the channel and mutual information
DESCRIPTION
Information Through a Channel symbols can’t be swallowed A a1 : : aq b1 : : bs B alphabet of symbols sent alphabet of symbols received P(bj|ai) or randomly generated For example, in an error correcting code over a noisy channel, s ≥ q. If two symbols sent are indistinguishable when received, s < q. Characterize a stationary channel by a matrix of conditional probabilities: Compare: noise (randomness) versus distortion (permutation) received row column sent s Pi,j Pi,j = P(bj | ai) P = q 7.1, 7.2, 7.3TRANSCRIPT
Chapter 7
The Channel and Mutual
Information
Information Through a Channel
alphabet of
symbols sent
a1 :
:
aq
A
b1 :
:
bs
B
alphabet of
symbols
receivedP(bj|ai)
symbols can’t be swallowed
or randomly generated
For example, in an error correcting code over a noisy channel, s ≥ q.
If two symbols sent are indistinguishable when received, s < q.
Characterize a stationary channel by a matrix of conditional probabilities:
Pi,j = P(bj | ai)
rowcolumn
received
sent
P = Pi,jq
s
out comemust some ,input each for 1)|(11
, ji
s
jij
s
jji baabPP
7.1, 7.2, 7.3
1)()|( then sent, being ofy probabilit )( if 1 1
q
i
s
jiijii apabPaap
For p(ai) = probability of source symbols, let p(bj) = probability of being received
[p(a1) … p(aq)]P = [p(b1) … p(bs)]
sjabPapbpq
iijij ,1)|()()(
1
no noise: Pi,j = I; p(bj) = p(aj)
all noise: Pi,j = 1/s; p(bj) = 1/s
The probability that ai was sent and bj was received is:
P(ai, bj) = p(ai) ∙ P(bj | ai) = p(bj) ∙ P(ai | bj). [coincidental probability]
Baye’s Theorem
So if p(bj) ≠ 0, the backwards conditional probabilities are:
.)()|(
)()|()(
)()|()|(
1
q
iiij
iij
j
iijji
apabP
apabPbp
apabPbaP
1)|(:over sum Now,1
q
iji baPi
7.1, 7.2, 7.3
Binary symmetric Channel
p(a = 0) = p
p(a = 1) = 1 − p
a = 0
a = 1
b = 0
b = 1
P0,0 = P1,1 = P
P0,1 = P1,0 = Q
P0,0
P1,1
P1,0
P0,1
P Q
Q P
(p 1−p)
a = 0 a = 1
( pP + (1 − p)Q pQ + (1 − p)P )
=
p(b = 0) p(b = 1)
7.4
Backwards conditional probabilities No noise All noise
P(a = 0 | b = 0) =Pp
1 pPp + Q(1−p)
P(a = 1 | b = 0) =Q(1−p)
0 1 − pPp + Q(1−p)
P(a = 0 | b = 1) =Qp
0 pQp + P(1−p)
P(a = 1 | b = 1) =P(1−p)
1 1 − pQp + P(1−p)
P = 1 Q = 0 P = Q = ½
If p = 1 − p = ½ (equiprobable) then:
P(a = 1 | b = 0) = P(a = 0 | b = 1) = Q
P(a = 0 | b = 0) = P(a = 1 | b = 1) = P
P
P
7.4
System Entropies
q
i ii ap
apAH1 )(
1log)()(
H(A)
Input entropy
s
j jj bp
bpBH1 )(
1log)()(
H(B)
Output entropy
condition on bj
q
i jijij baP
baPbAH1 )|(
1log)|()|(
average over all bj
s
j
q
i jiji
s
jjj baP
baPbAHbpBAH1 11 )|(
1log),()|()()|(
Similarly
q
i
s
j ijji abP
baPABH1 1 )|(
1log),()|(
The information loss in the channel, called equivocation (or noise entropy). The average uncertainty about the symbol
sent or received.
7.5
H(A| B) H(B| A)
Intuition: taking snapshots
A B
q
i
s
j jiji baP
baPBAH1 1 ),(
1log),(),(
q
i
s
j ijji
q
i
s
j iji abP
baPap
baP1 11 1 )|(
1log),()(
1log),(
H(B | A)
q
i
s
j jiij
q
j
s
i jji baP
abPbp
baP1 11 1 )|(
1log),()(
1log),(
H(B)
H(A, B) =
)|()(
),(
iji
ji
abPap
baP
)|()(
),(
jij
ij
baPbp
abP
7.5
Define :
Joint Entropy
H(A | B)
H(A)
H(A, B)
H(A| B) H(B| A)
Mutual Information
The amount of information they are sharing corresponds to Information gain upon receiving bj : I(ai) − I(ai | bj) .
)()|(
log)|(
1log)(
1log);(i
ji
jiiji ap
baPbapap
baI
By symmetry:
Theorem sBaye'by );()(
)|(log);( ji
j
ijij baI
bpabP
abI
If ai and bj are independent (all noise), then P(ai , bj) = p(ai) ∙ p(bj) and hence P(ai | bj) = p(ai) I(ai ; bj) = 0. No
information gained in channel.
7.6
p(ai) P(ai | bj)
a priori a posteriori
H(A| B) H(B| A)
H(A, B)
I(A; B)
shared
joint
Average over all ai:
)(
)|(log)|();()|();(
i
jiji
ijijij ap
baPbaPbaIbaPbAI
Similarly: j j
ijiji bp
abPabPBaI
)()|(
log)|();(
);()()(
),(log),();()();( ABI
bpapbaP
baPBaIapBAIi j ji
jiji
iii
from symmetry
)(1log
)(
),()(
1log
)(
),(),(
1log),(jj
j
iji
ii
i
jji
i j jiji bp
bp
baPap
ap
baPbaP
baP
i j
jijiji bpapbaPbaPBAI )](log)(log),()[log,();(
= H(A) + H(B) − H(A, B) ≥ 0 H(A, B) H(A) + H(B)
We know H(A, B) = H(A) + H(B | A) = H(B) + H(A | B). I(A ; B) = H(A) − H(A | B) = H(B) −
H(B | A) ≥ 0
H(A | B) H(A) and H(B | A) H(B).
By Gibbs I(A; B) ≥ 0. Equality only if P(ai, bj) = p(ai)∙p(bj) [independence].
7.6