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1 2
2 : n =1, 2,,
3 weak law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ),i=1, 2,,n 3. σi 2 = V (X i ) σ 2,i=1, 2,,n ɛ>0 ( lim P X 1 + X X n n n ) µ ɛ =0 X µ converge in probability 3
4 : Y = X 1 + X X n n E(Y )=µ V (Y )= σ σ2 n n 2 σ2 + + σ 2 n 2 P ( Y E(Y ) ɛ) V (Y ) ɛ 2 P ( Y µ ɛ) σ2 nɛ 2 ( lim P X 1 + X X n n n = σ2 n ) µ ɛ =0 4
5 : n =
6 X µ = E(X),. ν 2k = E(X µ) 2k ɛ>0 P ( X µ ɛ) ν 2k ɛ 2k : D = {x x µ ɛ} ν 2k = (x µ)2k f(x) dx (x D µ)2k f(x) dx D ɛ2k f(x) dx ɛ 2k f(x) dx D ɛ 2k P ( X µ ɛ) 6
7 4 X 1,X 2,,X n : E(X i )=µ, V (X i )=σ 2,E(X i µ) 4 = ν 4,i=1,,n E( X µ) 4 = 1 [ 1 n 2 n ν 4 + 3(1 1 ] n ) σ4 : ( X µ) 4 = 1 n = 1 n 4 n i=1 n (X i µ) 4 i µ) i=1(x 4 + 4C 2 (X i µ) 2 (X j µ) 2 i j 4C 1 (X i µ)(x j µ)(x k µ)(x l µ) + i j k l + i j 4C 1 (X i µ) 3 (X j µ), E( X µ) 4 = 1 n 4 (nν 4 + n C 2 4 C 2 σ 4 ) 7
8 strong law of large numbers 1. X 1,X 2,,X n 2. µ = E(X i ) σ 2 = V (X i ) i =1, 2,,n ν 4 = E(X i µ) 4 P ( lim n X 1 + X X n n ) = µ =1 X µ converge almost surely (converge with probability 1) 8
9 lim X = µ n ɛ>0 N ɛ, n>n ɛ X µ <ɛ, ɛ >0, n>n ɛ ɛ ɛ N, n>n X µ ɛ, for some n>n 9
10 n >N>ν 4 P n = P ( X µ ɛ ) E( X µ) 4 = 1 n 2 ɛ 4 ɛ 4 < 1+3σ4 ɛ 4 1 n 2 [ 1 n ν 4 + 3(1 1 n ) σ4 ], ɛ>0 N P ( lim X = µ n ) = 1 P ( X µ ɛ for some n>n ) 1 (P N + P N+1 + ) 1 1+3σ4 1 ɛ 4 N (N +1) N 1 i=1 1/N 2 10
11 3 11
12 : X 1,,X n : p =1/2 Bernoulli n i=1 X i 0.35 n=
13 : X 1,,X n : p =1/2 Bernoulli n i=1 X i 0.25 n=
14 : X 1,,X n : p =1/2 Bernoulli n i=1 X i n=
15 : X 1,,X n : p =1/2 Bernoulli n i=1 X i n=
16 1 Central Limit Theorem 1. X 1,,X n 2. E(X i )=µ, V (X i )=σ 2,i=1,,n Y = n( X µ) σ lim P n( X µ) n σ y y = 1 2π e x2 2 dx (1) 16
17 : (1/2) Y X i Y i = X i µ σ E(Y i )=0,V(Y i )=1,Y = 1 n n i=1 Y i Y i M(t) M(0) = E ( e 0 Y i) =1 M (0) = E(Y i )=0 M (0) = E(Yi 2 )=V(Y i)+(ey i ) 2 =1 ψ(t) = log M(t) ψ (t) = M (t) M(t), ψ (t) = M (t)m(t) [M (t)] 2 [M(t)] 2 ψ(0) = 0, ψ (0) = 0, ψ (0) = 1 17
18 : (2/2), θ < t ψ(t) ψ(t) =ψ(0) + ψ (0) 1! t + ψ (0) 2! Y E ( e Yt) = E (e 1 n n i=1 Y it ) = n ) E (e t n Yi i=1 = [ M ( t/ n )] n t 2 + ψ (θ) 3! = exp { n log M ( t/ n )} = exp n ( t/ n ) 2 + ψ (θ) 6 t 3 = exp 2 t2 + ψ (θ) 6 n { } 1 exp 2 t2 n t 3 = 1 2 t2 + ψ (θ) t 3 6 ( t/ n ) 3 18
19 2 (De Moivre-Laplace Limit Theorem) X lim P a n X Bi(n, p) X np b =Φ(b) Φ(a) (2) np(1 p) Φ(x) Φ(x) = x 1 2π e 1 2 y 2 dy 19
20 : X 1,,X n p Bernoulli, X = n X i Bi(n, p) E(X i ) = p i=1 V (X i ) = p(1 p) X np np(1 p) = = ni=1 X i np np(1 p) n( X p) p(1 p) N(0, 1) 20
21 4 χ 2 F 21
22 χ 2 X 1,,X n N(0, 1) X 1,,X n : Y = X X 2 n f(y) 3 f(y) f(y) = ( ) n n y 2 1 e y 2 I(0, ) (y) (3) Γ(n/2) 2 1 (3) n χ 2 22
23 X 1,,X n χ 2 : (1/2) M Y (t) = E [exp(yt)] n = E exp = (Ee X2 1 t) n i=1 X 2 i t Ee X2 1 t = = = t 1 ex2 2π e x2 2 dx 1 2π e x2 2 (1 2t) dx 1 1 2t (1 2t >0) M Y (t) =(1 2t) n 2 23
24 χ 2 : 2/2 0 e yt f(y) dy = (x = y 2 yt ) = = = = 0 0 ( M Y (t) ( ) n e yt n y 2 1 e y 2 dy Γ(n/2) 2 ( ) n n y 2 1 e y 2 +yt dy 0 Γ(n/2) 2 ( ) n 1 1 ( ) n 2 2 Γ(n/2) 2 1 2t x ) n x n 2 1 e x dx 1 2t Γ(n/2) 0 ) n t ( 2 1 e x ( ) 2 1 2t dx 24
25 χ 2 M(t) =(1 2t) n 2 M (t) = n(1 2t) n 2 1 M (t) = n(n + 2)(1 2t) n 2 2 E(Y )=M (0) = n V (Y )=E(X 2 ) (EX) 2 = M (0) n 2 =2n 25
26 χ 2 χ n= n=2 n=4 n=5 n=
27 χ 2 : χ n= n= n=
28 χ 2 Z 1 = X 2 1,,Z n = X 2 n 1 χ2 E(Z i )=1,V(Z i )=2 Z = 1 n ni=1 X 2 i = Y/n n n(y/n 1) n( Z 1) = 2 2 N(0, 1) = Y N(n, 2n) 28
29 F X: m χ2 Y : n χ2 X, Y : Z = X/m Y/n f(z) 4 f(z) f(z) = ( ) m 1 m 2 m z B( m 2, n 2 ) 2 1 (1 + m m+n z) 2 I (0, (z) (4) n n 2 (4) (m, n) F 29
30 F : (1/3) X, Y X, Y f(x, y) = 2 m 2 Γ(m/2) x m 2 1 e x 2 Ix>0 (x) 2 n 2 Γ(n/2) y n 2 1 e y 2 Iy>0 (y) 1 1 z = x/m y/n = nx my w = y x = m n zw y = w Jacobian (x, y) (z, w) = x z y z x w y w = m n w 0 1 m n z = m n w 30
31 F : (2/3) f(z) = = = = g(z,w) dw f( m n zw, w)m n wdzdw 2 m+n 2 ( ) m m Γ( m 2 )Γ( n 2 ) n zw 2 1 e m 2n zw w n 2 1 e w 2 ( m n ) m 2 z m m+n 2 Γ( m 2 )Γ( n 2 ) 0 w m+n 2 1 e w(mz+n) 2n dw m n wdwdz t = w(mz+n) 2n w = 2nt 2n, dw = mz + n mz + n dt 31
32 F : (3/3) 0 w m+n 2 1 e w(mz+n) 2n dw = = = 0 ( ( ( 2nt mz + n 2n mz + n 2n mz + n ) m+n 2 1 e t 2n ) m+n 2 ) m+n 2 0 t m+n 2 e t dt ( m + n Γ 2 ) mz + n dt f(z) = m m 2 n n 2 = ( m n ) m 2 Γ( m+n 2 ) z Γ( m 2 )Γ( n 2 ) B( m 2, n 2 )z m 2 1 (1 + m n m 2 1 (mz + n) m+n 2 m+n z) 2 This completes the proof. 32
33 (m, n) F F n E(Z) = (n >2) n 2 V (Z) = 2n2 (m + n 2) m(n 2) 2 (n 4) n>4 ( 1 E Y ) ( y 2 = w ) = E(Z) =E 0 1 y ( 1 1 Γ(n/2) 2 ) n/2 ) n/2 y n 2 1 e y 2 dy ( 1 1 = y n e y 2 dy Γ(n/2) 2 0 ( ) 1 1 n/2 = 2 n 2 2 w n e w 2 dw Γ(n/2) 2 0 = 1 Γ((n 2)/2) = 1 1 n 2 Γ(n/2) = 1 n 2 ( ) X/m Y/n = n m E(X) E ( ) 1 Y = n m m 1 n 2 = 33 n n 2
34 Z = X/m Y/n, n =1 : F m=1 m=3 m=5 m=7 m=
35 Z = X/m Y/n, m =1 : F n=1 n=3 n=5 n=7 n=
36 Z = X/m Y/n, m = n : F H2,2L H4,4L H6,6L H10,10L H20,20L
37 5 : t 37
38 t X N(0, 1) Y χ 2 (n) X, Y : T = X f(t) Y/n 5 f(t) f(t) = 1 nb( n 2, 1 2 ) 1+ t2 n n+1 2 (5) 3 (5) n t 38
39 t : (1/3) X Y X, Y f(x, y) =f X (x) f Y (y) = 1 ( ) n e x π 2 Γ(n/2) y n 2 1 e y t = x y/n w = y x = w y = w n t x t y t x w y w = w n t 2 wn 0 1 w = n 39
40 t : (2/3) f(t) = = = g(t, w) dw f( w n t, w) w n dw ( ) n 1 e 1 w 1 2 n t π 2 = 1 ( ) n π 2 n Γ(n/2) 0 s = w 2 + w 2n t2, w = Γ(n/2) w n 2 1 e w 2 w n dw w n e w 2 w 2n t2 dw s 1/2+t 2 /(2n), dw = t2 2n 1 ds 40
41 f(t) = = t : (3/3) ( )n 1 2π n Γ(n/2) 2 + t2 2n ( )n 1 2π n Γ(n/2) 2 + t2 2n 1 2 n 2 0 s n e s ds 1 2 n 2 Γ ( n ) ( n B 2 2), 1 = Γ ( n Γ ( n , t f(t) = = ) ) ( 2 Γ 1 2) = πγ ( n 2 ) Γ ( n ) 2 ( 1 2 2)n ( nb n 2, 1 2 2) + t2 2n 1 ( nb n 2, 1 ) 1+ t2 n+1 2 n n 2 41
42 t : 1+ t2 n n+1 2 = 1+ t2 n n t 2 t 2 n ( n+1 2 ) f(t) = n e t2 2 1 ( nb n 2, 1 ) 1+ t2 n 2 n 1 e t2 2 2π n t. 42
43 1. t : E(T ) = E X Y/n = E(X) E = 0 E 1 Y/n 1 =0 Y/n 2. E(T )=0 V (T ) = E(T 2 ) = ne X 2 Y ( ) 1 = ne(x 2 ) E Y 1 = n 1 n 2 = n n 2 43
44 t n=1 n=3 n=5 n=10 n=
45 7 X 1,X 2,,X n : N(0, 1) 1. X = n 1 n i=1 X i N(0, 1/n) 2. S 2 = n i=1 (X i X) 2 χ 2 (n 1) 3. X S 2 : (1) X M(t) M(t) = E = E = n i=1 (e Xt ) ( e 1 n n i=1 X it ) E ( e t n X ) i = [ exp ( t 2 /2n 2)] n = exp ( t 2 /2n ) N(0, 1/n) 45
46 X S 2 (1/3) (2), (3) Y 1 Y 2. Y n 1 Y n = n(n 1) 1 n(n 1) 1 n(n 1) n 1 n(n 1) 1 n 1 n 1 n 1 n X 1 X 2. X n 1 X n = a 1 a 2. a n 1 a n X 1 X 2. X n 1 X n = A X 1 X 2. X n 1 X n a i a t i =1, i =1,,n a i a t j =0, i j =1,,n 46
47 X S 2 (2/3) Y 1,,Y n : E(Y i )=0,V(Y i )=1,i =1,,n E(Y i Y j )=0,i j =1,,n Y 1,,Y n n i=1 X 2 i = n i=1 Y 2 i : n i=1 Y 2 i = [Y 1,Y 2,,Y n ][Y 1,Y 2,,Y n ] t = ( [X 1,X 2,,X n ] A t)( A [X 1,X 2,,X n ] t) = [X 1,X 2,,X n ] ( A t A ) [X 1,X 2,,X n ] t = [X 1,X 2,,X n ] I n n [X 1,X 2,,X n ] t = n Xi 2 i=1 47
48 X S 2 (3/3) n 1 i=1 Yi 2 χ 2 (n 1) Y n Y n = n X n 1 i=1 Y 2 i = n i=1 = n Xi 2 i=1 = n i=1 = S 2 X 2 i Y 2 n n( X) 2 (X i X) 2 48
49 6 : 49
50 confidence interval θ f(x θ) X = {X 1,X 2,,X n }: X 1,X 2,,X n i.i.d. f(x θ) 0 <α<1 P [L(X) θ U(X)] = 1 α (6) 4 (6) [L, U] =[L(X) θ U(X)] θ 1 α confidence interval L, U lower confidence limit upper confidence limit 1 α [L, U] confidence coefficient 1 α =0.9,
51 1 one sample problem 1. 1 : X 1,,X n i.i.d. N(µ, σ 2 ) 2. µ 3. σ 2 2 two sample problem µ x µ y 3. σ 2 x /σ2 y i.i.d. X 1,,X m N(µ x,σx 2 ) i.i.d. Y 1,,Y n N(µ y,σy 2 ) 51
52 µ (σ 2 ) α z α zα 1 2π e x2 2 dx = α 8 1. : X 1,,X n i.i.d. N(µ, σ 2 ) 2. σ 2 : 3. X = n 1 n i=1 X i µ 1 2α σ P X n z 1 α µ X + σ z 1 α =1 2α n 52
53 µ (σ 2 ): X Y = E( X) =µ, V ( X) = σ2 n X N(µ, σ2 n ) n( X µ) N(0, 1) σ 1 2α = P [ Y z 1 α ] n( X µ) = P z 1 α z 1 α σ σ = P X n z 1 α µ X + σ z 1 α n 53
54 µ (σ 2 ) 9 1. tα n 1 : n 1 t α 2. : i.i.d. X 1,,X n N(µ, σ 2 ) 3. µ, σ 2 : 4. X = n 1 n i=1 X i,s 2 = 1 n 1 n i=1 (X i X) 2 µ 1 2α S P X n t1 α n 1 µ X + S t n 1 n 1 α =1 2α 54
55 Y = Z = µ (σ 2 ): n( X µ) σ (n 1)S2 σ 2 N(0, 1) = n i=1 X i X σ 2 χ 2 (n 1) Y,Z T = Y Z/(n 1) = n( X µ) S t(n 1) 1 2α = P [ T t n 1 1 α n( = P t n 1 1 α X µ) t n 1 1 α S ] S = P X n t n 1 1 α µ X + S t n 1 n 1 α 55
56 µ (σ 2 ): , s 2 x = N(µ, σ 2 ) µ 1 2α = 95%, 90% [ x s n t n 1 1 α, x + s 50.0,α =0.025, 0.05 s = 1 n 1 (xi x) 2 = n n 1 s2 x = n t n 1 ] 1 α n = 100, x = = t = t =1.98, t = t = % : [ , ] = [48.0, 52.0] 90% : [ , ] = [48.3, 51.7] 56
57 7 : 57
58 σ χα n 1 : n 1 χ 2 α 2. : i.i.d. X 1,,X n N(µ, σ 2 ) 3. X = n 1 n i=1 X i,s 2 = n (X i X) 2 µ 1 2α P S2 χ1 α n 1 σ 2 S2 χ n 1 α i=1 =1 2α 58
59 S 2 σ 2 = n σ 2 : i=1 1 2α = P (X i X) 2 = P σ 2 χ 2 (n 1) χ n 1 α S2 χ n 1 1 α S2 σ 2 χn 1 1 α σ 2 S2 χ n 1 α 59
60 σ 2 : , s 2 x = N(µ, σ 2 ) σ 2 1 2α = 95%, 90% [ S 2 χ1 α n 1, ] S 2 χ n 1 α n = 100, x =50.0,s 2 = (x i x) 2 = ns 2 x = 100.2,α=0.025, 0.05 χ = , χ =73.361, χ = , χ = % : [ / , /73.361] = [77.9, 136.3] 90% : [ / , /77.046] = [81.2, 129.8] 60
61 σ 2 x /σ2 y Fα n 1;m 1 : (n 1,m 1) F α 2. : X 1,,X m i.i.d. N(µ x,σ 2 x), Y 1,,Y n i.i.d. N(µ y,σ 2 y) 3. X i,y j : X = 1 m Ȳ = 1 n m i=1 n j=1 X i,sx 2 = 1 m 1 Y i,sy 2 = 1 n 1 m i=1 n j=1 (X i X) 2 (Y i Ȳ )2 σ 2 x /σ2 y P 1 2α S2 x Sy 2 Fα n 1;m 1 σ2 x σy 2 S2 x S 2 y F1 α n 1;m 1 =1 2α 61
62 σ 2 x /σ2 y : (m 1) S2 x σx 2 (n 1) S2 y σy 2 = 1 σx 2 = 1 σy 2 m i=1 n j=1 (X i X) 2 χ 2 (m 1) (Y i Ȳ )2 χ 2 (n 1) Z = S2 y /σ2 y S 2 x /σ2 x = S2 y S 2 x σ2 x σ 2 y F (n 1,m 1) 1 2α = P = P Fα n 1;m 1 S2 y Fα n 1;m 1 Sx 2 Sy 2 62 S 2 x σ2 x σ 2 y σ2 x σ 2 y F n 1;m 1 1 α F n 1;m 1 1 α Sx 2 Sy 2
63 µ x µ y ( ) (σ 2 x,σ 2 y: ): X 1,,X m i.i.d. N(µ x,σ 2 x), Y 1,,Y n i.i.d. N(µ y,σ 2 y) 2. X i,y j : m X = 1 X i, m Ȳ = 1 i=1 n µ x µ y 1 2α P X Ȳ z 1 α σ 2 x m + σ2 y n j=1 Y i, n µ x µ y X Ȳ + z 1 α σx 2 m + σ2 y =1 2α n 63
64 µ x µ y ( ): X Ȳ E( X Ȳ ) = E( X) E(Ȳ )=µ x µ y V ( X Ȳ ) = V ( X)+V (Ȳ )=σ2 x m + σ2 y n Z = ( X Ȳ ) (µ x µ y ) σ 2 x /m + σ 2 y /n N(0, 1) 1 2α = P [ Z z 1 α ] = P z 1 α ( X Ȳ ) (µ x µ y ) σ 2 x /m + σy 2/n z 1 α 1 2α = P X Ȳ z 1 α σ 2 x m + σ2 y n µ x µ y X Ȳ + z 1 α 64 σx 2 m + σ2 y n
65 µ x µ y ( ) ( ): X 1,,X m i.i.d. N(µ x,σ 2 ), Y 1,,Y n i.i.d. N(µ y,σ 2 ) 2. X i,y j : X = 1 m Ȳ = 1 n m i=1 n j=1 ˆσ 2 = S2 x + S 2 y m + n 2 X i,sx 2 = m (X i X) 2 i=1 Y i,sy 2 = n (Y i Ȳ )2 µ x µ y 1 2α I = X Ȳ t m+n 2 1 m + 1 n ˆσ 1 α i=1 X Ȳ + t m+n 2 1 α 1 m + 1 n ˆσ 65
66 µ x µ y ( ): U = X Ȳ N (µ x µ y, ( 1 m + 1 n) σ 2 ) V = S2 x σ 2 + S2 y σ 2 χ2 (m + n 2) X,Ȳ,S2 x,s2 y : W =(U E(U))/ Var(U) T = W/ V/(m + n 2) = [ ( X Ȳ ) (µ x µ y ) ] / ( 1 m + 1 n )σ2 (S 2 x /σ 2 + S 2 y /σ2) /(m + n 2) = ( X Ȳ ) (µ x µ y ) 1 m + 1 n ˆσ t(m + n 2) 1 2α = P ( X Ȳ ) (µ x µ y ) 1 m + 1 n ˆσ t m+n 2 1 α 66
67 8 : Introduction 67
68 5( ) statistical hypothesis simple hypothesis composite hypothesis 1 X 1,,X n N(θ, 25) H : θ =17 H : θ 17 68
69 6( ) test of a statistical hypothesis reject 2 X 1,,X n N(θ, 25) H : θ 17 γ: X > n H 69
70 7( ) sample space n X 1,,X n X = {(x 1,,x n ); (x 1,,x n ) (X 1,,X n ) } 8( ) critical region X C X X (x 1,,x n ) C = C 3 X 1,,X n N(θ, 25) H : θ 17 γ: X > / n H : C γ = {(x 1,,x n ); x> / n} 70
71 2 9( ) null hypothesis, alternative hypothesis H 0 H 1 10 (2 ) two types of error H 0 H 0 1 type I error 1 1 size of type I error H 1 H 1 2 type error 2 2 size of type error 71
72 11 ( ) power function f(x θ) H 0 γ π γ (θ) =P [H 0 X 1,,X n f(x θ)] θ π γ (θ) [0, 1] 4 X 1,,X n N(θ, 25) H 0 : θ 17 X > / n H 0 π γ (θ) = P [ X > / n θ R] = P n( X θ) > / n θ 5 5/ n = 1 Φ / n θ 5/ n 72
73 π γ (θ) =1 Φ ( 17+10/ n θ 5/ n ) with n =
74 : H 0 : θ 17. π γ (θ) θ =
75 12 ( ) parameter space X f(x θ) θ Θ Θ ={θ θ } 5 X Bi(n, θ) P (X = x) = n C x θ n (1 θ) n x Θ={θ 0 θ 1} X N(θ, 1) f(x) = 1 (x θ)2 exp{ } 2π 2 Θ={θ <θ< } 75
76 13 ( ) size of test (significance level) : f(x θ) H 0 : θ Θ 0 γ π γ (θ) = sup [π γ (θ)] θ Θ 0 6 X 1,,X n N(θ, 25) H 0 : θ Θ 0 = {θ θ 17} γ: X > / n H0 : π γ (θ) =1 Φ ( ) 17+10/ n θ 5/ n = sup 1 Φ / n θ θ 17 5/ n = 1 Φ(2)
77 9 : 77
78 14 ( ) likelihood ratio test X 1,,X n f(x θ) H 0 : θ = θ 0 H 1 : θ = θ 1 λ = f(x 1 θ 0 ) f(x n θ 0 ) f(x 1 θ 1 ) f(x n θ 1 ) ni=1 f(x i θ 0 ) = ni=1 f(x i θ 1 ) k>0 λ k : H 0 λ>k : H 0 78
79 15 ( ) most powerful test H 0 : θ = θ 0 ; H 1 : θ = θ 1 α (0 <α<1) : γ π γ (θ 0 ) α γ α 1. π γ (θ 0 )=α 2. π γ (θ 1 ) π γ (θ 1 ) 79
80 Neyman-Pearson 14 (Neyman-Pearson ) Neyman-Pearson Lemma X 1,,X n f(x θ) H 0 : θ = θ 0 H 1 : θ = θ 1 : λ = n i=1 f(x i θ 0 )/ n i=1 f(x i θ 1 ) α(0 <α<1), 1. C = {(x 1,,x n ) λ k } 2. α = P [(X 1,,X n ) C θ = θ 0 ] γ α 80
81 Neyman-Pearson : α γ γ C P [(X 1,,X n ) C θ = θ 0 ] α A L 0 = A L 1 = n A i=1 n A i=1 f(x i θ 0 ) dx 1 dx n f(x i θ 1 ) dx 1 dx n π γ (θ 1 )= C L 1,π γ (θ 1 )= C L 1 C L 1 L 1 C 81
82 C L 1 Neyman-Pearson : ( ) C L 1 = C C L 1 1 k C C L 0 1 k = 1 k ( C C L 0 + = 1 k ( C L 0 This complets the proof. C C L 1 ) L 0 C = 1 (α size of γ) k 0 C C L 0 C C L 0 C C L 0 ) C C L 0 82
83 X 1,,X n N(θ, 1) H 0 : θ = θ 0 vs. H 1 : θ = θ 1 (θ 1 >θ 0 ) α λ = n i=1 1 2π e (X i θ 0 )2 2 / n n i=1 1 e (X i θ 1 )2 2 2π = exp 1 [ (Xi θ 0 ) 2 (X i θ 1 ) 2] 2 i=1 = exp [n(θ 0 θ 1 ) X n ] 2 (θ20 θ21 ) λ k X k 83
84 k α = P [ X k θ = θ 0 ] = P [ n( X θ 0 ) n(k θ 0 )] = 1 Φ [ n(k θ0 ) ] k = θ n Φ 1 (1 α) X θ n Φ 1 (1 α) H 0 α =0.025 Φ 1 (1 α) =Φ 1 (0.975) 1.96 α =0.05 Φ 1 (1 α) =Φ 1 (0.95) 1.65 α =0.10 Φ 1 (1 α) =Φ 1 (0.90)
85 10 : 85
86 f(x θ) X 1,,X n θ 1 α (L, U) =(L(X 1,,X n ),U(X 1,,X n )) H 0 : θ = θ 0 γ: θ 0 (L, U) H 0 θ 0 (L, U) H 0 P [H 0 H 0 ] = P [θ 0 (L, U) H 0 ] = 1 P [θ 0 (L, U) H 0 ] = α 86
87 15 ( ) 1. tα n 1 : n 1 t α 2. : i.i.d. X 1,,X n N(µ, σ 2 ) 3. µ, σ 2 : 4. X = n 1 n i=1 X i,s 2 = 1 n 1 n i=1 (X i X) 2 µ 1 2α S P X n t1 α n 1 µ X + S t n 1 n 1 α =1 2α 87
88 ( ) H 0 : µ = µ 0 C = { (x 1,,x n ); T >t n 1 1 α = (x 1,,x n ); n( X µ0 ) S ] >tn 1 1 α = P [H 0 H 0 ] = P n( X µ0 ) S = 1 P n( X µ0 ) S = 1 (1 2α) = 2α >tn 1 1 α tn 1 1 α 88
89 ( ) 20 ( kg) α =0.1, kg H0 : µ = µ 0 =49.5 n =20, x =51.80, s 2 = (x i x) 2 /(n 1) = T = n( X µ0 ) S = 20( ) =1.90 =0.1 t n = t =1.73 <T H 0 =0.05 t n = t =2.09 >T H 0 89
90 16 ( ) 1. χα n 1 : n 1 χ 2 α 2. : i.i.d. X 1,,X n N(µ, σ 2 ) 3. X = n 1 n i=1 X i,s 2 = n (X i X) 2 µ 1 2α P S2 χ1 α n 1 σ 2 S2 χ n 1 α i=1 =1 2α 90
91 ( ) H 0 : σ 2 = σ 2 0 C = { (x 1,,x n ); S 2 <σ 2 0 χn 1 α and S 2 >σ0 2 } χn 1 1 α = P [H 0 H 0 ] = P { S 2 <σ0 2 χ n 1 α and S 2 >σ0 2 χ n 1 1 α = 1 P { σ0 2 χn 1 S 2 σ0 2 χn 1 = 1 P α S 2 χ n 1 1 α = 1 (1 2α) = 2α σ 2 0 S2 χ n 1 α 1 α } } 91
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