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魁训较阎
7 years ago
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3 Y =1 0 =, L, ) Y Y ( 1 m f (.) Y = f, L, ) Y ( 1 m Y, L, ) ( k < m) ( 1 k ( k + 1, L, m ) Y ) ( 1, L, k, L, ) ( 1 k Y, L, ) ( 1 k Y Y Y Y Y 1

5 (Flush straght) (Four of a knd) (Full house) (Flush) (Straght) (Three of a knd) (Two pars) (Par) (one) 5 = :1 : :78 (Almost surely) 15.6 (=78/5) Y Y Ω Y = {,,,,,,,, } 40 : 64 : 3,744 : 5,108 : 10,00 : 54,91 : 13,55 : 1,098,40 : 1,30, ) 3

6 { }) = ; { }) = ; { }) = ; { }) = ; { }) = ; { }) = ; { }) = ; { }) = ; { }) =, { P }) ({ }) 40 ( = ) ) ) Y A A = {,, } = A = {,, } Ω Y ΩY φ φ 1. ΩY Y. A Ω Y 3. P [0,1] 4. 4

7 1. A Ω Y 0. A, B Ω Y A = φ) = 0 P ( A = + 3. A = + A 4. A B 5. Y ) = 1 Ω Y << >> Y = (,, ) () Ω () ω Ω ω) () A A (v) A Y Y Y = (,, ) () ΩY () Ω Y () A A 5

8 (Senstvty ) (Relablty ) Y = 1, 1, ( Y1, Y ), Y1 = Y = 0, 0, Ω Y = {(0,0), (0,1), (1,0), (1,1)} {( a, b)}), a = 0,1; b = 0,1 Y 00 P ({( 0,0)}) =, ) Y = ( a, b 10 ({( 0,1)}) = 01, {(1,0)}) = {(1,1 )}) = ab P 11 A = {(1,0),(1,1 )} B = {(0,1),(1,1)} 6

9 = = A B A Y ( 1,0) (1,1 ) 11 A B B = (B + 11 ) A B A = A B = A B c c c c A B ) B A ) = c A ) A Y Ω A A Y = A Y ) ( 1,0) (1,1 A ( 1,0) (1,1 ) A A 1 AB Ω > 0 A Y A B B = A A B B c c B A ) A c c B B A ) A ( ) 7

10 B Ω = Y B P B ) = P ( B Ω ) A A c Ω ( Y c c = B ( A A )) = B + B A ) A A B A = = = c c c A + A B + B A ) A ) c c c c B A ) = 1 B A ) A ) = 1 c c B B A ) A ) A A Y A B Ω > 0 A c A B A = c c B + B A ) A ) Y :19 A B c c 敍 B = 0.99, B A ) = 0.98P ( = * A = * * B A 8

11 Y = Y 1, Y ) ( Y = I( ), = 1, 0.5 = {(0,0),(0,1),(1,0),(1,1 )} Ω Y ω ΩP ( ω) = 0.5 A = {(0,1), (1,0 )} Y B = {(1,0),(1,1)} = 0.5 = 0. 5 A 0.5 A = 0.5 P ( A = = = 0.5 B A A B A P ( B = P ( A = A B Ω P ( A = Y 9

12 /0 1/ 絶寛 / 3 1/ / 3 10

13 敍 ΩY 00 = 01 = 10 = 11 = / 4 ) 1 {, } Ω Y = {(, ),(, ),(, ),(, )} { H, T} Ω = Y {( H, H ),( H, T ),( T, H ),( T, T )} Ω Y = {(0,0),(0,1),(1,0),(1,1 )} ( Y ) = ( Y1, Y ) = Y1 + Y = 0 = 1 = {0,1,,3} Ω Y Y ΩY ΩY ( : Ω R ) Y 敍 Y Y 11

14 n Y = I( ) n Y = 1 = Ω = { 0,1, L, n} =1, L,9 (Flush straght) (Four of a knd) (Full house) (Flush) (Straght) (Three of a knd) (Two pars) (Par) (one) ( ) { 4} = {1,,3,4} { 4}) 4) = {,3, L,1} 7 Ω 7) = 1/36 Ω (Probablty mass functon, p.m.f.) Ω x Ω f = = x) x Ω f = 0 f ( ) p.m.f. 1

15 x > 0 f Ω f ( ) f (x) = { 1,, L} = k k 1 Ω 1 x Ω, f = Ω ) = 1 x f x= f = = = 1 x x= 1 x= 1 Ω Ω p.m.f. f ( ) ) f ( x Ω, f > 0 x Ω, f = 0 f ( x ) = 1 x Ω (Cumulatve densty functon, c.d.f.) F ( ) 13

16 c.d.f. Ω = { x x R f > 0} ) f ( p.m.f. x R F = x) = f ( y) y Ω, y x F ( ) F ( ) = 1 F ( ) = 0 F f ( ) F ( ) ( ) f ( ) } Ω = {1,,3,4 f 3) = F (3) F () ( f ( ) F ( ) f ( ) F ( ) f ( ) F ( ) 14

17 (005) (0.7>0.1) }, = { 10,0,40 f Ω ( 10) = 0.7, f (0) = 0. f (40) = ε O( ) = ( ε 1 ε + O ) = O( ) = 40( 0.5 1:1 15

18 + + + : 1: 1 1 ε O( ) 1 ε O ( ) / ) [0.7 + O( )]*( 10) + [0.1 + O( 1 ε 1 ε )]*(40) 10* f ( 10) + 0* f (0) + 40* f (40) = 3 3 (Expectaton) ) E( Ω = { x x R f > 0} f ( ) p.m.f. E ( ) = x* f x Ω E( ) = 10* * *0.1 = 0 0 E( ) = E( ) ( 1, L, ) 16

19 1 = 1 ) E( E( ) = 49950* * = 5 Y Z T W E( Y ) = E( Z) = E( T ) = E( W ) = 6 E( ) 17

20 1 6 3 (1) () (3) 敍, Y, Z ), L, Y, L, Y ) Z, L, Z ), L, ) ( 1 ( 1 ( 1 ( 1 Z, L, Z ) ( 1 (Varance) ) Var( Ω = { x x R f > 0} f ( ) p.m.f. Var ( ) = ( x µ ) * f x Ω µ = E( ) σ ( ) Var( ) = 10 * *0.1 = 900 Var( Y ) = 75 Var( Z) = 40 x µ * f ( x x, x j Ω x j ) * f ( x )* f 18 x Ω ( x )

21 ( x µ ) * f = ( x µ x + µ )* f = x * f µ x Ω x Ω x Ω Var( ) = E( ) [ E( )] (Standard devaton) ) ( SD Var( ) SD ( ) = Var( ) σ ( ) (p.m.f) p.m.f. ( ) 19

22 1,L, 1 ~ D -U (1, ) p.m.f. f =, x = 1, L, 1 > E( ) = Var( ) = 1 p ( 1 p) = 1 = 0 ~ Ber( p) p.m.f f ( 0) = 1 p f (1) p 0 < p < 1 = E( ) = p Var( ) = p(1 p) p ( 1 p) n n ) ~ B( n, p p.m.f f n x n x = p (1 p), x = 0,1, L n x, n 1 0 < p < 1 E( ) = np Var( ) = np(1 p) n 1, L, n = 0 n = 1

23 p ( 1 p) ) ~ Ge( p p.m.f. f = p(1 p) x 1, x = 1,, L 1 (1 p) 0 < p <1 E( ) = Var( ) = p p p ( 1 p) r x + r 1 r x ) ~ B( r, p p.m.f. f = p (1 p), x = 0,1, L x r(1 p) r(1 p) r 1 0 < p < 1 E( ) = Var( ) = p p r = 1 r, L 1, r = D n n 1

24 D D ~ H (, D, n) p.m.f. f = /, x n x n max{ 0, n + D} x mn{ n, D}, D, n 1 E ( ) = nd( D)( n) Var( ) = ( 1) nd

絶 2 47

Part II 1 47 絶 2 47 3 47 4 47 5 47 6 47 7 47 8 47 9 47 10 47 11 47 12 47 13 47 14 47 15 47 16 47 敍 17 47 18 47 19 47 20 47 21 47 22 47 敍 23 47 24 47 25 47 敍 26 47 27 47 28 47 29 47 30 47 31 47 32 47 33