ii 2. F. ( ), ,,. 5. G., L., D. ( ) ( ), 2005.,. 6.,,. 7.,. 8. ( ), , (20 ). 1. (75% ) (25% ). 60.,. 2. =8 5, =8 4 (. 1.) 1.,,
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1 (1 C205) (2015) , P. G., 1995.,. 3., , , 2007,.
2 ii 2. F. ( ), ,,. 5. G., L., D. ( ) ( ), 2005.,. 6.,,. 7.,. 8. ( ), , (20 ). 1. (75% ) (25% ). 60.,. 2. =8 5, =8 4 (. 1.) 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0.
3 (1). (2).,. (3),,. 1.2 ( ), Ω., E., E Ω. ( ), E P (E) = E Ω..,.., Ω. 1.1 ( ) ( ). ( )., , 2 (K,Q,J). [11/221] 1 10, 1. [1023/1024] , ? [1/221, 1/270725]
4 ( ) , 1, 2. [2/10] 1.4 ( ) A,B 2. A 2/5, B 3/5. 3, A 2, B 1.? [.] 1.5 ( ) ( ). 3, ( ) 2.,. 1.,, 1 ( ).? , 3., 10. [2/10] 4 A,B 2. A p, B q = 1 p. 4, A 2, B 1., A, B,. 1.3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ).. ( ) ( ) ( ) ( ), ( ), ( ) ( )
5 : Ω: ( ) = (, F: ( ) P : Ω,. ω Ω E F (E = F. a < b.) Ω E c E E, E 1 E 2 E n E F, E 1 E 2 E n E F = 2.1 ( ), Ω., E P (E) = E Ω,. 2.2 (Ω ( ) ),,., P (X = k) = λk k! e λ, k = 0, 1, 2,...., λ > 0. λ.
6 4 2 5 (, ) λ.? [ e λ e λ ] 2.3 (Ω ). 2. [2/3] 1, ( ), ( )?, 3 1 : 2 : 3. [30] , 3. [1/2] 7, 30cm 40cm, 5cm. [1/2] Ω E. P (E) = E Ω,..,,,, , 10.,. 2 (, ). [9/25]
7 E P (E), 3, P Ω., P (E) E. (i) 0 P (E) 1. (ii) P (Ω) = 1. (iii) [ ] E 1, E 2, F (, i j E i E j = ), ( ) P E n = P (E n ). n=1, 3 (Ω, F, P ). n=1 2.3 AB 3. (, 3 3.) B A O 1/3, 1/2, 1/4] A C O C O B [ ], ( )?..
8 A, B, C,. A, B, C., ( ) 1. (1) 9 2. (2) 0, 1,..., (3) 5. 3 ( ) , , 4, 5 1 P, P ( 5 ) P, 2 P. 1/3. 6 A, B, C A B = A (A c B), A B C = A (A c B) (A c B c C),. 7 E, F P (E) = 1, P (F ) = 0., A. P (A E) = P (A F ) = P (A). 8 ( ) A, B, C, P (A B C) = P (A) + P (B) + P (C). 9 A 1, A 2,..., A n, ( n ) P A k 1. P (A B) P (B C) P (C A) + P (A B C) k=1 P (A c k) k=1
9 (1 ): x 1, x 2,..., x n ( ):, 2 (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ) (A) ( ) 120 (A) 120 (B) ,.
10 (A) (B) n x 1, x 2,..., x n 1,. :,. x = 1 x i n ( ): x 1, x 2,..., x n,. ( ): x 1, x 2,..., x n,., ( ). 2. (box plot): i=1 x : σ 2 = 1 n (x i x) 2 = 1 n i=1 x 2 i x 2 i=1 : σ = σ 2 = 1 n (x i x) 2 i=1 x, σ 2 x, σ x.
11 x, y (x, y) 3.2 (x) (y). (A) (B) (A) (B) n 2 (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ), x = 1 n ȳ = 1 n x i, i=1 y i, i=1 σ 2 x = 1 n σ 2 y = 1 n (x i x) 2, i=1 (y i ȳ) 2 i=1 : σ xy = 1 n (x i x)(y i ȳ) = 1 n i=1 x i y i xȳ i=1 ( ) σ xy = σ yx. σ xx = σ 2 x (, σ xx ). ( ) r xy = r yx. 3.3 ( ( )) r = r xy = σ xy σ x σ y = σ xy σxx σyy x i = x i x σ x, ỹ i = y i ȳ σ y x, y, x, ỹ, r xy = σ xỹ = r xỹ (3.1)., x, y, x, ỹ.
12 r xy (A) (B) A B σ xy = 1 n (x i x)(y i ȳ) = 1 n i=1 x i y i xȳ i= (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ) σ x > 0, σ y > 0., r = 1., r = (x i, y i ) y = f(x) (x, y )., 1 y = ax + b y x. 1 y = ax + b, x = x i y i, (x i, y i ) ϵ i y i = ax i + b + ϵ i
13 Q = ϵ 2 i = i=1 (y i ax i b) 2 i=1 a, b. Q a, b 2,.,. Q a = Q b Q a = 2an(σ2 x + x 2 ) 2n(σ xy + xȳ) + 2bn x, Q = 2bn 2nȳ + 2an x b = 0, 1, a 0 = σ xy σ 2 x y = a 0 x + b 0., b 0 = ȳ a 0 x (3.2) (x 1, y 1 ), (x 2, y 2 ),..., (x n, y n ), x, y y ȳ = σ xy (x x) = σ y r(x x) (3.3) σx 2 σ x., y, x., r. x x = σ xy (y ȳ) = σ x r(y ȳ) (3.4) σy 2 σ y ( ) 2, ( x, ȳ), ( ). 3.8 A,B (x) (y). A, x = , ȳ = 63.59, σ 2 x = , σ 2 y = , σ xy = , x,., y y = 0.73x (3.5) x = 0.27y (3.6)
14 12 3. (3.6) 1/ , (3.5)., B,, x,, y. x = , ȳ = 51.05, σ 2 X = , σ 2 Y = , σ XY = y = 0.72x x = 0.58y (A) (B) (0, 1), (1, 3), (3, 6), (4, 6) x. [y 4 = 1.3(x 2)] 10 5,. 11 x, y, σ xy σ x σ y x, y r xy. a, b, x = ax + b., a 0.,. r x y = { rxy, a > 0, r xy, a < 0
15 (1) 1, 0. (2) 5. (3). (4) 1,. ( )., x, y, z, t,...., 0 x 1, x 0 1.,,.,,.., X, Y, Z, T, ( ) 4.1 3, X. X {0, 1, 2, 3}., P (X = 0) = 1 8, P (X = 1) = 3 8, P (X = 2) = 3 8, P (X = 3) = 1 8,. X,, X ( ). X {a 1, a 2,..., a i,... }, P (X = a i ) = p i, i = 1, 2,...,
16 14 4, X., ( ) X., p i 0, p i = 1. (p i = 0 a i, p i = 0.) 4.2 X {a 1, a 2,..., }, p i = P (X = a i ). X m σ : i m = m X = E[X] = i a i p i = x xp (X = x), σ 2 = σx 2 = V[X] = E[(X m) 2 ] = E[X 2 ] m 2 = (a i m) 2 p i = a 2 i p i m 2 i i = x (x m) 2 P (X = x) = x x 2 P (X = x) m 2. σ X = σ 2 X = E[(X m) 2 ] 4.3 3, 3 100, 2 50, 1 10, 80. 1,. [m = 25, σ 2 = 2400, σ = 20 6] 4.3 ( ) X, P (X = a) = X ( ), F (x) = F X (x) = P (X x), x R, X. ( ). 4.5 L, X. X, 0, x L/2, 2x L F (x) =, L/2 x L, L 1, x L,
17 4.3. ( ) X, F ( x) = x f(t)dt F (x) = f(x) f(x) = f X (x) X. (F (x).), P (a X b) = b a f(x)dx. (,,.) f (x) f(x). a b x f(x) 0, + f(x)dx = ( 4.5 ) L, X. X. 4.8 f(x) ( X) : m = m X = E[X] = + xf(x) dx, σ 2 = σ 2 X = V[X] = E[(X m) 2 ] = E[X 2 ] m 2 = + σ = σ X. (x m) 2 f(x) dx = + x 2 f(x) dx m ( 4.7 ) L, X. X,,. [m = 3L/4, σ 2 = L 2 /48, σ = L/4 3] 12 L Y. Y,,,. 13 1, X. X,,,.
18 X, Y, σ XY = Cov (X, Y ) = E[(X E[X])(Y E[Y ])] = E[XY ] E[X]E[Y ]., : r XY = σ XY σ X σ Y = r XY 1.. σ XY σxx σy Y ( ) X, ( ) Y. X, Y. E[X] = X\Y / /36 2 2/36 1/ /36 3 2/36 2/36 1/ /36 4 2/36 2/36 2/36 1/ /36 5 2/36 2/36 2/36 2/36 1/36 0 9/36 6 2/36 2/36 2/36 2/36 2/36 1/36 11/36 11/36 9/36 7/36 5/36 3/36 1/36 1, E[Y ] = , V[X] = V[Y ] =, Cov (X, Y ) = , r = , 1 X, 6 Y. X, Y. [r XY = 1/5] 13 2 ( ) X, ( ) Y. X Y. [ 4.12] 14 O R 1, O X. X,,,. 15 X 1, X 2,..., X n, : [ ] V X k = Cov (X j, X k ). k=1 k=1 V[X k ] + j k
19 17 第 5 章 基本的な離散分布 5.1 二項分布 表が出る確率が p であるコインを n 回投げたとき, 表の出る回数 X の分布 ( ) n k P (X = k) = p (1 p)n k, k = 0, 1, 2,..., k を二項分布といい, B(n, p) で表す. 特に, B(1, p) を成功確率 p のベルヌーイ分布という. 例 題 5.1 B(4, 1/2) と B(4, 1/4) を図示せよ. 5.2 k k P (X = k) P (X = k) 幾何分布 表が出る確率が p であるコインを投げ続けるとき, 表が初めて出るまでに出た裏の回数 X の 分布は P (X = k) = p(1 p)k, k = 0, 1, 2,.... この分布をパラメータ p の幾何分布という. (待ち時間の分布として重要) 補注 文献によっては, 表が出る確率が p であるコインを投げ続けるとき, 表が初めて出るまで に要したコイン投げの回数 (表が出た回も 1 回と数える) Y の分布を幾何分布といっている. P (Y = k) = p(1 p)k 1, 5.3 k = 1, 2,.... ポアソン分布 確率変数 X がパラメータ λ > 0 のポアソン分布に従うとは, P (X = k) = λk λ e, k! k = 0, 1, 2,....
20 λ = 2. λ = 0.5, λ = 1? k P (X = k) ( ) 1 3,., 1, (1) 1. [0.05] (2) 5. [0.18] 5.4 ( ) B(n, p) np = λ ( ), n, p 0, λ ? 1 365,, 5 5 X B(50, 1/365)., P (X = k) (k = 0, 1, 2, 3, 4). [ : , , , , ] 5.4 (m) (σ 2 ) (2 ) B(1, p) p p(1 p) B(n, p) np np(1 p) ( p) (1 p)/p (1 p)/p 2 ( λ) λ λ 5.5 ( ) {0, 1, 2,... } X, G(z) = z k P (X = k) X ( X )., k=0 E(X) = G (1), E(X 2 ) = G (1) + G (1), V(X) = G (1) + G (1) G (1) 2., ? X, E[X]. 17,.
![19 第 6 章 基本的な連続分布 6.1 一様分布 区間 [a, b] からどの点も同等な確からしさで 1 点を選ぶときのモデルとして現れる. 1, a x b f (x) = b a 0, その他 6.2 指数分布 ランダム到着の待ち時間をモデル化するときに現れる.](/docs-images/92/110008826/images/21-0.jpg "λ > 0 を定数として { λe λx, x 0 f (x) = 0, x<0 6.")
21 19 第 6 章 基本的な連続分布 6.1 一様分布 区間 [a, b] からどの点も同等な確からしさで 1 点を選ぶときのモデルとして現れる. 1, a x b f (x) = b a 0, その他 6.2 指数分布 ランダム到着の待ち時間をモデル化するときに現れる. λ > 0 を定数として { λe λx, x 0 f (x) = 0, x<0 6.3 正規分布 (ガウス分布) N (m, σ 2 ): 平均 m, 分散 σ 2 の正規分布 (またはガウス分布) { } 1 (x m)2 f (x) = exp 2σ 2 2πσ 2 N (0, 1): 標準正規分布 他に, χ2 -(カイスクエア) 分布, t-分布, F -分布 (後出)
22 (m) (σ 2 ) [a, b] (a + b)/2 (b a) 2 /12 ( λ) 1/λ 1/λ 2 N(m, σ 2 ) m σ 2 18,.,, ( ). + e x2 dx = π Z ( Z N(0, 1) )., (1) : P (Z 1.15), P (Z 1.23), P ( Z < 2.4) (2) a. P (Z a) = 0.33, P (Z < a) = 0.75, P ( Z a) = ( ) X N(m, σ 2 ), ax + b N(am + b, a 2 σ 2 ),, Z = X m σ N(0, 1) 6.3 X N(2, 5 2 ), P (X 0), P ( X 4). 19 X N(20, 4 2 ), P (X > 17.8). [0.7088] 20 X N(50, 10 2 ), P (X > a) = a. [28.3] 21, 5%., 68, 8.,. [ ] ( )., x = x 1 y = y 1, x = x 2 y = y 2, x 1 < x < x 2 y : y = y 2 y 1 x 2 x 1 (x x 1 ) + y 1
23 B(100, 0.4) 6.4,. B(n, p) N(np, np(1 p)), 0 < p < 1, n , 225 ( ( ) ). 22, 4% ,. [0.0901] ( ) B(n, p) X, P (X = k) k. [P (X = k)/p (X = k 1).] 17 1,,, X. X, , 1 12., 600, 1 120,. 19 X N(0, 1), X 2 F (x) = P (X 2 x)., F (x), X 2 1 x 1/2 e x/2, x > 0, f(x) = 2π 0, x 0,.
24 22 6 I(z) = 1 2π z 0 e x2 /2 dx z
25 A, B 2. P (A) > 0, P (B A) = P (A B) P (A) A B. A, B. 7.2 ( ) 10, ,,? X, Y ( X = Y ). P (X 5 Y = 2) P (X + Y 8 X 4). [4/9, 5/9] 23 2 E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) A, B, P (A B) = P (A)P (B). A 1, A 2,..., A i1, A i2,..., A in (i 1 < i 2 < < i n ). P (A i1 A i2 A in ) = P (A i1 )P (A i2 ) P (A in ) 7.5 P (A) > 0, 2 A, B P (B) = P (B A).
26 , 121, 211, , 1 1 A 1, 10 1 A 2, A 3. A 1, A 2, A 3 2, A, B, C, P (A) = a, P (B) = b, P (C) = c. a, b, c. P (A B c ), P (A B), P (A B C), P (A B C) ( ) Ω = A 1 A 2, A 1 A 2 =, B, P (A 1 B) =. P (A 1 )P (B A 1 ) P (A 1 )P (B A 1 ) + P (A 2 )P (B A 2 ) 7.8 (1), A B, 95%, 2%... (2), 100p %,. p. 25, A B, 90%, 5%. (1). [ ] (2). [ ] 20 ( ) (1) T, P (T m + n T m) = P (T n), m, n = 0, 1, 2,.... (2) T, P (T a + b T a) = P (T b), a, b ( ) (1) 1, 6. [2/3] (2) 1, 6. [4/5]
27 25 第 8 章 母数の推定 I 二項母集団の母比率 8.1 視聴率調査 テレビ局では視聴率の獲得にしのぎを削っているようである. 果たして, コンマ以下の数字に 意味はあるのだろうか? 2015 年 5 月 25 日 (月) 5 月 31 日 (日) ドラマ (関東地区) 視聴率ベスト 10 番組名 放送局 連続テレビ小説 まれ 天皇の料理番 ようこそ わが家へ 木曜ドラマ アイムホーム Dr. 倫太郎 警視庁捜査一課９係 花燃ゆ 土曜ワイド劇場 事件 16 火曜ドラマ マザー ゲーム 木曜劇場 医師たちの恋愛事情 ＮＨＫ総合 ＴＢＳ フジテレビ テレビ朝日 日本テレビ テレビ朝日 ＮＨＫ総合 テレビ朝日 ＴＢＳ フジテレビ 放送日 放送開始時刻 分数 15/05/26(火) 8: /05/31(日) 21: /05/25(月) 21: /05/28(木) 21: /05/27(水) 22: /05/27(水) 21: /05/31(日) 20: /05/30(土) 21: /05/26(火) 22: /05/28(木) 22:00-54 視聴率 (%) ビデオリサーチ社による番組平均世帯視聴率 日本の放送エリアは全部で 32 ありますが, それぞれの放送エリアごとに視聴率調査が行な われています. ビデオリサーチでは, 関東地区をはじめ全国 27 地区の調査エリアで, PM シ ステムによる調査とオンラインメータシステムによる調査を実施しています. 日本全国を ひとつの調査エリアとした視聴率調査は実施していません また, 調査対象世帯数は, PM システムによる調査の関東地区 関西地区 名古屋地区で 600 世帯, それ以外のオンライン メータシステムによる調査地区は 200 世帯です. (ビデオリサーチ社のウェッブページから 現在) 参考: 藤平芳紀 視聴率の正しい使い方 (朝日新書) 8.2 標本抽出 調査対象の集団 (母集団) に対して, 全数調査が不可能である場合に, その一部分 (標本) を調 査して全体の性質を推定することが重要である. 標本を 1 個取り出せば, 観測値 x が 1 個得られる. 観測値は取り出された標本ごとに違った数 値となるが, 母集団をよくかき混ぜて無作為に標本を選ぶのなら, 観測値 x の現れ方に母集団
28 26 8 I., X, x X. 1,.,, 1 X 1, 2 X 2,..., n X n., X 1, X 2,..., X n n ( ).,., n, n.,..,, E 2, E p.. E 1, 0. n X 1, X 2,..., X n. k, X k = { 1, k E, 0, k E,, P (X k = 1) = p, P (X k = 0) = 1 p., X 1, X 2,..., X n., f(x 1, X 2,..., X n )., ˆp = 1 X k n. : k=1 (i) E[ˆp] = p ( ) (ii) P lim ˆp = p = 1 [ ] n
29 8.4. ˆp 27, ˆp ( ) (!)., ˆp p., ˆp,. 8.4 ˆp (1) X k B(n, p). k=1 (2) n, B(n, p) N(np, np(1 p)). pn 5, n(1 p) 5. (3), n ( ) p(1 p) ˆp N p, n ˆp p p(1 p)/n N(0, 1). 8.5 α = α/2 α, Z N(0, 1) ( ) P ( z Z z) = 1 α z N(0, 1) α. z α α N α z z p 1 α [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp z, ˆp + z n n
30 28 8 I. 90% (α = 0.1, z = 1.64) 95% (α = 0.05, z = 1.96) 99% (α = 0.01, z = 2.58). 2 ( ) p(1 p) ˆp p z n ˆp p z ˆp(1 ˆp) α 1 0 (1 α) 0% 100% 0 ( ) ( ) n ( ), x 1..., x n (, x k = 0 = 1). ˆp,.,.., 1 α, α.,. 8.1 ( ) %. 95%, 0.141( ) ± ± , 95% 0.01,? [38416] , 51% (NHK )., 90%. 27, 90% 0.01,? , 12.. [ ] 90%, 0.12(1 0.12) 0.12 ± ± ,,.
31 29 9 II ( ), 1, 0., x 1, x 2,... t n = 1 x k n. t n n,. k=1 9.2 ( ) X 1, X 2,..., m., ( ) P X k = m = 1 lim n 1 n k=1 X 1, X 2,..., (iid).,. 9.3 ( ) n X, ( ) P X = m = 1 lim n X. ( ): E[ X] = m
32 30 9 II 9.2 (CLT) 9.4 ( ) X 1, X 2,..., m = 0, σ 2 = 1., ( ) lim P 1 X k x = 1 x e t2 /2 dt. n n 2π, n, k=1 1 n X k N(0, 1). k=1 9.5 m, σ 2 X 1, X 2,..., X n, X, X m σ/ n = 1 n k=1 X k m σ N(0, 1) n., X = 1 n k=1 X k N ) (m, σ2 n n. 28 B(n, p) N(np, np(1 p)) ( - ). 9.3 ( ) m ( ), σ 2 X 1, X 2,..., X n : n ( (iid) ) 1 : X = X k n k=1 m 1 α, [ X z σ n, X + z σ n ], z N(0, 1) α 29, 200, 2.2 g., 1.5 g., g?. [1.992, 2.408]
33 9.4. ( ) ( ) m ( ), σ 2 X 1, X 2,..., X n : n ( (iid) ) 1 : X = X k n k=1 U 2 = 1 n 1 (X i X) 2, S 2 = 1 n i=1 (X i X) 2,. (,, ) 9.6 U 2 : E(U 2 ) = σ 2.,., n, S 2 U N(m, σ 2 ) n X 1,..., X n, i=1 T = X m U/ n t n 1 (n 1) t-,. n t- 1 n B ( n 2, 1 2) ( ) n t2 2 n = Γ( n+1 2 ) n Γ( n 2 )Γ( 1 2 ) ( ) n t2 2 n n n n (1) Γ. Γ(x) = 0 t x 1 e t dt, x > 0.
34 32 9 II (2) B. B(x, y) = 1 0 t x 1 (1 t) y 1 dt = Γ(x)Γ(y), x > 0, y > 0. Γ(x + y) (3) n = t- N(0, 1). (4), n 30 N(0, 1). m 1 α, [ X t U n, X + t U n ], t t n 1 α 9.8 8,. 90% [ x = , u 2 = = , t 7 = ± 0.375] 24,. 95% [33 ± 4.17] g., 8g. 1. [95% 156 ± 2.48] 26 25, 95% 1g? [984] 27 ( ) m, σ, ( ) = x m σ,., 20 80,.
35 9.4. ( ) 33 t P ( T t n (α)) = α n\α
36
37 Sir Ronald Aylmer Fisher ( ) 1. H 0 H T ( ), H 0, < α < 1., H 0., 10%, 5%, 1%., T, T α (P (T W ) = α). ( H 1. ),. 4. T t, W (t W ). t W ( T, H 0 ). α, H 0, H 1. t W. T, α., H , 223.? 1. p. H 0 : p = 1 2 H 1 : p X. H 0, X B(400, 1/2) N(200, 10 2 )., Z = X 200 N(0, 1) 10.
38 α = 0.05., 5% ( ). 5% (= 2.5% ) 1.96, W : z x = 223 Z z = = 2.3., H 0., 5% H 0.,. 5. 1%, 1% 2.58, z = % H 0. H 1 α α α W W W W N(0, 1) α α z α ( ) m, σ 2 n, X = 1 n ) X k N (m, σ2 n k=1 X m σ/ n N(0, 1),, n (. N(m, σ 2 ) ).
39 10.3. P ( ) ( ) 25 mm.,.,, 0.8 mm mm.? [ 5% H 0 : m = 25 ( )] 10.3 ( ). 120,., , [ m. H 0 : m = 120 H 1 : m > 120] 30 ( ), ( ), m = 60 (g).,, m 50 70, σ = 3 ( )., 25,, m = 60? 10.3 P ( ), α H 0.,, H 0. t, H 0, P = t, t P.,,. 32 A P. 33 ( ) 250.,, ? P.
40 H 0, 4. \ H 0 H 0 H 0 2 H 0 1 α: 1 = β: 2 1 = = 2 = = , 215.? 2., α β. θ θ β α c c, H 0,. H 0, ( 2 β). H ( ) ,. 4. ( ),. 5.,.,.
41 39 11 William Sealy Gosset ( ) 11.1 ( ) 11.1 N(m, σ 2 ) n X 1,..., X n, U 2 = 1 (X i n 1 X) 2,. X, i=1 T = X m U/ n t n 1 (n 1) t g g, 10 2 g.,? 11.3 ( ),. 50kg, 50kg. 12 (kg), x = 48.6, u 2 = [ 5% H 0 : m = 50 ( )] (kg), kg, A , A. A. [ 5% ]
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3 : 203 : 0. H = 0 0 2 6 0 () = 6, 2 = 2, 3 = 3 3 H 6 2 3 ϵ,2,3 (2) ψ = (, 2, 3 ) ψ Hψ H (3) P i = i i P P 2 = P 2 P 3 = P 3 P = O, P 2 i = P i (4) P + P 2 + P 3 = E 3 (5) i ϵ ip i H 0 0 (6) R = 0 0 [H,
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
1 1.1 R(x) = 0 y + P (x)y + Q(x)y = R(x)...(1) y + P (x)y + Q(x)y = 0...(2) 1 2 u(x) v(x) c 1 u(x)+ c 2 v(x) = 0 c 1 = c 2 = 0 c 1 = c 2 = 0 2 0 2 u(x) v(x) u(x) u (x) W (u, v)(x) = v(x) v (x) 0 1 1.2
n=1 1 n 2 = π = π f(z) f(z) 2 f(z) = u(z) + iv(z) *1 f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x
n= n 2 = π2 6 3 2 28 + 4 + 9 + = π2 6 2 f(z) f(z) 2 f(z) = u(z) + iv(z) * f (z) u(x, y), v(x, y) f(z) f (z) = f/ x u x = v y, u y = v x f x = i f y * u, v 3 3. 3 f(t) = u(t) + v(t) [, b] f(t)dt = u(t)dt
第７章
5. 推定と検定母集団分布の母数を推定する方法と仮説検定の方法を解説する まず 母数を一つの値で推定する点推定について 推定精度としての標準誤差を説明する また 母数が区間に存在することを推定する信頼区間も取り扱う 後半は統計的仮説検定について述べる 検定法の基本的な考え方と正規分布および二項確率についての検定法を解説する 5.1. 点推定先に述べた統計量は対応する母数の推定値である このように母数を一つの値およびベクトルで推定する場合を点推定
基礎統計
基礎統計 第 11 回講義資料 6.4.2 標本平均の差の標本分布 母平均の差 標本平均の差をみれば良い ただし, 母分散に依存するため場合分けをする 1 2 3 分散が既知分散が未知であるが等しい分散が未知であり等しいとは限らない 1 母分散が既知のとき が既知 標準化変量 2 母分散が未知であり, 等しいとき 分散が未知であるが, 等しいということは分かっているとき 標準化変量 自由度 の t
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yoshi@image.med.osaka-u.ac.jp http://www.image.med.osaka-u.ac.jp/member/yoshi/ II Excel, Mathematica Mathematica Osaka Electro-Communication University (2007 Apr) 09849-31503-64015-30704-18799-390 http://www.image.med.osaka-u.ac.jp/member/yoshi/
1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 +
1.3 1.4. (pp.14-27) 1 1 : f(z = re iθ ) = u(r, θ) + iv(r, θ). (re iθ ) 2 = r 2 e 2iθ = r 2 cos 2θ + ir 2 sin 2θ r f(z = x + iy) = u(x, y) + iv(x, y). (x + iy) 2 = x 2 y 2 + i2xy x = 1 y (1 + iy) 2 = 1
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.
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. ttp://www.morikita.co.jp/books/mid/00571 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. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)
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統計学 第 17 回 講義 母平均の区間推定 Part-1 014 年 6 17 ( )6-7 限 担当教員 : 唐渡 広志 ( からと こうじ ) 研究室 : 経済学研究棟 4 階 43 号室 email: kkarato@eco.u-toyama.ac.j website: htt://www3.u-toyama.ac.j/kkarato/ 1 講義の目的 標本平均は正規分布に従うという性質を
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