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network information theory chap5 课后题答案
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network information theory (网络信息论)第五章课后题答案
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Solutions to Selected Problems in Chapter 5 1
SOLUTIONS TO SELECTED PROBLEMS IN CHAPTER 5
.. Verify that the characterizations of the capacity region for the BS-BC and the
AWGN-BC in (??) and (??), respectively, are convex.
.. Show that the converse for the degraded DM-BC can be proved with the auxiliary
random variable identication U
i
=(M
2
, Y
i−1
2
)or U
i
=(M
2
, Y
i−1
1
, Y
i−1
2
).
.. Prove the converse for the capacity region of the AWGN-BC by starting from the
single-letter characterization
R
1
≤I(X; Y
1
|U ),
R
2
≤I(U ; Y
2
)
for some F(u, x)with E(X
2
)≤P.
.. Show that if a DM-BC is degraded, then it is also less noisy.
.. Show that if a DM-BC is less noisy, then it is also more capable.
.. Given a DM-BC p(y
1
, y
2
|x), let D(p(x)) = I(X; Y
1
)−I(X; Y
2
), where p(x)is the
pmf of X. Show that if D(p(x))is convex in p(x), then Y
1
is less noisy than Y
2
.
.. Prove the classication of the BSC–BEC broadcast channel in Example ??.
.. Prove the converse for the capacity region of the less noisy DM-BC p(y
1
, y
2
|x).
(Hint: Follow the steps of the converse for the more capable DM-BC.)
.. Show that the t wo characterizations of the capacity region for the more capable
DM-BC in Section ?? are equal.
.. Another simple outer bound. Consider the DM-BC with three messages.
(a) Show that a rate triple (R
0
, R
1
, R
2
)is in the capacity region if
R
0
+R
1
≤I(X; Y
1
) ,
R
0
+R
2
≤I(X; Y
2
)
for some p(x).
(b) Show that this outer bound on the capacity region is tighter than the simple
outer bound in Figure ??. (Hint: Consider two antisymmetric Z-channels.)
(c) S et R
1
= R
2
= 0 in the outer bound in part (a) to show that the common-
message capacity is
C
0
=max
p(x)
min{I(X; Y
1
) , I(X; Y
2
)}.
Argue that C
0
is in general strictly smaller than min{C
1
, C
2
}.
2 Solutions to Selected Problems in Chapter 5
.. Alternative decoding for degraded broadcast channels. Consider the following de-
coding rule for receiver Y
1
—declare that (
m
1
,
m
21
)is sent if it is the unique mes-
sage pair such that (x
n
(
m
1
,
m
21
) , y
n
1
) ∈ T
(n)
є
. Show that t h is modied rule gives
the same contraints as in eorem ??. (Hint: Assume (1, 1)is sent. First deal with
the case of (X
n
(1 , 1), Y
n
1
)∉T
(n)
є
and then break up the remaining error into t hree
types and bound each. e only tricky part is to nd p(x
n
(m
1
, 1), y
n
1
)for m
1
=1.)
.. Product of two degraded broadcast channels. Consider two degraded DM-BCs
p(y
11
|x
1
)p(y
21
|y
11
)and p(y
12
|x
2
)p(y
22
|y
12
). e product of these two degraded
DM-BCs depicted in Figure . is a DM-BC with X = (X
1
, X
2
), Y
1
= (Y
11
, Y
12
),
Y
2
= (Y
21
, Y
22
), and p(y
1
, y
2
|x) = p(y
11
|x
1
)p(y
21
|y
11
)p(y
12
|x
2
)p(y
22
|y
12
). Show
that the private-message c apacity region for the product DM-BC is the set of rate
pairs ( R
1
, R
2
)such t hat
R
1
≤I(X
1
; Y
11
|U
1
)+I(X
2
; Y
12
|U
2
) ,
R
2
≤I(U
1
; Y
21
)+I(U
2
; Y
22
)
for some p(u
1
, x
1
)p(u
2
, x
2
).
X
1
X
2
Y
11
Y
21
Y
12
Y
22
p(y
11
|x
1
)
p(y
12
|x
2
)
p(y
21
|y
11
)
p(y
22
|y
12
)
Figure .. Product of two degraded broadcast channels.
Remark: Poltyrev. CHECK!
Solution: Consider an auxiliary random variables U = (U
1
, U
2
). e channel
X → Y
1
→Y
2
is a degraded broadcast channel. erefore, the capacity region is
the set of all rate pairs (R
1
, R
2
)such t hat
R
1
≤I(X; Y |U)=I(X
1
, X
2
; Y
11
, Y
12
|U
1
, U
2
) ,
R
2
≤I(U ; Z)=I(U
1
, U
2
; Y
21
, Y
22
)
for some p(u
1
, u
2
)p(x
1
, x
2
|u
1
, u
2
). As a result of the independence of the two
channels, it seems reasonable that the above region is not changed if we restrict
ourselves to p(u
1
, u
2
)p(x
1
, x
2
|u
1
, u
2
)= p(u
1
)p(x
1
|u
1
)p(u
2
)p(x
2
|u
2
)(see converse
below). is simplies the above characterization of the capacity region to the set
of all rate pairs (R
1
, R
2
)such t hat
R
1
≤I(X
1
; Y
11
|U
1
)+I(X
2
; Y
12
|U
2
) ,
R
2
≤I(U
1
; Y
21
)+I(U
2
; Y
22
)
Solutions to Selected Problems in Chapter 5 3
for some p(u
1
)p(u
2
)p(x
1
|u
1
)p(x
2
|u
2
). Achievability of this region follows from the
achievability for the degraded broadcast channel. We prove the converse. Con-
sider
nR
2
=H(W
2
)
≤I(W
2
; Z
n
1
, Z
n
2
)+nє
n
=I(W
2
; Z
n
1
)+I(W
2
; Z
n
2
|Z
n
1
)+nє
n
=
n
i=1
(I(W
2
; Z
1i
|Z
i−1
1
)+I(W
2
; Z
2i
|Z
i−1
2
, Z
n
1
))+nє
n
=
n
i=1
(H(Z
1i
|Z
i−1
1
)−H(Z
1i
|Z
i−1
1
, W
2
)+H(Z
2i
|Z
i−1
2
, Z
n
1
)
−H(Z
2i
|Z
i−1
2
, Z
n
1
, W
2
))+nє
n
≤
n
i=1
(H(Z
1i
)−H(Z
1i
|Z
i−1
1
, W
2
)+H(Z
2i
)−H(Z
2i
|Z
i−1
2
, Z
n
1
, W
2
))+nє
n
≤
n
i=1
(H(Z
1i
)−H(Z
1i
|Y
i−1
1
, Z
i−1
1
, W
2
)+H(Z
2i
)−H(Z
2i
|Y
i−1
2
, Z
i−1
2
, Z
n
1
, W
2
))+nє
n
=
n
i=1
(H(Z
1i
)−H(Z
1i
|Y
i−1
1
, W
2
)+H(Z
2i
)−H(Z
2i
|Y
i−1
2
, Z
n
1
, W
2
))+nє
n
=
n
i=1
(I(Y
i−1
1
, W
2
; Z
1i
)+I(Y
i−1
2
, Z
n
1
, W
2
; Z
2i
)+nє
n
=
n
i=1
(I(U
1i
; Z
1i
)+I(U
2i
; Z
2i
))+nє
n
≤n(I(U
1
; Z
1
)+I(U
2
; Z
2
))+nє
n
,
where U
1i
=(Y
i−1
1
, W
2
)and U
2i
=(Y
i−1
2
, Z
n
1
, W
2
). Note that U
1i
→ X
1i
→Y
1i
→
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