情報理論 第5回 情報量とエントロピー

Similar documents
CD口頭目次.indd

卓球の試合への興味度に関する確率論的分析

: : : : ) ) 1. d ij f i e i x i v j m a ij m f ij n x i =

II

BIT -2-

A A = a 41 a 42 a 43 a 44 A (7) 1 (3) A = M 12 = = a 41 (8) a 41 a 43 a 44 (3) n n A, B a i AB = A B ii aa

II Time-stamp: <05/09/30 17:14:06 waki> ii

( 28 ) ( ) ( ) 0 This note is c 2016, 2017 by Setsuo Taniguchi. It may be used for personal or classroom purposes, but not for commercial purp

0 (18) /12/13 (19) n Z (n Z ) 5 30 (5 30 ) (mod 5) (20) ( ) (12, 8) = 4


PDF


FX ) 2

FX自己アフリエイトマニュアル

ax 2 + bx + c = n 8 (n ) a n x n + a n 1 x n a 1 x + a 0 = 0 ( a n, a n 1,, a 1, a 0 a n 0) n n ( ) ( ) ax 3 + bx 2 + cx + d = 0 4

( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a

α = 2 2 α 2 = ( 2) 2 = 2 x = α, y = 2 x, y X 0, X 1.X 2,... x 0 X 0, x 1 X 1, x 2 X 2.. Zorn A, B A B A B A B A B B A A B N 2

x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)


060407人材派遣業界における適正な健康保険年金制度の適用.doc

タ 縺29135 タ 縺5 [ y 1 x i R 8 x j 1 7,5 2 x , チ7192, (2) チ41299 f 675


資料5:聖ウルスラ学院英智小・中学校 提出資料(1)

特別支援1~8ページ.PDF


2 3 2


untitled

(1) (2) (3) (4) (5) (6) (7) (8)


14 12 ( ) 2004

第1 予算編成の基本的な考え方

untitled


2

0.表紙


,

-2-


300 10


0

総務委員会会議録

I y = f(x) a I a x I x = a + x 1 f(x) f(a) x a = f(a + x) f(a) x (11.1) x a x 0 f(x) f(a) f(a + x) f(a) lim = lim x a x a x 0 x (11.2) f(x) x

untitled

untitled


, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx


文庫●注文一覧表2016c(7月)/岩波文庫


PowerPoint プレゼンテーション

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

EPSON エプソンプリンタ共通 取扱説明書 ネットワーク編

untitled

ありがとうございました

EPSON エプソンプリンタ共通 取扱説明書 ネットワーク編

公務員人件費のシミュレーション分析


橡hashik-f.PDF

198

ネットショップ・オーナー2 ユーザーマニュアル


1

新婚世帯家賃あらまし

05[ ]戸田(責)村.indd

/9/ ) 1) 1 2 2) 4) ) ) 2x + y 42x + y + 1) 4) : 6 = x 5) : x 2) x ) x 2 8x + 10 = 0

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

数理.indd

「産業上利用することができる発明」の審査の運用指針(案)

1 1.1 Excel Excel Excel log 1, log 2, log 3,, log 10 e = ln 10 log cm 1mm 1 10 =0.1mm = f(x) f(x) = n

大野川水系中流圏域

10:30 12:00 P.G. vs vs vs 2

2009 IA 5 I 22, 23, 24, 25, 26, (1) Arcsin 1 ( 2 (4) Arccos 1 ) 2 3 (2) Arcsin( 1) (3) Arccos 2 (5) Arctan 1 (6) Arctan ( 3 ) 3 2. n (1) ta

.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(

Microsoft PowerPoint - 7.pptx

(1) + b = b +, (2) b = b, (3) + 0 =, (4) 1 =, (5) ( + b) + c = + (b + c), (6) ( b) c = (b c), (7) (b + c) = b + c, (8) ( + b)c = c + bc (9

CALCULUS II (Hiroshi SUZUKI ) f(x, y) A(a, b) 1. P (x, y) A(a, b) A(a, b) f(x, y) c f(x, y) A(a, b) c f(x, y) c f(x, y) c (x a, y b)

DVIOUT

II (1) log(1 + r/100) n = log 2 n log(1 + r/100) = log 2 n = log 2 log(1 + r/100) (2) y = f(x) = log(1 + x) x = 0 1 f (x) = 1/(1 + x) f (0) = 1


2


271124【議運】レジュメ


untitled


Taro12-第4回意見募集結果(改訂


14号A4indd


平成16年度外務省事後評価実施計画策定について



newmain.dvi

201_P1_P24(2)

katagami No.65

Transcription:

5 ()

( ) ( ) ( ) p(a) a I(a) p(a) p(a) I(a) p(a) I(a) (2)

(self information) p(a) = I(a) = 0 I(a) = 0 I(a) a I(a) = log 2 p(a) = log 2 p(a) bit 2 (log 2 ) (3)

I(a) 7 6 5 4 3 2 0 0.5 p(a) p(a) = /2 I(a) = p(a) = I(a) = 0 (4)

: boy girl : I(boy) = log 2 2 = bit I(girl) = log 2 2 = bit 2 /8 I( ) = log 2 2 3 = 3 bit I( ) = log 2 7 8 = log 2 7 + log 2 8 = 2.807 + 3 = 0.93 bit (5)

E E E 2 E I(E) = I(E ) + I(E 2 ) 52 A I( A) = log 2 52 5.7 bit I( ) = log 2 4 = 2 bit A I(A) = log 2 3 3.7 bit I( A) = I( ) + I(A) (6)

( )..... log a b = c b = a c ( ) 2 log a b = log 0 b log 0 a 3 log a (xy) = log a (x) + log a (y) x 4 log a y = log a(x) log a (y) 5 log a x y = y log a x 6 log a x = log a x ( 5 y = ) log 2 x 2 log 2 x = log 0 x/ log 0 2 = log 0 x/0.300 3.3223 log 0 x. (7)

(average information) ( ) A A = {a, a 2,..., a n } n p(a i ) ( ) I(a i ) H(A) H(A) = n p(a i )I(a i ) = i= p(a i ) p i n p(a i ) log 2 p(a i ) i= n H(A) = p i log 2 p i i= bit (8)

0 H(A) log 2 n bit A a i p(a i ) = 0 H(A) = 0 0 A p(a i ) = /n H(A) = log 2 n (9)

8 : p( ) = 4, p( ) = 2 H(A) = p( ) = 4, p( ) = 0 4 p i log 2 p i i= = 4 log 2 4 2 log 2 2 4 log 2 4 0 log 2 0 = 2 4 + 2 + 2 0 =.5 bit 4 x 0 x log 2 x 0 (0)

(entropy) H = K k n k ln n k K n k k H = i p i log 2 p i ( ) ()

K 40% 30% 30% H = 0.4 log 2 0.4 0.3 log 2 0.3 0.3 log 2 0.3 =.57 bit 00% H =.0 log 2.0 0 log 2 0 0 log 2 0 = 0 bit 00% 0 (2)

(maximum entropy) (2 ) : ( ) a a 2 A = p p 2 (p + p 2 = ) H = p log 2 p p 2 log 2 p 2 p + p 2 = H : L = p log 2 p p 2 log 2 p 2 + λ( p p 2 ) L/ p i = log 2 p i + λ = 0 L/ λ = p p 2 = 0 log 2 p = log 2 p 2 H max = log 2 = bit 2 (3)

g(x) = 0 f(x) λ L = f(x) λg(x) x L = f λ g = 0, L λ = 0 d + x,...,x d, λ d + (4)

f ( x) = const. f g(x) = 0 g g(x) = 0 f(x) f = λ g (5)

n 2 n ( ) a a 2 a n A = p p 2 p n H = n p i log 2 p i i= 2 : ( ) n n L = p i log 2 p i + λ p i i= L/ p i = log 2 p i + λ = 0 L/ λ = n i= pi = 0 p = p 2 = = p n H max = log 2 bit n i= (6)

H max = 6 i= 6 log 2 6 = log 2 6 = 2.585 bit (A Z 27 ) 27 H max = i= 27 log 2 27 = log 2 27 = 4.755 bit (945 ) 945 H max = i= 945 log 2 945 = log 2 945 = 0.925 bit (7)

(entropy function) 2 H = p log 2 p p 2 log 2 p 2 p = p p 2 = p H(p) = p log 2 p ( p) log 2 ( p) H(p) H(p) p (8)

A 0.6 ( 0.4) B 0.9 ( 0.) H(A ) = H(0.6) = H(0.4) 0.97 bit H(B ) = H(0.9) = H(0.) 0.496 bit B A (9)

(joint entropy) : ( ) a a A = 2 p(a ) p(a 2 ) B = ( ) b b 2 p(b ) p(b 2 ) A B A B AB : ( ) (a, b AB = ) (a, b 2 ) (a 2, b ) (a 2, b 2 ) p p 2 p 2 p 22 (a i, b j ) = a i b j p ij = p(a i b j ) AB H(AB) = p ij log 2 p ij i j (20)

(conditional entropy) H(AB) H(AB) = p(a i b j) log 2 p(a i b j) i j = p(a i )p(b j a i ) log 2 p(a i )p(b j a i ) i j = p(a i)p(b j a i){log 2 p(a i) + log 2 p(b j a i)} i j = p(a i )p(b j a i ) log 2 p(a i ) i j p(a i)p(b j a i) log 2 p(b j a i) i j = p(a i ) log 2 p(a i ) p(b j a i ) i j i p(a i) p(b j a i) log 2 p(b j a i) j j p(b j a i ) = H(A) (2)

2 j p(b j a i ) log 2 p(b j a i ) a i b j 2 a i H(B A) H(B A) = i p(a i ) j p(b j a i ) log 2 p(b j a i ) H(B A) H(AB) H(AB) = H(A) + H(B A) H(AB) = H(B) + H(A B) H(AB) = H(BA) (22)

(Shannon s fundamental inequality) : H(A B) H(A), H(B A) H(B) ( ) A: B: B A H(AB) = H(A) + H(B A) H(AB) = H(A) + H(B A) H(A) + H(B) A B (23)

H(AB) = H(A) + H(B A) H(A) + H(B) A B A: B: A B 0 H(A B) H(A) H(AB) (24)

H(AB) H(A B) H(B A) H(A) H(B) H(AB) = H(A) + H(B A) = H(B) + H(A B) (25)

5. 45% 35% 2% 8% H 2 5.2 48 3 5.3 A 3 75% A 3 30% (26)