Topic 5: Non-Linear Regression
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1 Topc 5: Non-Lnear Regresson The models we ve worked wth so far have been lnear n the parameters. They ve been of the form: y = Xβ + ε Many models based on economc theory are actually non-lnear n the parameters. CES Producton functon: Y = γ[δk ρ + (1 δ)l ρ ] v/ρ exp (ε ) or, ln(y ) = ln(γ) ( v ρ ) ln[δk ρ + (1 δ)l ρ ] +ε Lnear Expendture System: (Stone, 1954) Max. U(q) = β ln(q γ ) (Stone-Geary /Klen-Rubn) s.t. p q = M Yelds the followng system of demand equatons: p q = γ p + β (M j γ j p j ) ; = 1, 2,., n The β s are the Margnal Budget Shares. So, we requre that 0 < β < 1 ; = 1, 2,., n. Engel aggregaton mples that 1. γ = β = 1. In general, suppose we have a sngle non-lnear equaton: y = f(x 1, x 2,, x k ; θ 1, θ 2,, θ p ) + ε We can stll consder a Least Squares approach. The Non-Lnear Least Squares estmator s the vector, θ, that mnmzes the quantty: S(X, θ) = [y f (X, θ )] 2. Clearly the usual LS estmator s just a specal case of ths. To obtan the estmator, we dfferentate S wth respect to each element of θ ; set up the p frst-order condtons and solve. 1
2 Dffculty usually, the frst-order condtons are themselves non-lnear n the unknowns (the parameters). Ths means there s (generally) no exact, closed-form, soluton. Can t wrte down an explct formula for the estmators of parameters. Example y = θ 1 + θ 2 x 2 + θ 3 x 3 + (θ 2 θ 3 )x 4 + ε S = [y θ 1 θ 2 x 2 θ 3 x 3 (θ 2 θ 3 )x 4 ] 2 S θ 1 = 2 [y θ 1 θ 2 x 2 θ 3 x 3 (θ 2 θ 3 )x 4 ] S θ 2 = 2 [(θ 3 x 4 + x 2 )(y θ 1 θ 2 x 2 θ 3 x 3 θ 2 θ 3 x 4 )] S θ 3 = 2 [(θ 2 x 4 + x 3 )(y θ 1 θ 2 x 2 θ 3 x 3 θ 2 θ 3 x 4 )] Settng these 3 equatons to zero, we can t solve analytcally for the estmators of the three parameters. In stuatons such as ths, we need to use a numercal algorthm to obtan a soluton to the frst-order condtons. Lots of methods for dong ths one possblty s Newton s algorthm (the Newton- Raphson algorthm). Methods of Descent θ = θ 0 + s d(θ 0 ) θ 0 = ntal (vector) value. s = step-length (postve scalar) d(. ) = drecton vector 2
3 Usually, d(. ) Depends on the gradent vector at θ 0. It may also depend on the change n the gradent (the Hessan matrx) at θ 0. Some specfc algorthms n the famly make the step-length a functon of the Hessan. One very useful, specfc member of the famly of Descent Methods s the Newton- Raphson algorthm: Suppose we want to mnmze some functon, f(θ). Approxmate the functon usng a Taylor s seres expanson about θ, the vector value that mnmzes f(θ): f(θ) f(θ ) + (θ θ ) ( f θ ) + 1 θ 2! (θ θ ) [ 2 f θ θ ] (θ θ ) θ Or: f(θ) f(θ ) + (θ θ ) g(θ ) + 1 2! (θ θ ) H(θ )(θ θ ) So, f(θ) θ 0 + (θ θ ) g(θ ) + 1 2! 2H(θ )(θ θ ) However, g(θ ) = 0 ; as θ locates a mnmum. So, (θ θ ) H 1 (θ ) ( f(θ) θ ) ; or, θ θ H 1 (θ )g(θ) Ths suggests a numercal algorthm: 3
4 Set θ = θ 0 to begn, and then terate θ 1 = θ 0 H 1 (θ 1 )g(θ 0 ) θ 2 = θ 1 H 1 (θ 2 )g(θ 1 ) θ n+1 = θ n H 1 (θ n+1 )g(θ n ) or, approxmately: θ n+1 = θ n H 1 (θ n )g(θ n ) Stop f (θ () () n+1 θn ) () < ε () ; = 1, 2,, p θ n Note: 1. s = d(θ n ) = H 1 (θ n )g(θ n ). 3. Algorthm fals f H ever becomes sngular at any teraton. 4. Acheve a mnmum of f (.) f H s postve defnte. 5. Algorthm may locate only a local mnmum. 6. Algorthm may oscllate. The algorthm can be gven a nce geometrc nterpretaton scalar θ. To fnd an extremum of f (.), solve f(θ) θ = g(θ) = 0. 4
5 5
6 If f(θ) s quadratc n θ, then the algorthm converges n one teraton: 6
7 In general, dfferent choces of θ 0 may lead to dfferent solutons, or no soluton at all. 7
8 Example f(θ) = 3θ 4 4θ (Where we actually know the answer) locate mnmum Analytcally: g(θ) = 12θ 3 12θ 2 = 12θ 2 (θ 1) H(θ) = 36θ 2 24θ = 12θ(3θ 2) Turnng ponts at = 0, 0, 1. H(0) = 0 H(1) = 12 saddlepont mnmum Algorthm θ n+1 = θ n H 1 (θ n )g(θ n ) θ 0 = 2 (say) θ 1 = 2 ( ) = 1.5 θ 2 = 1.5 ( ) = 1.2 θ 3 = 1.2 ( ) = 1.05 etc. Try: θ 0 = 2; θ 0 = 0.5 8
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