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数理統計学 - web.econ.keio.ac.jp

20040607 13;15;20 30;0701 02;04 06;0710;13;0911;28;20050815( );.. 2006 . web .. ~ .. 1 4. 2 .. 4.. 5. 2 .. 6. 2 8. JR Y 1990 .. 8.. 8.. 9.. 10. Y .. 12.. 12. 3 13.. 13.. 15.. 17.. 18.. 19.. 20. 4 20.. 20.. 22.. 23.. 24. LLN, CLT .. 24. 5 26.. 26.. 28.. 29. 13 .. 29. 2 .. 31. 6 32.. 32.. 33. 2 .. 34. 3 .. 35. 7 1 1 2 t 37. 2.. 37.. 39. t .. 40.. 41.. 42. 28 .. 43. 32 .. 44. 8 2 F t 44. F .. 44.. 46.. 47.. 47. 2 .. 48. 9 49.. 49. 2. F .. 51.. 54. 10 55.. 55.. 56.. 57.. 57.. 59.. 60.. 61. 11 62. 64. 4. 1 . 2 . 2 . n . p 0 < p < 1 p = 0 . p = 1 .. 1 .. (1) .. 3.. n . 1 0 0, 1 n . = {(s1 , s2 , , sn ) {0, 1}n} . 1 p . 1 p 1 . 1 p.

1.1 4 1 不公平な硬貨— 離散分布の確率論. 硬貨投げ(2項分布)を例として,離散分布について,確率測度,平均,分散の定義を復習する.

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Transcription of 数理統計学 - web.econ.keio.ac.jp

1 20040607 13;15;20 30;0701 02;04 06;0710;13;0911;28;20050815( );.. 2006 . web .. ~ .. 1 4. 2 .. 4.. 5. 2 .. 6. 2 8. JR Y 1990 .. 8.. 8.. 9.. 10. Y .. 12.. 12. 3 13.. 13.. 15.. 17.. 18.. 19.. 20. 4 20.. 20.. 22.. 23.. 24. LLN, CLT .. 24. 5 26.. 26.. 28.. 29. 13 .. 29. 2 .. 31. 6 32.. 32.. 33. 2 .. 34. 3 .. 35. 7 1 1 2 t 37. 2.. 37.. 39. t .. 40.. 41.. 42. 28 .. 43. 32 .. 44. 8 2 F t 44. F .. 44.. 46.. 47.. 47. 2 .. 48. 9 49.. 49. 2. F .. 51.. 54. 10 55.. 55.. 56.. 57.. 57.. 59.. 60.. 61. 11 62. 64. 4. 1 . 2 . 2 . n . p 0 < p < 1 p = 0 . p = 1 .. 1 .. (1) .. 3.. n . 1 0 0, 1 n . = {(s1 , s2 , , sn ) {0, 1}n} . 1 p . 1 p 1 . 1 p.

2 P[ (s1 , s2 , , sn ) ] = ps1 + +sn (1 p)n (s1 + +sn ) . (1). n . n k . P w . Nn : {0, 1, , n} P Nn . P N P Nn 1 . Qn = P Nn 1 k . Qn ({k}) = P[ Nn 1 (k) ] = P[ Nn = k ]. (2). P Nn = k k .. {w = (s1 , , sn ) | Nn (w) = k} = {Nn = k}. P[ Nn = k ] .. N k .. Nn . Nn ((s1 , , sn )) = s1 + + sn (1) .. Qn ({k}) = P[ Nn = k ] = P[ {(s1 , , sn )} ] = pk (1 p)n k s1 , ,sn {0,1}; s1 , ,sn {0,1}; (3). s1 + +sn =k s1 + +sn =k k n k = p (1 p) {s1 , ,sn {0,1}|s1 + +sn =k}. 5. n k n C k . Qn ({k}) = n C k pk (1 p)n k , k = 0, 1, , n 1, n, (4). 2 Bn,p . n p n p .. n k 2 (4) 2 k . n + 1 Q({k}) = Qn ({k}) . A {0, 1, , n} = .. Q(A) = Q({k}) (5). k A.

3 (2) (5) Q = Qn .. 2 = {0, 1, , n} . Q({k}), k , Q( ) = 1. n . Q({k}) = 1 (6). k=0. A Q(A) (5) Q . {k} . 2 n p . (4) 2 .. Q({k}) .. = k Q({k}) (7). k .. n .. 6.. v= (k )2 Q({k}). (8). k . (8) 2 .. v= k 2 Q({k}) 2 (9). k .. 2 (3) . 2 . 2 x, y n . 2 . n . (x + y)n = k n k nC k x y (10). k=0. x, y x = p, y = 1 p . n . k nC k p (1 p)n k = 1. k=0. 2 Bn,p (3) Qn ( ) = 1 . (10) y x x x . n . nx(x + y)n 1 = k n k n C k kx y . (11). k=0. x, y x = p, y = 1 p . n . k nC k k p (1 p)n k = np. k=0. (7) = np (11) x x . n . nx(x + y)n 1 + n(n 1)x2 (x + y)n 2 = nC k k 2 k n k x y . (12). k=0. x, y x = p, y = 1 p . n . 2. nC k k pk (1 p)n k = np + n(n 1)p2.

4 K=0. = np (9) v = np + n(n 1)p2 (np)2 = np(1 p) . 2 Bn,p = np v = np(1 p) . 2 . 2 n n .. 7. 2 (2 ) 0 < p < 1 Qn Bn,p y [y] y .. kn (x) = [np + x np(1 p)] .. ( lim np(1 p) Qn ({kn (x)}) =). n . 1 2 (13). lim np(1 p) nC kn (x) pkn (x) (1 p)n kn (x) = e x /2 , x R. n 2 . 3. [y] y y . Qn ({k}) |k| n |x| . n k < 0 k > n Qn ({k}) = 0. = Z .. 2 Stirling . n! lim = 1, (14). n 2 nnn e n e n 1. lim 1 + =e n n n x 2. x n lim e 1+ = e x /2. n n Stirling . n = 1 n! (14) . [ , ] . (13) N (0, 1) . 2 . Qn ({kn (x)}) 2 . x .. secunormaldistri . (Nn np). 3 (2 ) p n Nn . np(1 p). n N (0, 1) . [a, b] . b Nn np 1 2. lim Qn (a b) = e x /2 dx n np(1 p) a 2.

5 3. (2) . 1 . lim np(1 p) Qn ({k}). n np(1 p) k np k; a b np(1 p). (13) 3 3 . 2 .. 8. 2 .. 3 . 1 .. JR Y 1990 . JR ATM . 1 . 2 .. 1980 1990 .. 1990 .. JR Y . 1990 . Y . (i) . (ii) Y . (iii) . 2 3 1 .. 2 .. 2 .. R R+ . [a, b] = Rn 1 1995 .. 2 .. 9. n n .. (5) = R, R+ , [a, b] . 1 : R+ .. Q(A) = (x) dx (15). A.. (x) 0, x , (16).. (x) dx = 1 (17).. n n 2 . (x, y) 0, (x, y) R, (18).. (x, y) dx dy = 1 (19).. 2 : R2 R+ .. Q(A) = (x, y) dx dy (20). A.. (15) A = {x} . 1 0 . x Q({x}) = (y) dy = 0. (21). x . v . (7) (8) (9) .. = x (x) dx (22).. v= (x )2 (x) dx = x2 (x) dx 2 . (23).. = v . |x | .. E[ X ], V[ X ]. (2) X : R . Q(A) = (P X 1 )(A) = P[ X 1 (A) ] = P[ X A ] (24).

6 10. X Q X P . X . X X Q X X Q .. X E[ X ], V[ X ] X X Q =. P X 1 {0, 1, 2, , n} . n n . E[ X ] = kP[ X = k ] = kQ({k}) (25). k=0 k=0. X. Q = P X 1 .. E[ X ] = x (x) dx (26). R. (25) .. (7), (22) E[ X ] X .. V[ X ] X . V[ X ] = E[ (X E[ X ])2 ] (27). X Q = P X 1 (25) Q = E[ X ] (25). X (X )2 . n . V[ X ] = (k )2 Q({k}). k=0. (8) V[ X ] X Q v X Q = P X 1 . (26), (22), (23) V[ X ] X Q v . a . E[ aX ] = aE[ X ], V[ aX ] = a2 V[ X ]. (28). X, Y . E[ X + Y ] = E[ X ] + E[ Y ]. (29). a2 .. x = V[ X ] (29) .. E[ 1 ] = 1, V[ 1 ] = 0. (30).. Y .. 11. (i) M N N 1 . M N .. (ii) N M . M .. N/M 3 . (iii) i (i = 1, 2, , N ) Si 4 . (iv) {Si }.

7 = E[ Si ] .. 5 . (v) .. N . N . N M . N. T T = . M. Y . T 6 T .. t0 . T (t0 > T ) . T T . T < t0 T > t0 .. P[ T > t0 ] . T T P[ T > t0 ]. Si . t0 > T P[ T > t0 ] .. 3 .. 4 (2) . Si Si ( , P). Si Si : . sample .. n 5 S1 , S2 , , Sn P[ S1 A1 , S2 A2 , , Sn An ] = P[ Si Ai ] . i=1. Ai , i = 1, , n, i = j E[ Si Sj ] = E[ Si ]E[ Sj ] .. 6 . 12. 7 . N. 1 . T (1) = Si , M i=1. N/M (31).. (2). T = Sji . i=1. j1 , j2 , , jN/M , E[ T (1) ] =. E[ T (2) ] = T . V[ Si ] i 2 = V[ Si ] . N 2. V[ T (1) ] = , . M2. (32). N 2. V[ T (2) ]= , . M. (43) (43) V[ T (1) ] (32) .. N. V[ T (1) ] = 2 2 (43) {Si | i = 1, , N } V[ S1 + S2 +. M. + Sn ] = V[ S1 ] + V[ S2 ] + + V[ Sn ] (28) (31).

8 V[ T (1) ] = V[ M 1 S1 + M 1 S2 + + M 1 SN ] = V[ M 1 S1 ] + V[ M 1 S2 ] + + V[ M 1 SN ]. N. = M 2 (V[ S1 ] + + V[ SN ]) = 2 2 . M. M > 1 V[ T (1) ] < V[ T (2) ] .. JR .. Y . Y 2000 .. (16) 2 (18) 1 . Q(A) 0, A (33). (17) 1 . Q( ) = 1. (34). (17) .. A B = Q(A B) = Q(A) + Q(B), (35). 7 P[ T > t0 ] . P[ T > t0 ] . P[ T > t0 ] . 13. n . Ai Aj = , i = j, 1 i, j n, Q(. n Ak ) =. Q(A ). n i (36). k=1 k=1.. Ai Aj = , i = j, i, j , Q(.. Ak ) =. Q(A ).. i (37). k=1 k=1. (33) (34) . Ak . (21) (37) . Q( ) = 0. (38).. (33) (34) (37) 3 Q .. (33) (34) (37) . 3 1.. n .. Q (15) (20) . A . Q F F .. F 1 . , [ , ), ( , ], [ , ] F, , .. Ak F, k , Ak F.

9 K=1. (39). (37) . A F Ac F A Ac 1 .. (15) F (39) . Borel B .. 3 .. 1 .. n k (4) . 0 1 k . k . v 2 . 14. p 1 .. (random sampling) .. 8 .. w . P P .. 7 8 . 2 .. 9 .. 2 .. w . w . (1) (0) .. Xk , k = 1, 2, , n, p .. Xk k . {Xk } .. n m = m(w) (1). (0) Xk , k = 1, 2, , n, m 2 Bn,p . 8 .. 9 . 2 . 15. Xk p Xk P Xk 1 .. P[ Xk = 1 ] = p (40). P[ Xk = 0 ] = 1 p n m = X1 +. + Xn 2 . 2 . (1) . 6 .. X P X 1 E[ X ] V[ X ] . Xk , k = . 1, , n, . n . P[ Xk Ak , k = 1, , n ] = P[ Xk Ak ] (41). k=1. Ak Xk . Xk (41) . n . P[ Xk = ak , k = 1, , n ] = P[ Xk = ak ]. k=1. {ak | k = 1, , n} . (42) .. 4 Xk , k = 1, , n, fk : R R, k = 1, , n, .. n n . E[ fk Xk ] = E[ fk Xk ] (42).

10 K=1 k=1. 3. (f g)(w) = f (g(w)) (f g)(w) =. f (w) g(w) .. X Y . V[ X + Y ] = V[ X ] + V[ Y ] + 2 E[ X E[ X ] ] E[ Y E[ Y ] ] = V[ X ] + V[ Y ]. (43).. (29) . 16.. X1 X2 (X1 , X2 ) X1 X2 . 1 (24) . Q(A) = (P (X1 , X2 ) 1 )(A) = P[ (X1 , X2 ) 1 (A) ] = P[ (X1 , X2 ) A ] (44). = R2 A R2 2 . A 2 . A = [a, b] [c, d] = {(x, y) | a x b, c y d}. 2 (44) . Q([a, b] [c, d]) = P[ (X1 , X2 ) A ] = P[ a X1 b, c X2 d ]. 2 (18) .. Q(A) = P[ (X1 , X2 ) A ] = (x, y) dx dy (45). A. 2 .. (X1 , X2 ) 2 . X1 X2 .. X2 . X1 2 Xi . Qi = P Xi 1 X1 (45) .. P[ X1 A1 ] = P[ (X1 , X2 ) A1 R ] = (x, y) dy dx A1 R. X1 . 1 (x) = (x, y) dy (46).. X2 .. 2 (x) = (x, y) dx.