4 Solutions to Selected Problems in Chapter 3
Solution:
(a) An index
m will be chosen by MLD if for ever index m =
m, p(y
n
|x
n
(
m))>
p(y
n
|x
n
(m)), or equivalently,
p
d(x
(
m), y
)
(1 −p)
n−d(x
(
m), y
)
>p
d(x
(m) , y
)
(1 −p)
n−d(x
(m) , y
)
,
or equivalently, d(x
n
(
m), y
n
)<d(x
n
(m), y
n
)considering the fact that p <1/2.
(b) Let E ={d(X
n
(1) , Y
n
)>n(p +є)}. en, t he probablity of error averaged over
codebooks is upper bounded as
P
(n)
e
=P{
M =1 |M =1}
=P{d(X
n
(m), Y
n
)<d(X
n
(1) , Y
n
)for some m =1 |M =1}
=P{E , d(X
n
(m), Y
n
)<d(X
n
(1) , Y
n
)for some m =1 |M =1}
+P{E
c
, d(X
n
(m), Y
n
)<d(X
n
(1) , Y
n
)for some m =1 |M =1}
≤P(E |M =1}
+P{E
c
and d(X
n
(m), Y
n
)<d(X
n
(1) , Y
n
)for some m =1 |M =1}
≤P{d(X
n
(1) , Y
n
)>n(p +є)|M =1}
+(2
nR
−1)P{d(X
n
(2) , Y
n
)≤n(p +є)|M =1}.
(c) We know that d(X
n
(1) , Y
n
)is a binomial random variable with mean np and
variance np(1 −p)given {M =1}. erefore, by the Chebyshev inequality,
P{d(X
n
(1) , Y
n
)−np >nє |M =1}≤
p(1 −p)
nє
2
,
which tends to zero as n →∞. Also, by the Cherno bound, the se cond term
tends to zero as n → ∞, provided R < 1 −H(p +є). erefore, by taking
є →0, any rate R <1 −H(p)=C is achievable.
.. Randomized code. Suppose in the denition of the (2
nR
, n)code for the DMC
p(y|x)we allow the encoder and decoder to use random mappings. Specically, let
W be an arbitrary random variable independent of the message M and the channel,
i.e., p(y
i
|x
i
, y
i−1
, m, )=p
Y|X
(y
i
|x
i
)for i ∈[1 : n]. e encoder generates a code-
word x
n
(m, W), m ∈[1 : 2
nR
], and the decoder generates an estimate
m(y
n
, W).
Show that this randomization does not increase the capacity of the DMC.
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