卓球の試合への興味度に関する確率論的分析
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- るるみ もてぎ
- 5 years ago
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10 (2004) 80 K. 2 2 A x 1 x B B B y y A B 1 y A A 1 2 A 1 a B b a b 2 A B 50 B A 55 B 45 B B b =0.6 A 50
11 b =30 30 B B 3 A B 1 2 A x B y A y ax B y B 1 x b(1 y) A x y = ax, 1 x = b(1 y) x y A B A 1 A x B A x B 1 x = b(1 a)/(1 ab) ( )
12
13 p p F (p) F (1) = 0 F (p) p F (p) F (p) 2.1: F (p) y = log x x 0 1 y F (p) log p
14 6 2 p F (p) = log p(0 p 1) p F (p) p 1 p I(p) = pf (p)+(1 p)f(1 p) = p log p (1 p) log(1 p) A = {a 1,a 2,,a k } a 1,a 2,,a k P (a 1 )=p 1,P(a 2 )= p 2,,P(a k )=p k L= L=999 a 1 = a 1 A, B a 1,a 2 A log 2 a 1 B log 2 a 2 2 a 1 log 2 a 1 a 2 log 2 a 2 2
15 2.2. (1) : 2.2 (1) 2 A, B 2 1 A m, B 1 m A p = m 2 +2m 2 (1 m) = m 2 (3 2m) B q = (1 m) 2 +2m(1 m) 2 = (1+2m)(1 m) 2 1 A A p 1 = m + m(1 m) = m(2 m)
16 8 2 B q 1 =(1 m) 2 B 1 A p 2 = m 2 B q 2 = (1 m)+m(1 m) = 1 m A p 3 = m B p 3 =1 m m m =1.0 I p = 1.0( ) = 1.0 q = (0+2 0)(1 0) 2 = 0 I = log = 0 (2.1) 1, 2 I A m =0.1I p = 0.01(3 0.2) = q = ( )(1 0.1) 2 = 0.972
17 2.2. (1) 9 I = log log = (2.2) 1 I 1 A p 1 = = 0.19 q 1 = = 0.81 I 1 = 0.19 log log? = B p 2 = = 0.01 q 2 = = 0.99 I 2 = 0.01 log log = I I = = (2.3) 2 I 1 1 I = 0.1 log log = I ( ) = (2.4)
18 10 2 m =0.3 I p = 0.09(3 0.6) = q = (1+0.6)(1 0.3) 2 = I = log log = (2.5) 1 I 1 A p 1 = = 0.51 q 1 = (1 0.3) 2 = 0.49 I 1 = 0.51 log log = B p 2 = = 0.09 q 2 = = 0.91 I 2 = 0.09 log log = I I = = (2.6)
19 2.2. (1) 11 2 I 1 1 I = 0.3 log log = I ( ) = (2.7) m =0.5 I p = ( ) = 0.5 q = ( )(1 0.5) 2 = 0.5 I = log = 1 (2.8) 1 I 1 A p 1 = 0.5(2 0.5) = 0.75 q 1 = (1 0.5) 2 = 0.25 I 1 = 0.75 log log = B p 2 = = 0.25 q 2 = = 0.75
20 12 2 I 1 = 0.75 log log = I I = = (2.9) 2 I 1 1 I = log = 1 2 I =0.5 (2.10) (1) m (2) 4 A, B (3 3 ) 1 A a, B b A 2 A a 2 B b 2 1 a 2 b 2 A a 4 +4a 4 b +10a 4 b 2 +20a 5 b 3 {1+ (1 a 2 b 2 ) n } = a 4 +4a 4 b +10a 4 b 2 +20a 5 b 3 {1+(1 a 2 b 2 /a 2 + b 2 )}
21 2.3. (2) 13 B A 4 A, B (3 3 ) A a 4 +4a 4 b +20a 4 b 2 +20a 4 b 3 B A,B 1 a, b a = 0.6 b = 0.4 A {1+( / )} = = A I I d I nd I d = log log = (2.11) I nd = log log = (2.12) A, B 1 a, b a = 0.7 b = 0.3 A {1+( / )} = 0.901
22 = A I d, I nd I d = log log = (2.13) I nd = log log = (2.14) A, B 1 a, b a = 0.8 b = 0.2 A {1+( / )} = = A I d, I nd I d = log log = (2.15) I nd = log log = (2.16) A, B 1 a, b a = 0.9 b = 0.1
23 2.3. (2) 15 A {1+( / )} = = A I d, I nd I d = log log = (2.17) I nd = log log = (2.18) : m = m = m = m =
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24 11 10 24 12 10 30 1 0.45m1.00m 1.00m 1.00m 0.33m 0.33m 0.33m 0.45m 1.00m 2 23% 29% 71% 67% 6% 4% n=1525 n=1137 6% +6% -4% -2% 21% 30% 5% 35% 6% 6% 11% 40% 37% 36 172 166 371 213 226 177 54 382 704 216
情報理論 第5回 情報量とエントロピー
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漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,