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1 Chinese Journal of Applied Probability and Statistics Vol.25 No.3 Jun (,, 20024;,, 54004).,,., P,. :,,. : O22... (Credibility Theory) 20 20, 80. ( []).,.,,,.,,,,.,. Buhlmann Buhlmann-Straub ( [2])., 999 Frees, E.W., ( [3]).,,, ( [5]).,,,. [3],,,.,,, P, F. (066006) (200707MS33) ,

2 : 29 : n ( ),,. T, T +,. y ij i =, 2,, n, j =, 2,, T, y ij y ij = β + jβ 2 + α i + ɛ ij i =, 2,, n, j =, 2,, T, (.) β, β 2,. jβ 2,. β, β 2, α i i, ɛ ij, Var (α i ) = σα, 2 Var (ɛ ij ) = σɛ 2, α i N(0, σα), 2 ɛ ij N(0, σɛ 2 ), α i ɛ ij. i : Y i = X i β + U i α i + ɛ i i =, 2,, n, (.2) Y i = (y i, y i2,, y it ) 2, X i = (X i, X i2 ) =.., U i = T, T T, ɛ i = (ɛ i, ɛ i2,, ɛ it ), β = (β, β 2 ), Y i Cov (Y i ) = σα 2 T T + σɛ 2 I T = V i. (.3). 2., ( [3]). 2. Y N, E(Y ) = Xβ, Var (Y ) = V. W, E(W ) = λ t β, Var (W ) = σw 2. Cov (W, Y ) = Cov W Y, Y W W = E(W ) + Cov W Y V [Y E(Y )] = λ t β + Cov W Y V (Y Xβ).,.

3 292 : Y = Xβ + Uα + ɛ, (2.) Y =(y, y 2,, y T, y 2,, y nt ) =(Y, Y 2,, Y n), X = n X i, U =I n T, α = (α, α 2,, α n ), ɛ = (ɛ, ɛ 2,, ɛ T, ɛ 2,, ɛ nt ) = (ɛ, ɛ 2,, ɛ n ). T +, E(y i,t + α i ) = β + (T + )β 2 + α i, i =, 2,, n. (2.2) W i = β + (T + )β 2 + α i, i =, 2,, n, E(W i ) = β + (T + )β 2, j = i, Cov (W i, Y j ) = σ 2 α T, j i, Cov (W i, Y j ) = 0, [3] (3), W i W Y i = T i = E(W i ) + Cov Wi Y V i (Y i X i β) ( = T σ2 ) ( α T σ 2 ) α σɛ 2 + T σα 2 β + (T + ) 2(σɛ 2 + T σα) 2 β 2 + T σ2 α σɛ 2 + T σα 2 Y i, (2.3) 2. T y ij.. (.) ŷ i,t + = ( ξ) β + (T + )( ξ/2) β 2 + ξy i, i =, 2,, n, (2.4) ξ = T σ 2 α/( σ 2 ɛ + T σ 2 α), Y i = T T y ij..,,,., β, β 2, σ 2 α σ 2 ɛ,. 3.,. ( H T, H = T / ) T, H (T ) T, H HH = I T, H T = 0. H, H = (H 2,, H T ), H 2 = (/ 2, / 2, 0,, 0), H 3 = (/ 2 3, / 2 3, 2 2 3, 0,, 0),

4 : 293. Z i = H Y i, Z i = (z i, z i2,, z it ), Z i(2) = (z i2, z i3,, z it ). : ) Z i : Cov (Z i ) = diag(σɛ 2 + T σα, 2 σɛ 2,, σɛ 2 ), {z it : i =,, n, t =,, T }. 2) H T / T, z i = ( T ) / T y it, t= E(zi ) = T β + ( + T ) T 2 β 2. {z i : i =,, n},, ( ( N T β + + T ) β 2 ), σɛ 2 + T σα 2 = N(µ, στ 2 ) i =, 2,, n. 2 3) H T = 0, E(Z i(2) ) = HX i2 β 2, η = (η 2, η 3,, η T ) = HX 2 β 2, t = 2, 3,, T, {z it : i =,, n} N(η t, σ 2 ɛ ). {z i : i =,, n} T {β + [( + T )/2]β 2 } σ 2 ɛ + T σ 2 α µ = z = z i, σ τ 2 = SS = n n (z i z ) 2. (3.) t = 2, 3,, T, {z it : i =,, n} N(η t, σ 2 ɛ ), β 2, σ 2 ɛ : [ / T β 2 = (H t X i2 ) 2] T H t X i2 z it, (3.2) σ ɛ 2 T = (z it H t X i2 β2 ) 2. (3.3), 3. β, β 2, σ 2 α σ 2 ɛ β = z + T [ / T β 2, β2 = T 2 t=2 σ α 2 = T (SS σ2 ɛ ) = ( T n σ 2 ɛ = T (z i z ) 2 σ 2 ɛ (z it H t X i2 β2 ) 2, (H t X i2 ) 2] T ), SS = i z ) n (z 2, z = n z i, H t X i2 z it, H t H, H (T ) T, HH = I T, H T = 0.

5 β, β 2 β, β 2 ; { / [ ]} σ 2 ɛ, (/T ) {[n/(n )] σ 2 τ [/[ ]] σ 2 ɛ } σ 2 ɛ, σ 2 α. :,,, [6] 4..4 β 2 β 2, µ µ, µ = T {β +[( + T )/2]β 2 }, β β. {/[ ]} σ 2 ɛ σ 2 ɛ, [n/(n )] σ 2 τ σ 2 τ, σ 2 τ = σ 2 ɛ +T σ 2 α, (/T ){[n/(n )] σ 2 τ [n(t )/[n(t ) ]] σ 2 ɛ } σ 2 α.?,,. 4., P, ( [4]). 4. Z h E(Z h ) ( E(Z h q / ) = C b b j r ) h { r / j a b q } k k Γ[a k ( + h) + ξ k ] Γ[b j ( + h) + η j ], (4.) C, Z 0 E(Z 0 ) =, C = q / r Γ[b j + η j ] Γ[a k + ξ k ]. (4.2) a k =O(n) (k =, 2,, r) b j =O(n) (j =, 2,, q), ξ k (k =, 2,, r) η j (j =, 2,, q), q r n. r q a k = b j, [ r f = 2 ρ = f [ r ξ k q η j ], 2 (r q) (4.3) a ω 2 = 6ρ 2 { r B 2(ξ k ) a 2 k q B 3[( ρ)a k + ξ k ] B 2 (h) B 3 (h) Bernoulli : ] b B 2(η j ), (4.4) q B 2 (h) = h 2 h + 6, B 3(h) = h h2 + 2 h. } b 2 j B 3 [( ρ)b j + η j ], (4.5)

6 : 295 P( 2 ln Z x) = P(χ 2 (f) x) + O(n ), (4.6) P( 2ρ ln Z x) = P(χ 2 (f) x) + O(n 2 ), (4.7) P( 2ρ ln Z x) = P(χ 2 (f) x) + ω 2 [P(χ 2 (f + 4) x) P(χ 2 (f) x)] + O(n 3 ).(4.8) [4] , H 0 : σα 2 = 0 H : σα 2 0., σ 2 τ = σ 2 ɛ + T σ 2 α 3 z i. H 0 : σ 2 τ = σ 2 ɛ H : σ 2 τ σ 2 ɛ, σ 2 α = 0, β, β 2 σ 2 ɛ z = n n : Λ = = β = z + T β 2, T 2 [ / T β 2 = (H t X i2 ) 2] T H t X i2 z it, (4.9) σ ɛ 2 = [ (z i z ) 2 + T (z i H t X i2 β2 ) 2], (4.0) nt z i. 3. max L(β, β 2, σɛ 2 ) H 0 max L(β, β 2, σɛ 2, σα) 2 H 0 H ( [ (z i z ) 2 + T nt t=2 ( ) n/2 ( (z i z ) n 2 ]) nt/2 (z it H t X i2 β2 ) 2 T ). (4.) n(t )/2 (z it H t X i2 β2 ) 2 Λ,. 4, 3, χ 2, 2 ln Λ L χ 2 (), n. P P(χ 2 () 2 ln Λ).

7 296 P : P = n (z i z ) 2, P 2 = T (z it H t X i2 β2 ) 2, Λ = (nt ) nt/2 ( n n/2 () n(t )/2 P = P /(P + P 2 ), Λ = P ) n/2( P + P 2 P 2 (nt ) nt/2 n n/2 () n(t )/2 (P )n/2 ( P ) n(t )/2. P + P 2 ) n(t )/2. (4.2), P Beta ( (n )/2, [ ]/2 ). Λ h,, E(Λ h ) = r = 2, a = n 2, a 2 = T nt h/2 P. f, ρ ω 2 f = [ 2 ( 2 ) 2 [ 2 ρ = f ω 2 = 6ρ 2 { 2 = 4 6n 2 ρ 2 { T 2 B 3 = 4 6n 2 ρ 2 {( h ( h + Γ((nT 2)/2) (T ) n(t )h/2 Γ((n )/2)Γ([ ]/2) Γ([n(h + ) ]/2)Γ([(h + ) ]/2). (4.3) Γ([nT (h + ) 2]/2), ξ = ξ 2 = 2 2, q =, b = nt 2, η =, a B 2( /2) ( ) 2 (2 ) ] =, ] b j B 2 ( ) = T 2 26T + 26, 6T } a 2 k B 3[( ρ)a k + ξ k ] b 2 B 3[( ρ)b j + η j ] [ T 2 32T + 26 [ 5T 2 20T + 26 B 3 2T (T ) [ T 2 38T + 38 ] 2(T ) ] + (T ) 2 h3 2 (T ) 2 h 2 T 2 h 3 T 2 h3 3 )}, (T ) 2 B 3 2T ]} ) 3 ( h (T ) 2 h2 2 ) T 2 h2 3, h = 5T 2 20T T (T ), h 2 = T 2 32T T, h 3 = T 2 38T (T )

8 : 297, O(n ), O(n 2 ) O(n 3 ) P P(χ 2 () 2 ln Λ)+O(n ), P(χ 2 () 2ρ ln Λ)+O(n 2 ), P(χ 2 () 2ρ ln Λ)+ ω 2 [P(χ 2 (5) 2ρ ln Λ) P(χ 2 () 2ρ ln Λ)]+O(n 3 ). O(n 3 ) P., α = 0.05, T = 5, σɛ 2 =, n = 0 n = 30 n = 50 σ 2 α = σ 2 α = σ 2 α = ,,. : H 0 : β 2 = 0, 5. H 0 : β 2 = 0 H : β 2 0. y ij = β + α i + ɛ ij i =, 2,, n, j =, 2,, T. (5.) : β = n z i, σ ɛ 2 = T σ 2 = n n (z i T β ). : Λ = = = T zit, 2 σ α = T ( σ2 σ ɛ 2 ), max L(β, β 2, σɛ 2 ) H 0 max L(β, β 2, σɛ 2, σα) 2 H 0 H ( ) n/2 ( T ) n(t )/2 (z i z ) 2 zit 2 n nt ( ) n/2 ( (z i z ) n 2 T ) n(t )/2 (z it H t X i2 β2 ) 2 ( T ) n(t )/2 T n(t )/2 zit 2 ( T ). (5.2) n(t )/2 (T ) n(t )/2 (z it H t X i2 β2 ) 2 t=2

9 298 Q = T zit 2, Q 2 = T, Q σ 2 ɛ χ 2 (). (z it H t X i2 β2 ) 2, Q 2 σɛ 2 χ 2 ( ). Λ, Λ, Q /Q 2,. Q Q 2 = Q Q 2 Q 2 +. (5.3) Q /Q 2, (Q Q 2 )/Q 2,. Q Q 2 σ 2 ɛ χ 2 (), Q 2, F =, F F (, ), P (Q Q 2 )/ Q 2 /[ ]. (5.4) P = P(F (, ) F ). (5.5), n = 0, 30, 50, α = 0.05, T = 5, β =, 2 n = 0 n = 30 n = 50 β 2 = β 2 = β 2 = ,,. 6. Hachemeister (975) [7] [3],, : y ij = β + jβ 2 + α i + ɛ ij i =, 2,, 5, j =, 2,, 2.,,. : H 0 : β 2 = 0 H : β 2 0.

10 : 299 (5.5) F, P :,. P = P(F (, 54) F ) < , H 0 : σα 2 = 0 H : σα P, P(χ 2 () 2 ln(λ)) < 0.00.,.,. 3 2, 5 3, : ŷ i,3, i =, 2,, 5, 4,,

11 300. [],, :, [2] Buhlmann, H. and Gisler, A., A Course in Credibility Theory and Its Application, Springer, [3] Frees, E.W., Young, V.R. and Luo Y., A longitudinal date analysis interpretation of credibility models, Insurance: Mathematics and Economics, 24(999), [4] Anderson, T.W., An Introduction to Multivariate Statistical Analysis (Second Edition), New York: John Wiley, 984. [5],,, :,, 998. [6],, :, [7] Hachemeister, C.A., Credibility for regression models with application to trend in credibility theory and application, Proceedings of the Berkeley Actuarial Research Conference on Credibility, Academic Press, New York, 975, [8] Lo, C.H., Fung, W.K. and Zhu, Z.Y., Generalized estimating equations for variance and covariance parameters in regression credibility models, Insurance: Mathematics and Economics 39(2006), [9] Antonio, K. and Beirlant, J., Actuarial statistics with generalized linear mixed models, Insurance: Mathematics and Economics, 40(2007), Estimating and Testing Parameter for Regression Credibility Model with Linear Trend Tang Guoqiang (School of Finance and Statistics, East China Normal University, Shanghai, ) (Department of Mathematics and Physics, Guilin University of Technology, Guilin, ) In this paper, the parameters of regression credibility model with linear trend are estimated and tested. Orthogonal transformation is used to estimate parameter and unbiased estimate of parameters are obtained. Likelihood ratio test is used to test randomness and linear trend. The better P-value of likelihood ratio test is got and Monte-Carlo simulation is performed. Keywords: Regression credibility model, orthogonal transformation, likelihood ratio test. AMS Subject Classification: 62P05.

[9] R Ã : (1) x 0 R A(x 0 ) = 1; (2) α [0 1] Ã α = {x A(x) α} = [A α A α ]. A(x) Ã. R R. Ã 1 m x m α x m α > 0; α A(x) = 1 x m m x m +

[9] R Ã : (1) x 0 R A(x 0 ) = 1; (2) α [0 1] Ã α = {x A(x) α} = [A α A α ]. A(x) Ã. R R. Ã 1 m x m α x m α > 0; α A(x) = 1 x m m x m + 2012 12 Chinese Journal of Applied Probability and Statistics Vol.28 No.6 Dec. 2012 ( 224002) Euclidean Lebesgue... :. : O212.2 O159. 1.. Zadeh [1 2]. Tanaa (1982) ; Diamond (1988) (FLS) FLS LS ; Savic

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