solutionJIS.dvi

Similar documents
Microsoft Word - 表紙.docx

1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2

6.1 (P (P (P (P (P (P (, P (, P.

renshumondai-kaito.dvi

6.1 (P (P (P (P (P (P (, P (, P.101

y = x x R = 0. 9, R = σ $ = y x w = x y x x w = x y α ε = + β + x x x y α ε = + β + γ x + x x x x' = / x y' = y/ x y' =

日本内科学会雑誌第102巻第4号

29

Part () () Γ Part ,

Ł\”ƒ-2005

第90回日本感染症学会学術講演会抄録(I)

n Y 1 (x),..., Y n (x) 1 W (Y 1 (x),..., Y n (x)) 0 W (Y 1 (x),..., Y n (x)) = Y 1 (x)... Y n (x) Y 1(x)... Y n(x) (x)... Y n (n 1) (x) Y (n 1)

kou05.dvi

i

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4

v er.1/ c /(21)

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

all.dvi

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

68 A mm 1/10 A. (a) (b) A.: (a) A.3 A.4 1 1

y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4


10:30 12:00 P.G. vs vs vs 2

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

1 (1) () (3) I 0 3 I I d θ = L () dt θ L L θ I d θ = L = κθ (3) dt κ T I T = π κ (4) T I κ κ κ L l a θ L r δr δl L θ ϕ ϕ = rθ (5) l

st.dvi

2011de.dvi

( )/2 hara/lectures/lectures-j.html 2, {H} {T } S = {H, T } {(H, H), (H, T )} {(H, T ), (T, T )} {(H, H), (T, T )} {1

untitled

日本内科学会雑誌第98巻第4号

日本内科学会雑誌第97巻第7号

tnbp59-21_Web:P2/ky132379509610002944

抄録/抄録1    (1)V

untitled

x (x, ) x y (, y) iy x y z = x + iy (x, y) (r, θ) r = x + y, θ = tan ( y ), π < θ π x r = z, θ = arg z z = x + iy = r cos θ + ir sin θ = r(cos θ + i s

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

DVIOUT-HYOU

Z: Q: R: C: 3. Green Cauchy

医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.

パーキンソン病治療ガイドライン2002

研修コーナー

I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%

151021slide.dvi

変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,

Akito Tsuboi June 22, T ϕ T M M ϕ M M ϕ T ϕ 2 Definition 1 X, Y, Z,... 1


Morse ( ) 2014

A B P (A B) = P (A)P (B) (3) A B A B P (B A) A B A B P (A B) = P (B A)P (A) (4) P (B A) = P (A B) P (A) (5) P (A B) P (B A) P (A B) A B P

1 12 *1 *2 (1991) (1992) (2002) (1991) (1992) (2002) 13 (1991) (1992) (2002) *1 (2003) *2 (1997) 1

1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =

B ver B

newmain.dvi

DVIOUT

chap10.dvi

9. 05 L x P(x) P(0) P(x) u(x) u(x) (0 < = x < = L) P(x) E(x) A(x) P(L) f ( d EA du ) = 0 (9.) dx dx u(0) = 0 (9.2) E(L)A(L) du (L) = f (9.3) dx (9.) P

: , 2.0, 3.0, 2.0, (%) ( 2.

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

A

( z = x 3 y + y ( z = cos(x y ( 8 ( s8.7 y = xe x ( 8 ( s83.8 ( ( + xdx ( cos 3 xdx t = sin x ( 8 ( s84 ( 8 ( s85. C : y = x + 4, l : y = x + a,

I A A441 : April 21, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) Google

N cos s s cos ψ e e e e 3 3 e e 3 e 3 e

waseda2010a-jukaiki1-main.dvi


II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

x x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {

, = = 7 6 = 42, =

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

液晶の物理1:連続体理論(弾性,粘性)

A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

f(x) = f(x ) + α(x)(x x ) α(x) x = x. x = f (y), x = f (y ) y = f f (y) = f f (y ) + α(f (y))(f (y) f (y )) f (y) = f (y ) + α(f (y)) (y y ) ( (2) ) f

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

nsg02-13/ky045059301600033210

i

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

untitled


本文/目次(裏白)

untitled

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

A 2 3. m S m = {x R m+1 x = 1} U + k = {x S m x k > 0}, U k = {x S m x k < 0}, ϕ ± k (x) = (x 0,..., ˆx k,... x m ) 1. {(U ± k, ϕ± k ) 0 k m} S m 1.2.

201711grade1ouyou.pdf

n (1.6) i j=1 1 n a ij x j = b i (1.7) (1.7) (1.4) (1.5) (1.4) (1.7) u, v, w ε x, ε y, ε x, γ yz, γ zx, γ xy (1.8) ε x = u x ε y = v y ε z = w z γ yz

ii

73

2012 IA 8 I p.3, 2 p.19, 3 p.19, 4 p.22, 5 p.27, 6 p.27, 7 p

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

() Remrk I = [0, ] [x i, x i ]. (x : ) f(x) = 0 (x : ) ξ i, (f) = f(ξ i )(x i x i ) = (x i x i ) = ξ i, (f) = f(ξ i )(x i x i ) = 0 (f) 0.

『共形場理論』

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

Transcription:

May 0, 006 6 morimune@econ.kyoto-u.ac.jp /9/005 (7 0/5/006 1 1.1 (a) (b) (c) c + c + + c = nc (x 1 x)+(x x)+ +(x n x) =(x 1 + x + + x n ) nx = nx nx =0 c(x 1 x)+c(x x)+ + c(x n x) =c (x i x) =0 y i (x i x) = y i x i y i x = = = y i x i (y 1 x + y x + + y n x) y i x i x(y 1 + y + + y n ) y i x i x(ny) 1

x i (y i y) = y i x i x i y = y i x i y(nx) (x i x)(y i y) = y i x i nyx n 1 (x i x)(y i y) = 1 y i x i yx n n 1. (a) x = 1 3 (1 1+ +3 3+4 4 4.5.5) = 5 3, 5 x 3, y = 1 3 (1 4+ 3+3 +4 1 4.5.5) = 5 3, 5 3 /( 5 3 5 3 )= 1. (b) (a) = 1 3 (1 1+ +3 3+4 4 4.5.5) = 5 3, (c) x y = 1 3 (1 1+ +( 3) ( 3) + ( 4) ( 4) 4 ( 1) ( 1)) = 6 3, = 1 3 (1 ( 4) + ( 3) + 3 +4 1 4 ( 1).5) = 10 3 10 3 /( 6 3 5 3 )= 10 13 y =( 4, 3, 3, 4) -3.5 = 1 3 (1 ( 4) + ( 3) + 3 ( 3) + 4 ( 4) 4 ( 3.5).5) = 0

1.3 x : (1,),(,4),(-1,),(-,4) y : (1,),(,4),(1,-),(,-4) (1,),(,4),(-1,-),(-,-4), 1.4 1.5 t = 0 55.18 1685.36 = 5 110.36 113 5. t -.191 19 t 5% -1.73-1.65 t = 33.71 17 = 17 33.71 1849.79 135 17 1 4 (t 1.089) 16 t 1.75 1.65 1.5 t = 0.1 17 16 ( 0.1) 0.4 1 ( 0.1) 15 t t 0.389.1 y x x xy y 1 1 1 1 1 3+ 9 1=1.5 1 1.5 = 0.5 7 4 14 49 3+ 9 =6 7 6=1 10 3 9 30 100 3+ 9 3=10.5 10 10.5 = 0.5 18 6 14 45 150 18 0 3

b = 45 6 18/3 14 (6 6/3) = 9 a = (18/3) 9 (6/3) = 3. y x x xy y -5-1 1 5 5 1 0 0 0 1 4 1 1 4 16 0 0 9 4 b = 9 0 0 = 9, a =0 9 0=0..3 b = 00 980 (t t)y t (t t) 00 980 t t t = t 1979 t = t 1979 t t =(t 1979) (t 1979) = t t b = b 00 980 (t t )y t 00 980 (t t ) = 00 980 (t t)y t (t t) 00 980 4

.4 β = (x t x)(y t y) (x t x), γ = (y t y)(x t x) (y, t y) r = (y t y)(x t x) (y t y) n (x t x) r = β γ.5 r =0.567 1.31 = 0.698. 100 β = (x t x)(y t y) (x t x) α = y βx β = (x t x )(yt y n ) n = 100(x t x)10(y t y) (x t x ) n (100(x = 1 t x)) 10 β α = y β x =10y 1 10 β 100x =10 α ŷ i =1+x i 5

x 100 10ŷ i =10+0x i 10ŷ i =10+0 1 100 (100x i) =10+0. (100x i ).6 y i = µ + u i µ (y i µ) = = µ (y i µ) (y i µ) µ (y i µ) = =0 (y i µ) (y i µ) =0 ( ) µ y i n µ =0 µ = 1 n y i = y ( ) ŷ i = µ 6

ŷ i µ, RSS = = (y i µ) (y i y) TSS ESS = (ŷ i y) =0..7 y i = βx i + u i β (y i βx i ) = β (y i βx i ) = (y i βx i ) β (y i βx i ) = (y i βx i )x i ( ) = { =0 y i x i β x i x i } β = y ix i n. x i y i = βx i y i βx i, ( ) x i 0 7

.8 8.1 ln(439/479) = 0.087. 8. R ntt =0.0001(0.11) + 0.71699(0.6)R nikkei 0.00 0.% 8.3 0.005 0.087 0.005 = 0.09 (R ntt 0.005) = 0.0006(0.034) + 0.71699(0.6)(R nikkei 0.005). t P 0.97t P 0.047 3 3.1 x t =0, z t =0, x =0, z =0. z tx t =0, y t = n α y t x t = β x t y t z t = γ 3 z t 1 = 6 α =4 β 8

9=6 γ. y y c x z 7 1 1 1 1 1 + 1 ( 1) + 3 ( 1) = 3 1 1 4 7 4 1 1 1 + 1 (1) + 3 ( 1) = 5 1 1 4 7 3 9 1 0 1 + 1 (0) + 3 ( 1) = 4 1 1 7 4 16 1 1 1 + 1 ( 1) + 3 (1) = 9 1 1 4 7 5 5 1 1 1 + 1 (1) + 3 11 (1) = 1 1 4 7 6 36 1 0 1 + 1 (0) + 3 10 (1) = 1 1 1 91 0 0 1 0 3 TSS =91 1 6 1 1 = 35 R =1 35 3=9 35. y 7 1 x y 1 7 3 7 x y 1 7 3 7 (3.7) 0 y 3.5 3. x 1 z x,y 1 z y β = y t x t (x t ) θ = y t x t (y t ) β θ = ( y t x t ) (x t ) (y t ) 1.66 0.4 = 0.63 9

3.3 RES RES t =0 3.4 RES RESi = = RES t x t =0 RES t z t =0 RES i (y i α βx i γz i ) RES i y i α RES i β RES i x i γ RES i z i RES iy i RES i y i = 3.5 Φ (y i α βx i γz i )y i Φ α = y βx γz y i y = β(x i x)+ γ(z i z)+error (x i x) (z i z) (x i x) =δ(z i z)+error 10

δ {(x i x) δ(z i z)} ( ) δ x i =(x i x) δ(z i z) x i (x i ĉ ζz i ) ( ) ĉ ζ ζ ĉ ĉ = x ζz ( ) ( ) 4 15 (4.8) β K β K β K 4.1 x y y y x x xy 0 0-4 0 3 9-1 1-3 8 64 3 9 4 sum 11 73 0 14 1 y =3.6(5.9) + 1.5(5.)x, R =0.96,RSS =1.17,ESS =31.5 11

4. x z y,x,z y x z y 5/ 1/4 3/4 x 1/4 3/ 0 z 3/4 0 53/ y =0.4565(0.08446)x +0.0830(0.05564)z. t 5.41 0.51 b 0.967 0.9016 t 5% 1.71 4.3 K= K-1=5 DW=1.8 5% U=1.86 L=1.0 U L U F f = 6 3 0.88 0.660 =0.066 1 6 0.660 f = 6 3 0.955 0.94 =0.066 1 6 1 0.955 F 3 1 5% 3.34 3 16 5% 3.4 3 1 5% 3.0 3 4.4 118 log(l) b log(k) c b+c=1+d, b=1+d-c, log Y log L = a + d log L + c{log K log L} + e log H4+error log(y/l)=a + d log L + c log(k/l)+e log H4+error 1

log(y/l) log(k/l) log(y/l) =.01 + 0.045(4.73) log L + 0.31(7.74) log(k/l) + 0.16(6.1) log H4, R =0.9119,RSS=0.05380, t d d 4.5 y i = a + bx i + error x =(x 1 + x )/ (x 1 x) +(x x) = (x 1 x ) (x 1 x)y 1 +(x x)y = (x 1 x ) y 1 + (x x 1 ) y = (x 1 x ) (y 1 y ) b = (y 1 y ) (x 1 x ) y = c + (y 1 y ) (x 1 x ) x c y 1 = c + (y 1 y ) (x 1 x ) x 1 y = c + (y 1 y ) (x 1 x ) x ĉ = y (y 1 y ) (x 1 x ) x 13

4.6 (4.15) (4.5) m (4.5) x mi =1 {y i ( β 1 x 1i + + β K x Ki )} =0 (y i ŷ i )=0 y i = n 4.7 (4.5) ŷ i (y i ŷ i )x mi =0 4.8 ŷ i = β 1 x 1i + + β K x Ki (4.5) (y i ŷ i )ŷ i = β n 1 (y i ŷ i )x 1i + + β n K (y i ŷ i )x Ki =0 14

(y i ŷ i )ŷ i = y i ŷ i y i ŷ i = (ŷ i ) =0, (ŷ i ). n(y) y i = ŷi (y i y)(ŷ i y) = (ŷ i y). ± n (ŷ i y) 4.9 {y i ( β 1 x 1i + + β K x Ki )} β 1,, β K x 1i x i k l {y i ( β 1 x 1i + + β K x Ki )},i m,i l + {y m ( β 1 x 1m + β x m + + β K x Km )} + {y l ( β 1 x 1l + β x m + + β K x Kl )} x 1l =1,x m =0, x 1l =0,x m = 1, x 1l =0,x m =0,,i m,i l {y i ( β 3 x 3i + + β K x Ki )} + {y m ( β 1 + β 3 x 3m + + β K x Km )} + {y l ( β + β 3 x 3l + + β K x Kl )} 15

β 3 β K β 1 β β 1 = y m ( β 3 x 3m + + β K x Km ) β = y l ( β 3 x 3l + + β K x Kl ) 4.10 b=c d=1 b c = e d 1=f b = c + e d =1+f b d y t = a + ex t + c(x t + z t )+z t + fz t + u t (y t z t )=a + ex t + c(x t + z t )+fz t + u t e =0,f =0 F (y t z t )=a + c(x t + z t )+u t (y t z t )=a + c(x t + z t )+u t F 16

4.11 t V (b c) =V (b)+v (c) Cov(b, c) b c V (b)+v (c) Cov(b, c) t = 0.65 0.44 0.6 +0.0 ( 0.0075) =0.6 5% 5 5.1 1.1 Y 1 t 0.58 0.79 1 1. n=5 K=5 k=4 5% L=1.04,U=1.77 DW 1.7 L L 1.3 1.4 TSS=1350 1.5 r =1 dw =0.15. 0.8 =1 70 TSS f = 5 5 4 0.8 1 0.8 =0 4,0 F 1% 4.43 1.6 17

5..1 y x y = y x = x y x (x ) 16 1 (1/4) = 3 1 (1/4) 16 = 8 3 4 19 4 10 4 0.5 =3 10 0.5 16 = 6 4 6 4 6 0.5 4=4 4 0.5 10 = 1-4 1 7+ 3 1+8 3 6 197.1 (, 3), ( 1, 4) -1/3 (,3) 3=a + 1 3 a=11/3. b = 6 (7 + 3)(1 + 8 3)/3 197 (1 + 8 = 0.13969 3) /3 a = 7+ 3 3 +0.13969 1+8 3 3 5.3 3.1 y = 3 1 x, RSS =0 3 E(y t )=E(a + bx t + u t ) = a + bx t + E(u t ) = a + bx t =3.604 V (y t )=E{(a + bx t + u t ) E(a + bx t + u t )} = E{(a + bx t + u t ) (a + bx t )} = E(u t ) = V (u t ) t 1, t 3 3. 18

y x y/σ x/σ x*y* x* y* 14 1 7 7 49 1 4 10 5 10 5 4 6 4 6 4 4 16 36 sum 9 16 41 90 41 b = 41 9 16/3 90 16 16/3 = 1.5 a = a =3 ( 1.5) 16/3 =11. 3.3 y t = a + bx t + u t a = b y t = b(1 + x t )+u t b = (1 + x t )y t (1 + x t ) y t + x t yt = 3+ x t + x t a=0.8 3.4 = 9+41 3+ 16 + 90 =0.4 RSS a = (y t ) a y t b x t y t =41 (11) 9 ( 1.5) 41 = 3.5 RSS 0 = (y t ) b (1 + x t )y t =41 (0.4) (41 + 9) = 1 19

f = 3 1 1 3.5 =6 1,1 F 5% 1 t 3.8 14.44 t d=0 a = b + d y t = d + b(1 + x t )+u t d t 6 6.1 y t = α + β 0 x t + β 1 x t 1 + β x t + ε t β 0 = γ 0 β 1 = γ 0 + γ 1 β = γ 0 +γ 1 3β γ 3β β β β 0 β 1 γ 1 = β 1 β 0 γ 1 = β β 1 β β 1 = β 1 β 0 β =β 1 β 0 0

P=3 q= y t = α + β 0 x t + β 1 x t 1 + β x t + β 3 x t 3 + ε t β 0 = γ 0 β 1 = γ 0 + γ 1 + γ β = γ 0 +γ 1 +4γ β 3 = γ 0 +3γ 1 +9γ 1 4β 1 3β β 1 β 0 = γ 1 + γ β β 1 = γ 1 +3γ β 3 β = γ 1 +5γ (β β 1 ) (β 1 β 0 )=γ (β 3 β ) (β β 1 )=γ (β 3 β ) (β β 1 )=(β β 1 ) (β 1 β 0 ) 6. y t = α + β 0 x t + β 1 x t 1 + β x t + ε t, β 0 = γ 0 β 1 = γ 0 + γ 1 1

β = γ 0 +γ 1 β 1 = γ 0 γ 1 =0 γ 0 = γ 1 β 0 = γ 0 β 1 = γ 0 + γ 1 =β 0 β = γ 0 +γ 1 =3β 0 y t = α + β 0 (x t +x t 1 +3x t )+ε t, 6.3 6.4 6.5 6.6 6.7 6.8 SUR

6.9 6.10 6.11 1 1 W trend tax W g G X 1 C W p X P 0 0-1 0 0 0 0 0 1 W 0 0 0 1 0 0 0 1 0 C α 3 0 0 0 0 0-1 0 0 W p 0 γ 3 0 0 0 γ 0-1 γ 1 X 0 0 0 0 1 1 1 0-1 5 1 5 7 [ 1] (0,1) 1 P ( 1 3 < X< 3 ) 7.1 1/ 1/1 P ( 1 1 3 < X< 3 )=P{ 1 1 = P { 1 1 6 <Z<1 1 6 } = P { <Z<} 0.05 ( 1 3 1 1 ) < 1 1 (X 1 1 ) < ( 1 3 1 )} 1 X (a,b) 3

0.0454 1 [ ] 8 α=4 P (m <X<m) E(X) = = E(X )= m m = α α 1 m = m m = α α m x αm α x α 1 dx αm α x α dx x αm α x α 1 dx αm α x 1 α dx V (X) = α α α m ( α 1 m) α = (α )(α 1) m. P (m <X<m+ m ) = P (m α α 1 m<x α α 1 m<m + m α α 1 m) = P ( 1 α 1 m<x α α 1 m< α (α 1) m) = P ( n( 1 (α 1) (α ) m) <Z< n α = P ( n α 1 (α ) α m α <Z< n(α ) (α 1) (α ) α ) m = P ( <Z< 4 ) = P ( <Z<) (α 1) (α ) m m ) α 4

95 7.1 X,Y 7. X = 1 0 1 Y = 1 1 4 4 81 4 0 81 4 1 81 81 16 81 16 81 81 16 81 16 81 Z X = 1 0 1 Y = 1 1 0.5 0 0 0.5 0 0.5 1 0 0.5 1 P (Z = 1) = 1 81, P (Z = 0.5) = 8 81, P (Z =0)= 4 81, P (Z =0.5) = 3 81, P (Z =1)= 16 81 n=3 X 1 1 1 ( 1 3 )3, 1 3( 1 3 ) ( 3 ), 1 1 3( 1 3 )( 3 ), 1 0 ( 3 )3, E(Z) = ( 1 3 )3 +3( 1 3 ) ( 3 ) 3(1 3 )( 3 ) +( 3 )3 =( 1 3 )3 +3( 1 3 ) ( 3 )+3( 1 3 )( 3 ) +( 3 )3 =(( 1 3 )+( 3 ))3 =( 1 3 )3 5

E(Z )=( 1 3 )3 +3( 1 3 ) ( 3 )+3(1 3 )( 3 ) +( 3 )3 =(( 1 3 )+( 3 ))3 = (1) 3 =1 V (Z) =E(Z ) E(Z) =1 ( 1 3 )6 E(Z) =( 1 3 )n 7.3 V (Z) =1 ( 1 3 )n x 1 E(X) = 0 x xdx = 3 E(X )= x xdx =1 0 V (X) =1 ( 3 ) = 1 9 6

7.4 P(-1)=1/3,P(1)=/3. {X 1,X,X 3 } s = 1 3 1 {(X 1 X) +(X X) +(X 3 X) } = 1 {X 1 + X + X 3 3X } 1. -1 s =0, ( 1 3 )3.. 1-1 s = 4 3, 3( 1 3 ) ( 3 ). 3. 1 1-1 s = 4 3, 3( 3 ) ( 1 3 ). 4. 1 s =0, ( 3 )3. s 0 4 3 s =0, ( 1 3 )3 +( 3 )3 = 9 7 = 1 3. s = 4 3, 3( 3 ) ( 1 3 )+3(1 3 ) ( 3 )= 3. 4 3 3 = 8 9. 7.5 X s X s 1 (-1,-1) 3 1 3 = 1 9-1 0 (-1,1) or (1,-1) 9 + 9 = 4 9 0 (1,1) 3 3 = 4 9 1 0 X s X = 1 X =0 X =1 s s 1 4 5 =0 9 0 9 9 s 4 = 0 X 1 9 4 9 0 9 4 4 9 9 1 X s E(X) =( 1) 1 9 +0 4 9 + (1) 4 9 = 1 3 E(s )=0 5 9 + () 4 9 = 8 9 E(X s )=0. Cov(X,s )= 1 3 8 9 = 8 7. 7

7.6 A= A= B B= 0.5 0 0.5 B= 0 0.5 0.5 A 0.5 0.5 1 A= A= B B= 0 0.5 0.5 B= 0.5 0 0.5 A 0.5 0.5 1 A= A= B B= 0.4 0.1 0.5 B= 0.1 0.4 0.5 A 0.5 0.5 1 A= A= B B= 0.1 0.4 0.5 B= 0.4 0.1 0.5 A 0.5 0.5 1 A= A= B B= 0.5 0.5 0.5 B= 0.5 0.5 0.5 A 0.5 0.5 1 8

7.7 Z=XY Z Z 0 1 4 1 P 15 + 4 15 + 1 15 = 11 4 15 15 0+0 0 Z 0 1 P 11 15 4 15 7.8 3 1. V=0 ( 11 15 )3.. V=1 3( 11 15 ) ( 4 15 ). 3. V= 3( 11 15 )( 4 15 ). 4. V=3 ( 4 15 )3. X s X s 1 (-1,-1,-1) 7-1 0 (-1,-1,1) 3 7 = 6 1 4 7 3 3 (-1,1,1) 3 4 7 = 1 1 4 7 3 3 8 (1,1,1) 7 1 0 X s X = 1 X = 1 3 X = 1 3 X =1 s s 1 8 9 =0 7 0 0 7 7 s = 4 6 1 3 0 X 1 7 7 6 7 18 7 0 7 1 8 7 7 1 X s 9