization problem. The second option is equivalent to the first because it yields linear first-order conditions around the steady state. To pursue the second option recall that xt kt i kt At 0. Write the period utility function in the unconstrained optimization problem as R Xt U g Xt . Let Rj dR Xt BXjt be the partial derivative of R xt with respect to the j th element of xt and Rjj dỉ R xt dXitdXjt be the second cross-partial derivative. Since Rjj Rjj the relevant derivatives are Ri Ucgi Rỉ Ucgỉ R3 Ucg3 Rii Uccgiỉ Rỉỉ Uccgỉ Ucgỉỉ R33 Uccg32 Ri2 Uccgig2 Ri3 Uccgig3 Rỉ3 Uccgỉg3 Ucgỉ3. The second-order Taylor expansion of the period utility function is 1 R R x R1 kt 1 k Rỉ kt k R3 At A - Rii kt 1 k 2 146 CHAPTER 5. INTERNATIONAL REAL BUSINESS CYCLES 1 1 2Ryiikt - k 2R33 At - A R12 kt 1 - k kt k R13 kt 1 - k At - A R23 kt - k At - A - Suppose we let q R1 R2 R3Ỵ be the 3 X 1 row vector of partial derivatives the gradient of R and Q be the 3 X 3 matrix of second partial derivatives the Hessian multiplied by 1 2 where Qij Rj 2. Then the approximate period utility function can be compactly written in matrix form as R At R A q At - A 0Q At - A . 5-21 The problem is now to maximize EtE A R At j . j 0 The first order conditions are for all t 0 AR2 R1 AR12 kt 2 - k R11 AR22 kt 1 - k R12 kt - k AR23 At 1 - 1 R13 At - 1 . If you compare to you ll see that a0 AR2 R1 1 AR12 a2 R11 AR22 a3 R12 4 AR23 as R13. This verifies that the two approaches are indeed equivalent. Now to solve the linearized first-order conditions work with . Since the data that we want to explain are in logarithms