foundations of econometrics phần 1 potx

... inverse of the product of an arbitrary number of factors. 1. 17 Consider the following example of multiplying partitioned matrices:  A 11 A 12 A 21 A 22  B 11 B 12 B 21 B 22  =  A 11 B 11 + A 12 B 21 A 11 B 12 + ... A 12 B 21 A 11 B 12 + A 12 B 22 A 21 B 11 + A 22 B 21 A 21 B 12 + A 22 B 22  . Check all the expressions on the right-ha...

Ngày tải lên: 14/08/2014, 22:21

... equation η 2 2 (y 1  y 2 y 2  M ∆Y y 2 − y 1  M ∆Y y 2 y 2  y 2 ) + η(y 2  y 2 y 1  M ∆Y y 1 − y 1  y 1 y 2  M ∆Y y 2 ) + (y 1  y 1 y 1  M ∆Y y 2 − y 1  M ∆Y y 1 y 1  y 2 ) = 0. (14 .78) 14 .24 ... matrices implies that x 21 x 11 + x 22 x 21 = 0, we obtain the error-correction model ∆y t1 = η 2 ∆y t2 + (λ 2 − 1) (y t 1, 1 − η 2 y t 1, 2 ) + e  t2 ,...

Ngày tải lên: 14/08/2014, 22:21

Ngày tải lên: 24/07/2014, 08:20

... X 2 β 2 + u;e) M 1 y = X 2 β 2 + u;f) M 1 y = M 1 X 2 β 2 + u;g) M 1 y = X 1 β 1 + M 1 X 2 β 2 + u;h) M 1 y = M 1 X 1 β 1 + M 1 X 2 β 2 + u;i) P X y = M 1 X 2 β 2 + u.j) Here P 1 projects orthogonally ... sensible estimators are the following: ˆµ 1 = 1 n + 1 n  t =1 y t , ˆµ 2 = 1. 01 n n  t =1 y t , and ˆµ 3 = 0.01y 1 + 0.99 n − 1 n  t=2 y t . The...

Ngày tải lên: 14/08/2014, 22:21

... be σ 2 0 [ −S 21 S 1 11 1 ]  S 11 S 12 S 21 S 22  −S 1 11 S 12 1  = σ 2 0  S 22 − S 21 S 1 11 S 12  . (4.57) Copyright c  19 99, Russell Davidson and James G. MacKinnon 4.7 The Power of Hypothesis ... square of the second factor is n 1 x 2  M 1 x 2 = n 1 x 2  x 2 − n 1 x 2  P 1 x 2 = n 1 x 2  x 2 − n 1 x 2  X 1  n 1 X 1  X 1 ...

Ngày tải lên: 14/08/2014, 22:21

... β 2 ) + u t  n 1  n t =1  1 / t (β 0 1 − β 1 ) + 1 / t 2 (β 0 2 − β 2 ) + 1 / t u t   . (6 .13 ) It is known that the deterministic sums n 1  n t =1 (1/ t) and n 1  n t =1 (1/ t 2 ) both tend ... X 1 and X 2 denote X 1 (β 0 ) and X 2 (β 0 ), respectively, we see that n 1 ´ X 2  M ´ X 1 ´ X 2 = n 1 ´ X 2  ´ X 2 − n 1 ´ X 2  ´ X 1 (n 1 ´ X 1  ´ X...

Ngày tải lên: 14/08/2014, 22:21

... ˆρX 1 )  (X − ˆρX 1 ) (X − ˆρX 1 )  ˆ u 1 ˆ u 1  (X − ˆρX 1 ) ˆ u 1  ˆ u 1  1 , (7.63) where the n×k matrix X 1 has typical row X t 1 , and the vector ˆ u 1 has typical element y t 1 − ... equations n  t=2 (X t − ˆρX t 1 )   y t − X t ˆ β − ˆρ(y t 1 − X t 1 ˆ β)  + (1 − ˆρ 2 )X 1  (y 1 − X 1 ˆ β) = 0, and n  t=2 (y t 1 − X t 1 ˆ β)  y t − X t ˆ...

Ngày tải lên: 14/08/2014, 22:21

... elements of θ. For the estimating equations (9 .11 1), the iterative step (9 .11 4) becomes  µ (j +1) σ 2 (j +1)  =  µ (j) σ 2 (j)  −     ∂ ¯m 1 ∂µ ∂ ¯m 1 ∂σ 2 ∂ ¯m 2 ∂µ ∂ ¯m 2 ∂σ 2      ¯m 1 (µ (j) , ... then our estimate of the covariance matrix of ˆµ and ˆσ 2 is  Var  ˆµ ˆσ 2  = (W  ˆ F ) 1 W  ˆ Ω W ( ˆ F  W ) 1 , (9 .11 6) with W given by (9 .11 2) and...

Ngày tải lên: 14/08/2014, 22:21

... −I 0 21 plim n→∞ n 1/ 2 ( ˜ θ 1 − θ 0 1 ) = plim n→∞  n 1/ 2 g 2 (θ 0 ) −I 0 21 (I 0 11 ) 1 n 1/ 2 g 1 (θ 0 )  = [ −I 0 21 (I 0 11 ) 1 I ] plim n→∞  n 1/ 2 g 1 (θ 0 ) n 1/ 2 g 2 (θ 0 )  , (10 .66) where ... −I 0 21 (I 0 11 ) 1 I ]  I 0 11 I 0 12 I 0 21 I 0 22  −(I 0 11 ) 1 I 0 12 I  = I 0 22 − I 0 21 (I 0 11 ) 1 I 0 12 . (10 .67) I...

Ngày tải lên: 14/08/2014, 22:21

... N(0, 1) . Generate 500 samples of 20 observations on (x t , y t ) pairs, 10 0 assuming that β 1 = 0 and β 2 = 1, 10 0 assuming that β 1 = 1 and β 2 = 1, 10 0 assuming that β 1 = 1 and β 2 = 1, 10 0 ... (12 .13 ), we see that (12 .10 ) can be expressed as Var( ˆ β GLS • ) =    σ 11 X 1  X 1 · · · σ 1g X 1  X g . . . . . . . . . σ g1 X g  X 1 · · · σ gg X g ...

Ngày tải lên: 14/08/2014, 22:21