2 1 Introduction
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- けんじ いんそん
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1 toyoizumi@waseda.jp
2 2 1 Introduction Bayes
3 7 Poisson Poisson Poisson χ 2 ( t F t
4
5 5 1 Introduction Methods & Evaluation ( Problem Requirements 1 Text book Business Statistics (Barron s Business Review Series) Douglas Downing ( ), Jeff Clark ( ) Barrons Educational Series Inc ; 4th (2003/09) 1.
6 6 1 Introduction 1.1: Business Statistics (Barron s Business Review Series)[1]
7 Example 1.1 ( ) Problem 1.1.
8 Definition 2.1 (average). Given n samples of data, the quantity below is called by average. x = x 1 + x x n n = n i=1 x i. (2.1) n Definition 2.2 (median). Given n samples of data, the center of the ordered samples is called by median. Definition 2.3 (mode). Given n samples of data, the value which is appeared most in the data is called by mode. Problem average median mode Problem 2.2.
9 Definition 2.4 (variance). Given n samples of data, the quantity below is called by variance of data. The sample variance is also defined by Var(x) = σ 2 = n i=1 (x i x) 2 n (2.2) = x 2 x 2. (2.3) s 2 2 = n i=1 (x i x) 2 n 1 Problem 2.3. (2.3) (2.4) x 2 Definition 2.5 (standard deviation). Given n samples of data, the square-root of the variance Var(x) is called by standard deviation, and given by σ = Var(x) (2.5) = x 2 x 2. (2.6) 2.3 Example 2.1 ( ). Problem 2.4. Excel
10 Remark 3.1. Definition 3.1. Ω A A P{A or B} = P{A B} P{A}, (3.1) = P(A) + P(B) P(A B). (3.2) P{A and B} = P{A B} (3.3) P{ not A} = P{A c } = 1 P{A}. (3.4)
11 Problem n j np j = n! (n j)!, (3.5) n j ( ) n n! = n C j = j (n j)! j!, (3.6) Problem
12 n h P{ n h } = = ( )( n 1 h 2 n! h!(n h)! ) n (4.1) ( ) 1 n 2 (4.2) Problem 4.1. Problem (null hypothesis) (alternative hypothesis) false positive)
13 error) Problem 4.3. Remark % 10% Example 4.1. (null hypothesis) p = 1/2. (alternative hypothesis) p 1/2. Problem 4.4. false positive) error)
14 P{ } = (5.1) MRI 5% 5% P{A B} B B P{ } = 1 20, P{ } = (5.2) 1 [3, p.207]
15 P{A B} P{A B} = P{A B} P{B}. (5.3) P{ }. (5.4) (5.2) (5.4) Problem 5.1 (False positives 2 ). Answer the followings: 1. Suppose there are illegal acts in one in companies on the average. You as a accountant audit companies. The auditing contains some uncertainty. There is a 1% chance that a normal company is declared to have some problem. Find the probability that the company declared to have a problem is actually illegal. 2. Suppose you are tested by a disease that strikes 1/1000 population. This test has 5% false positives, that mean even if you are not affected by this disease, you have 5% chance to be diagnosed to be suffered by it. A medical operation will cure the disease, but of course there is a misoperation. Given that your result is positive, what can you say about your situation? 1/6 2 Modified from [3, p.207].
16 Definition 5.1. B P{A B} = P{A B} P{B}. (5.5) Example 5.1. P{( ) ( )} P{ } = P{ } P{ } = P{ } P{ }P{ 3} = P{ } = P{ } = 1 6. Problem 5.2. A K Problem 5.3. A
17 5.4. Bayes 17 Problem 5.4. Definition 5.2. A B Theorem 5.1 ( ). A B Problem 5.5. Theorem 5.1 P{A B} = P{A} (5.6) P{A B} = P{A}P{B}. (5.7) 5.4 Bayes Bayes Theorem 5.2 (Bayes). P{B A} = Problem 5.6. Theorem 5.2 P{A B}P{B} P{A B}P{B} + P{A B c }P{B c }. (5.8)
18 Definition 6.1. Example 6.1. W: 6.2 Definition 6.2 (). X f (a) = P{X = a}, (6.1) Problem 6.1. X X Definition 6.3 (). X F(x) = P{X x}, (6.2) f (x) Theorem 6.1. f (x) = df(x) dx = dp(x x). (6.3) dx lim F(x) = 1, (6.4) x lim F(x) = 0 (6.5) x
19 Definition 6.4 ( ). E[X] = xdp{x x} (6.6) Remark 6.1. X E[X] = xdp{x x} = ap{x = a} (6.7) a 6.4 Example 6.2. U,V 1/2 U 1/2, (6.8) V = { 1 with probability 1/2, 0 with probability 1/2. (6.9) Problem 6.2. E[U] = E[V ] = 1/2 Definition 6.5 ( ). Var[X] = E [ (X E[X]) 2]. (6.10)
20 20 6 Theorem 6.2. Var[X] = E[X 2 ] (E[X]) 2. (6.11) Var[aX + b] = a 2 Var[X]. (6.12) Remark 6.2. X Definition 6.6 (). σ = Var[X]. (6.13) Problem 6.3. Example 6.2 Theorem 6.3. X,Y E[XY ] = E[X]E[Y ], (6.14) Var[X +Y ] = Var[X] +Var[Y ]. (6.15) 6.5 A Definition 6.7 ( ). A = { 1 with probability p, 0 with probability 1 p. (6.16) Theorem 6.4. E[A] = p, (6.17) Var[A] = p(1 p). (6.18)
21 Definition 6.8. X 1,X 2,...,X n, (6.19) P{X i x} = F(x), (6.20) X 1,X 2,...,X n i.i.d: independent and identically distributed) X Theorem 6.5. X 1,X 2,...,X n, (6.21) X = 1 n n i=1 X i, (6.22) E[ X] = E[X], (6.23) Var[ X] = Var[X]. (6.24) n
22 22 7 Poisson 7.1 Definition 7.1 (). p X X Theorem 7.1. X P{X = i} = (1 p) i 1 p, (7.1) E[X] = 1 p, (7.2) Var[X] = 1 p p 2. (7.3) Proof. X = { 1 with probability p, 1 + X with probability 1 p, (7.4) X X E[X] = p 1 + (1 p)e[1 + X], (7.5) E[X] = 1/p
23 Problem 7.1. E[X 2 ] = p (1 p)e[(1 + X) 2 ], (7.6) Var[X] Problem / Definition 7.2 (). p n X X P{X = i} = ( ) n p i (1 p) n i, (7.7) i Theorem 7.2. X Bernouilli A i P{A i = 1} = p X = n i=1 Theorem 7.3. X A i, (7.8) E[X] = np, (7.9) Var[X] = np(1 p). (7.10)
24 24 7 Poisson 7.3 Problem 7.3 ( ) P_n n 7.1: : Problem 7.4 ( ). 7% 5% 7.4 Poisson Poisson Definition 7.3 (Poisson ). N N λ Poisson
25 7.4. Poisson 25 P_n n 7.2: P_n n 7.3: risk n 7.4:
26 26 7 Poisson P{N = n} = λ n Theorem 7.4. Poisson N n! e λ. (7.11) E[N] = λ, (7.12) Var[N] = λ. (7.13) Theorem 7.5. n np λ = np Poisson 7.5 Poisson Problem 7.5 ( ) P_n n 7.5: : Problem 7.6 ( )
27 Definition 7.4 ( ). M A N M B n A X X ( M )( N M ) i n i P{X = i} = ( N. (7.14) n) Theorem 7.6. N E[N] = nm N, (7.15) ( )( M Var[N] = n 1 M )( ) N n. (7.16) N N N 1
28 Definition 8.1. X a b X [a,b] a c d b P{c X d} = d c b a, (8.1) Remark 8.1. P{X = x} = 0. (8.2) 8.2 Definition 8.2. Example 8.1. H H Definition 8.3 ( ). X (Cumulative Distribution Function: CDF or Probability Density Function: PDF) Theorem 8.1. F(x) = P{X x}. (8.3) P{X > a} = 1 F(a), (8.4) P{b < X < c} = F(c) F(b). (8.5)
29 Problem 8.1. [a,b] Definition 8.4 ( ). F(x) (probability density function:pdf) f (x) = df(x) dx. (8.6) Problem 8.2. [a,b] Theorem 8.2. Theorem 8.3. P{a < X b} = E[X] = E[X 2 ] = b a f (x)dx = F(b) F(a). (8.7) x f (x)dx, (8.8) x 2 f (x)dx, (8.9) Var[X] = E[X 2 ] E[X] 2. (8.10) Problem 8.3. [a,b] X 8.3 µ σ 2 Definition 8.5 (). X µ σ 2 f (x) f (x) = 1 2πσ e [(x µ)/σ]2 /2. (8.11) X N(µ,σ 2 ) Definition 8.6 ( ). Z µ = 0 σ 2 = 1 Z N(0,1) Φ(x) = P{Z x}. (8.12)
30 30 8 Theorem % P{µ 2σ X µ + 2σ} (8.13) Theorem 8.5. µ σ 2 Y Z Y = µ + σz. (8.14) Z = Y µ σ. (8.15) Theorem 8.6. ( ) a µ P{X a} = Φ. (8.16) σ Proof. { X µ P{X a} = P a µ } σ σ { = P Z a µ } σ ( ) a µ = Φ. σ Remark 8.2. Excel Example 8.2 (). X POS 2.5% Theorem 8.5 Z = X µ σ (8.17)
31 N[0,1] P{ 2 Z 2} (8.18) P{Z 2} + P{ 2 Z 2} + P{Z 2} = 1. (8.19) P{Z 2} = P{Z 2} X P{Z 2} = 1/2P{ 2 Z 2} = (8.20) = P{Z 2} { X µ = P σ } 2 = P{X µ + 2σ} = P{X 440} % Problem 8.4. Sony Sony Sony Excel Theorem 8.7. X i µ i σi 2 ) X = n i=1 X i N ( n i=1 µ i, n i=1 σ 2 i. (8.21)
32 Definition 8.7 ( ). log(y ) Y X Y = e X, (8.22) Theorem 8.8 ( ). X N[µ,σ 2 ] Y = e X E[Y ] = e µ+σ 2 /2, (8.23) Var[Y ] = e 2µ+2σ 2 e 2µ+σ 2. (8.24) Problem 8.5. E[Y ] e µ 8.5 Theorem 8.9 ( ). Figure 8.1 Remark χ 2 ( χ 2 Definition 8.8 (χ 2 ). Z Z χ 2 χ = Z 2 (8.25)
33 8.6. χ 2 ( : The detailed histgram of the sample average A = 1 n n i=1 X i when n = 10, where X i is a Bernouilli random variable with E[X i ] = 1/2. The solid line is the corresponding Normal distribution. Theorem Proof. E[χ] = 1, (8.26) Var[χ] = 2. (8.27) E[χ] = E[Z 2 ] = 1. (8.28) Definition 8.9 ( n χ 2 ). Z i χ n = n χ 2 n i=1 Z 2 i, (8.29) Remark 8.4. n Z i Remark 8.5. χ 2 Theorem E[χ n ] = n, (8.30) Var[χ n ] = 2n. (8.31)
34 t Definition 8.10 (t ). Z Y m χ 2 T = Z Y /m, (8.32) m student t Remark 8.6. student t 8.8 F Definition 8.11 (t ). X Y m n χ 2 m n F F = X/m Y /n, (8.33)
35 Definition 9.1 ( ). X Y F(x,y) = P{X x,y y}. (9.1) Theorem 9.1. f (x,y) = d2 F(x,y). (9.2) dxdy E[XY ] = xy f (x, y)dxdy. (9.3) Problem 9.1. X Y 9.2 Definition 9.2 (). P{X x} = F X (x) = F(x, ). (9.4) f X (x) = y= f (x,y)dy. (9.5) Problem 9.2. X Y
36 Definition 9.3 ( ). P{X x Y = y}. (9.6) f (x Y = y) = f (x,y) f Y (y). (9.7) Problem 9.3. X Y f (x Y = 2), (9.8) 9.4 X Y (9.7) Theorem 9.2. X Y f (x Y = y) = f X (x). (9.9) f (x,y) = f X (x) f Y (y). (9.10) E[XY ] = E[X]E[Y ]. (9.11)
37 Definition 9.4. X Y Cov(X,Y ) = E [(X E[X])(Y E[Y ])] (9.12) = E [XY ] E[X]E[Y ]. (9.13) Problem 9.4. Cov(X,Y ) Problem 9.5. X Y Definition 9.5. ρ(x,y ) = Cov(X,Y ) Var(X)Var(Y ). (9.14) Problem 9.6. ρ(x,y ) 1-1 Problem 9.7. X Y 9.6 Theorem 9.3. E[X +Y ] = E[X] + E[Y ], (9.15) Var[X +Y ] = Var[X] +Var[Y ] + 2Cov[X,Y ]. (9.16) Remark 9.1. X Y Problem 9.8. Var[X +Y ] = Var[X] +Var[Y ] + 2Cov[X,Y ]. (9.17)
38 Example 9.1. Worldwide Fastburgers, Inc W E[W] = 1000, (9.18) Var[W] = 400, (9.19) HaveItYourWay Burgers, Inc H E[H] = 1000, (9.20) Var[H] = 400. (9.21) FunGoodTimes Pizza, Inc F E[F] = 1000, (9.22) Var[F] = 400. (9.23) Problem 9.9. Problem Theorem 9.3 Cov(W, H) = 380, (9.24) Cov(W, F) = 200, (9.25)
39 Problem HaveItYourWay Burgers, Inc Var[W + H] = Var[W] +Var[H] + 2Cov(W,H) (9.26) = (9.27) = 1,560. (9.28) Problem FunGoodTimes Pizza, Inc Problem 9.13.
40 X µ n X 1,X 2,...,X n µ ˆµ x ˆµ = x = 1 n n i=0 X i. (10.1) x Definition Example ,7,4,10,12, (10.2) µ = E[X] ˆµ ˆµ = x = 1 ( ) = 7. (10.3) 5 Remark ˆµ
41 Definition 10.2 ( ). P{X 1 = x 1,X 2 = x 2,...,X n = x n µ}, (10.4) µ Remark Example 10.2 () {1,0,0,0,1,0,1,0,0,0}. (10.5) p p = 1/2 P{X 1 = 1,X 2 = 0,...,X 10 = 0} = , (10.6) 210 p p = 1/3 P{X 1 = 1,X 2 = 0,...,X 10 = 0} = , (10.7) 310 Problem p P{X 1 = 1,X 2 = 0,...,X 10 = 0} = p 3 (1 p) 7 (10.8) p log log logp{x 1 = 1,X 2 = 0,...,X 10 = 0} = 3log p + 7log(1 p), (10.9)
42 42 10 p p 0 3 p + 7 = 0, (10.10) 1 p p = 3/10 n p = 1 n n i=1 x i (10.11) (x 1,x 2,...,x n ) Example X ˆµ = x = 1 n µ = E[X] n i=0 Example X ˆ σ 2 = 1 n n i=1 σ 2 = Var[X] X i, (10.12) (X i x) 2, (10.13) Example 10.5 (). (Definition 7.2 ( ) n P{X = i} = p i (1 p) n i, (10.14) i (10.15) p
43 Definition consistent estimator) Remark Problem Example x µ [ ] n 1 Var[ x] = Var X i n i = 1 n 2 nvar[x] = 1 Var[X] 0 as n 0. n 10.4 Definition Problem x Example ˆ σ 2 E[X] = 0, (10.16) σ 2 = Var[X] = E[X 2 ]. (10.17) ˆ σ 2 = 1 n n i=0 (X i x) 2. (10.18)
44 44 10 [ n ] E[ σ ˆ2 ] = 1 n E (X i x) 2 i=0 (10.19) = 1 n n i=0 E [ X 2 i 2 xx i + x 2] (10.20) (n 1)σ 2 =. (10.21) n E[X 2 i ] = σ 2, (10.22) E[ xx i ] = 1 n σ 2, (10.23) E[ x 2 ] = 1 n σ 2, (10.24) E[ σ ˆ2 (n 1)σ 2 ] =. (10.25) n σ ˆ2 s ˆ2 sˆ 2 = 1 n 1 n i=0 (X i x) 2 (10.26) E[ ˆ s 2 ] = σ 2, (10.27) Example
45 Example µ (X 1,X 2 ) ˆµ 2 = X 1 + X 2 2 (10.28) 1000 ˆµ 1000 = 1 n n i X i, (10.29) Problem ˆµ 1000 Problem 10.5.
46 46 11 Chapter 10 Problem 11.1 ( ). = 1, (11.1) Problem = 1, (11.2) 11.1 X 1,X 2,...,X n µ = E[X] x x = 1 n n i=0 X i. (11.3)
47 Theorem x Proof. Theorem 8.7 x x x µ c µ [ x c, x + c] µ [ x c, x + c] 95% c Definition 11.1 (). X 1,X 2,...,X n µ x [ x c, x + c] CL µ x P{ x c < µ < x + c} = CL. (11.4) Theorem 11.2 (). X 1,X 2,...,X n c P{ x c < µ < x + c} = 0.95, (11.5) (11.6) c = 1.96σ n, (11.7) 95% [ [ x c, x + c] = x 1.96σ, x σ ], (11.8) n n
48 48 11 Lemma Lemma Z = x µ σ 2 /n = n x µ σ, (11.9) N[0,1] Proof. x µ, σ 2 /n Theorem 8.5 Proof of Theorem Lemma 11.1 P{ x c < µ < x + c} = P{ c < x µ < c} = P{ c n σ < n x µ σ < c n σ } = P{ c n σ < Z < c n σ }. P{ 1.95 < Z < 1.95} = (11.10) c = 1.96σ n. (11.11) Problem t Section 11.1 c = 1.96σ n, (11.12)
49 11.2. t 49 X σ 2 Problem σ 2 1. σ 2 ˆσ 2 2. ˆσ 2 c Problem Lemma 11.1 Lemma Lemma ŝ 2 = 1 n 1 n i=0 (X i x) 2. (11.13) n( x µ) T =, (11.14) ŝ n 1 t Section 8.7 Proof. Section 8.7 Z χ 2 Y T = Z Y /m, (11.15) T t (11.14) n( x µ)/σ T = (11.16) ŝ/σ n( x µ)/σ = n i=0 (X i x) 2 /[(n 1)σ 2 ]. (11.17)
50 50 11 Lemma 11.1 n i=0 (X i x) 2 σ 2, (11.18) x n 1 χ 2 [2]P84-P87 Lemma 11.2 Theorem 11.3 ( ). X 1,X 2,...,X n n P{ x c < µ < x + c} = 0.95, (11.19) (11.20) c c = a ˆσ n, (11.21) a t P{ a < T < a} = 0.95, (11.22) n = 7 a = % [ [ x c, x + c] = x a ˆσ, x + a ˆσ ], (11.23) n n Proof. P{ x c < µ < x + c} = P{ c < x µ < c} = P{ c n ˆσ < T < c n ˆσ }. t P{ c n ˆσ < T < c n ˆσ } (11.24) Remark t t
51 Normal t n : t [1]
52 Problem Example 12.1 ( ) :
53 Problem N M p = M N, (12.1) n n X ˆp ˆp = X n, (12.2) ˆp p Problem Lemma n X (n, p) n X np np(1 p)
54 54 12 Proof. n X n Theorem n X ˆp = X n, (12.3) N(p, p(1 p)/n) Proof. Lemma 12.1 X N(np,np(1 p)) ˆp = X n, (12.4) X [ ] X E[ ˆp] = E = np = p. (12.5) n n [ ] X Var[ ˆp] = Var = 1 p(1 p) np(1 p) =. (12.6) n n2 n Remark ˆp E[ ˆp] = p. (12.7) n ˆp Var[ ˆp] = p(1 p) n 0, (12.8) ˆp
55 Theorem n ˆp 95% [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp 1.96, ˆp (12.9) n n Proof. Theorem 12.1 ˆp σ 2 = Var[ ˆp] = ˆp(1 ˆp). (12.10) n Theorem 11.3 Remark [1] P259 Example n = 2018 ˆp = % [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp 1.96, ˆp = [ , ], n n (12.11) 12.3 Example Literacy Digest Roosevelt Roosevelt Problem Literacy Digest Example 12.4 (). 12.2
56 Problem 12.5 ( ). : ( ) Problem 12.6.
57 :
58 : : Problem (YES or NO)
59 Section 4.2 (null hypothesis) (alternative hypothesis) error false positive) false negative) ( 5%
60 60 13 Example Problem % 20% 5% 13.2 Definition 13.1 ( ). Example X X X n = 100 p = 1/2
61 X = 90 n = 100 p = 1/2 Problem example 13.3 Definition X 1,X 2,...,X n µ = E[X] µ H 0 : µ = µ. (13.1) Example 13.3 ( ). N[µ,σ 2 ] σ 2 =
62 62 13 x {9,11,6,10,7,4,0,7,8,6,8,2,18}. (13.2) x = 7.38, (13.3) Problem x = 7.38 µ = 7 H 0 H 0 : µ = 7. (13.4) X 1,X 2,...,X 13 N[7,16.16] Lemma 11.1 x N[7,16.16/n] Z Z = n x µ σ = = 0.341, N[0,1] Z N[0,1] N[0,1] Z P{ 1.96 < Z < 1.96} = 0.95, (13.5) Z = 0.341, 7 5%
63 Problem Remark 13.1 ( ). 95% 5% 13.4 Example 13.4 () : {7,16,19,12,15,9,6,16,14,7,2,15,23,15,12,18,9}. (13.6) Problem
64 64 13 Section 13.3 H 0 : µ = µ. (13.7) Section H 0 : µ < µ. (13.8) Definition Example 13.5 ( ). 11 σ T T = n x µ ˆσ = = n 1 t P{T < 1.75} = 0.95, (13.9) T = 1.26 n 1 t 5% Problem {3,2,7,5,4,8,7,8} (13.10)
65 Problem ( % Problem 13.8.
66 66 [1] Douglas Downing and Jeffrey Clark. Business Statistics. Barrons s Educational Series, Inc., [2] Shingo Shirahata. Toukei Kaiseki Nyumon. Kyouritu, [3] Nassim Nicholas Taleb. Fooled by Randomness: The Hidden Role of Chance in the Markets and in Life. Random House Trade Paperbacks, 2005.
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〈論文〉興行データベースから「古典芸能」の定義を考える
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