0 2 4 6 8 10 12 14 16
0 2 4 6 8 10 12 14 16
µ
σ2 σ2
has to be restricted to the following region:
Λ= µ
R+×[0,σ2max]∩
½ λ: µ
σ2 ≥∂σ∂µ2¯¯¯
σ2max
¾¶
∪ {∞} .
In Figure4.3, the setΛis depicted by the shaded region. Notice that for a fixed value ofσ2, the slopeµ′=∂µ/∂σ2is equal for allµalong any ¯xcontour. Further, this slope is maximized at the upper bound ofσ2defined byσ2max. If the lines connectingλandλ′to the origin have slopes that are larger than the maximumµ′, then the sum is guaranteed to lie in a higher contour.
In the following, we will prove that the constrained AWGN satisfies all the compatibility properties for the GDA. We also prove that addition of a vectorλ′′to any pair ofλand λ′will not change the ordering between them.
P1 Except for closure, the other monoid properties are obvious because⊕is the stan- dard vector addition. The main reason forΛnot occupying the full non-negative octant, but rather being bounded byσ2≤σ2max =ǫ−max2 /(2π), is to satisfy the clo- sure property. Inside the shaded area, every vector has a slope of at leastµ′. The sum of two such vectors must also have a slope of at leastµ′. The valueσ2max is given by:
σ2max=max
π∈Π
nX
i:vi∈Vπσ2io
(4.15)
This value ofσ2max can be obtained by running a regular DA once onG, with a time complexity cost ofO(V2). If one (or both) of the vectors is∞, then by the definition of∞, the⊕sum must be∞.
P2 The proof is derived from closure on∞.
P3 The proof follows from the definition ofΛand.
P4 Both terms in equation (4.13) are minimized when they are zero, i.e.,µ=0 and
eitherσ2=0 orσ2=ǫ−2/(2π). However, sinceλ=(0, 0)∈Λ, then (0, 0) is indeed the 0 element inΛ.
P5 From Figure4.3and the definition ofΛ, the addition of (µ, 0) followed by (0,σ2) to
any two points inΛpreserves their ordering.
The following proposition proves that the presence of error- and erasure-correcting nodes in the network can have practical benefit for routing information, allowing a very high level of reliability to be achieved by relatively unreliable edges. Mathematically, we say that if the path includesu∈U, then for somej∈[0,J], we can haveβ(πj)=0 even with pj 6=0 for all j.
Proposition 3. IfVπJ∩U =∅, thenβ(πj) is an non-decreasing function ofj. The mini- mumβ(πj)=0 is only possible ifpj =0 for all j =0 . . .J.
PROOF: If we apply the isotonicity propertyP5with 0=a≺b=pj, andc=pπ
j−1, then we will haveβ(πj−1)=pπ
j−1≺pπj−1⊕ pj=β(πj). This proves that thatβ(πj) is a non- decreasing function of j. The second part of the proof can be derived directly fromP1.
Having proven the GDA compatibility, we will now prove that the ¯xvalues in examples 1–3can be defined as non-decreasing functions ofλs. If this is true, then the path with minimumλis also the path with minimum ¯x.
For AWGN, this monotonicity property automatically follows from its definition of λ λ′. In contrast, the proof of monotonicity for q-SC and q-EC is quite involved.
Luckily, the proof for both channels is identical. First, we claim — and prove — that within an admissible range of ǫ∈[0, ¯ǫ], the WCE function ¯x(p,ǫ) is a non-decreasing (albeit discontinuous) function ofp.
Lemma 4. For a given n and a fixed x, the probability function P(x,p) is maximized atp = x
n. Furthermore,P∗=P(⌊n2⌋,12) minimizesP(x,nx) over all possible values ofx∈ [0,n].
PROOF: Let us start from the definition ofP(x,p) in equation (4.7). For a givenn, the lemma is true atx=0 andx=nbecause the value ofP(x,p) reaches the maximum of δ(0)=1. To prove the lemma for other values ofp, we first compute the derivative of P(x,p) with respect topto find the extrema:
∂P(x,p)
∂p =P(x,p) (x
p −n1−−px)=0 . (4.16) On the right-hand side of Equation (4.16), the definition ofP(x,p) shows that it does not have any root with respect top. Therefore the root, if there is any, must be contained in the factor (xp−n1−−px)=0. Indeed, solving the factor forpgives usp=x
n, which maximizes the functionP(x,p) for any givenx∈(0, 1).
Having identified the value ofp that minimizesP(x,p), the next question is, for 0<
x <n, whichxminimizesP(x,x
n)? Unlike withp, we cannot differentiateP(x,p) with respect toxbecause it is a discrete variable. We choose not to take the easiest shortcut of approximatingP(x,p) with a Gaussian distribution and taking the derivative of this approximation because this approach is only valid for certain values ofnandp. Instead, we find the location of the minimum by using the upper and lower bounds of¡n
x
¢given in [29] :
λnx<¡n
x
¢<µnx (4.17)
λnx=λ(n,x)=ξ¡ 1
12n−12x1 −12(n1−x)¢ µnx=µ(n,x)=ξ¡ 1
12n−12x1+1−12(n−1x)+1
¢
ξ(A)= eAnn
+12 (2π)12(n−x)n−x+12 xx+12
.
From Equation (4.17), we can then conclude thatP(x,p) is bounded from below and above by two continuous and differentiable functions ofx.
λnxpx(1−p)n−x<P(x,p)<µnxpx(1−p)n−x
Sincep= x
n, we substitutex=np into the equation above and solve thep roots of the p derivatives of both the lower and upper bounds ofP(x,p) to find the minima with respect top. The lower and upper bounds are minimized atp= 1
2. Sincep= x
n, then x=n
2, and the inequality becomes:
q 2
nπe−12n(6n18n−+11) <P(⌊n2⌋,12)<
q 2
nπe−4n1 . (4.18)
For large values ofn the lower and upper bounds converge. In fact, in (4.17),λnx con- verge toµnx for allx, including at the discrete points 0<x<n ∈N. Thus the minima forλnxandµnxover the continuousxmust also be the minimum forP(x,x/n) over the
discretex, denoted byP∗=P(⌊n2⌋,12).
Lemma 5. P(x,p) is unimodal overxandp. A functionf(x) isunimodaloverx∈[a,b]
if there exists a single valuex0such that f(x) is monotonically increasing forx<x0and monotonically decreasing forx>x0.
PROOF: To prove unimodality overx, solve the inequalityP(x,p)<P(x+1,p) for x, which gives usx<np−(1−p). Since 0≤(1−p)≤1, thenx<⌊np⌋. In other words, x0=⌊np⌋. FromP(x,p)>P(x+1,p) we also concludex>x0. To prove unimodality overp:
P′(x,p)= ∂
∂pP(x,p)=P(x,p)(x
p −n1−−px) . We have calculated thatP′(x,p0)=0 at p0 = x
n. Therefore, p0is an extremum and a candidate for a global maximum. Indeed, forp<p0, we can easily show thatP′(x,p)>0 and vice versa forp>p0, we haveP′(x,p)<0.
Corrolary 6. The setǫof admissibleǫis defined by the interval [0, ¯ǫ=P∗].
PROOF: First, recall the definition ¯x(p,ǫ)=max{x|P(x,p)≥ǫ}. To ensure that ¯x(p,ǫ) is properly defined for allp∈ P, for each value ofpwe must haveP(x,p)≥ǫfor some x. This condition is trivially met ifǫ≤ 0 because P(x,p) ≥0. However, we are only
interested inǫ≥0. Ifǫ>P∗, then P(x,p) lies completely beneathǫand the set{x | P(x,12)≥ǫ}=∅. Consequently, ¯x(p,12) is invalid. Thus, the set of admissibleǫis given
by [0,P∗].
Consider the equationP(x,p)=ǫfor somen andǫ. Define the set of all the roots of this equation byP ={p∈ P |P(x,p)=ǫ}. Thei-th root of this equation is denoted by pi(x,ǫ)=pi(x)=Pi. From Lemma5, if we can guaranteeǫ∈ǫ, then there is at least one root, i.e.,P6=∅.
In three special cases,P only contains one root. Atx=0, it is denoted byp0(0)=0, and atx =n, by p1(n)=1. Finally, whenǫ=P∗, at the midpointxm =⌊n2⌋the root is p0(xm)=p1(xm)= 1
2. In general, however, there are two distinct roots p0(x),p1(x)∈ [0, 1], withp0(x)< x
n <p1(x). Ifǫgoes toward 0, thenp0(x)→0 andp1(x)→1, except p1(0)=0 andp0(1)=1. Ifǫgoes towardP∗, thenp0(x) increases, whilep1(x) decreases.
Define the setsP0andP1, each having then+1 values of the first and second roots p0(x) andp1(x) ofP. Each of them corresponds to then+1 different values of 0≤x≤n.
Using the equationP(x,p)=ǫ, these roots correspond to the setsx0(p) andx1(p), for 0≤p= x
n ≤n.
Lemma 7. The roots p0(x) and p1(x) are non-decreasing functions of x with p0(x)= p1(x) only atx=0 andx=n(orx=xm for the case whereǫ=P∗).
PROOF: First, observe that P(x,p) = P(x+1,p) only has one root at p = p× = x+1
n+1
between the maxima ofP(x,p) andP(x+1,p), i.e., x
n <p×< x+1
n . From lemma5, this implies that ifp <p×thenP(x,p)>P(x+1,p), and ifp >p×, P(x,p)<P(x+1,p).
Hence, p0(x)≤ p0(x+1) and p1(x) ≤p1(x+1). Therefore, both p0(x) and p1(x) are
non-decreasing functions ofx.
Theorem 8. For q-SCs and q-ECs, the worst case function values
¯
x(p,ǫ)=maxx=np{x0(p),x1(p)}are non-decreasing functions ofp.
PROOF: First, the maxxfunction is used because it is possible to have identical values of
p that solvesP(x,p)=ǫ, that correspond to different values ofx. In other words, it is possible to havep0(x)=p0(x′)∈P0forx6=x′, whereP(x,p0(x))=P(x′,p0(x′))=0. For example, ifǫ=0,p0(0)=···=p0(n−1)=0. The exact same argument applies to the roots in the setP1.
It is interesting to note that in our definition of ¯x(p,ǫ), we also have used maxx. The function maxxin ¯x(p,ǫ) isolates the largestxsatisfyingP(x,p)≥ǫ, while maxxx0(p),x1(p) isolates the largestxthat satisfiesP(x,p)=ǫ. Finally, we claim thatP(x,p)=ǫimplic- itly defines a functionp(x) that links each value ofp to a value ofx. The proof for this theorem then follows directly from the monotonicity ofp(x) already proven in Lemma
7.
Theorem8, proves that the worst case functions ¯x(p,ǫ) of theq-SC andq-EC are non- decreasing functions ofp. Therefore, as we stated before, a network path that minimizes pπalso minimizes ¯x. The constrained AWGN channel also satisfies this due to its defini- tion ofthat directly utilizes ¯x.
0 2 4 6 8 10 12 0
2 4 6 8 10 12 14 16
µ
σ2 σ2
0 2 4 6 8 10 12 14 16 0 2 4 6 8 10 12