Copyright 1996 Lawrence C. Marsh 0 PowerPoint Slides for Undergraduate Econometrics by
Download 0.54 Mb. Pdf ko'rish
|
allChap6
|
this model with
least squares : 1. One observation is used up in creating the transformed (lagged) variables leaving only (T − 1) observations for estimating the model. 2. The value of ρ is not known . We must find some way to estimate it. 11.11
Copyright 1996 Lawrence C. Marsh Recovering the 1st Observation Dropping the 1st observation and applying least squares is
not the
best linear unbiased estimation method. Efficiency is lost because the variance of the error associated with the 1st observation is not equal to that of the other errors. This is a special case of the heteroskedasticity problem except that here all errors are assumed to have equal variance except the 1st error. 11.12
Recovering the 1st Observation y 1
β 1 + β 2 x 1 + e
1 The 1st observation should fit the original model as: We could include this as the 1st observation for our estimation procedure but we must first transform it so that it has the same error variance as the other observations. with error variance: var(e 1 ) =
σ e 2 = σ ν 2 /(1- ρ 2
Note: The other observations all have error variance σ ν 2 . 11.13 51 Copyright 1996 Lawrence C. Marsh y 1 = β 1 + β 2 x 1
+ e 1 with error variance: var(e 1 ) =
σ e 2 = σ ν 2 /(1- ρ 2
The other observations all have error variance σ ν 2 . Given any constant c : var(ce 1 ) = c
2 var(e
1 ). If c = 1- ρ 2
, then var( 1- ρ 2 e 1 ) = (1- ρ 2 ) var(e 1 ).
= (1- ρ 2 ) σ e 2 = (1-
ρ 2 ) σ ν 2 /(1- ρ 2 ) =
σ ν 2 The transformation ν 1
= 1- ρ 2 e 1 has variance σ ν 2 . 11.14 Copyright 1996 Lawrence C. Marsh y 1 = β 1 + β 2 x 1
+ e 1 The transformed error ν 1
= 1- ρ 2 e 1 has variance σ ν 2 . Multiply through by 1- ρ 2 to get: 1- ρ 2 y 1 = 1- ρ 2
β 1 +
1- ρ 2 β 2 x 1
+ 1- ρ 2 e 1 This transformed first observation may now be added to the other (T-1) observations to obtain the fully restored set of T observations. 11.15
Copyright 1996 Lawrence C. Marsh Estimating Unknown ρ Value
e t
= ρ e t − 1 + ν t First, use least squares to estimate the model: If we had values for the e t
y t = β 1 + β 2 x t + e
t The residuals from this estimation are: e t
t - b
1 - b
2 x t ^ 11.16
Copyright 1996 Lawrence C. Marsh e t = y t - b 1 - b
2 x t ^ e t = ρ e t − 1 + ν t ^ ^ ^ Next, estimate the following by least squares: The least squares solution is: Σ e
e t-1
Σ e t-1 T T t = 2 t = 2 2 ^ ^
^ ρ
= ^ 11.17 Copyright 1996 Lawrence C. Marsh Durbin-Watson Test H o : ρ = 0 vs. H 1 : ρ
≠ 0 ,
ρ > 0, or ρ < 0
Σ ( e t − e t-1 ) Σ e t T T t = 2 t = 1 2 ^ ^ ^ d
= 2 The Durbin-Watson Test statistic, d, is : 11.18
Testing for Autocorrelation The test statistic, d, is approximately related to ρ as: ^ d ≈ 2(1 −ρ ) ^ When
ρ = 0 , the Durbin-Watson statistic is d ≈ 2.
^ When
ρ = 1 , the Durbin-Watson statistic is d ≈ 0.
^ Tables for critical values for d are not always readily available so it is easier to use the p-value that most computer programs provide for d. Reject H o if p-value < α , the significance level. 11.19
52 Copyright 1996 Lawrence C. Marsh Prediction with AR(1) Errors When errors are autocorrelated, the previous period’s error may help us predict next period’s error. The best predictor, y T+1 , for next period is: y T+1 = β 1 + β 2 x T+1
+ ρ e T ^ ^ ^ ^ ~ where β 1 and
β 2 are generalized least squares estimates and e T is given by: ~ ^ ^ e T = y T
−
β 1
− β 2 x T ^ ^ ~ 11.20 Copyright 1996 Lawrence C. Marsh y T+h = β 1 + β 2 x T+h
+ ρ h e T ^ ^ ^ ^ ~ For h periods ahead, the best predictor is: Assuming | ρ | < 1, the influence of ρ h e T
diminishes the further we go into the future (the larger h becomes). ^ ^ ~ 11.21
Copyright 1996 Lawrence C. Marsh Pooling Time-Series and Cross-Sectional Data Chapter 12 Copyright © 1997 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Copyright Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein. 12.1
Copyright 1996 Lawrence C. Marsh Pooling Time and Cross Sections y it
β 1it
+ β 2it x 2it
+ β 3it x 3it
+ e it If left unrestricted, this model requires different equations for each firm in each time period. for the i th firm in the t th time period 12.2
Seemingly Unrelated Regressions y it
β 1i + β 2i x 2it +
β 3i x 3it + e
it SUR models impose the restrictions: β 1it
= β 1i β 2it
= β 2i β 3it
= β 3i Each firm gets its own coefficients: β 1i , β 2i and β 3i but those coefficients are constant over time . 12.3
Copyright 1996 Lawrence C. Marsh The investment expenditures (INV) of General Electric (G) and Westinghouse(W) may be related to their stock market value (V) and actual capital stock (K) as follows: INV Gt
β 1G + β 2G V Gt +
β 3G K Gt + e
Gt INV
Wt =
β 1W + β 2W V Wt + β 3W K Wt + e Wt i = G, W t = 1, . . . , 20 Two-Equation SUR Model 12.4
53 Copyright 1996 Lawrence C. Marsh Estimating Separate Equations For now make the assumption of no correlation between the error terms across equations: We make the usual error term assumptions: cov(
e Gt , e Gs ) = 0 cov( e Wt , e Ws ) = 0 var(
e Gt ) = σ G
var( e
) = σ W 2 E( e Gt ) = 0
E( e Wt ) = 0 cov(
e Gt , e Wt ) = 0 cov( e Gt , e Ws ) = 0 12.5
Copyright 1996 Lawrence C. Marsh homoskedasticity assumption: σ G
σ W
2 INV
t =
β 1G + δ 1 D t + β 2G V t + δ 2 D t V t + β 3G K t + δ 3 D t K t + e
t Dummy variable model assumes that : σ G
σ W
2 For Westinghouse observations D t = 1; otherwise D t = 0.
β 1W
= β 1G + δ 1 β 2W
= β 2G + δ 2 β 3W
= β 3G + δ 3 12.6 Copyright 1996 Lawrence C. Marsh Problem with OLS on Each Equation The first assumption of the Gauss-Markov Theorem concerns the model specification .
If the model is not fully and correctly specified the Gauss-Markov properties might not hold. Any
of error terms across equations must be part of model specification. 12.7
Copyright 1996 Lawrence C. Marsh Any
correlation between the dependent variables of two or more equations that is not due to their explanatory variables is by default due to correlated error terms . Correlated Error Terms 12.8 Copyright 1996 Lawrence C. Marsh 1. Sales of Pepsi vs. sales of Coke. (uncontrolled factor: outdoor temperature) 2. Investments in bonds vs. investments in stocks . (uncontrolled factor: computer/appliance sales) 3. Movie admissions vs. Golf Course admissions.
(uncontrolled factor: weather conditions) 4. Sales of butter vs. sales of bread.
(uncontrolled factor: bagels and cream cheese) Which of the following models would be likely to produce positively correlated errors and which would produce
12.9
Copyright 1996 Lawrence C. Marsh Joint Estimation of the Equations INV Gt
β 1G + β 2G V Gt +
β 3G K Gt + e
Gt INV
Wt =
β 1W + β 2W V Wt + β 3W K Wt + e Wt cov( e G t , e W t ) =
σ GW 12.10 54 Copyright 1996 Lawrence C. Marsh Seemingly Unrelated Regressions When the error terms of two or more equations are
correlated , efficient estimation requires the use of a
(SUR)
type estimator to take the correlation into account. Be sure to use the Seemingly Unrelated Regressions (SUR) procedure in your regression software program to estimate any equations that you believe might have
. 12.11 Copyright 1996 Lawrence C. Marsh Separate vs. Joint Estimation SUR will give exactly the same results as estimating each equation separately with OLS if either or both of the following two conditions are true: 1. Every equation has exactly the same set of explanatory variables with exactly the same values.
2. There is no correlation between the error terms of any of the equations. 12.12
Test for Correlation Η ο
GW = 0
Test the null hypothesis of zero correlation σ GW σ G σ W ^ ^ ^ r GW = 2 2 2 2 λ = T r GW 2 λ ∼ χ 2 (1) asy. 12.13
Copyright 1996 Lawrence C. Marsh Start with the residuals e Gt and e Wt from each equation estimated separately. ^ ^
GW σ G σ W ^ ^ ^ r GW = 2 2 2 2 λ = T r
GW 2 λ ∼ χ 2 (1)
asy. σ GW = Σ
e Gt e Wt 1 T ^ ^ ^ σ G = Σ
e Gt 1 T ^ ^ 2 2 σ W = Σ
e Wt 1 T ^ ^ 2 2 12.14 Copyright 1996 Lawrence C. Marsh Fixed Effects Model y it
β 1it
+ β 2it x 2it
+ β 3it x 3it
+ e it y it =
β 1i + β 2 x 2it +
β 3 x 3it + e
it Fixed effects models impose the restrictions: β 1it
= β 1i β 2it
= β 2 β 3it
= β 3 For each i th cross section in the t th time period: Each i th
β 1i
intercept. 12.15
Copyright 1996 Lawrence C. Marsh The
Fixed Effects Model is conveniently represented using dummy variables: y it = β 11 D 1i
+ β 12 D 2i
+ β 13 D 3i
+ β 14 D 4 i
+ β 2 x 2it
+ β 3 x 3it
+ e it D 1i =1 if North D 1i
D 2i =1 if East D 2i =0 if not E D 3i =1 if South D 3i =0 if not S D 4i =1 if West D 4i =0 if not W y it
= millions of bushels of corn produced x 2it = price of corn in dollars per bushel x 3it = price of soybeans in dollars per bushel Each cross-sectional unit gets its own intercept, but each cross-sectional intercept is constant over time. 12.16
55 Copyright 1996 Lawrence C. Marsh H o : β 11 = β 12 = β 13 = β 14 Test for Equality of Fixed Effects H 1 : H o not true The H o joint null hypothesis may be tested with F-statistic: (SSE
R
− SSE U ) / J SSE U / (NT − K)
F = ~ F
(NT − K) J SSE
R is the restricted error sum of squares (one intercept) SSE U
N is the number of cross-sectional units (N = 4) K is the number of parameters in the model (K = 6) J is the number of restrictions being tested (J = N − 1 = 3) T is the number of time periods 12.17
Copyright 1996 Lawrence C. Marsh Random Effects Model y it
β 1i + β 2 x 2it +
β 3 x 3it + e
it β 1i = β 1 + µ i β 1 is the population mean intercept. µ i is an unobservable random error that accounts for the cross-sectional differences. 12.18
Copyright 1996 Lawrence C. Marsh β 1i = β 1 + µ i µ i are independent of one another and of e it E(
i ) = 0
var( µ i ) = σ µ 2 where
i = 1, ... ,N Consequently, E( β
) = β 1 var( β 1i ) = σ µ 2 Random Intercept Term 12.19
y it = β 1i + β 2 x 2it
+ β 3 x 3it
+ e it y it = (
β 1 + µ i ) + β 2 x 2it +
β 3 x 3it + e
it y it = β 1 + β 2 x 2it
+ β 3 x 3it
+ ( µ i +e it ) y it = β 1 + β 2 x 2it +
β 3 x 3it +
ν it
Random Effects Model 12.20
Copyright 1996 Lawrence C. Marsh ν it = ( µ i +e it ) y it = β 1 + β 2 x 2it +
β 3 x 3it +
ν it
ν it
has zero mean : E( ν it ) = 0 ν it
is homoskedastic : var( ν it ) = σ µ
+ σ e 2 2 The errors from the same firm in different time periods are correlated: The errors from different firms are always uncorrelated: cov( ν
, ν is ) = σ µ 2 cov(
ν it , ν js ) = 0 t ≠ s i ≠ j 12.21
Copyright 1996 Lawrence C. Marsh Simultaneous Equations Models Chapter 13 Copyright © 1997 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Copyright Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein. 13.1
56 Copyright 1996 Lawrence C. Marsh Keynesian Macro Model Assumptions of Simple Keynesian Model 1. Consumption, c, is function of income, y. 2. Total expenditures = consumption + investment. 3. Investment assumed independent of income. 13.2
consumption is a function of income: income is either consumed or invested: c =
β 1 + β 2 y y = c + i The Structural Equations 13.3
The Statistical Model c t
β 1 + β 2 y t + e
t y t = c t + i t The consumption equation: The income identity: 13.4
Copyright 1996 Lawrence C. Marsh The Simultaneous Nature of Simultaneous Equations c t = β 1 + β 2 y t
+ e t y t = c
t + i
t Since y
t contains e t they are
correlated 2. 1. 3. 4. 5. 13.5 Copyright 1996 Lawrence C. Marsh The Failure of Least Squares The least squares estimators of parameters in a structural simul- taneous equation is biased
and inconsistent because of the cor- relation between the random error and the endogenous variables on the right-hand side of the equation. 13.6
Single Equation: Simultaneous Equations: Single vs. Simultaneous Equations c t
t e t c t y t i t e t 13.7 57 Copyright 1996 Lawrence C. Marsh Deriving the Reduced Form c t
β 1 + β 2 y t + e
t y t = c t + i t c t = β 1 + β 2 (c t + i t ) + e
t (1
−
β 2 )c t = β 1 + β 2 i t + e t 13.8
Copyright 1996 Lawrence C. Marsh Deriving the Reduced Form (1 −
β 2 )c t =
β 1 + β 2 i t + e
t c t = + i t + e t (1 −β 2 ) (1 −β 2 ) (1 −β 2 ) 1 β 1 β 2 c t = π 11 + π 21 i t +
ν t The Reduced Form Equation 13.9 Copyright 1996 Lawrence C. Marsh Reduced Form Equation c t
π 11 + π 21 i t +
ν t (1 −β 2 ) β 1 π 11 = (1 −β 2 ) β 2 π 21 = (1 −β 2 ) 1 ν t = + e t and 13.10 Copyright 1996 Lawrence C. Marsh y t = c t + i t where
c t = π 11 + π 21 i t +
ν t y t =
π 12 + π 22 i t
+ ν t It is sometimes useful to give this equation its own reduced form parameters as follows: y t
π 11 + ( 1 + π 21 ) i
t
+ ν t 13.11 Copyright 1996 Lawrence C. Marsh y t = π 12 + π 22 i t
+ ν t c t = π 11 + π 21 i t + ν t Since c t and y
t are related through the identity: y t
t + i
t , the error term, ν t
equations is the same, and it is easy to show that: (1 −β
) β 1 π 11
= π 12 = (1 −β 2 ) π 22 = (
1 −π 21 )
= 1 13.12
Copyright 1996 Lawrence C. Marsh Identification The structural parameters are β 1 and β 2 . The reduced form parameters are π 11
and π 21. Once the reduced form parameters are estimated, the identification problem is to determine if the orginal structural parameters can be expressed uniquely in terms of the reduced form parameters. (1 +
21 ) β 2
= π 21 ^ ^ ^ (1 + π 21 ) β 1
= π 11 ^ ^ ^ 13.13 58 Copyright 1996 Lawrence C. Marsh Identification An equation is exactly identified if its structural (behavorial) parameters can be uniquely expres- sed in terms of the reduced form parameters. An equation is over-identified if there is more than one solution for expressing its structural (behavorial) parameters in terms of the reduced form parameters. An equation is under-identified if its structural (behavorial) parameters cannot be expressed in terms of the reduced form parameters. 13.14
The Identification Problem A system of M equations containing M endogenous variables must exclude at least M − 1 variables from a given equation in order for the parameters of that equation to be identified and to be able to be consistently estimated. 13.15
Copyright 1996 Lawrence C. Marsh Two Stage Least Squares Problem: right-hand endogenous variables y t2 and y t1 are correlated with the error terms. y t1 = β 1 + β 2 y t2 + β 3 x t1 + e t1 y t2 = α 1 + α 2 y t1 + α 3 x t2 + e t2 13.16
Copyright 1996 Lawrence C. Marsh Problem: right-hand endogenous variables y t2
y t1 are correlated with the error terms. Solution: First, derive the reduced form equations. y t1 = β 1 + β 2 y t2 + β 3 x t1 + e t1 y t2 = α 1 + α 2 y t1 + α 3 x t2 + e t2 y t1 = π 11 + π 21 x t1 + π 31 x t2 +
ν t1 y t2 =
π 12 + π 22 x t1 +
π 32 x t2 +
ν t2 Solve two equations for two unknowns, y t1 , y
t2 : 13.17 Copyright 1996 Lawrence C. Marsh y t1 = π 11 + π 21 x t1 + π 31 x t2 +
ν t1 y t2 =
π 12 + π 22 x t1 +
π 32 x t2 +
ν t2 Use least squares to get fitted values: 2SLS: Stage I y t1 = π 11 + π 21 x t1 + π 31 x t2 ^ ^ ^ ^ y t2 =
π 12 + π 22 x t1 +
π 32 x t2 ^ ^ ^ ^ y t2 = y
t2 +
ν t2 ^ ^ y t1 = y t1 + ν t1 ^ ^ 13.18
Copyright 1996 Lawrence C. Marsh 2SLS: Stage II y t2
t2 +
ν t2 ^ ^ y t1 = y t1 + ν t1 ^ ^ and
y t1 = β 1 + β 2
y t2 + β 3 x t1 + e
t1 y t2 = α 1 + α 2 y t1 + α 3 x t2 + e t2 Substitue in for y
, y t2 y t1 = β 1 + β 2 (y t2 +
ν t2 ) + β 3 x t1 + e
t1 ^ ^ y t2 = α 1 + α 2 (y t1 +
ν t1 ) + α 3 x t2 + e
t2 ^ ^ 13.19 59 Copyright 1996 Lawrence C. Marsh 2SLS: Stage II (continued) y t1
β 1 + β 2 y t2 +
β 3 x t1 + u
t1 ^ y t2 =
α 1 + α 2 y t1 +
α 3 x t2 + u
t2 ^ ^ u t1
= β 2 ν t2
+ e t1 u t2 = α 2 ν t1 + e t2 ^ where and
Run least squares on each of the above equations to get 2SLS estimates: β 1
β 2 , β 3 , α 1 , α 2
and
α 3 ~ ~ ~ ~ ~ ~ 13.20
Copyright 1996 Lawrence C. Marsh Nonlinear Least Squares Chapter 14 Copyright © 1997 John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that permitted in Section 117 of the 1976 United States Copyright Act without the express written permission of the copyright owner is unlawful. Request for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. The purchaser may make back-up copies for his/her own use only and not for distribution or resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein. 14.1
Copyright 1996 Lawrence C. Marsh (A.) “Regression” model with only an intercept term: Review of Least Squares Principle y t = α + e t e t = y t
−
α Σ e t = Σ (y t
− α ) 2 2 SSE = Σ (y t −
α ) 2 ∂ SSE
∂ α = − 2
Σ (y t −
α ) = 0
^ Σ
y t
−
Σ α = 0 ^ Σ
y t
−
Τ α = 0 ^ α = Σ
y t
= y ^ 1 T (minimize the sum of squared errors) Yields an exact analytical solution: 14.2
Copyright 1996 Lawrence C. Marsh Review of Least Squares (B.) Regression model without an intercept term: y t = β x t + e
t e t = y t
−
β x t Σ e t = Σ (y t −
β x t ) 2 2 SSE = Σ (y t
− β x t ) 2 ∂ SSE
∂ α = − 2
Σ x t (y t
− β x t )= 0 ^ Σ
x t y t
− Σ
β x t = 0
^ 2 Σ x t
y t
−
β Σ x t
= 0 ^ 2 β Σ x t = Σ
x t y t ^ 2 β = ^ Σ
x t y t 2 Σ x t This yields an exact analytical solution: 14.3
Copyright 1996 Lawrence C. Marsh Review of Least Squares (C.) Regression model with both an intercept and a slope: y t = α + β x t + e t SSE = Σ (y t −
α
− β x t ) 2 ∂ SSE
∂ α = − 2
Σ (y t − α
− β x t ) = 0 ^ ^ ∂ SSE
∂ β = − 2
Σ x t (y t
− α
− β x t ) = 0 ^ ^ y
− α
− β x = 0
^ ^ β = ^ Σ (x t − x)(y
t − y) Σ (x t − x) 2 Σ x t y t
−
αΣ x t
− βΣ x t = 0 ^ ^ 2 This yields an exact analytical solution: α = y − β
x ^ ^ 14.4 Copyright 1996 Lawrence C. Marsh Nonlinear Least Squares (D.) Nonlinear Regression model: y t = x t β + e t SSE = Σ (y t − x t β ) 2 PROBLEM: An exact analytical solution to this does not exist. ∂ SSE
∂ β = − 2
Σ x t β ln (x t )(y t − x t β ) = 0
^ ^ Σ [ x t β ln (x t )y t ] − Σ [
x t 2β ln (x t )] = 0
^ ^ Must use numerical search algorithm to try to find value of β to satisfy this. 14.5 60 Copyright 1996 Lawrence C. Marsh Find Minimum of Nonlinear SSE β β
SSE SSE =
Σ (y t − x t β ) 2
14.6 Copyright 1996 Lawrence C. Marsh The
least squares principle is still appropriate when the model is
, but it is harder to find the solution. Conclusion 14.7
Nonlinear least squares optimization methods: The Gauss-Newton Method Optional Appendix 14.8
Copyright 1996 Lawrence C. Marsh The Gauss-Newton Algorithm 1. Apply the Taylor Series Expansion to the nonlinear model around some initial b
. 2. Run Ordinary Least Squares (OLS) on the linear part of the Taylor Series to get b
. 3. Perform a Taylor Series around the new b (m)
to get b (m+1) . 4. Relabel b (m+1) as
b (m) and rerun steps 2.-4. 5. Stop when ( b
−
(m) ) becomes very small. 14.9
ε
t for
t = 1, . . . , n. Do a Taylor Series Expansion around the vector b = b (o) as follows: y t = f(X t ,b (ο)
) + f’(X t ,b (ο)
)(b - b (ο)
) + ε
∗
ε
∗ ≡
(o) ) T f’’(X t ,b (ο)
)(b - b (ο)
) + R t + ε
t f(X t ,b) = f(X t ,b (ο)
) + f’(X t ,b (ο)
)(b - b (ο)
)
(ο)
) T f’’(X t ,b (ο)
)(b - b (ο)
) + R t 14.10
Copyright 1996 Lawrence C. Marsh y t = f(X t ,b (ο)
) + f’(X t ,b (ο)
)(b - b (ο)
) + ε
∗
(ο)
) = f’(X t ,b (ο)
)b - f’(X t ,b (ο)
) b (ο)
+ ε
∗
(ο)
) + f’(X t ,b (ο)
) b (ο)
= f’(X t ,b (ο)
)b + ε
∗
∗ (ο) = f’(X t ,b (ο)
)b + ε
∗
∗ (ο) ≡
y t - f(X t ,b (ο)
) + f’(X t ,b (ο)
) b (ο)
This is linear in b . Gauss-Newton just runs OLS on this transformed truncated Taylor series. 14.11
61 Copyright 1996 Lawrence C. Marsh y t ∗ (ο) = f’(X t ,b (ο)
)b + ε
∗
∗ (ο) = f’(X,b (ο)
)b + ∈ ∗ for t = 1, . . . , n in matrix terms b =
f’(X,b (ο)
) T f’(X,b (ο)
) ] -1 f’(X,b (ο)
) T y ∗ (ο) ^ This is analogous to linear OLS where y = Xb + ∈
b = (
X T X) −1
T y ^ except that X is replaced with the matrix of first Download 0.54 Mb. Do'stlaringiz bilan baham: 
|