Embed Size (px)
DESCRIPTION
78
Citation preview
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 1/15
pL (1 − p)L
p
p
L
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 2/15
ητρωπη
B() L
p
(1 − p)
B()
0 20 40 60 80 100
0.0
0.2
0.4
0.6
0.8
1.0
p = 0.2
(1 − p) = 0.8
L
pL
(1−
p)L
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 3/15
pL (1 − p)L
n p(m)
m0 1000 2000 3000 4000
0
5
10
15
20
L = 5000
p
L
m
L
m/L = p
B()
L
(1 − p)log2
1
1− p + p log2
1
p
L
2L
H B( p) = (1 − p)log2
1
1 − p + p log2
1
p,
B()
m L − m
n p(m) = L!
m!(L − m)!(1 − p)L−m pm,
p
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 4/15
n p(m)/2L
m
0 200 400 600 800 1000
0.01
0.03
L = 1000
p m
L
p×L
Lp(1− p)
m/L
mL = p
m! √
2πmm
e
m,
n p(m)
n p(m) L
2πm(L−
m) L
e L
e(1 − p)
L−
m L−m
ep
mm
,
n p(m)
1
2πLmL (1 − m
L )
1 − p
1 − mL
L(1−m
L)
pmL
LmL
.
m
L = q,
n
p(q ) = n p(m) 1
2πLq (1 − q )
1 − p
1 − q
L(1−q) p
q
Lq,
loge n
p(q ) loge n p(m) = loge1
2πLq (1 − q )
+ L [(1 − q )(loge(1 − p) − loge(1 − q )) + q (loge( p) − loge(q ))] .
Lp
Lp(1 − p) q p + ε
n
p(ε) = loge n
p(q ) −1
2 loge [2πL( p + ε)(1 − p − ε)]
+ L [(1−
p−
ε) (loge
(1−
p)−
loge
(1−
p−
ε)) + ( p + ε) (loge
( p)−
loge
( p + ε))] ,
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 5/15
n
p(ε) −1
2 loge (2πLp(1 − p)) +
− ε
2(1 − p) +
ε
2 p
+ L
(1 − p − ε)
ε
1 − p − ( p + ε)
ε
p
.
L
n
p(ε) −1
2 loge (2πLp(1 − p)) − L
ε2
2 p(1 − p).
mL
log[n p(m)]
m
0 200 400 600 800 1000
-40
-30
-20
-10
0
L = 1000
p
1
2 loge
L
2πp(1 − p) − L
mL − p
22 p(1 − p)
.
L
L( p − δ )
L( p + δ ) δ
L
m
m
L − p
> s
< p(1 − p)
s2L .
L
m
1/L
e−2(1−2 p)L
pL
(1 − p)L
pL
pL
(1 − p)L
n1/2( p) = L!
( pL)![(1 − p)L]!
,
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 6/15
m
m/L0.15 0.20 0.25 0.30 0.35
30
70
L = 1000
L = 5000 p
m
m/L0.0 0.2 0.4 0.6 0.8 1.0
-30
-20
-10
0
10
L = 1000
L = 5000 p
− L2 p(1− p)
2L
L pL=0
n1/2( p) =L
pL=0
L!
( pL)![(1 − p)L]! = 2L.
n1/2( p) =
√ 2πL
Le
L√
2πLpLpe
Lp 2πL(1 − p)
L(1− p)
e
L(1− p) .
n1/2( p) = 1
2πLp(1 − p) pLp(1 − p)L(1− p),
n1/2( p) = 1
2πLp(1 − p)2
−L( p log2 p+(1− p) log2(1− p)),
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 7/15
n1/2( p) = 2−L( p log2 p+(1− p) log2(1− p))− 12 log2(2πLp(1− p)).
n1/2( p)
p0.0 0.2 0.4 0.6 0.8 1.0
-150
-100
-50
L
m
2L(H B−1)
L
LH B( p) = L (− p log2 p − (1 − p)log2(1 − p)).
H B( p)
p0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
Lp
2L(H B( p)+δ)
δ
L(H B( p) + δ )
p
LpLx=0
n1/2(x) =
LpLx=0
L!
Lx!L(1 − x)!,
x
[0, 1] Lx
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 8/15
LpLx=0
n1/2(x) ≤ n1/2( p)(1 + δ ),
δ
L
L
n1/2(x)
2L
π e−2L(x− 1
2 )2
.
x
p
x < p
p
n1/2(x)
2L
π e−2L( p− 1
2 )2
e−2L
(x− 1
2 )2−( p− 1
2 )2
,
n1/2(x)
n1/2( p)e−2L[x− p](x−1+ p) < n1/2( p)e2(1−2 p)L[x− p].
LpLx=0
n1/2(x) < 1 − e−2Lp(1−2 p)
1 − e−2(1−2 p)L)n1/2( p).
L
LpLx=0
n1/2(x) < (1 + e−2(1−2 p)L)n1/2( p).
p
p e−2L(x− p)2 (x − p)
12L
LpLx=0 n1/2(x)
p
0.0 0.1 0.2 0.3 0.4 0.5
-300-270
-240
-210
-180
-150
-120
-90
-60
-30
0
p
p
L = 1000
p 2LH ( p)
p
p + δ
2LH B( p)
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 9/15
12L
LpLx=0 n1/2(x)
L
0 500 1000 1500 2000 2500 3000
-220
-180
-140
-100
-60
-20
20
p = 0.2
L
p
m
L0.0 0.2 0.4 0.6 0.8 1.0
-150
-100
-50
p − δ p + δ
L m
p + δ
p − δ
LH B( p)
p
p
LH B( p) = Lp log2
L
Lp − L(1 − p)log2 e loge(1 − p),
LH B( p) Lp
log2
L
Lp + log2 e
.
Lp
L
Lp log
2e
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 10/15
( pM = 1/2) M
L
L
M 2M 2L
H M = M/L 2M H M
2M 2L
2M = 2LH M .
1 − p
1 − p
p
p
p
p
H B = − p log2 p − (1 − p)log2(1 − p)
M
B
H M + H B
H M + H B < 1,
M
M
L − M
L − M > LH B ,
M
L < 1 − H B .
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 11/15
M
LH B
L
B
M
L
Lp
2M
2M 2L
M L
Lp
Lp
2M
2L
(2L)!(2L−M )!
2L
2L−M 2L2L−(L−M )2L−M
e2L−2L−M .
2L
2M
2L
2L 2M L M
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 12/15
p
2M
Lp
Lp
L
p
L 2L(1−H B( p))
p 2M 2LH B
M
2M 2LH B≤
2L.
p = 0, H B( p) = 0 2L
p = 1
2 H B( p) = 1
2L−LH B( p) 2LH B( p)
2L
L 2M
L2
2L
n1/2(x) = L!
(xL)![(1 − x)L]!.
L2
2M (2M − 1)
(2M − 1)2LH B M
L( p + ε)
Lp
Q =
LpxL=0
1
2Ln1/2(x) =
LpxL=0
L!
xL!(L − xL)!
1
2L
,
Q =
LpxL=0
1 2πL(1 − x)x
1
(1 − x)L(1−x)
1
(x)Lx
1
2L
.
Q
n1/2( p)
2L 2LH B
2L .
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 13/15
LpLx=0
12L
n1/2(x)
p
1L×
0.0 0.2 0.4 0.6 0.8 1.0
-150
-100
-50
L = 5000
L = 1000
L
2M L
M L
2M − 1
Lp
S =
2M −11
Q 2M Q 2M +LH B−L.
M + LH B − L < 0.
L
M L
α
S 2M +LH B−L < 2α,
M
L < 1 − H B +
α
L.
L
H M = M
L < 1 − H B ,
L
M =
1
1 − H B ,
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 14/15
p0.00 0.05 0.10 0.15 0.20 0.25 0.30
0
1
2
3
4
5
6
7
8
9
7/21/2019 Coursentropie4 Vf
http://slidepdf.com/reader/full/coursentropie4-vf 15/15
π
