On Stochastic Orders and Fast Fading Multiuser
Channels with Statistical CSIT
Pin-Hsun Lin, Member, IEEE, Eduard A. Jorswieck, Senior Member, IEEE,

arXiv:1712.03692v1 [] 11 Dec 2017

Rafael F. Schaefer, Senior Member, IEEE, Martin Mittelbach, and
Carsten R. Janda, Student Member, IEEE
Abstract
In this paper, we investigate the ergodic capacity of fast fading Gaussian multiuser channels when only the
statistics of the channel state are known at the transmitter. In general, the characterization of capacity regions
of multiuser channels with only statistical channel state information at the transmitter (CSIT) is open. Instead of
directly matching achievable rate regions and the corresponding outer bounds, in this work we resort to classifying
the random channels through their probability distributions. To be more precise, in order to attain capacity results,
we first derive sufficient conditions to attain some information theoretic channel orders such as degraded and
strong/very strong interference by applying the usual stochastic order and exploiting the same marginal property
such that the capacity regions of the memoryless Gaussian multiuser channels can be characterized. These include
Gaussian interference channels, Gaussian broadcast channels, and Gaussian wiretap channels/secret key generation.
We also extend the framework to channels with a specific memory structure, namely, channels with finite-state,
wherein the Markov fading channel is discussed as a special case. Several practical examples such as Rayleigh
fading and Nakagami-m fading, etc., illustrate the application of the derived results.
Index Terms
Stochastic orders, Same marginal property, Imperfect CSIT, Coupling, Maximal coupling, Copula, Multi-user
channels, Capacity regions, Channel with memory
Parts of the work were presented in ITW 2016, Cambridge, UK and SCC 2017, Hamburg, Germany. Pin-Hsun Lin, Eduard A. Jorswieck,
Martin Mittelbach and Carsten R. Janda are with the Communications Laboratory, Department of Electrical Engineering and Information
Technology, Technische Universität Dresden, Germany. Rafael F. Schaefer is with the Information Theory and Applications Chair, Technische
Universität Berlin, Germany Emails: {pin-hsun.lin, eduard.jorswieck, martin.mittelbach, carsten.janda}@tu-dresden.de, rafael.schaefer@tuberlin.de. Part of this work is funded by FastCloud 03ZZ0517A and FastSecure 03ZZ0522A.

2

I. I NTRODUCTION
For Gaussian multiuser (GMU) channels with perfect channel state information at the transmitter (CSIT),
due to the capability of ordering the channels of different users, capacity regions/secrecy capacity are
known for the degraded broadcast channel (BC) [1] [2] and wiretap channel (WTC) [3], [4], [5], and also
the sum capacity of low-interference regime [6] and the capacity region for some cases of interference
channel (IC) such as strong IC [7], [8], [9] and very strong IC [8]. When fading effects of wireless
channels are taken into account, if there is perfect CSIT, some of the above capacity results still hold with
an additional operation of taking the average with respect to the fading channels. For example, in [10], the
ergodic secrecy capacity of Gaussian WTC is derived; in [11], the ergodic capacity regions are derived for
ergodic very strong and uniformly strong (each realization of the fading process is a strong interference
channel) Gaussian IC. Because of limited feedback bandwidth and the delay caused by channel estimation,
the transmitter may not be able to track channel realizations instantaneously if they vary rapidly. Thus,
for fast fading channels, it is more practical to consider the case with only partial CSIT of the legitimate
channel. However, when there is only statistical CSIT, there are only few known capacity results, such as
the layered BC [12], the binary fading interference channel [13], the one-sided layered IC [14], Gaussian
WTC [15], [16], and layered WTC [16], etc. The main difficulty is that, without the knowledge of the
instantaneous CSIT, we may not be able to directly compare the channels in a simple form and manner.
In particular, channel orders including degraded, less noisy, and more capable [17] [18] in Gaussian BC
and Gaussian WTC or the strong and very strong in IC depend on the knowledge of CSIT. Note that
these channel orders can help to simplify the functional optimization with respect to the channel input
distribution and/or channel prefixing.
To consider a GMU-channel in which the transmitters only know the distributions of the channels
but not the realizations, some immediate questions are: How to compare channel qualities only by their
distributions? How to derive the capacity region by exploiting such comparison of channel qualities? From
an information-theoretic point of view, to deal with this problem, a better achievable scheme and tighter

3

outer bound shall be developed. In contrast, in this work we resort to identifying that whether the random
channels are stochastically orderable or not, to find the capacity region. In particular, a GMU channel with
orderable random channels means that there exists an equivalent GMU channel in which we can reorder
channel realizations among different transmitter-receiver pairs in the desired manner1 . Taking the Gaussian
BC (GBC) as an example, an orderable two-user GBC means that under the same noise distributions at
the two receivers, in the equivalent GBC, one channel strength is always stronger or weaker than the
other for all realizations within a codeword length. We attain this goal mainly by the following elements:
stochastic orders [19], coupling [20], and the same marginal property [21]. The stochastic orders have
been widely used in the last several decades in diverse areas of probability and statistics such as reliability
theory, queueing theory, and operations research, etc., see [19] and references therein. Different stochastic
orders such as the usual stochastic order, the convex order, and the increasing convex order can help us
to identify the location, dispersion, or both location and dispersion of random variables, respectively [19].
Choosing a proper stochastic order to compare the unknown channels allows us to form an equivalent
channel in which realizations of channel gains are ordered in the desired manner. Then, we are able to
derive the capacity regions of the equivalent MU channel, which is simpler than directly considering the
original channel.
We also investigate the applicability of the developed framework to MU channels with memory, e.g.,
the finite-state BC (FSBC) [22]. In the FSBC model, the channels from the transmitter to the receivers
are governed by a state sequence that depends on the channel input, outputs, and previous states, which
are described by the transition function. Some examples for FSC include multiple access channels [23],
degraded BC [22], etc. More detailed discussion on FSC without and with feedback can be found in [22],
[24], respectively, and references therein.
The main contributions of this paper, which are all novel in the literature to the best of our knowledge,
are as follows:
1 The

desired manner is related to the intrinsic property of the channel to be analyzed, which will be shown later.

4

•

We use maximal coupling to illustrate the insight of the proposed framework and we use coupling to
derive sufficient conditions to attain the capacity results. In addition to the coupling scheme, we also
identify that another alternatively explicit construction to derive the sufficient conditions is indeed a
copula.

•

By coupling, we classify memoryless fast fading GMU channels such that we can obtain the capacity
results of them under statistical CSIT. To attain this goal, we integrate the concepts of usual stochastic
order and the same marginal property, i.e., an intrinsic property of MU channels without receivers
cooperation, such that we can align the realizations of the fading channel gains between different
users in an identical trichotomy order over time to obtain an equivalent channel.

•

We then connect the trichotomy order of channel strengths in the equivalent channels and different
information theoretic orders to characterize the capacity regions of: Gaussian IC (GIC), GBC, GWTC,
and also the secret key capacity of the secret key generation with a Gaussian-Maurer’s satellite model
[25].

•

We further extend the framework to time-varying channels with memory. In particular, we consider
the finite-state channel (FSC) model introduced by [26]. We use the finite-state BC (FSBC) [22] as
an example. The Markov fading channel, which is commonly used to model the memory effect in
wireless channels, is also considered in this paper.

•

Several examples with practical channel distributions are illustrated to show the usage scenarios of
the developed framework.

Notation: Upper case normal/bold letters denote random variables/random vectors (or matrices), which
will be defined when they are first mentioned; lower case bold letters denote vectors. The expectation is
denoted by E[.]. We denote the probability mass function (PMF) and probability density function (PDF) of
a random variable X by PX and fX , respectively. The cumulative distribution function (CDF) is denoted by
FX (x), where F̄X (x) = 1 − FX (x) is the complementary CDF (CCDF) of X. X ∼ F denotes that the random
variable X follows the distribution with CDF F. The mutual information between two random variables X

5

and Y is denoted by I(X;Y ) while the conditional mutual information given Z is denoted by I(X;Y |Z). The
differential and conditional differential entropies are denoted by h(·) and h(·|·), respectively. A Markov
chain relation between X, Y , and Z is described by X −Y − Z. Unif(a, b) denotes the uniform distribution
between a and b and N0 = {0, N}. Bernoulli distribution with probability p is defined by Bern(p). The
indicator function is denoted by 1{.} , e.g., 1{(1)} = 1 means the expression in (1) is valid. The support
/ The
of a random variable X is denoted by either supp(X) or supp( fX ). The null set is denoted by 0.
logarithms used in the paper are all with respect to base 2. We define C(x) , 12 log(1 + x). We denote
the equality in distribution by =d . The convolution of functions f and g is denoted by f ∗ g. Calligraphic
alphabets in capital denote sets. The convex hull of a set A is denoted by co(A ).
The remainder of the paper is organized as follows. In Section II, we introduce our motivation and
important preliminaries. In Section III, we develop our framework from maximal coupling, coupling, and
copulas. In Section IV we apply the developed framework to fast fading Gaussian interference channels,
broadcast channels, wiretap channels, and source model secret key generation with statistical CSIT. In
Section V we discuss the finite state broadcast channel as an example of applying the developed framework
for channels with memory. Finally, Section VI concludes the paper.
II. M OTIVATION AND P RELIMINARIES
In this section, we first explain our motivation and then review several key properties and definitions
including the same marginal property, degradedness, and the usual stochastic orders, which are crucial
for deriving the main results of this work. We assume that each node in the considered MU channel is
equipped with a single antenna.
A. Motivation
With perfect CSIT, the transmitter can easily compare the strengths of instantaneous channels among
time and/or users and find the optimal strategy for the transmission [18], including the designs of the
codebook, channel input distribution, resource allocation, etc., simply according to the trichotomy law. In

6

contrast, when the transmitter has imperfect CSIT, e.g., only the statistics of the channels, the comparison
is non-trivial due to the random strength of the fading channels. In the following, we use two simple
examples, e.g., fast fading channels with additive white Gaussian noises (AWGN) to illustrate the difficulty
of comparing channels in such a scenario. For simplicity, we consider a two receiver case where we
assume that the noise variances at different receivers are identical, without loss of generality. Denote
the square of the magnitudes of the two real random channels by H1 and H2 with PDF’s f1 and f2 ,
respectively. In Fig. 1 (a), the supports of f1 and f2 are non-overlapping. Therefore, even when there is
only statistical CSIT, we still can know that the second channel is always stronger than the first channel.
However, due to the lacked knowledge of CSI, the capacity results degrade. In contrast, in Fig. 1 (b),
the supports of the two channels are overlapping. Intuitively, the transmitter is not able to distinguish
the stronger channel just based on the channel statistics. This is because, due to the overlapping part
of the PDF’s, the trichotomy order of the channel realizations H1 = h1 and H2 = h2 may alter over
time. As an example, from Fig. 1 (b) we may have two vectors of absolute square of channel gains
as h 1 = [3.1, 2.3, 0.7, 8.5, · · · ] and h 2 = [1.6, 4.9, 2.8, 3.1, · · · ], where each entry is independently and
identically generated by the corresponding distributions at a specific sample time. It can be easily seen
that there is no fixed trichotomy order between h1 and h2 within the whole codeword length. Therefore,
we may not be able to claim which channel is better in this scenario. Based on the above observation,
we would like to ask the following question: can the transmitter compare the strengths of channels to
different receivers, only based on the knowledge of the distributions of the fading channels in general,
just like the comparison of f1 and f2 in Fig. 1(a)? In addition, can we extend the scenarios of channel
comparison from the easy case as Fig. 1(a) to Fig. 1(b)? In the following, we partly answer this question.
B. Preliminaries
In the following, we review important background knowledge that will be necessary in the following.
1) The Same Marginal Property: The same marginal property is crucial in the sense that it provides us
the degree of freedom to construct an equivalent channel in which the realizations of all random channel

7

PDF

PDF
f2

f1

f1

f2

h
(a)

Fig. 1.

h
(b)

Two examples of relations between two fading channels.

tuples are aligned in the desired manner. For channels with multiple transmitters and multiple receivers,
e.g., the interference channel, we can extend from the following building block:
Theorem 1 (The Same Marginal Property for Two Transmitters [27, Theorem 16.6]). The capacity
region of a discrete memoryless multiuser channel including two transmitters and two non-cooperative
receivers with input and output alphabets X1 × X2 and Y1 × Y2 , respectively, depends only on the conditional marginal distributions PY1 |X1 ,X2 and PY2 |X1 ,X2 and not on the joint conditional distribution PY1 ,Y2 |X1 ,X2 ,
where (X1 , X2 ) ∈ X1 × X2 and (Y1 , Y2 ) ∈ Y1 × Y2 are the transmit and receive signal pairs, respectively.
For channels with a single transmitter and multiple receivers, e.g., BC or WTC, we can specialize from
Theorem 1 as follows by removing X1 or X2 .
Corollary 1 (The Same Marginal Property for One Transmitter [27, Theorem 13.9]). The capacity
region of a discrete memoryless multiuser channel including one transmitter and two non-cooperative
receivers with input and output alphabets X and Y1 × Y2 , respectively, depends only on the conditional

marginal distributions PY1 |X and PY2 |X and not on the joint conditional distribution PY1 ,Y2 |X , where X ∈ X
and (Y1 , Y2 ) ∈ Y1 × Y2 are the transmit signal and receive signal pair, respectively.

2) Information-Theoretic Orders for Memoryless Channels and Stochastic Orders: Based on the same
marginal property, we introduce the following definitions describing the relation of reception qualities
among different receivers.
Definition 1. A discrete memoryless channel with two non-cooperative receivers and one transmitter is
physically degraded if the transition probability satisfies PY1Y2 |X (y1 , y2 |x) = PY1 |X (y1 |x)PY2 |Y1 (y2 |y1 ), for all

x ∈ X , y1 ∈ Y1 , y2 ∈ Y2 , i.e., X, Y1 , and Y2 form a Markov chain X −Y1 −Y2 . The channel is stochastically

degraded if its conditional marginal distribution is the same as that of a physically degraded channel,

8

i.e.,
∃ P̃Y2 |Y1 (y2 |y1 ) : PY2 |X (y2 |x) = ∑ PY1 |X (y1 |x)P̃Y2 |Y1 (y2 |y1 ), ∀ x, y1 , y2 .

(1)

y1

By applying the quantization scheme used in [18, Proof of Theorem 3.3 and Remark 3.8], we can
extend the discussion from discrete alphabets to continuous ones. In the following we consider fast fading
GMU channels. Denote the square of the fading channels from the transmitter to the first and second
receivers considered in Definition 1 by the random variables H1 and H2 , respectively. We then define two
sets of pairs of the random channels as H1 = {(H1 , H2 )} and H10 = {(H1 , H2 ) : 1{(1)} = 1}, i.e., (H1 , H2 )
in H10 must satisfy (1). In the following, we call a stochastically degraded channel simply a degraded
channel due to the same marginal property. Note that discussions on the relation between degradedness
and other information theoretic channel orders can be referred to [17], [27], [28].
Definition 2. Denote the square of channels from the i-th transmitter to the first and second receivers by
H1i and H2i , respectively, i = 1, 2. A discrete memoryless-IC is said to have strong interference if
I(X1 ;Y1 |X2 , H11 , H12 ) ≤ I(X1 ;Y2 |X2 , H21 , H22 ),

(2)

I(X2 ;Y2 |X1 , H21 , H22 ) ≤ I(X2 ;Y1 |X1 , H11 , H12 ),

(3)

for all fX1 · fX2 . Define the sets H2 = {(H11 , H12 , H21 , H22 )} and H20 = {(H11 , H12 , H21 , H22 ) : 1{(2)} =
1 and 1{(3)} = 1}.
Definition 3. A discrete memoryless-IC is said to have very strong interference if
I(X1 ;Y1 |X2 , H11 , H12 ) ≤ I(X1 ;Y2 |H21 , H22 ),

(4)

I(X2 ;Y2 |X1 , H21 , H22 ) ≤ I(X2 ;Y1 |H11 , H12 ),

(5)

for all fX1 · fX2 . Define the set H30 = {(H11 , H12 , H21 , H22 ) : 1{(4)} = 1 and 1{(5)} = 1}.
In the following, we introduce the definition of the usual stochastic order, which is the underlying tool
to derive the main results in this paper.
Definition 4. [19, (1.A.3)] For random variables X and Y , X is smaller than Y in the usual stochastic
order, namely, X ≤st Y , if and only if F̄X (x) ≤ F̄Y (x) for all x ∈ (−∞, ∞).
Note that the definition is applicable to both discrete or continuous random variables.

9

III. M AIN R ESULTS
In this section, we develop a general framework to classify fading channels such that we are able to
characterize the corresponding capacity results under statistical CSIT.
A. Problem Formulation and the Proposed Framework
According to the motivation described in Section II-A, we aim to formulate a problem in a tractable
way. We first define a set A ⊆ {H1 , H2 }, which is a subset of all tuples of random channels of the
aforementioned MU channels. We also define a set B , which includes all above MU channel orders from
Definitions 1, 2, and 3 as

B , {H10 , H20 , H30 }.

(6)

Intuitively, we want to find a subset of all fading channel tuples, namely, A , which should possess the
following properties:
•

encompasses a constructive way to find the transformation f : A 7→ B ;

•

allows the existence of a corresponding set B in which the channel qualities follow a certain order;

•

by this channel order, capacity results are attainable.

The considered problem in this work is formulated as follows, also illustrated in Fig. 6.

P1: Find a set of tuples ( f , A ), namely, S , such that

S = {( f , A ) :
f : A 7→ B ,

(7)

a ∈ A and b ∈ B have the same marginal distributions.}

(8)

Note that (7) is due to the fact that in the desired equivalent channel, the tractability of the derivation
of the capacity region can be guaranteed, while (8) ensures that after the transformation f , the capacity
result is not changed by Theorem 1.

10

All random
fading channels

Fig. 2.

f(a)

a

b

A

B

For each realization of b,
the capacity results are known

The proposed scheme identifies the ergodic capacity regions under statistical CSIT.

Remark 1. The optimal classification shall be finding the three elements of a tuple ( f , A , B ), simultane-

ously, instead of fixing B and then finding ( f , A ). However, as mentioned previously, the goal of this work
is not to match new capacity inner and outer bounds for the open scenarios, which is out of the scope of
this work.
Note that under different assumptions on different MU channel models, ways to identify A and B may
be different, not to mention the corresponding capacity results. Therefore, we will introduce those results
case by case.
In the following we summarize three feasible schemes for P1 to compare channel strengths for the case
in Fig. 1 (b), when the transmitter has only statistical CSIT. The first two schemes are related to coupling
and the third one is related to copula [29]. At the end we will show that the schemes by coupling and
copula are equivalent. In brevity, the trick of all schemes is that under fixed marginal distributions, to
find a special structure on the dependence among fading channels corresponding to different users to fit
our purpose, e.g., to attain the capacity results. Note that the three schemes can be easily extended to an
arbitrary number of receivers. We first give the following definition.
Definition 5. [30, Definition 2.1] The pair (X̃, Ỹ ) is a coupling of the random variables (X,Y ) if X̃ =d X
and Ỹ =d Y .
It will be clarified that coupling plays an important role of the function f in P1 and the proposed
framework.
1) A Construction from Maximal Coupling: For the random variables whose PDF’s partially overlap as
in Fig. 1 (b), a straightforward idea to align H1 and H2 for each realization is as follows: we try to form
an equivalent channel pair (H̃1 , H̃2 ) with PDF’s f˜1 and f˜2 , respectively, which is a coupling of (H1 , H2 ),

11

to reorder H̃1 and H̃2 in the sense that H̃1 = H̃2 = h̃1 = h̃2 if h̃1 , h̃2 ∈ S , supp{ f˜1 } ∩ supp{ f˜2 }. Otherwise,
H̃1 and H̃2 follow PDF’s f˜1 = f1 and f˜2 = f2 , respectively. If this alignment is possible, we can transform
the remaining parts of the PDF’s into two non-overlapped ones. Then we know that one of the fading
channel is always equivalently no weaker than the other one, even if the statistics of the channels are
known, only. However, we may ask whether such an alignment exists or not? Our answer is yes, which
relies on the concept of maximal coupling [30] by reordering part of the channel realizations as defined
as follows.
Definition 6. [30, Section 2.2] For the random variables (X, Y ), the coupling (X̃, Ỹ ) is called a maximal
coupling if P(X̃ = Ỹ ) gets its maximal value among all the couplings of (X, Y ).
To proceed, we introduce a result of maximal coupling from [30].
Proposition 1. [30, Proposition 2.5] Suppose X and Y are random variables with respective piecewise
continuous density functions fX and fY . The maximal coupling (X̃, Ỹ ) for (X, Y ) results in
P(X̃ = Ỹ ) =

Z ∞
−∞

min( fX (x), fY (x))dx.

(9)

Based on Proposition 1, we derive the following result, which reorders the realizations of a class of
(H1 , H2 ) in the same trichotomy order.
Theorem 2. For a single-transmitter two-receiver AWGN channel, assume the PDF’s of the two channels,
namely, f1 and f2 , are continuous or with finite discontinuities. Then the selection
(
)
/ sup(supp( f2 )) < sup(supp( f1 )) , i ∈ {1, 2},
A = (H1 , H2 ) : fi − fInt 6= 0,

(10)

B = {(H̃1 , H̃2 ) ∈ f (A )},

(11)

(

f : A 7→ B , f (a) = (H̃1 , H̃2 ) :
fk (h) − min( f1 (h), f2 (h))
, k = 1, 2, with probability 1 − p;
H̃k follows the PDF f˜k (h) ,
1− p
)
min(
f
(h),
f
(h))
1
2
H̃1 = H̃2 follow the PDF f˜0 (h) ,
, with probability p ,
(12)
p
where fInt (h) , min( f1 (h), f2 (h)), h ∈ supp(H1 ) ∪ supp(H2 ), p ,
feasible to P1.

R∞

−∞ min( f 1 (h), f 2 (h))dh

and a ∈ A , is

12

Proof: The explicit construction of f in (12) is from the proof of maximal coupling [30, Proposition
2.5], while (10) is a sufficient condition to guarantee that the realizations of the two equivalent channels
follow the desired order after maximal coupling. To be self-contained, we firstly restate the important
steps of the proof of [30, Proposition 2.5] in the following. Define H = {h : f1 (h) < f2 (h)}. Any coupling
(H̃1 , H̃2 ) of (H1 , H2 ) should satisfy
P(H̃1 = H̃2 ) = P(H̃1 = H̃2 , H̃1 ∈ H ) + P(H̃1 = H̃2 , H̃2 ∈ H c )
≤ P(H̃1 ∈ H ) + P(H̃2 ∈ H c )
(a)

= P(H1 ∈ H ) + P(H2 ∈ H c )

=
(b)

=

Z
H

f1 (h)dh +

Z ∞
−∞

Z
Hc

f2 (h)dh

min( f1 (h), f2 (h))dh

(c)

= p,

(13)

where (a) is by Definition 5, (b) is by the definition of H and (c) is by the definition of p.
On the other hand, we can use (12) as an explicit construction of the maximal coupling. Define three
independent random variables G0 , G1 , and G2 , following the PDF’s f˜0 , f˜1 , and f˜2 , respectively, as shown
in (12). We can show that (12) is a coupling as follows:
(a)

P(H̃k ≤ h0 ) = pP(H̃k = G̃0 ≤ h0 ) + (1 − p)P(H̃k = G̃k ≤ h0 )
=p
=

Z h0
min( f1 (h), f2 (h))

∞
Z h0
∞

p

dh + (1 − p)

Z h0
fk (h) − min( f1 (h), f2 (h))
∞

fk (h)dh = P(Hk ≤ h0 ), for k = 1, 2,

1− p

dh
(14)

where (a) is by construction, i.e., by assigning G0 and Gk to H̃k with probability p and 1 − p, respectively.
Hence, it is clear that (14) fulfills the definition of coupling in Definition 5. On the other hand, it is clear
that P(H̃1 = H̃2 ) ≥ P(H̃1 = H̃2 = G0 ) = p. Therefore, from (13) and (14), we know that (12) can achieve
maximal coupling.
In the following, we will prove that (10) and (12) form a feasible solution for P1. By (9) it is clear that

13

P(H̃1 = H̃2 ) is the area of the intersection of f1 and f2 . Then from (12), we know that f˜1 (x) and f˜2 (x)
do not overlap with probability 1 − p. In addition, from (10) we know that h1 < h2 if h1 ∈ supp{ f˜1 } and
h2 ∈ supp{ f˜1 }. That is, H̃1 < H̃2 with probability 1 − p. Furthermore, from (12) (or from (9)), we know
that H̃1 = H̃2 is with probability p. In other words, by the maximal coupling (12), we can construct an
equivalent channel where there are only two relations between the fading channels realizations h̃1 and h̃2 :
1) h̃1 = h̃2 ; 2) h̃1 < h̃2 , which completes the proof.
Remark 2. Note that by applying the maximal coupling, we can transform the two PDF’s like the ones

in Fig. 1(b) into equivalent ones where one of the PDF’s can generate channel realizations always no
worse than those of the other. In the following we use two examples to illustrate the feasibility of the
selection of A and f in Theorem 2. Assume supp( f1 ) ⊆supp( f2 ) and sup(supp( f1 )) ⊆sup(supp( f2 )), there

may exist R21 ⊂supp( f˜2 ) and R22 ⊂supp( f˜2 ), such that h̃21 < h̃1 < h̃22 , where h̃21 ∈ R21 and h̃22 ∈ R22 ,

and h̃1 ∈ supp( f˜1 ) for the case h̃1 6= h̃2 . Then h̃1 can be larger or smaller than h̃2 . An example is shown
in Fig. 3(a). On the other hand, if supp( f1 ) ⊆supp( f2 ) and the maximum value of the alphabets of H1 is

the same as that of H2 , it is possible that h̃1 > h̃2 , for all h̃1 ∈ supp{ f˜1 } and h̃2 ∈ supp{ f˜2 } for the case

h̃1 6= h̃2 . An example is shown in Fig. 3(b). This example shows that (10) is sufficient but not necessary.
Note that the latter example fulfills the definition of the usual stochastic order and further discussion can
be seen in the next method.
~
f2

f1

= (1-p)

f2

+p

~
f1

h

~ ~
f 1= f 2

h

h

(a)

f2

f1

= (1-p)
h

Fig. 3.

(b)

~
f2

~
f1

+p
h

~ ~
f1 = f2

h

Examples of different PDF’s f1 , f2 and also f˜1 and f˜2 constructed by Theorem 2, where the bold curves denote f2 and f˜2 .

Remark 3. Note that from [30, Proposition 2.7] we know that the probability of H̃1 6= H̃2 from the coupling

R

14

(H̃1 , H̃2 ) of (H1 , H2 ) can be described by the total variation distance between H1 and H2 as
P(H̃1 6= H̃2 ) = (1 − p)

Z ∞
−∞

( f˜1 (h) + f˜2 (h))dh = 1 −

Z ∞
−∞

min( f1 (h), f2 (h))dh = dTV (H1 , H2 ),

(15)

where dTV (X,Y ) = sup|P(X ∈ A ) − P(Y ∈ A )|. By this way we can observe the relationship between the
A

closeness of two random variables in distributions and the closeness of the two random variables can be
coupled.
In fact, the condition described in Theorem 2 corresponds to H1 ≤st H2 . The reason that H1 ≤st H2 also
results in H̃1 ≤ H̃2 with probability 1, will be explained by the next method, namely, coupling.
2) A Construction from Coupling: In this method we resort to constructing an explicit coupling such
that each realization of channel pair has the same trichotomy order.
Theorem 3. For a single-transmitter two-receiver AWGN channel, the selection

A = {(H1 , H2 ) : H1 ≤st H2 },

(16)

B = {(H̃1 , H̃2 ) ∈ f (A )},

(17)

f : A 7→ B , f (a) = {(H̃1 , H̃2 ) : H̃1 = FH−1
(U), H̃2 = FH−1
(U)},
1
2

(18)

where a ∈ A , U ∼ Unif(0, 1), is feasible to P1.
Proof: From the coupling theorem [20] we know that, H1 ≤st H2 if and only if there exist random
variables H̃1 =d H1 and H̃2 =d H2 such that H̃1 ≤ H̃2 with probability 1. Therefore, by construction we
know that distributions of H̃1 and H̃2 fulfill the same marginal condition (8) in P1. In addition, the
trichotomy order H̃1 ≤ H̃2 fulfills (H̃1 , H̃2 ) ∈ B , i.e., the 1st channel is degraded from the 2nd one by (1)
under this trichotomy order for AWGN channel. The proof of the coupling theorem [30, Ch. 2] provides
us a constructive way to find f , which is restated as follows for a self contained proof. If H1 ≤st H2 ,
and if the generalized inverses FH−1
and FH−1
exist, where the generalized inverse of F is defined by
1
2
F −1 (u) = inf{x ∈ R : F(x) ≥ u}, u ∈ [0, 1], then the equivalent channels H̃1 and H̃2 can be constructed by
H̃1 = FH−1
(U) and H̃2 = FH−1
(U), respectively, where U ∼ Unif(0, 1). This is because
1
2
P(H̃1 ≤ h) = P(FH−1
(U) ≤ h) = P(U ≤ FH1 (h)) = FH1 (h),
1

(19)

15

i.e., H̃1 ∼ FH1 . Similarly, H̃2 ∼ FH2 . Since H1 ≤st H2 , from Definition 4 we know that FH2 (h) ≤ FH1 (h),
(u) ≤ FH−1
(u), ∀u, such that P(H̃1 ≤ H̃2 ) = 1. Therefore, we attain (18),
for all h. Then it is clear that FH−1
1
2
which completes the proof.
Remark 4. Originally, even though we know H1 ≤st H2 , the order of the channel realizations H1 = h1 and

H2 = h2 may vary for each realization. Then we are not able to claim which channel is stronger for the
duration of a codeword. However, from Theorem 3 we know that we can virtually explicitly align all the
channel realizations within a codeword length such that each channel gain realization of H1 is no worse
than that of H2 , if H1 ≤st H2 . Then we can claim that there exists an equivalently degraded channel.
Note that the usual stochastic order aligns random fading gains in the equivalent channel, which makes
it simple to adopt the MU channel orders, e.g., in Sec. II-B2. However, the alignment operation done in
the equivalent channel may be too strict for feasible scenarios. This is because the considered channel
orders, e.g., the degradedness, among MU channels are in fact described by just statistics but not to the
realizations of individual channels.
3) A Construction from Copula: In this method, we first explicitly construct a joint distribution between
the fading channels, such that each realization of channel pairs has also the same trichotomy order. We
then identify this construction is indeed the copula [29]. The concepts of coupling and copula are similar
in the sense that, in both cases the marginal distributions are given, and we want to form joint distributions
to achieve certain goals. In the former case, we can introduce any relation between the random channels
such that, for example, we can order the channel pairs for all realizations in the desired trichotomy
order. However, we do not construct an explicit joint distribution. In the latter case, we may try to find a
constructive way to attain an explicit joint distribution from the marginal ones, then identify that this joint
distribution fulfill our requirements. Note that the existence of the copula is proved by Sklar’s theorem
[29].
Theorem 4. For a single-transmitter two-receiver AWGN channel, the selection of

A = {(H1 , H2 ) : H1 ≤st H2 },

(20)

B = {(H̃1 , H̃2 ) ∈ f (A )},

(21)

f : A 7→ B , f (a) = {(H̃1 , H̃2 ) : H̃1 and H̃2 follow the joint CCDF F̄H̃1 , H̃2 (h̃1 , h̃2 ) = min{F̄H1 (h̃1 ), F̄H2 (h̃2 )}},
(22)

16

where F̄X,Y (x, y) , P(X ≥ x, Y ≥ y), a ∈ A , is feasible to P1.
Proof: By the definition of F̄X,Y (x, y) it is clear that the marginal distributions are unchanged, i.e.,
F̄H̃1 (h̃1 ) = F̄H̃1 ,H̃2 (h̃1 , 0) = F̄H1 (h̃1 ) and F̄H̃2 (h̃1 ) = F̄H̃1 ,H̃2 (0, h̃2 ) = F̄H̃2 (h̃2 ). Note that we consider the square
of channel gains, so we substitute 0 into F̄H̃1 , H̃2 (h̃1 , h̃2 ) to marginalize it. With the selection A = {(H1 , H2 ) :
H1 ≤st H2 }, we prove that h̃1 ≤ h̃2 , ∀(H̃1 , H̃2 ) ∈ B = f (A ) in the following.
Assume that h̃1 > h̃2 + ε, ε > 0, we can prove that
P(h̃1 ≤ H̃1 , h̃2 ≤ H̃2 ≤ h̃2 + ε) =P(h̃1 ≤ H̃1 , h̃2 ≤ H̃2 ) − P(h̃1 ≤ H̃1 , h̃2 + ε ≤ H̃2 )
(a)

= F̄H̃1 ,H̃2 (h̃1 , h̃2 ) − F̄H̃1 ,H̃2 (h̃1 , h̃2 + ε)

(b)

= F̄H̃1 (h̃1 ) − F̄H̃1 (h̃1 )
=0,

(23)

where (a) follows from the definition of the joint CCDF and (b) follows from (22) with the given property
H1 ≤st H2 , which implies that F̄H̃2 (h̃2 ) ≥ F̄H̃2 (h̃2 + ε) ≥ F̄H̃1 (h̃2 + ε) ≥ F̄H̃1 (h̃1 ). To ensure that H̃1 ≤ H̃2
for all random samples, we let ε → 0. Thus, as long as H1 ≤st H2 , we can form an equivalent joint
distribution as (22) such that it has the same marginal distribution as the original channel, then the
capacity is unchanged.
We can re-express Theorem 4 in terms of the joint CDF instead of the joint CCDF (or the joint survival
function [29]) as follows.
Corollary 2. For a single-transmitter two-receiver AWGN channel, the selection of

A = {(H1 , H2 ) : H1 ≤st H2 },
f (a) = {(H̃1 , H̃2 ) : H̃1 and H̃2 follow the joint CDF FH̃1 , H̃2 (h̃1 , h̃2 ) = min{FH1 (h̃1 ), FH2 (h̃2 )}},

B = {(H̃1 , H̃2 ) ∈ f (A )},

(24)
(25)
(26)

where a ∈ A , solves P1.
Proof: Here we construct the function f by showing that (25) is identical to the Fréchet-Hoeffdinglike upper bound for the survival copulas. From Fréchet-Hoeffding bounds [29, Sec. 2.5] we know that a

17

joint CDF FXY (x, y) can be upper and lower bounded by the marginals FX (x) and FY (y) as follows:
max{FX (x) + FY (y) − 1, 0} ≤ FXY (x, y) ≤ min{FX (x), FY (y)}.

(27)

On the other hand, by definition of the joint CCDF F̄XY (x, y) and the joint CDF FXY (x, y), we can easily
see:
F̄XY (x, y) = 1 − FX (x) − FY (y) + FXY (x, y).

(28)

After substituting the upper bound in (27) into (28), we can get:
(a)

F̄XY (x, y) ≤ 1 − FX (x) − FY (y) + min{FX (x), FY (y)} = 1 + min{−FX (x), −FY (y)} = min{F̄X (x), F̄Y (y)},
where (a) is by the definition of CCDF, which completes the proof.
In fact, we can prove that (18) and (25) are equivalent by showing that:
P(FH−1
(U) ≤ h̃1 , FH−1
(U) ≤ h̃2 ) = P(U ≤ FH1 (h̃1 ), U ≤ FH2 (h̃2 ))
1
2
= P(U ≤ min{FH−1
(h̃1 ), FH−1
(h̃2 )})
1
2
(a)

= min{FH−1
(h̃1 ), FH−1
(h̃2 )},
1
2

(29)

where (a) is due to the assumption that U ∼ Unif(0, 1).
Note that copula and coupling introduce different expressions in the joint distributions in the sense
that, in the latter case, the joint distribution is implicitly indicated by the random variable U and also the
marginal distributions in (18) while the joint distribution is explicitly constructed as (22) in copula.
Remark 5. We can directly identity that (22) is a two-dimensional copula [29, Definition 2.2.2] by definition

without the aid of Fréchet-Hoeffding-like upper bound and the derivation in the proof of Corollary 2. An
equivalent definition of a two-dimensional copula [29, (2.2.2a), (2.2.2b), (2.2.3)] is a function C0 : [0, 1]2 7→
[0, 1] with the following properties:
1) For every u, v in [0, 1],
C0 (u, 0) = 0 = C0 (0, v),

(30)

C0 (u, 1) = u and C0 (1, v) = v;

(31)

18

2) For every u1 , u2 , v1 , v2 in [0, 1] such that u1 ≤ u2 and v1 ≤ v2 ,
C0 (u2 , v2 ) −C0 (u2 , v1 ) −C0 (u1 , v2 ) +C0 (u1 , v1 ) ≥ 0.

(32)

The identification that F̄H̃1 , H̃2 (h̃1 , h̃2 ) = min{F̄H1 (h̃1 ), F̄H2 (h̃2 )} in (22) is a copula is derived in Appendix
I.
IV. A PPLICATION ON FAST FADING M EMORYLESS MU C HANNELS
In the following, we will use the coupling scheme introduced in Theorem 3 for memoryless MU channels
due to its intuitive characteristics. We first consider the binary symmetric memoryless MU channels as
examples for discrete memoryless MU channels and then extend to memoryless GMU channels.
A. Binary and Binary Fading MU Channels with Statistical CSIT
In the following we take the binary symmetric channel with and without fading as examples.
1) Binary Symmetric MU Channels with Statistical CSIT: For binary symmetric channels (BSC), we
can model the random bit flip by a switch S ∼ Bern(p), i.e., S = 1 if it switches to a bit flip with probability
p. Consider a binary symmetric MU channel with one transmitter and two receivers where S1 ∼ Bern(p1 )
and S2 ∼ Bern(p2 ) for the first and second BSC’s, respectively. Define the statistical CSIT as that the
transmitter only knows the probabilities p1 and p2 , p1 ≥ p2 and define the perfect CSIT as that the
transmitter knows the instantaneous positions of S1 and S2 . For statistical CSIT, we can use Theorem 3 to
compare the two sub-BSC’s, i.e., the second channel is stronger than the first one if S1 ≥st S2 . Note that
the direction of the inequality in usual stochastic order is in the reverse direction compared to the three
schemes since here higher probability of S = 1 means the channel is worse. Note also that the definition
of the usual stochastic order in Definition (4) is also valid for discrete random variables.
2) Binary Erasure MU Channels with Statistical CSIT: Similar to the BSC case, we can use a switch S to
change the input bit to an erasure. Then for two binary erasure channels S1 ∼ Bern(p1 ) and S2 ∼ Bern(p2 ),
p1 ≥ p2 , similar to the BSC case, we can claim that the second channel is stronger than the first one if
S1 ≥st S2 .

19

3) Binary Fading MU Channels with Statistical CSIT: For BSC’s, similar to the AWGN cases, we can
consider a binary fading [13] as following, where the receive signal at the k-th receiver is:
Yk = Hk X ⊕ Nk ,

(33)

where Hk ∼ Bern(pk ) and Nk ∼ Bern(pN ) denote the fading channel and the equivalent additive noise,
respectively. Then for a two-receiver case, from Theorem 3 we can easily see that if p1 ≥ p2 , i.e., H1 ≥st H2 ,
the first channel is stronger than the second one.
B. Fading Gaussian Interference Channels with Statistical CSIT
From this section we consider the Gaussian MU channels. When there is only statistical CSIT and full
CSI at the receiver, the ergodic capacity region of GIC is unknown in general. In this section, we identify
a sufficient condition based on the results in Section III-A to attain the capacity regions of Gaussian
interference channel with strong and very strong interference. Examples illustrate the results.
1) Preliminaries and Results: We assume that each receiver perfectly knows the two channels to it. .
Therefore, the considered received signals of a two-user fast fading Gaussian interference channel can be
stated as
Y1 =

√
√
H11 e jΦ11 X1 + H12 e jΦ12 X2 + N1 , H̃11 X1 + H̃12 X2 + N1 ,

(34)

Y2 =

√
√
H21 e jΦ21 X1 + H22 e jΦ22 X2 + N2 , H̃21 X1 + H̃22 X2 + N2 ,

(35)

where Hk j and Φk j ∈ [0, 2π] are real-valued non-negative independent random variables denoting the
absolute square and the phase of the fading channel between the j-th transmitter to the k-th receiver,
respectively, where k, j ∈ {1, 2}. The CCDF of Hk j is denoted by F̄Hk j . The channel inputs at the
transmitters 1 and 2 are denoted by X1 and X2 , respectively. We consider the channel input power
constraint as E[|X1 |2 ] ≤ P1 and E[|X2 |2 ] ≤ P2 , respectively. Noises N1 and N2 at the corresponding receivers
are independent circularly symmetric AWGN with zero mean and unit variance. We assume that the
transmitters only know the statistics but not the instantaneous realizations of {Hk j } and {Φk j }. Hence, the
channel input signals {X j } are not functions of the channel realizations and therefore are independent to

20

{Hk j } and {Φk j }. In addition, without loss of generality, we assume {Hk j }, {Φk j }, and {Nk } are mutually
independent.
In the following derivation we do not exploit the commonly used standard form of GIC. This is
because that the normalization operation in the standard form results in a ratio of random variables,
whose distribution may not be easy to derive and thus hinders us on identifying all channels easily.
We first extend the sufficient condition to have a strong interference channel to the case with only
statistical CSIT. In the sequel we illustrate our first main result.
Theorem 5. If we consider the set A as

A = {(H11 , H12 , H21 , H22 ) : H21 ≥st H11 and H12 ≥st H22 },

(36)

then the following ergodic capacity region of a strong GIC with statistical CSIT can be achieved

C (P1 , P2 ) = C1 (P1 ) ∩ C2 (P2 ),

(37)

where


C j (Pj ) , (R1 , R2 ) : R1 ≤ E[C(H j1 P1 )],
R2 ≤ E[C(H j2 P2 )],
R1 + R2 ≤ E[C(H j1 P1 + H j2 P2 )] .

(38)

Please refer to Appendix II for the proof.
Remark 6. A recent work [13] proves the capacity region of a two user binary fading interference channel

without CSIT under both weak and strong interferences, where the receive signals are modeled by
Yk = Gkk Xk ⊕ Gk̄k Xk̄ , k = 1, 2,

(39)

where k̄ = 3−k, Gkk , Gk̄k ∈ {0, 1} and all algebraic operations are in F2 . Our result in Theorem 5 and also
the discussion in Section III are valid for the binary fading IC (39). More specifically, in [13], the strong
interference channel is defined as pd ≤ pc ≤ 1, where H11 ∼ Bern(pd ), H22 ∼ Bern(pd ), H12 ∼ Bern(pc ),
and H21 ∼ Bern(pc ). It is easy to see that the complementary cumulative distribution function of H21 is no
less than that of H11 , i.e., H21 ≥st H11 . Similarly, we can deduce H12 ≥st H22 . This fact manifests that (36)
is a more general expression of the strong interference condition, i.e., it can be used for either discrete
or real valued random channels. We can also extend (39) into cases with noises. Cases in [13] achieving
capacity regions but not covered in our current paper can be considered as future works. Note that (36)

21

can also be specialized to the full CSIT case. For strong binary fading ICs, the capacity region achieved
by interference-decoding at each receiver [13], is the same as in our results in Theorem 5.
Remark 7. With the technique of stochastic orders, we can consider a fading IC with a stricter condition,

i.e., the uniformly strong IC [11], which is introduced in Appendix II, such that we can more easily derive
the capacity region of the IC. In contrast, directly dealing with the hybrid IC2 may be too difficult3 .
The ergodic capacity region result of a very strong interference channel is as follows.
Theorem 6. If we consider the set A as

A = (H11 , H12 , H21 , H22 ) :


H12
H21
≥st H11 and
≥st H22 ,
1 + P2 H22
1 + P1 H11

(40)

then the following gives the ergodic capacity region of a very strong GIC with statistical CSIT


C (P1 , P2 ) = (R1 , R2 ) : R1 ≤ E[C(H11 P1 )],
R2 ≤ E[C(H22 P2 )] .

(41)

Please refer to Appendix III for the proof.
By the considered scenarios, i.e., strong/very strong interference, we can decouple the effect of the
other user to the single user capacity constraint, such that we are able to prove the optimality of Gaussian
input. Further discussion is presented in Remark 9 in Sec. IV-D.
2) Examples: In this subsection we provide examples to show the scenarios that the sufficient conditions
in Theorem 5 and Theorem 6 are feasible.

Example of (36): Assume that the squares of the four channel gains follow exponential distributions,
i.e., H jk ∼ Exp(1/σ2jk ), j, k = 1, 2. By the CCDF of exponential distributions, it is easy to see that if
σ221 ≥ σ211 and σ212 ≥ σ222 , then the two constraints in (36) are fulfilled.
When there is perfect CSIT, constraints in (36) can still be utilized by bestowing the density functions
of the channels as Dirac delta functions.
2 It

is defined in [11] as: at least one strong and one weak fading state happen when averaged over all fading states.
may treat the problem considered in [13], which is only for binary alphabet, as the evidence of the concern of the difficulty.

3 We

22

Example of (40): In this case, to proceed, we find the distributions of the two ratios of random variables
in (40). We can first rearrange H21 /(1 + P2 H22 ) into a ratio of quadratic form as
H H B1 H
H21
=
, Z,
1 + P2 H22 1 + P2 H H B 2 H

(42)

where we assume H , [Hw,21 Hw,22 ]T , Hw,21 ∼ N(0, 1), Hw,22 ∼ N(0, 1), Hw,21 is independent of Hw,22 ,
B 1 , diag{[σ221 , 0]} and B 2 , diag{[0, σ221 ]}. Hence, H21 , (σ221 Hw,21 )2 , H22 , (σ222 Hw,22 )2 . From [31,
(16)] we know that the CDF of the RHS of (42) can be calculated by the following
 
λ2i
1 −t
t
λ
FZ (t) =
e iu
Πl6=i (λi − λl ) |λi |
λi
 −t 



1
1
t
−1
(b)
σ221
P2 σ222
e u
+e
,
u
= u(t) − 2
σ21 + tP2 σ222
σ221
P2 σ222
(a)

u(t) − Σ2i=1

(43)

where in (a), u(t) is the unit step function, {λi} are the eigenvalues of B1 − tP2 B2 , i.e., (λ1 , λ2 ) =
(σ221 , −tP2 σ222 ); in (b) we substitute the eigenvalues into the RHS of (a). Then we evaluate the first
−t
2

constraint in (40) by checking the difference of CCDFs of Z and H11 , i.e., 1 − FZ (t) − e σ11 numerically. In
the first comparison, we fix the variances of the cross channels as c , σ212 = σ221 = 1 and the transmit powers
P , P1 = P2 = 1, and scan the variances of the dedicated channels by a , σ211 = σ222 = 0.1, 0.3, 0.5, 0.7.
Since the conditions in (40) are symmetric and the considered settings are symmetric, once the first
condition in (40) is valid, the second one will be automatically valid. The results are shown in Fig. 4,
from which we can observe that when a increases, the difference of the CCDFs decreases. This is because
the support of the CCDF of H11 increases with increasing a and in the considered case when a = 0.7,
the distribution of H11 concentrates on larger channel values than H21 /(1 + P2 H22 ), which reflects that
the CCDFs that, H11 has a larger CCDF than H21 /(1 + P2 H22 ) at smaller h, which violates the condition
in (40). In contrast, the values a = 0.1, 0.3, 0.5 result in (H11 , H12 , H21 , H22 ) satisfying (40). We also
investigate the effect of the transmit power constraints to the validity of the sufficient condition. We
consider the case with a = 0.1, c = 1 with P = 1, 10, 50, 100 (in linear scale). From Fig. 5 we can observe
that the differences of CCDFs decrease with increasing P, which is intuitive from (40), since then the
distribution of H21 /(1 + P2 H22 ) will be more concentrated in a smaller range. We can also see that, when

23

P = 100, the condition in (40) is not valid anymore. In contrast, the choices P = 1, 10, 50 satisfy (40).
c=1, P=1
0.7
0.6
1 − FZ (t) − e−t/σ11

0.5

2

Very strong GIC

0.4
0.3
a = 0.1, 0.3, 0.5, 0.7

0.2
0.1
0

−0.1
0

1

2

3

4

5

6

7

H
Fig. 4.

Identification of (40) under different variances of channel gains of the dedicated channels with c = 1 and P = 1.

a = 0.1, c = 1
0.7
0.6

2

1 − FZ (t) − e−t/σ11

0.5
Very strong GIC

0.4
0.3
0.2

P = 1, 10, 50, 100

0.1
0

−0.1
0

1

2

3

4

5

6

7

H

Fig. 5.

Identification of (40) under different transmit power constraints with a = 0.1 and c = 1.

C. Fading Gaussian Broadcast Channels with Statistical CSIT
1) Preliminaries and Results: When a 2-receiver BC is degraded, we know that the capacity region
[18] is the union over all V , X satisfying the Markov chain V − X − Y1 − Y2 of rate pairs (R1 , R2 ) such

24

that




 R1 ≤ I(X;Y1 |V ),


 R2 ≤ I(V ;Y2 ),

for some p(v, x), where the cardinality of the random variable V satisfies |V | ≤ min{|X |, |Y1 |, |Y2 |} + 1.
Note that for non-degraded BC’s, only the inner and outer bounds are known, e.g., Marton’s inner bound
[32] and Nair-El Gamal’s outer bound [33]. Therefore, it shall be easier to characterize the capacity region
of a broadcast channel if we can identify that it is degraded.
The capacity region of a GBC is known for both fixed and fading channels when there is perfect
CSIT as well as perfect CSI at the receiver (CSIR). For multiple antennas GBC with perfect CSIT and
CSIR, the channel enhancement technique is used to form an equivalent degraded broadcast channel [34].
Immense endeavors have been made to find the optimal input covariance matrix of the Gaussian input,
e.g., [35]–[37]. However, the problem is open when there is imperfect CSIT in general and only limited
cases are known [12]. The fading GBC with only perfect CSIR but imperfect CSIT in general lacks the
degraded structure for arbitrary fading distributions, which makes it a challenging problem.
We assume that there is full CSIR such that they can compensate the phase rotation of their own
channels, respectively, without changing the capacity to form real channels. Therefore, the signal of
receiver k of the considered L-receiver fast fading GBC can be stated as
Yk =

p

Hk X + Nk , k = 1 · · · L,

(44)

where Hk is a real-valued non-negative independent random variable denoting the square of receiver k’s
fading channel with CCDF F̄Hk . The channel input is denoted by X. We consider the channel input power
constraint as E[X 2 ] ≤ PT . The noises {Nk } at the corresponding receivers are independent AWGN with zero
mean and unit variance. We assume that the transmitter only knows the statistics but not the instantaneous
realizations of {Hk }. In the following we will first consider the two-receiver case then extend to the
L-receiver case.
The following result can be easily derived from Theorem 5.

25

Corollary 3. For an L-receiver fast fading Gaussian broadcast channel, if H1 ≥st H2 ≥st · · · ≥st HL , then
it is degraded.
Remark 8. An important note is that the degradedness does not guarantee the optimality of Gaussian input

for fading GBC’s with statistical CSIT. In [38], it is shown that with statistical CSIT, a local perturbation
of Gaussian channel input can be better than the Gaussian one. Gaussian input is optimal only proved
for very few cases, e.g., [39].
Now we can further generalize Corollary 3 to the case in which fading channels are formed by clusters
of scatterers [40]. The received signals at receivers j from M clusters of scatters can be expressed as
M

Yj =

∑ H̃ j,k X + N j ,
k=1

where H̃ j,k are mutually independent, for all j, k. In particular, we consider the case in which we are only
provided the information of each cluster but not the superimposed result as

√
Hk in (44). Therefore, the

phases of each channel cluster should be taken into account, i.e., we consider the k-th clusters of the first
and second user as H̃1k = H̃1k,Re + i · H̃1k,Im =

√
√
H1k e−iφ1k and H̃2k = H̃2k,Re + i · H̃2k,Im = H2k e−iφ2k .

Proposition 2. Let M be the number of clusters of scatterers for both users 1 and 2, and denote the
channels of users 1’s and 2’s kth clusters as H̃1k and H̃2k , respectively, where k = 1, · · · , M. The broadcast
channel is degraded if user 1’s scatterers are stronger than those of user 2 in the sense that H̃π1k ,Re H̃π1 j ,Re ≥st
H̃π2k ,Re H̃π2 j ,Re , and H̃π1k ,Im H̃π1 j ,Im ≥st H̃π2k ,Im H̃π2 j ,Im , ∀k, j ∈ {1, · · · , M}, for some permutation π1 and π2
of users 1’s and 2’s clusters.
Proof: We attain the result by the fact that the usual stochastic order is closed under convolution
[19, Theorem 1.A.3], which claims that for two sets of independent random variables {Ak } and {Bk },
if Ak ≥st Bk , ∀k ∈ {1, · · · , M} then for any increasing function φ : RM → R, one has φ(A1 , · · · , AM ) ≥st
M
φ(B1 , · · · , BM ). In particular, here φ is the summation and we have ∑M
k=1 Ak ≥st ∑k=1 Bk . In our problem,

from Corollary 1 we know that
M

∑ H̃1k
k=1

2

M

≥st

∑ H̃2k

2

(45)

k=1

is sufficient to attain a degraded GBC. After expanding the left and right hand sides of (45) in addition
with [19, Theorem 1.A.3], we can get Proposition 2.

26

To encompass a broader scenario of fading channels, we can relax the conditions of the number of
clusters of channels 1 and 2 in Lemma 2 to two non-negative integer-valued random variables N and M,
M
2
2
respectively, and the result is still valid. In particular, if | ∑N
k=1 H̃1k | ≥st | ∑k=1 H̃2k | and if N ≥st M, then

receiver 2 is a degraded version of receiver 1, which can be proved with the aid of [19, Theorem 1.A.4].
2) Example: In the following, we show an example with practical fading distributions based on
Corollary 3. Assume the magnitudes of the three channels are independent Nakagami-m random variables
with shape parameters m1 , m2 , and m3 , and spread parameters w1 , w2 , and w3 , respectively. From
Corollary 3 we know that the fading GBC is degraded if






γ m2 , mw22x
γ m3 , mw33x
γ m1 , mw11x
≥
≥
, ∀x,
Γ(m1 )
Γ(m2 )
Γ(m3 )
where γ(s, x) =

R x s−1 −t
R
e dt is the incomplete gamma function and Γ(s) = 0∞ t s−1 e−t dt is the ordinary
0t

gamma function [41, p. 255]. An example satisfying the above inequality is (m1 , w1 ) = (1, 3), (m2 , w2 ) =
(1, 2) and (m3 , w3 ) = (0.5, 1).
D. Physical Layer Security with Fading Gaussian Channels and Statistical CSIT
In this section we introduce two important notions in physical layer security: the wiretap channel for
secure data transmission and the secret key generation (SKG), to show how to apply the developed
framework to them. For SKG, we focus on the source-model SKG to compare the random source
stochastically.
1) Wiretap channels: For WTC, compared to [16], in this paper we provide a more intuitive and
complete derivation. In addition, we discuss the secret key capacity of SKG with source model. In
particular, we aim at characterizing the ergodic secret key capacity of a fast fading Gaussian Maurer’s
satellite model, which is a very practical model for SKG through wireless channels.
We consider a Gaussian WTC as:
Y=

√
HX + NY ,

Z=

√
GX + NZ ,

27

where H and G are channel gains from Alice to Bob and Eve, respectively, while NY and NZ are AWGN’s
at Bob and Eve respectively. Without loss of generality, we assume NY and NZ following the same
distribution, i.e., both with zero means and unit variances. Assume that Alice only knows the statistics of
both channels H and G, while Bob knows perfectly the realization of his own channel H but Eve knows
perfectly both H and her own channel G.
Theorem 7. If we consider the set A as

A = {(H, G) : H ≥st G},

(46)

then the ergodic secrecy capacity with statistical CSIT of both H and G is
CS (PT ) = E[C(HPT ) −C(GPT )].

(47)

Please refer to Appendix IV for the proof.
Remark 9. Recall that for a degraded GBC with only statistical CSIT, we cannot claim that the capacity

region is achievable by a Gaussian input. On the contrary, for a degraded Gaussian WTC (GWTC) with
only statistical CSIT, we can prove that a Gaussian input is optimal. The difference can be explained
by the following. For a GWTC, there is only one message set to be transmitted to the single dedicated
receiver and the Markov chain U − X − Y − Z reduces to X − Y − Z [42] when it is degraded, i.e., no
prefixing is needed. The simplification in solving the optimal channel input distribution from two random
variables to only one makes it easier to prove that Gaussian input is the optimal one. On the other hand,
for a two user degraded GBC there are two message sets and one of the user signals is interference to
the other, i.e., the chain U − X −Y − Z does not degenerate as that of a WTC. In particular, the boundary
point of a BC with statistical CSIT is expressed as
max

I(X;Y1 |U, H1 ) + µI(U;Y2 |H2 )

max

h(Y1 |U, X, H1 )−h(N1 )+µh(Y2 |H2 )−µh(Y2 |U, H2 ),

(U, X):U−X−(Y1 ,Y2 ),
E[X 2 ]≤PT

=

(U, X):U−X−(Y1 ,Y2 ),
E[X 2 ]≤PT

(48)

where µ > 0. Note that Gaussian is optimal for each term on the RHS of (48) but the negative sign
prevents to show the exact optimality. More precisely, h(Y2 |U, H2 ) comprises the interference from the
first user such that this term cannot be reduced to h(N2 ) as the counter part h(Y1 |U, X, H1 ) = h(N1 ) [43].

28

2) Source-model SKG: To invoke the developed scheme, we first prove a sufficient condition to achieve
the secret key capacity CSK = I(X;Y ) − I(X; Z), i.e., the common randomness forming a Markov chain
X −Y − Z, which is physically degraded, can be relaxed to stochastic degradedness. The motivation is as
follows. To achieve the secret key capacity, the original analysis [44, Corollary 1] requires the common
randomness forming a Markov chain. However, this requirement is quite stringent on classifying random
sources which can attain the secret key capacity. From extending the physically degraded broadcast/wiretap
channels to the stochastic one we know that, the latter is equivalent to the former one, but it has broader
classes of channels. Therefore, in the following we define the stochastically degraded source first. We then
show that the stochastically degraded condition can be fulfilled by, e.g., the fast fading Gaussian Maurer’s
(satellite) model. More specifically, we assume that there is a central random source S observed through
fast fading AWGN channels as X, Y , and Z at Alice, Bob, and Eve, respectively. We apply the usual
stochastic order to derive the sufficient condition on the fading channels such that CSK is achieved. The
sufficient condition provides a simple way to verify the stochastic degradedness and thereby to identify
the effective SK capacity. We first give a definition on the relation between the common randomness
followed by our result.
Definition 7. A source of common randomness (X , Y , Z , PX Ỹ Z̃ ) is called stochastically degraded if the
conditional marginal distributions PỸ |X and PZ̃|X are identical to those of another source of common

randomness (X , Y , Z , PXY Z ) following X − Y − Z. A source of common randomness (X , Y , Z , PXY Z ) is
called physically degraded if X, Y, and Z form a Markov chain X −Y − Z.
Theorem 8. If a source of common randomness (X , Y , Z , PX Ỹ Z̃ ) is stochastically degraded such that
PỸ |X = PY |X and PZ̃|X = PZ|X , where X − Y − Z, then the SK capacity of the source (X, Ỹ , Z̃) is CSK =
I(X;Y ) − I(X; Z).

Please refer to Appendix V for the proof. The proof is sketched as follows and summarized in Fig. .
The main idea of the proof is to show that the stochastically degraded source (X, Ỹ , Z̃) implies that the
conceptual WTC (CWTC) [45] is also stochastically degraded, which then has the same CSK as that of the
CWTC from a physically degraded source (X,Y, Z). There are three main steps. First is to construct the

29

CWTC of the source (X,Y, Z) and prove that, if X −Y − Z, then U −Y 0 − Z 0 , i.e., the corresponding CWTC
is also a physically degraded one, where U is the conceptual code symbol transmitted in the CWTC,
which is uniformly distribution in X and independent to (X,Y, Z); the equivalently received signals at
Bob and Eve in the CWTC are Y 0 , (Y,U ⊕ X) and Z 0 , (Z,U ⊕ X), respectively, while U ⊕ X is the
signal transmitted through the public channel. The second step is to construct a stochastically degraded
source (X, Ỹ , Z̃) from (X,Y, Z) and then construct the corresponding CWTC of (X, Ỹ , Z̃) as (U,Y 00 , Z 00 ),
where Y 00 , (Ỹ ,U ⊕ X) and Z 00 , (Z̃,U ⊕ X) and (U,Y 00 , Z 00 ). The third step is to show that the CWTC
described by (U,Y 00 , Z 00 ) is a stochastically degraded version of the CWTC described by (U,Y 0 , Z 0 ), i.e.,
to prove these two CWTC’s have the same marginals. Then from same marginal property we know the
two CWTC’s have the same secrecy capacity. As a result, the two common random sources (X,Y, Z) and
(X, Ỹ , Z̃) have the same CSK , which completes the proof.

Common random source

Conceptual WTC

1)

Physically degraded:

2)
Stochastically degraded:

2)
3)

Fig. 6. Key steps in the proof of Theorem 8. First, to show the CWTC is also physically degraded if the source is physically degraded. Then
to show the CWTC constructed from a stochastically degraded source corresponding to the physically degraded source, is a stochastically
degraded CWTC with the same marginal as that physically degraded CWTC. Finally to show the secrecy capacity of the second CWTC is
indeed the secret key capacity of the stochastically degraded source. The key and secrecy capacities of the physically degraded source and
the corresponding CWTC are denoted by CSK and CS , respectively, while the key (rate) and secrecy capacities of the stochastic source and
0 (R0 ) and C0 , respectively.
the corresponding CWTC are denoted by CSK
SK
S

3) Example: In the following we give an example to show the usage scenario of Theorem 8. Consider
the following fast fading Gaussian Maurer’s (satellite) model

4

[25]: X = AX S + NX , Y = AY S + NY , and

Z = AZ S + NZ , where NX , NY and NZ are independent AWGNs at Alice, Bob, and Eve, respectively, while

30

all noises are with zero mean and unit variance; AX , AY , and AZ follow CDFs FX , FY , and FZ , respectively,
are the i.i.d. fast fading channel gains from the source S to Alice, Bob, and Eve, respectively. Thus we can
extend Theorem 3 to the sufficient condition for the source to be stochastically degraded and the example
shown in Sec. IV-C can be treated as a practical one for the fast fading Gaussian Maurer’s model.
V. C HANNELS WITH M EMORY
In this section, we discuss channels with memory which possesses a specific structure, i.e., channels
with finite-state [26]. In particular, we investigate the relationship between the stochastic orders for random
processes and the ergodic capacity of BC as a representative example. Due to the memory, the information
theoretic order of degradedness has to be revisited. Therefore, we first introduce some important properties
extended to multiuser channels with memory such as the same marginal property and degradedness. Then
we discuss the usual stochastic order for random processes.
A. Preliminaries
Definition 8 (Finite state broadcast channels (FSBC) [22]). The discrete finite-state broadcast channel
is defined by the triplet {X × S , p(y, z, s|x, s0 ), Y × Z × S }, where X is the input symbol, Y and Z are

the output symbols, S0 and S are the channel states at the end of the previous and the current symbol
transmissions, respectively. S , X , Y , and Z are finite sets. The PMF of the FSBC satisfies
p(yi , zi , si |yi−1 , zi−1 , si−1 , xi , s0 ) = p(yi , zi , si |xi , si−1 ),

(49)

where s0 is the initial channel state.
For FSBC, a single letter expression of the condition for degradedness is in general not possible.
Therefore we introduce the following definitions of physical and stochastic degradedness for FSBC.
Definition 9 (Degradedness for FSBC [22]). A finite state BC is called physically degraded if for every
s0 ∈ S and symbol time i its PMF satisfies

4 Since

p(yi |xi , yi−1 , zi−1 , s0 ) = p(yi |xi , yi−1 , s0 )

(50)

p(zi |xi , yi , zi−1 , s0 ) = p(zi |yi , zi−1 , s0 ).

(51)

the SKG capacity can be achieved by one time discussion, i.e., without feedback/iteration, we can resort to the quantization scheme

as shown in [46, Appendix 1] to extend the result in Theorem 8 to continuous alphabets.

31

The FSBC is called stochastically degraded if there exists a PMF p̃(z|y) such that for every block length
n and initial state s0 ∈ S

n
n

n

n

n

n

n

n

p(z |x , s0 ) = ∑ p(y , z |x , s0 ) = ∑ p(y |x , s0 ) ∏ p̃(zi |yi ).
Yn

Yn

(52)

i=1

Note that Definition 9 can be specialized to the degradedness of the memoryless case easily as p(y, z|x) =
p(y|x)p(z|y), i.e., X, Y , and Z form a Markov chain X −Y − Z for physically degraded BC and p(z|x) =
∑y p(y|x) p̃(z|y) for stochastically degraded BC.
Remark 10. Two important facts of channels with memory considered in [22] are: 1) The channel output

at the weaker receiver does not contain more information than that of the stronger receiver. 2) When the
degraded outputs are given causally, the stronger/better output up to the current time results in the case
where the degraded current output is independent to the channel input up to the current time. In addition,
the definition of stochastic degradedness (52), i.e., P̃Z|Y,S0 considered in [22] is a strictly equivalent channel
from Y to Z, which is memoryless, compared to the expression P̃Z n |Y n ,S0 . This is because the authors aim to
show the explicit causal operation of the degraded channel. More importantly, with this definition, we can
identify the degradedness more easily by adopting the stochastic ordering to align the random channels,
which will be shown in the next section.
In this work, we consider the indecomposable FSBC (IFSBC), where the concept of indecomposability
is defined as [47]: if there exists an N0 (ε) ∈ N for every ε > 0 such that for all n > N0 (ε), then we have
|p(sn |xn , s0 ) − p(sn |xn , s00 )| ≤ ε, for all sn , xn , and different initial states s0 and s00 . This means that the
effect of the initial channel state s0 on the state transition probabilities diminishes over time.
To apply the stochastic orders, similar to the aforementioned memoryless channels, we need the same
marginal property. In the following, to take the memory effect into account, we need to consider a multiletter version of the same marginal property.
Corollary 4. The capacity region of an FSBC depends only on the conditional marginal distributions
pY n |X n , S0 and pZ n |X n , S0 and not on the joint conditional distribution pY n ,Z n |X n , S0 .
This corollary can be easily proved by removing the memoryless property of [27, Theorem 13.9].

32

B. Sufficient Conditions for an IFSBC to be Degraded
In this subsection, we discuss our results for the IFSBC. In particular, we investigate the relationship
between the stochastic orders for random processes and the ergodic capacities of IFSBC. In brief,
we identify whether the fading processes satisfy the usual stochastic order or not. Again by invoking
the coupling scheme with the same marginal property, we can form an equivalent channel with the
same capacity region, which has all fading states ordered in the same trichotomy order for all channel
realizations. Therefore, it results in the degradedness. The detailed proof is derived in the following.
When considering the memory property, the aforementioned stochastic order is not sufficient. On the
contrary, we need to resort to the stochastic order for a random process. We introduce the following
sufficient condition for the degraded IFSBC by the usual stochastic order for random processes. To describe
the memory effect to the current random variable, we define [X(m)|X(m − 1) = x(m − 1), · · · , X(0) = x(0)]
as the random variable X at time m, given the realizations of X prior to m.
Theorem 9. Let {HZ (m)} and {HY (m)} be two discrete-time random processes. If

A = ({HZ (m)}, {HY (m)}) :
HZ (0) ≤st HY (0),
[HZ (m)|HZ (m − 1) = hZ (m − 1), · · · , HZ (0) = hZ (0)]
≤st [HY (m)|HY (m − 1) = hY (m − 1), · · · , HY (0) = hY (0)],

whenever hZ ( j) ≤ hY ( j), j = 0, 1, · · · , m − 1, m ∈ N ,

(53)

then the IFSBC is degraded and the following capacity region of a degraded IFSBC can be achieved
(

C = lim co
n→∞

[
qn ∈Qn

(R1 , R2 ) : R1 ≥ 0, R2 ≥ 0
1
R1 ≤ I(X n ;Y n |U n , s00 )qn
n
)
1
R2 ≤ I(U n ; Z n |s000 )qn ,
n

where Qn is the set of all joint distributions p(un , xn ), s00 and s000 are arbitrary initial states.
The proof is sketched as follows. By definition of the usual stochastic order, it is clear that (53),
namely, the strong stochastic order [19, Theorem 6.B.31], results in {HZ (m)} ≤st {HY (m)}, which is

33

therefore a sufficient condition for the usual stochastic order. Note that this comparison is element-wise.
Then, from Theorem 3 we know that there exists {HZ0 (m)} =st {HZ (m)} and {HY0 (m)} =st {HY (m)} such
that Pr(HZ0 (m) ≤ HY0 (m)) = 1, ∀ m ∈ N0 . On the other hand, note that the relation between Yi and Zi of
the equivalent physically degraded channel in (52) is memoryless, i.e., described by ∏ni=1 p̃(zi |yi ). This
means considering zi and yi in an element-wise manner is sufficient, which is compatible to the usual
stochastic order of random processes. Therefore, as long as the conditions in (53) are fulfilled, there exists
∏ni=1 p̃(zi |yi ) as shown in (52). Then it is clear that {HZ (m)} ≤st {HY (m)} implies (52). By combining
the capacity region results in [22], this completes the proof.
Remark 11. For WTC’s with finite state and statistical CSIT, we can also apply the previous discussion

to identify the degradedness. From [48] it is clear that the identification of degradedness can simplify
the functional optimization, i.e., there is no need to optimize the channel prefixing p(xn |un ), if the WTC
is degraded. Based on the multi-letter expression of secrecy capacity with memory in [49], we know that
under statistical CSIT, if the conditions in Theorem 9 are valid, then the following secrecy capacity of a
degraded WTC with finite state can be achieved
o
1n n n 0
0
n n 00
00
CS = lim
max
I(X ;Y |S0 = s0 ) − I(X ; Z |S0 = s0 ) ,
n→∞ p(xn ):Pr( 1 c (X n )≤P)=1 n
n n

(54)

where {cn }n≥1 is a sequence of cost functions, cn : X → R+ , such that any transmit sequence X n ∈ X n

should satisfy 1n cn (X n ) ≤ P.

In the following, we specialize Theorem 9 to the time homogeneous5 finite-state Markov channels.
There are two main reasons to consider the Markov fading channels for memory cases. First, it is a
useful model for mobile wireless channels when the underlying channel state process is time-invariant
and indecomposable [50], [51], [52]. More detailed discussions can be found in [53]. Second, the structure
of a Markov chain simplifies the sufficient condition of usual stochastic order for random processes, which
increases the tractability and provides insights on the analysis of fading channel with memory. The latter
point can be observed in the derivation shown in the following.
Remark 12. According to [53] we know that the first order Markov chain can model slow fading relatively

accurately. For intermediate fading rate, higher orders from second to seventh order Markov chain should
5 The

transition probability matrix does not change over time.

34

be used. For cases in which the channel gains change over each code symbol, no memory will be used.
Independent to the fading speed, by our framework we can construct an equivalent degraded channel for
each code symbol.
Consider a k-th order N-alphabet time-homogeneous Markov process with a transition probability matrix
as P , which is fixed over time. Indices of each row and column of P represent a vector of the current
states s = (s j , s j+1 , · · · , s j+k−1 ) and the next state s 0 = (s j+1 , s j+2 , · · · , s j+k ), respectively, where j ∈ N.
Denote the j-row of P and Q as p j and q j , respectively. We name s and s 0 as the super states. Then
the transition probability from the super state s to s 0 is expressed as Psi+1 |ss =

Psi+1 ,ss
Ps ,

i = j, · · · , j + k − 1.

Define p i and q i as the i-th row vectors of the transition matrices P and Q , respectively. To simplify the
expression, we define an N-ary expansion of L, namely, an integer indicating the number of the row of
P to the vector s , by g : L 7→ s , where L ∈ {1, · · · , N k }.
We now derive the result for the Markov fading channels in the following.
Corollary 5. Consider two Markov fading channels both with k-th order, namely, {HZ (m)} and {HY (m)},
with transition matrices P and Q , respectively. A two-receiver finite state Markov fading BC is degraded
if

A ={({HY (m)}, {HZ (m)}) :
HZ (0) ≤st HY (0),

(55)

[HZ (m)|HZ (m − 1) = hZ (m − 1), · · · , HZ (0) = hZ (0)] ≤st
[HY (m)|HY (m − 1) = hY (m − 1), · · · , HY (0) = hY (0)],
whenever hZ ( j) ≤ hY ( j), j = 0, 1, · · · , m − 1, if 1 ≤ m ≤ k, and,

(56)

F̄p l (n) ≤ F̄q j (n), ∀ n, ∀ (l, j) = {l, j|g(l) ≤ g( j)} , if m > k},

(57)

where F̄p i (n) and F̄q i (n) are the CCDFs of the i-th state of the two receivers, respectively, and n is the
index of channel states, ≤ compares two vectors element-wise.
Please refer to Appendix VI for the proof. Note that conditions (55) and (56) should be identified
according to initial conditions which are additionally given, but cannot be derived from the transition
matrices P and Q .
The first order Markov fading channels can be further specialized from Corollary 5 shown as follows.

35

Corollary 6. Consider two first order Markov fading channels {HZ (m)} and {HY (m)} with fading states
arranged in an increasing manner with respect to the state-values. A two-receiver finite state Markov
fading broadcast channel is degraded if
HZ (0) ≤st HY (0), and

(58)

F̄p l (n) ≤ F̄q j (n), ∀ n, ∀ l, j, s.t. l ≤ j.

(59)

Proof: Following the same steps in the proof of Corollary 5, the condition hZ (m) ≤ hY (m) can be
simply replaced by (59) due to the ordered super states.
C. Examples
In this subsection, we illustrate two examples to show an application of our results to fading channels
with memory.

Example 1 of channels with memory (a 1st-order Markov chain with 3-state): Consider two threestate first order Markov chains. Given any HZ (0) ≤st HY (0), the following transition probability matrices
of HZ and HY , respectively, satisfy Corollary 6 and form a degraded channel







P=



1
2

1
4

1
4

3
4

1
8

1
8

5
8

1
4

1
8







, Q = 





1
4

3
8

3
8

1
8

2
8

5
8

1
2

1
8

3
8




.



(60)

To identify the degradedness, we resort to Corollary 6, and verify the condition for the corresponding
CCDF matrices as




F̄P = 



1
2

1
4

1
4

1
8

3
8

1
8





0 





,
F̄
=


Q
0 



0

3
4

3
8

7
8

5
8

1
2

3
8


0 


.
0 


0

(61)

After comparing the pairs of CCDFs: (F̄p 1 , F̄q 1 ), (F̄p 1 , F̄q 2 ), (F̄p 1 , F̄q 3 ), (F̄p 2 , F̄q 2 ), (F̄p 2 , F̄q 3 ), (F̄p 3 , F̄q 3 ), it
is clear that (59) is valid. Therefore, it is a degraded IFSBC.

36

Example 2 of channels with memory (a 2nd-order Markov chain with 2-state): Consider two
binary-valued second order Markov chains. The general transition matrix can be expressed as [54]
00

01

10

11



00 PP000
 00


01
 0


P100
10
 P10


11
0

P001
P00

0

0

P010
P01

P101
P10

0

0

P110
P11

0 


P011 

P01 
,

0 




(62)

P111
P11

where row and column indices show the current and next super states, respectively.
Consider the following transition matrices




1 1
1 2
 2 2 0 0 
 3 3 0 0 








 0 0 1 2 
 0 0 1 3 
3 3 
4 4 


P=
, Q = 
.


 1 3
 1 4
 4 4 0 0 
 5 5 0 0 








1 4
1 5
0 0 5 5
0 0 6 6
The corresponding CCDF matrices can be easily derived as



2
1
 3
 2 0 0 0 






2

 1
 1 1
3 0 


F̄P = 
 , F̄Q = 

 4
 3
 4 0 0 0 
 5






4
1 1 5 0
1

(63)


0 0 0 


3
1 4 0 

.

0 0 0 


5
1 6 0

(64)

By comparing the pairs of CCDFs: (F̄p 1 , F̄q 1 ), (F̄p 1 , F̄q 2 ), (F̄p 1 , F̄q 3 ), (F̄p 1 , F̄q 4 ), (F̄p 2 , F̄q 2 ), (F̄p 2 , F̄q 3 ),
(F̄p 2 , F̄q 4 ), (F̄p 3 , F̄q 3 ), (F̄p 3 , F̄q 4 ), (F̄p 4 , F̄q 4 ), we can verify (57). In addition, given any initial conditions
satisfying (55) and
[HZ (2)|HZ (1) = hZ (1), HZ (0) = hZ (0)] ≤st [HY (2)|HY (1) = hY (1), HY (0) = hY (0)],

(65)

[HZ (1)|HZ (0) = hZ (0)] ≤st [HY (1)|HY (0) = hY (0)],

(66)

are all fulfilled, we can identify it is a degraded IFSBC.

37

VI. C ONCLUSION
In this paper, we investigated the ergodic capacity of several fast fading Gaussian memoryless multiuser
channels, when only the statistics of the channel state information are known at the transmitter. To achieve
this goal, we resorted to classifying the random channels through their probability distributions by which
we are able to attain the capacity results. In particular, we derived sufficient conditions to attain some
information theoretic channel orders such as degraded and strong/very strong interference by combining
the usual stochastic order and the same marginal property, such that the capacity regions can be simply
characterized, which includes Gaussian interference channels, Gaussian broadcast channels, and Gaussian
wiretap channels/secret key generation. Extension of the framework to channels with finite-state memory
was also considered, wherein the Markov fading channel is discussed as a special case. Practical examples
were illustrated to show the successful application of the derived results.
A PPENDIX I. P ROOF OF R EMARK 5
We verify (30), (31), and (32), respectively as follows. It can be easily seen that
min{F̄H1 (h̃1 ), 0} = min{0, F̄H2 (h̃2 )} = 0,

(67)

then (30) is fulfilled. As previously mentioned, this property corresponds to the same marginal property.
We can also easily see that
min{F̄H1 (h̃1 ), 1} = F̄H1 (h̃1 ) and min{1, F̄H2 (h̃2 )} = F̄H2 (h̃2 ),

(68)

then (31) is fulfilled. To check (32), i.e., the 2-increasing property, we first define u j , F̄H1 (h̃1, j ) and
v j , F̄H2 (h̃2, j ), j = 1, 2, where the subscript j is to indicate the different realizations of H1 and H2 . The
condition {u1 ≤ u2 and v1 ≤ v2 } has the following cases: 1) v1 ≤ v2 ≤ u1 ≤ u2 , 2) v1 ≤ u1 ≤ v2 ≤ u2 ,
3) v1 ≤ u1 ≤ u2 ≤ v2 , 4) u1 ≤ v1 ≤ v2 ≤ u2 , 5) u1 ≤ v1 ≤ u2 ≤ v2 , 6) u1 ≤ u2 ≤ v1 ≤ v2 . We can further
combine the above cases into the following 4 classes:
Class 1 (Cases 1 and 2): u2 ≥ v2 , u1 ≥ v1 : the LHS of (32) can be expressed as
v2 − v1 − min(u1 , v2 ) + v1 = v2 − min(u1 , v2 ) ≥ 0.

(69)

38

Class 2 (Case 4): u2 ≥ v2 , u1 ≤ v1 : the LHS of (32) can be expressed as
v2 − v1 − u1 + u1 = v2 − v1 ≥ 0.

(70)

Class 3 (Case 3): u2 ≤ v2 , u1 ≥ v1 : the LHS of (32) can be expressed as
u2 − v1 − u1 + v1 = u2 − u1 ≥ 0.

(71)

Class 4 (Cases 5 and 6): u2 ≤ v2 , u1 ≤ v1 : the LHS of (32) can be expressed as
u2 − min(u2 , v1 ) − u1 + u1 = u2 − min(u2 , v1 ) ≥ 0.

(72)

Therefore, from (69), (70), (71), and (72), the 2-increasing property is fulfilled, which completes the proof.
A PPENDIX II. P ROOF OF T HEOREM 5
We first derive the ergodic capacity region of a strong GIC with statistical CSIT, which is not reported
in literature. Then we derive the sufficient condition to achieve this ergodic capacity region. To commence
the first part, we focus on the uniformly strong IC (US IC) [11], in which every fading state results in
strong interference, i.e., all realizations of channel gains over time satisfy h21 ≥ h11 and h12 ≥ h22 . Hinging
upon the US IC, we can smoothly connect the stochastic orders with the capacity region. To prove the
ergodic capacity region, we mirror the proof steps in [11, Theorem 3] with proper modifications to fit our
assumptions. For the achievable scheme, it is clear that allowing each receiver to decode both messages
from the two transmitters provides an inner bound, i.e., (37), of the capacity region.
Now we prove that (38) is an outer bound of the capacity region of the considered model, where a
genie bestows the information of the interference to only one of the receiver, e.g., the second receiver,
which is equivalent to setting h21 = 0. By this genie aided channel, we aim to prove6
R1 + R2 ≤ E [C (H11 P1 + H12 P2 )] .

6 The

bound.

(73)

single user capacity outer bounds of R1 and R2 can be easily derived. Therefore, here we only focus on the sum capacity outer

39

From Fano’s inequality we know that the sum rate must satisfy
(R1 + R2 − ε)
(a)

n

n

H ) + I(X2n ;Y2n |X1n , H̃
H )
≤I(X1n ;Y1n |H̃

(b)

n

n

n

n

H = h̃h ) + I(X2n ; h̃n22 X2n + N2n |H̃
H = h̃h )]
=E[I(X1n ; h̃n11 X1n + h̃n12 X2n + N1n |H̃
n

n

n

n

n

n

H = h̃h ) − h(h̃n12 X2n + N1n |H̃
H = h̃h ) + h(h̃n22 X2n + N2n |H̃
H = h̃h ) − h(N2n )]
=E[h(h̃n11 X1n + h̃n12 X2n + N1n |H̃
(c)


H = h̃h)−h(N2,k ) +h(h̃n22 X2n + N2n |H̃
H = h̃h)−h(h̃n12 X2n +N1n |H̃
H = h̃h)],
≤E[Σnk=1 h(h̃11,k X1,k +h̃12,k X2,k +N1,k |H̃
(74)
n

n , H̃ n ]; in
H , [H̃11
where in (a) the condition of the second term is due to the genie and we define H̃
12
n

n

H . To simplify the notation, we omit the subscript of H̃
H in expectation;
(b) the expectation is over H̃
(c) is by applying the chain rule of entropy and conditioning reduces entropy and i.i.d property to the
first and the fourth terms on the RHS of the second equality, respectively. Since the last term on the
RHS of (74) has a sign change and also it is not as simple as the term h(N2n ), we concentrate on
n X n + N n ) − h(H̃ n X n + N n ) together for the ease of single letterization. To proceed, we exploit the
h(H̃22
2
2
12 2
1

property |h̃12 | ≥ |h̃22 |7 by definition of US IC as
(a)

n n
n n
n
n
E[h(H̃22
X2 + N2n ) − h(H̃12
X2 + N1n )] =E[h(X2n + Ñ2n ) − h(X2n + Ñ1n ) + 2 log(|H̃22
|/|H̃12
|)]
(b)

n
n
=E[h(X2n + Ñ1n + N n ) − h(X2n + Ñ1n ) + 2 log(|H̃22
|/|H̃12
|)]
n
n
|/|H̃12
|)]
=E[h(N n + X2n + Ñ1n )−h(N n + X2n + Ñ1n |N n )+ 2 log(|H̃22

(c)

n
n
|/|H̃12
|)]
=E[I(N n ; N n + X2n + Ñ1n ) + 2 log(|H̃22

(d)

n
n
|/|H̃12
|)]
≤ E[I(N n ; N n + Ñ1n ) + 2 log(|H̃22

(e)

≤Σnk=1 E[h(N2,k ) − h(N1,k )],

(75)

where in (a), Ñ2,k ∼ C N (0, |H̃22,k |−2 ) and Ñ1,k ∼ C N (0, |H̃12,k |−2 ); in (b) we define Nk ∼ C N (0, |H̃22,k |−2 −
|H̃12,k |−2 ), while N n = [N1 , N2 , · · · Nn ]T and N n is independent to Ñ n ; in (c), we treat N n as the transmitted
signal and X2n + Ñ1n as an equivalent noise at the receiver; in (d) we apply data processing inequality with

40

the Markov chain: N n − Ỹ2n − Ỹ1n , where Ỹ1n , X2n + Ñ1n + N n and Ỹ2n , Ñ1n + N n ; in (e) we use the fact
that N1n and N2n are i.i.d., respectively, and also by the assumption |h̃12 | ≥ |h̃22 |. Note that even though
(48) and (74) both have the differences of differential entropies and both are to solve the optimal input
distributions, in (48) due to the asymmetric conditioned U and X in the differential entropies, (48) cannot
be reformed as (75).
After substituting (75) into (74), we obtain
1
H ) , I(X1 , X2 ;Y1 |H̃
H , Q),
R1 + R2 ≤ Σnk=1 I(X1k , X2k ;Y1k |H̃
n

(76)

where Q is the time sharing random variable. To proceed, we apply the result in [55], wherein the
capacity region of a MAC is derived for cases in which only the receiver can perfectly track the channel
information but the transmitter has no such information. For such channels, the optimal input distribution
is proved to be Gaussian. Note that since the transmitter has no instantaneous CSIT, it cannot do power
allocation across time. As a result, full powers P1 and P2 are always used. Then we obtain (73). Likewise,
when the genie provides the interference only to the first receiver, we can get
R1 + R2 ≤ E [C (H21 P1 + H22 P2 )] .

(77)

Finally, after comparing the outer bounds (73) and (77) to (37) and (38), we can observe that decoding
both messages at each receiver can achieve the capacity region outer bound and under the assumptions
on the knowledge of channel gains at each node.
In the ensuing part, we derive the sufficient condition to achieve the capacity region of GIC with strong
interference. In particular, we derive the condition to achieve US IC. From Definition 2 we can derive

7 Note

I(X1 ;Y1 |X2 , H11 = h11 , H12 = h12 ) = log(1 + h11 P1 ),

(78)

I(X1 ;Y2 |X2 , H21 = h21 , H22 = h22 ) = log(1 + h21 P1 ).

(79)

that without this property, we may not be able to rearrange the outer bound of the sum rate as (76). This seemly very strict condition

can be relaxed as long as the channel distributions follow proper stochastic orders, which will be explained in the latter part of this proof.

41

If we impose the following constraint
h21 ≥ h11 and h12 ≥ h22 ,

(80)

for all realizations, it is clear that (2) is fulfilled. Note that it is the uniformly strong IC [11], by which
we derive (73) and (77). To connect (80) to our considered model, we apply Theorem 3. Recall that we
assume that all random channels, signals, and noises are jointly independent. From (34) and (35), we can
find that
fY1 |X1 ,X2 = fH̃11 ∗ fH̃12 ∗ fN1 ,

(81)

fY2 |X1 ,X2 = fH̃21 ∗ fH̃22 ∗ fN2 ,

(82)

where the convolution ∗ is due to the mutual independence assumption, while for a given x j , H̃k j , x j Hk j .
0 , H̃ 0 ) and (H̃ 0 , H̃ 0 ) such
From (81) and (82) we know that as long as we can find equivalent channels (H̃11
12
22
21

that fH̃ 0 ∗ fH̃ 0 = fH̃11 ∗ fH̃12 and fH̃ 0 ∗ fH̃ 0 = fH̃22 ∗ fH̃21 , then the capacity regions of the two interference
11

12

22

21

channels are identical. To achieve this goal, we resort to finding equivalent {Hk0 j } which are jointly
independent such that8
fHk0 j = fHk j .

(83)

A constructive way to generate the equivalent {Hk0 j } is by (18), i.e.,
Hk0 j = FH−1
(U), k, j = {1, 2}.
kj

(84)

Then we know that Hk0 j has the same CDF as that of Hk j , namely, FHk j , which satisfies the condition (83).
0 ≥ H 0 ) = P(H 0 ≥ H 0 ) = 1 if H ≥ H
From Theorem 2, we also know that P(H21
21 st 11 and H12 ≥st H22 ,
11
12
22

respectively, which satisfies (80) a.s. and therefore completes the proof.
8 Finding

individually equivalent channels as (83) but not finding the compound distributions such that fH̃ 0 ∗ fH̃ 0 = fH̃11 ∗ fH̃12 may make
11

12

the sufficient condition in the theorem statement stricter. But this way is more tractable and therefore we adopt it.

42

A PPENDIX III. P ROOF OF T HEOREM 6
Similar to Appendix II, we first derive the ergodic capacity region of GIC with very strong interference
with statistical CSIT and then we derive the sufficient condition to achieve the derived capacity region.
For the first part, we can deduce the proof following the same steps as in Appendix II with proper
modifications. But here we proceed in an easier way since for the capacity region of a discrete memoryless IC with very strong interference, there is no sum rate constraint. We can solve the optimal input
distributions of GIC with statistical CSIT by considering
arg max

I(X1 ;Y1 |X2 , H11 , H12)+µI(X2 ;Y2 |X1 , H22 , H21),

pX1 , pX2 :
E[|X1 |2 ]≤P1 ,E[|X2 |2 ]≤P2

(85)

where µ ∈ R+ . Note that (85) can be further expressed as
I(X1 ; H11 X1 +Z1 |H11)+µI(X2 ; H22 X1 +Z2 |H22).
pX1 , pX2 :
E[|X1 |2 ]≤P1 ,E[|X2 |2 ]≤P2
arg max

(86)

It is clear that for each µ, (86) can be maximized by Gaussian inputs, i.e., X1 ∼ C N (0, P1 ) and X2 ∼

C N (0, P2 ). Then the capacity region can be delineated by (41).
In the second part, from (4) we can derive

I(X1 ;Y2 |H21 = h21 , H22 = h22 ) = log 1 +


h21 P1
.
1 + h22 P2

(87)

After comparing (87) and (78), we can find the condition
h11 ≤

h21
.
1 + P2 h22

(88)

h22 ≤

h12
.
1 + P1 h11

(89)

Similarly, from (5) we can derive

By treating H21 /(1 + P2 H22 ) and H12 /(1 + P1 H11 ) as new random variables, respectively, and follow the
same steps in the proof of Theorem 5, we can extend (88) and (89) to the stochastic case as
H21
≥st H11 ,
1 + P2 H22
H12
≥st H22 .
1 + P1 H11

(90)
(91)

43

Now we need to guarantee that the marginal distributions are intact after applying the construction by
0 /(1 + P H 0 )
Theorem 3. This can be simply checked by the fact that, because the equivalent channels H21
2 22

have the same distribution as that of H21 /(1 + P2 H22 ) after applying Theorem 3, we can exploit the
0 and H 0 whose distributions are the same as those of H
transform (84) to construct H21
21 and H22 ,
22

respectively. Therefore, the marginal distributions in (81) and (82) are not changed, which completes the
proof.
A PPENDIX IV. P ROOF OF T HEOREM 7
The achievability of (47) can be easily derived by substituting U = X ∼ N(0, PT ) into the secrecy
capacity in [42, (8)]. In the following, we focus on the derivation of the outer bound of the secrecy
capacity. In particular, we adopt the Sato-type outer bound combined with the coupling scheme to show
that Gaussian input is optimal for the outer bound and also show the outer bound matches the lower
bound. In the following, we first verify the validity of using the coupling scheme.
To construct an equivalent wiretap channel, we require the two channels to have the same: 1) error
probability P(W 6= Ŵ ) and 2) equivocation rate limn→∞ n1 h(W |Z n ), where W and Ŵ are the transmitted
and detected messages at Alice and Bob, respectively. The first requirement is valid because the coupling
scheme does not change the channel distribution. Checking the second requirement is more involved. The
reason is that we have asymmetric knowledge of the CSI at Bob and Eve. In general, to design for the
worst case we assume that Eve has more knowledge of the CSI than Bob. As mentioned previously, a
common assumption is that Bob knows perfectly the realization of his own channel H but Eve knows
perfectly both H and her own channel G. The equivocation rate then becomes
1
lim h(W |Z n , H n , Gn ),
n→∞ n

(92)

and the calculation of which depends on p(zn , hn , gn |xn ). If we directly use the coupling scheme, we will
end up with the equivocation rate as h(W |H̃ n , G̃n , Z̃ n ), whose calculation relies on p(z̃n , h̃n , g̃n |xn ), where
Z̃ is the channel output at Eve of the equivalent WTC when the coupling scheme is used. Note that

44

p(zn , hn , gn |xn ) may not be identical to p(z̃n , h̃n , g̃n |xn ) because same marginal property does not guarantee
the same joint distribution. More specifically, the correlation between H n and Gn can be arbitrary. In
contrast, the correlation between H̃ n and G̃n cannot be arbitrary, i.e., it is fixed by the marginal CDFs FH
and FG and also the property of U. To avoid this inconsistence, we can consider a new wiretap channel
where Eve only knows gn with equivocation rate as h(W |Gn , Z n ) calculated by p(zn , gn |xn ). Note that the
secrecy capacity of the new GWTC is no less than the original one, since Eve here knows less CSI than
in the original setting. After applying the coupling scheme, we have the equivocation rate as h(W |G̃n , Z̃ n ),
whose calculation relies on p(z̃n , g̃n |xn ). A sufficient condition to equal these two equivocation rates is
that p(z, g|x) = p(z̃, g̃|x), which can be verified as follows:
p(z, x|g)p(g)
p(x)
(a) p(z, x|g)p(g̃)
=
p(x)
(b) p(z̃, x|g̃)p(g̃)
=
p(x)

p(z, g|x) =

= p(z̃, g̃|x),

(93)

where (a) is due to the coupling, i.e., p(g) = p(g̃); (b) comes from the fact that g̃ and g have the same set
of alphabets and distribution while z is replaced by z̃ due to the equivalent channel z̃ =

√
g̃x + nz . Then

we know that the new WTC where Eve only knows gn before and after using the coupling scheme are
equivalent.
To verify the optimality of Gaussian input of the outer bound, we invoke the Sato-type outer bound by
letting Bob know what Eve knows, i.e., Bob receives Ỹ and Z̃. Then we can express the outer bound as
(a)

max I(U; Ỹ , Z̃, H̃, G̃) − I(U; Z̃, G̃) = max I(X; Ỹ , Z̃, H̃, G̃) − I(X; Z̃, G̃)

PU , PX|U

PX

(b)

= max I(X; Ỹ , Z̃|H̃, G̃) − I(X; Z̃|G̃)
PX

(c)

= max I(X; Ỹ , Z̃|H̃, G̃) − I(X; Z̃|H̃, G̃)
PX

= max I(X; Ỹ |Z̃, H̃, G̃),
PX

(94)

45

where (a) is because the new WTC is a degraded one; (b) is due to the assumption of statistical CSIT;
(c) comes from the fact that knowing G̃ causes H̃ also known. In particular, since Bob knows both FH
and FG in the new WTC, once he knows g̃ = FG−1 (u), he can simply get u and then derive h̃ = FH−1 (u).
In addition, because
arg max I(X; Ỹ |Z̃, H̃, G̃) = arg max h(Ỹ |Z̃, H̃, G̃),
PX

PX

(95)

we can extend [56, Lemma 1] to show that
h(Ỹ |Z̃, H̃, G̃) = h(Ỹ − αZ̃|Z̃, H̃, G̃)
≤ h(Ỹ − αZ̃|H̃, G̃)
= EH̃, G̃ [h(Ỹ − αZ̃|H̃ = h̃, G̃ = g̃)],
where α =

√ √
h̃ g̃/(1 + g̃) is the linear MMSE estimator of Ỹ from Z̃ and the inequality is due to the fact

that conditioning only reduces the differential entropy. And the equality is achieved by Gaussian X for
the outer bound.
We match the outer and lower bounds as follows. Due to the coupling scheme, we have the Markov
chain X − Ỹ − Z̃ for each channel use. Hence we can further express the RHS of (b) in (94) as
max I(X; Ỹ , Z̃|H̃, G̃) − I(X; Z̃|G̃) = max I(X; Ỹ |H̃, G̃) − I(X; Z̃|G̃).
PX

PX

(96)

After substituting Gaussian input into (96) followed by the full power usage since power allocation can
not be done due to statistical CSIT, we can get (47), which completes the proof.
A PPENDIX V. P ROOF OF T HEOREM 8
To prove that the stochastically degraded random source (X, Ỹ , Z̃) implies that the corresponding
conceptual WTC (CWTC) [45] is also stochastically degraded from the CWTC constructed by the corresponding physically degraded source (X,Y, Z), we start from constructing the CWTC of the random source
(X,Y, Z), where the equivalently received signals at Bob and Eve are Y 0 , (Y,U ⊕ X) and Z 0 , (Z,U ⊕ X),

46

respectively. If X −Y − Z, then U −Y 0 − Z 0 , i.e., the CWTC is also a physically degraded one, which can
be shown as follows:
(a)

I(U; Z 0 |Y 0 ) =H(Z,U ⊕ X|Y,U ⊕ X) − H(Z,U ⊕ X|Y,U ⊕ X,U)
(b)

=H(Z|Y ) − H(Z|Y,U ⊕ X,U)

(c)

=H(Z|Y ) − H(Z|Y, X,U)

(d)

= H(Z|Y ) − H(Z|Y, X)

(e)

=0,

(97)

where (a) is by definition of Y 0 and Z 0 ; (b) is by the crypto lemma [57] and U is selected to be independent
to Y and Z; (c) is from the fact that given U, we can know X from U ⊕ X; (d) is by the same reason as
(b); (e) is due to X −Y − Z and by definition of conditional mutual information.
Now we consider the stochastically degraded source of common randomness (X, Ỹ , Z̃) fulfilling PỸ |X =
PY |X and PZ̃|X = PZ|X . Similar to the first step, we construct a CWTC from (X, Ỹ , Z̃), namely, {(U,Y 00 , Z 00 ),
PU,Y 00 ,Z 00 = PU PY 00 ,Z 00 |U }, where Y 00 , (Ỹ ,U ⊕ X) and Z 00 , (Z̃,U ⊕ X) are the equivalent channel outputs at
Bob and Eve, respectively. To prove the two CWTC’s PY 0 ,Z 0 |U and PY 00 ,Z 00 |U are equivalent, inherited from the
WTC, we can invoke the same marginal property in Corollary 1 to prove PY 00 |U = PY 0 |U and PZ 00 |U = PZ 0 |U ,
which is shown in the following. By definition of Y 00 and Y 0 , we know that PY 00 |U=u = PỸ ,u⊕X , PỸ ,Xu and
PY 0 |U=u = PY,u⊕X , PY,Xu , respectively, where we define Xu as u ⊕ X. Note that due to the closed operation
⊕ in X , the distribution of Xu is just a left circular shift of that of X by u. Instead of directly proving
PỸ ,Xu = PY,Xu , we can equivalently prove PỸ |Xu = PY |Xu , ∀ u, as follows:
PỸ |Xu =x = PỸ |X=x = PY |X=x = PY |Xu =x ,

(98)

where the 1st and 3rd equality is the 2nd equality is due to the assumption of (X, Ỹ , Z̃) forming a
stochastically degraded source from the physically degraded one (X,Y, Z). Therefore, PY 00 |U = PY 0 |U .
Similarly, we can derive PZ 00 |U = PZ 0 |U . Hence, we prove that the CWTC of a stochastic degraded source
is also a degraded CWTC and the two CWTC’s have the same secrecy capacity.

47

Note that due to X − Y − Z, we can derive the key capacity CSK from the corresponding CWTC, but
not just an achievable key rate. For the source (X, Ỹ , Z̃), till now we may only claim the achievable
0 , is the same as C . That is because the
secret key rate but not the secret key capacity, namely, CSK
SK

CWTC is in general only an achievable scheme to derive the secret key rate. However, due to the fact
that the stochastic degradedness is more general than the physical degradedness, i.e., less stringent on
characterizing the relation between X, tildeY and Z̃, the former can not result in a larger secret key
0 =C
capacity than the latter one. Therefore, we attain that CSK
SK = I(X;Y ) − I(X; Z), which completes the

proof.
A PPENDIX VI. P ROOF OF C OROLLARY 5
To simplify the expression, we first define
H̃Z (hhZ (m)) =st [HZ (m)|HZ (m − 1) = hZ (m − 1), · · · , HZ (m − k) = hZ (m − k)]
as the channel state of the m-th sample, which is a random variable given k consecutively past channel
states h Z (m) = [hZ (m − 1), · · · , hZ (m − k)]. We also define H̃Y (hhY (m)), similarly. Due to the property of
the k-th order Markov chain, we can express (53) as
H̃Z (hhZ (m)) ≤st H̃Y (hhY (m)), ∀m ∈ N

(99)

with h Z (k) ≤ hY (k) for all k < m. In addition to Theorem 9, we know that from HZ (0) ≤st HY (0) and (99)
whenever h Z (k) ≤ hY (k) are valid, it follows that {HZ (m)} ≤st {HY (m)}, m ∈ N. Note that the inequality
operates element-wise.
Based on the given transition matrices P and Q , we can further simplify the constraint (99) for the case
m > k. It is clear that the j-th entries of p i and q i are the transition probabilities from the i-th super state
to the j-th super state of the Markov processes {HZ (m)} and {HY (m)}, respectively. Given p i and q i ,
the i-th rows of P and Q , respectively, i = 1, 2, · · · , N k , we can form the corresponding CCDF matrices,
respectively, as F̄P = [F̄pT1 , F̄pT2 , · · · , F̄pT k ]T and F̄Q = [F̄qT1 , F̄qT2 , · · · , F̄qT k ]T , where F̄p j and F̄q j are the CCDF
N

N

vectors derived by p j and q j , respectively. From Definition 4, we can equivalently express (99) for m > k by

48

F̄p l (n) ≤ F̄q j (n), ∀ n, with the constraint h Z (k) ≤ hY (k), where we need to properly choose (l, j) according
to (l, j) = {l, j|g(l) ≤ g( j)}, where g is an N-ary expansion of L defined as g : L 7→ s ∈ {1, · · · , N}k ,
L ∈ {1, · · · , N k }. Then we attain (56) and (57). Combining with (55), we obtain the sufficient conditions to
attain {HZ (m)} ≤st {HY (m)}, or equivalently, there exist two random processes {HZ0 (m)} and {HY0 (m)},
such that {HZ0 (m)} =st {HZ (m)}, {HY0 (m)} =st {HY (m)}, and Pr(HZ0 (m) ≤ HY0 (m), m ∈ N0 ) = 1, which
implies the degradedness and completes the proof.
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