Table Of ContentEstimation of RFID Tag Population Size by
Gaussian Estimator
Md Mahmudul Hasan, Shuangqing Wei, Ramachandran Vaidyanathan
Abstract—In this paper we propose a novel approach to the frame size of the forthcoming frame to the tags in its
estimating RFID tag population size. Unlike all previous {0,1} vicinity. Each tag randomly picks a time-slot in the frame
estimation schemes, we analysed our scheme under {0,1,e}
and replies during that slot. Previous {0,1} channel models
and presented results under both {0,1} and {0,1,e} channel
do not differentiate between a singleton and a collision slot,
models. Under both the channel models we used a well justified
Gaussianestimatorforestimation.Wehavenamedouralgorithm where a 0 represents an empty slot and a 1 represents a non-
7 ”Gaussian Estimation of RFID Tags,” namely, GERT. The most empty slot. In this {0,1,e} model [2] [3], 0 represents ’no
1 prominentfeatureofGERTisthequalitywithwhichitestimates reply’, 1 represents ’exactly one reply’ and e stands for ’any
0 a tag population size. We supported all the required approxi-
number of replies greater than 1’, to a slot in a frame. The
2 mations with detailed analytical work and counted for all the
n approximation errors when we considered the overall quality of reader receives a sequence of 0s, 1s and e’s. GERT calculates
a theestimation.GERTisshowntobemorecosteffectivethanthe Ne−fN1 for each round where, Ne represents the number of
J previous methods suggested so far, under the same performance e’s ( i.e. collision slots) and N1 is the number of 1’s in the
1 constraints. reader sequence. After n rounds of these measurements we
Index Terms—RFID, Collision Model, Gaussian Approxima-
2 take the average of all these values. This average is finally
tion
substituted for the true mean in the expected value equation
]
of the estimator to estimate the tag population size by an
T I. INTRODUCTION (cid:104) (cid:105)
I inverse function. Figure 1 demonstrates that E Ne−N1 is
A. The Problem Statement f
.
s a monotonic function of t for given f and p, except for a
c Tag estimation is useful in many everyday applications
singularity at the start. We have analyzed in this paper the
[ includingintagidentification,privacy-sensitiveRFIDsystems (cid:104) (cid:105)
1 and warehouse monitoring. In this paper, we propose a more conditionsunderwhichE Ne−fN1 isinvertibleand Ne−fN1 is
asymptoticallyGaussiandistributedforlargef,whilemeeting
v efficientmethodtoestimatethesizeofatagpopulation.There
2 have been a good number of estimation schemes so far, and the imposed estimation accuracy requirement.
5
we have presented in the paper the comparative efficacy of
9
our algorithm with some recently proposed approaches. The
5
0 problem at hand can be formally stated as follows: For a
. given reliability requirement α ∈ [0,1), a confidence interval
1 C. Related Work
β ∈ [0,1) a reader will have to estimate an unknown tag
0
7 population size t in a particular area. The estimation has to
1 maintaintheminimumaccuracyconditionP[|tˆ−t|≤βt]≥α, The first work on tag estimation was that of Kodialam and
v: wheretˆistheestimatedvalueoftheactualtagpopulationsize Nandagopal, titled Unified Probabilistic Estimator (UPE) [4].
i t. UPEwasbasedonthenumberofemptyslotsorthenumberof
X
collision slots in the frame. Kodialam et al. later proposed an
r B. Proposed Scheme Overview
a improved framed slotted Aloha protocol-based estimation in
In this paper, we present a new algorithm named Gaussian [5] called Enhanced Zero Based (EZB) estimator. EZB makes
Estimation of RFID Tags (GERT). At the beginning of the its estimation based on the total number of empty slots in
estimation process we run a probe using the Flazolet Martin a frame. The difference between EZB and UPE is that UPE
algorithm [1] to get a rough upper bound tm on the tag makes an estimation of the population size in each frame and
population size t. Based on tm and the accuracy requirements attheendaveragesoutalltheestimationresultswhereas,EZB
α, β we determine the critical parameters p, f and n, where findstheaverageofthenumberof0sineachframeandfinally
p is the persistence probability (the probability that a tag makestheestimationbasedonthisaveragevalue.Inthispaper
remains active to respond to a forthcoming frame), f is we compared the proposed method ( GERT) with a relatively
the frame size and n is the required number of rounds. recent tag estimation scheme (ART) by Shahzad and Liu [6],
A standardized framed slotted Aloha protocol is what we whichclaimstobebothfasterandmorecosteffectiveinterms
used as communication protocol. The reader first broadcasts of the number of slots required for estimation than the prior
algorithms. ART uses a {0,1} channel model and estimates
1MdMHasan,S.WeiandR.VaidyanathanarewiththeschoolofElectrical
the tag population size based on the average run size of 1s in
EngineeringandComputerScience,LouisianaStateUniversity,BatonRouge,
LA70803,USA(Email:[email protected],[email protected],[email protected]). a frame.
1
II. FORMULATIONOFTHEPROBLEM
(cid:104) (cid:105)
We are considering Ne−fN1 as the estimator in our pro- 0.8
posed model. We define two important variables in (1),
(cid:20) (cid:21) 0.6
N −N N −N
Z (t)(cid:44) e 1, g (t)(cid:44)E e 1 (1)
f f f f
0.4
For a given persistence probability p and frame size f, Z]f
E[
let X ∼ Bernoulli(p) be the variable that represents the 0.2
ij f
probability that the ith tag replies to the jth slot or not. So,
(cid:40) 0
1, with probability p A B
X = f (2)
ij 0, with probability (1− p) −0.2 D
f
Let Yj = (cid:80)ti=1Xij. It is straightforward to see that the −0.40 C E 500 1000 1500
random variable Yj has the following probability distribution, t
(cid:16) (cid:17)t Fig.1. gf versustagpopulationsizet.’x’representsthesimulatedresults.
(cid:16)1−(cid:17)fp(cid:16) =p(cid:17)0t,−1 y =0 (f =200)
pYj(y)= t fp 1− fp =p1, y =1
1−(cid:16)1− fp(cid:17)t−t(cid:16)fp(cid:17)(cid:16)1− fp(cid:17)t−1 =pe, y =e ofWtheeneoxtipceectiantioFnigucruerv1ethwaht,icthheprereissenatsdiupsawt tihthe abegcainsneinogf
singularity. We define, r (cid:44) tp. Proofs to the following two
Introduction of the following two indicators namely Y(1) f
j facts for f (cid:29) 1, are given in Appendix A and Appendix B
and Y(e) will make our analysis easier, p
j respectively,
Y(1) =(cid:40)1, when Yj =1, Y(e) =(cid:40)1, when Yj =e 1) atDinFigure1,tLM = 2fp orequivalentlyatrLM = 12.
j 0, when Yj (cid:54)=1 j 0, when Yj (cid:54)=e 2) gf(t) convex t < 32fp and concave for t ≥ 32fp for given
f and p.
(3)
At point B in Figure 1, the singularity ends . We numerically
Now using (1) we have,
calculated the value of r at B to be r =1.2564. Using the
min
Z = 1 (cid:88)f (Y(e)−Y(1)) (4) definition of r we get the corresponding frame size at B,
f f j j tp
j=1 f = (11)
max r
Next we define, Z (cid:44)Y(e)−Y(1). Then (4) gives, min
j,f j j Which gives us the maximum allowable frame size for given
f t and p for which g (t) will avoid the singularity region and
1 (cid:88) f
Zf = f Zj,f (5) hence be invertible.
j=1
III. GAUSSIANAPPROXIMATIONOFTHEESTIMATOR
Since Y(1) ∈ {0,1} and Y(e) ∈ {0,1} it is easy to see that
j j
Our estimation of the tag population size has to maintain
we have the following pdf for Z ,
j,f
the accuracy requirement given by the condition,
0, with probability p
0 P[|tˆ−t|≤βt]≥α (12)
Z = −1, with probability p (6)
j,f 1
1, with probability p SinceweareusingZ asourestimatortodeterminethevalue
e f
of tˆ, using (1) the condition in (12) can be written as,
LetthemeanandvarianceofZ beµˆ andσˆ2 respectively.
j,f f f
Simple algebraic manipulations give us, P[g {(1−β)t}≤Z ≤g {(1+β)t}]≥α (13)
f f f
E[Z ]=µˆ =p −p (7)
j,f f e 1 Now to perform our estimation of the tag population size
σˆ2 =p +p −(p −p )2 (8) maintaining the accuracy requirements given in (13), we need
f e 1 e 1
the following,
Let, µ and σ2 be the mean and variance of Z respectively.
f f f
Its easy to see µ =µˆ . Using (1), (5), (7) and (8) we have, 1) gf(t) has to be an invertible function.
f f
2) Analytical expression for the distribution of Z (t) .
(cid:18) p(cid:19)t (cid:18)p(cid:19)(cid:18) p(cid:19)t−1 f
g (t)=1− 1− −2t 1− (9) The previous section clearly analyzed the conditions under
f f f f
which g (t) is monotonic function and hence invertible. This
f
1
σ2 = [p +p −(p −p )2] (10) sectionisparticularlydevotedtotheanalysisoftheconditions
f f e 1 e 1 under which Z (t) has a Gaussian distribution. We resort to
f
a triangular array version of Central Limit Theorem [7], i.e.
Lindeberg Feller Theorem [8] for that. k(r)
The statement of Lindeberg Feller Theorem says, when f ≥ . (24)
(cid:15)2
{X } is an independent array of random variables with
E[Xn,i ] = 0 and E[X2 ] = σ2 , Z = (cid:80)n X and This serves as the lower bound on f for a given value of (cid:15).
B2 =n,iVar(Z ) = (cid:80)nn,iσ2 , thne,ni Zn→ N(i0=,1B2)n,iif the If we select a frame size that satisfies (24), the distribution of
n n i=1 n,i n n the estimator will be Z ∼N(µ ,σ2).
condition below holds for every (cid:15)>0, f f f
n A. Quality Considerations of Gaussian Approximation
1 (cid:88)E(cid:2)X2 1 (|X |>(cid:15)B )(cid:3)→0 (14)
B2 n,i A n,i n The quality of the above approximation depends on the
n i=1 value of the approximation error (cid:15). Exactly speaking, satis-
For our algorithm,we modify Z to get a new variable fying (16) means,
j,f
Z(cid:101)j,f, so that the new variable has zero mean and variance
pσˆrfo2b.aIbtiilsitystrdaiisgtrhitbfuotriwona,rd to show that, Z(cid:101)j,f has the following (cid:12)(cid:12)(cid:12)P (cid:20)l≤ Zf −µf ≤u(cid:21)−P [l≤θ ≤u](cid:12)(cid:12)(cid:12)≤(cid:15) (25)
(cid:12) σ (cid:12)
f
−µf, with probability p0 where, θ ∼ N(0,1). It is straightforward to see that, (25)
Z(cid:101)j,f = −1−µf, with probability p1 (15) implies P [l≤θ ≤u] ≥ α + (cid:15). Which means that, If we
1−µf, with probability pe approximate Zfσ−fµf as standard normal, to compensate for
the approximation error we will have to maintain the actual
Nowinlinewith(14)andthestatementofLindebergFeller reliabilityα+(cid:15)insteadofthegivenreliabilityα.Thefactthat
theorem, simple algebraic manipulations using (5), (8), (10) probabilitycannotbegreaterthan1rendersα+(cid:15)≤1,giving
and (15) we get, Zf ∼N(µf,σf2) if the following holds, the following upper bound on (cid:15),
(cid:15) =1−α (26)
f max
1 (cid:88) (cid:104) (cid:16) (cid:112) (cid:17)(cid:105)
fσˆ2 E Z(cid:101)j2,f1A |Z(cid:101)j,f|>(cid:15) fσˆf →0 (16) Equation (26) gives the maximum value of (cid:15) that we can
f i=1 operate on for a given reliability requirement α .
In the above condition given in (16) , the indicator function
(cid:16) √ (cid:17)
1A |Z(cid:101)j,f|>(cid:15) fσˆf plays a pivotal role. For the variable Algorithm 1: Estimate RFID Tag Population (α,β,nr)
Zi,f we have the following 3 cases of the indicator function, Input:
(cid:112) 1) Required reliability α
|1−µ |>(cid:15) fσˆ (17)
f f
(cid:112) 2) Required confidence interval β
|−1−µ |>(cid:15) fσˆ (18)
f f Output: Estimated tag population size tˆ
(cid:112)
|−µf|>(cid:15) fσˆf (19) Calculate tm := upper bound
Calculate persistence probability p using (27)
It is easy to see that if none of (17), (18) and (19) holds, then
obtain f and (cid:15) using (11) and (28) respectively.
(16)holds.IthasbeenprovedanalyticallyinAppendixCthat, max min
obtain (cid:15) and f using (26) and (29) respectively.
if the following condition holds, then none of the (17), (18), max min
make the array, f =[f ,f ]
(19) holds; that is, (16) holds, array min max
calculate, l =length(f )
f array
(cid:15)2f ≥k(r) (20) for i:=1:lf do
Evaluate n for given p and f (i)
i array
where k(r) is defined as k(r) (cid:44) max{k1(r),k2(r),k3(r)}. end
and the values for k1,k2 and k3 are given by the following Obtain fop, and nop such that
equations. (f +l)×n :=min {(f (i)+l)×n }
op op i array i
for j :=1:n do
op
1 Provide the reader with frame size fop, persistence
k1(r)= (cid:12) (cid:12) (21) probability p, and random seed R .
(cid:12)−er(1+4r)(cid:12)−1 j
(cid:12) (1+2r)2 (cid:12) Run Aloha on the jth frame.
k (r)=(cid:12)(cid:12)(cid:12)e2r+(1+2r)(cid:0)14 −er(cid:1)(cid:12)(cid:12)(cid:12) (22) Obtain Zf(j)= Nef−opN1 for the jth frame
2 (cid:12)(cid:12) 14(1+2r)2−rer (cid:12)(cid:12) eZ¯nd← 1 (cid:80)nopZ (j)
k (r)=(cid:12)(cid:12)(cid:12)e2r−2er(1+2r)+(1+2r)2(cid:12)(cid:12)(cid:12) (23) Seft gf(nto)p:=Zj¯f anfd solve (9) to get the estimated value
3 (cid:12)(cid:12) (1+2r)2−er(1+4r) (cid:12)(cid:12) tˆfor tag population size t.
return tˆ
Theargumentnowreducesto,if(20)holds,then(16)holds.
Solving (20) for frame size f, we can write,
3500
tags as low as 200. To be on the safer side and appreciating
α=87%
α=90% the fact that tm is a random quantity, we only estimate the
3000 α=92% tag population sizes greater than 300 under {0,1,e} channel
model. That ensures (cid:15) ≤(cid:15) holds.
min max
2500
C. Frame size
2000
n It is apparent from (20) that, corresponding to (cid:15) given
× max
(f+l) 1500 in (26) we have the minimum frame size fmin, as mentioned
in the Algorithm 1, that will ensure that Z is Gaussian
f
distributed. Using the equation (20) we get,
1000
k(f,p,t)
f =min (29)
500 min f (cid:15)2max
D. Number of rounds n
0
500 1000 1500 2000 2500 3000 3500 4000 For any f in f mentioned in Algorithm 1 g (t) is a
Tag population size t array f
monotonic function and Z ∼N(µ ,σ2) with an approxima-
f f f
Fig.2. Numberofslotsforestimationagainstthetagpopulationsizetfor tion error (cid:15) corresponding to a given f, and we know that
√
different accuracy requirements under {0,1,e} channel model. p=1,β = standard deviation gets scaled down n times if we take n
0.05.
rounds of the measurements, using (13) we have,
(cid:34)
g {(1−β)t}−µ Z −µ
IV. SELECTIONOFCRITICALPARAMETERS P f f ≤ f f
√σf √σf
Thissectionclarifiesthestepstoattaintheoptimumparam- n n
(cid:35)
eters mentioned in Algorithm 1. ≤ gf{(1+β)t}−µf ≥α+(cid:15) (30)
√σf
A. Persistence Probability p n
To decide the persistence probability p, we resort to the It is obvious that the left and right inequalities in (30) will
give us two different values of n, we name them n and
numerical evidences presented in Figure 2. Figure 2 shows left
n respectively. So, the required number of rounds for
that for different accuracy requirements the required number right
of slots for estimation (f+l)×n hits minimum for different given frame size f, can be given by
valuesoft.Sincewewillnotaimatanyreliabilitylowerthan
n=(cid:100)max{n ,n }(cid:101). (31)
left right
87%, the minimum that applies to α=87% curve will surely
encompass the minimum of the higher accuracy requirements V. PERFORMANCEEVALUATION
curves. So, we select t = 1500 as the point where we bring
We used MATLAB to get simulation results for GERT.
down all the higher tag population sizes. Since at the time of
Figures 3 and 4 illustrate the actual reliability of GERT for
estimation we do not know t, we substitute t for t and get
m different reliability requirements under {0,1} and {0,1,e}
the following equation for persistence probability,
channel models respectively. To get each point in Figures 3
(cid:40)1500, for t >1500 and4weran300to600trials.Weseethattheactualreliability
p= tm m (27) of GERT is much greater than the required reliability under
1, otherwise
both channel models. This higher level of quality can be
B. Selection of (cid:15) attributed to two distinct properties of GERT. Firstly, unlike
ART, because of the restrictions imposed by the Gaussian
We have already discussed the maximum value of (cid:15) that
approximation of Z , for a given number of tags t GERT
we can operate on, given by the equation (26). When we are f
always maintains a commensurate frame size which is highly
looking for the minimum value of (cid:15) we are actually looking
unlikely to get saturated. Secondly, we controlled quality
for the value of (cid:15) corresponding to point B in Figure 1 in
by taking all the approximation errors into account when
order to avoid singularity. We know from (11) at point B in
we calculated the overall estimation error. This gives us the
Figure 1, f =f and r =r . Now, using (20) we have,
max min advantage of being able to target a lower reliability than the
(cid:115)
required reliability. For example for {0,1} channel model we
k(r )
(cid:15)min = f min (28) see that, when we have a required reliability of 91%, we can
max actuallyaimat87%.Figure5and6presentthecorresponding
This would only work as long as (cid:15) ≤ (cid:15) . GERT number of slots required for the achieved reliability given in
min max
performance shows that if we aim at 92% reliability, the Figure 3 and 4 respectively. The required number of slots for
achieved reliability is around 97% which is top notch. To aim ART for the reliability requirements 91%, 95% and 97% are
at 92% our (cid:15) has to be less than 0.08. We numerically 1760, 2340 and 2880 respectively. As we see in Figure 5,
min
found that 0.08 is the value of epsilon for the number of except for the very small tag population sizes GERT under
1 3200
α=87% actual reliability ≥ 91%
α=90% 3000 actual reliability ≥ 95%
0.98 α=92% actual reliability ≥ 97%
2800
eved 0.96 2600
achi 2400
eliability 0.94 ×(f+l) n 2200
al r 2000
Actu 0.92 1800
1600
0.9
1400
0.88 1200
200 400 600 800 1000 1200 1400 1600 0 500 1000 1500 2000 2500 3000
Tag population size t Tag population size t
Fig.3. ActualreliabilityachievedbyGERTunder{0,1}channelmodelfor Fig. 5. Number of slots required for estimation by GERT under {0,1}
differenttagpopulationsizesfordifferentaccuracyrequirements.β=0.05 channelmodelagainstthetagpopulationsizet,fordifferentlevelsofachieved
reliability.(β=0.05)
1
α=87% 3500
α=90% actual reliability ≥ 90%
0.98 α=92% actual reliability ≥ 93%
3000 actual reliability ≥ 96%
d
eve 0.96 2500
hi
ac
bility 0.94 n 2000
al relia ×(f+l) 1500
Actu 0.92
1000
0.9
500
0.88
400 600 800 1000 1200 1400 1600
Tag population size t 0 500 1000 1500 2000 2500 3000
Tag population size t
Fig.4. ActualreliabilityachievedbyGERTunder{0,1,e}channelmodel
for different tag population sizes for different accuracy requirements. β = Fig. 6. Number of slots required for estimation by GERT under {0,1,e}
0.05 channelmodelagainstthetagpopulationsizet,fordifferentlevelsofachieved
reliability.(β=0.05)
{0,1} channel model takes 27%, 30% and 28% less number
of slots than ART, to achieve 91%, 95% and 97% levels of any other proposed algorithm so far under the same accuracy
reliability respectively. Comparing between Figures 5 and 6, constraints.
we notice that GERT under {0,1,e} requires fewer number
of slots than GERT under {0,1} to achieve similar level APPENDIXA
of accuracy. This shows the advantage of having more side
information. This is because under {0,1,e} each 1 is certain. Lemma 1. The local minimum of g (t) curve occurs at a tag
f
Because of this added certainty, we need fewer number of population size t = f or equivalently at r = 1, given
LM 2p LM 2
rounds to meet the accuracy requirements under {0,1,e} than frame size f and persistence probability p.
under {0,1} channel model.
Proof:Tofindthelocalminimumweneedtodifferentiate
the expectation curve and set the derivative to 0. Solving that
VI. CONCLUSION
equation we will get the local minimum of the dip. Using (9),
Thekeycontributionofthispaperisthatitproposesacom-
pletelynewandmoreeffectivetechniquefortheestimationof d d (cid:34) (cid:18) p(cid:19)t (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35)
g (t)]= 1− 1− −2t 1− =0
anRFIDtagpopulation.ThemostprominentfeatureofGERT dt f dt f f f
is the quality of estimation coming off rigorous analysis. (32)
All our analytical findings have been well supported by the
simulation results. GERT is much more cost effective than Simple algebraic calculations give us,
APPENDIXC
DERIVATIONOFLINDEBERGFELLERCONDITIONSFOR
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
−2 p − 1− p ln 1− p GERTUNDER{0,1,e}CHANNELMODEL
f f f
t = (33)
LM (cid:16) (cid:17) (cid:16) (cid:17) 1) Firstcondition: Fromequation(17)wehavethefollow-
2 p ln 1− p
f f ing,
(cid:112)
⇒1−µ >(cid:15) fσ
Here, t stands for the t value where the local minimum f j,f
LM (cid:112)
of the dip occurs (i.e. at point D in Figure:1). Now, since the ⇒1−(pe−p1)>(cid:15) f[pe+p1−(pe−p1)2]
(cid:16) (cid:17)
valueof p <<1wecanapproximateln 1− p as−p.This ⇒1−2(p −p )+(p −p )2 >(cid:15)2f[p +p −
f f f e 1 e 1 e 1
will give us the following, tLM ≈ 2fp. Which means the local (pe−p1)2(cid:3)
minimum for the dip occurs at a value of t = tLM which is ⇒1−(cid:0)2+(cid:15)2f(cid:1)pe+(cid:0)2−(cid:15)2f(cid:1)p1
supported by our simulation results. +(cid:0)1+(cid:15)2f(cid:1)(p −p )2 >0 (36)
e 1
As we defined earlier,
letting (cid:15)2f be represented by k, and inserting the expression
tp for p , p and p we have,
r (cid:44) (34) 0 1 e
f (cid:34) (cid:18) p(cid:19)t (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35)
⇒1−(2+k) 1− 1− −t 1− +(2
f f f
Substituting, t in (34) gives us r = 1.
LM LM 2 (cid:34) (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35) (cid:34) (cid:18) p(cid:19)2t
−k) t 1− +(1+k) 1+ 1−
f f f
APPENDIXB
(cid:18)p(cid:19)2(cid:18) p(cid:19)2(t−1) (cid:18) p(cid:19)t (cid:18)p(cid:19)
Lemma 2. gf(t) is a convex function of t for t< 32fp, and for +4t2 f 1− f −2 1− f +4t f
the rest of the t values the function is concave, given frame (cid:18) p(cid:19)2t−1 (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35)
size f and persistence probability p. 1− −4t 1− >0 (37)
f f f
Proof:Tofindaninverseweneedg (t)tobeamonotonic
f We know, for x (cid:28) 1 and y (cid:29) 1, (1 − x)y can be
function of t . To find which part of the g (t) demonstrates
f appriximated as eyln[1−x] and that in turn can be reduced to
monotonic behavior we need the second derivative of g (t)
and check for it’s convexity and concavity characterisftics. ey[−x−12x2] applying Taylor series. Applying this we get,
Again using (9)
(cid:20) (cid:18) (cid:19) (cid:21)
p
⇒1−(2+k) 1−e−t{fp+12(fp)2}−t e−(t−1){fp+12(fp)2}
f
d2 g (t)=(cid:18)1− p(cid:19)t−1ln(cid:18)1− p(cid:19)(cid:20)−2t(cid:18)p(cid:19)ln(cid:18)1− p(cid:19) +(2−k)(cid:20)t(cid:18)p(cid:19)e−(t−1){fp+12(fp)2}(cid:21)+(1+k)
dt2 f f f f f f
(cid:18)p(cid:19) (cid:18) p(cid:19) (cid:18) p(cid:19)(cid:21) (cid:34) (cid:18)p(cid:19)2
−4 − 1− ln 1− 1+e−2t{fp+12(fp)2}+4t2 e−2(t−1){fp+12(fp)2}−
f f f f
(35) (cid:105) (cid:20) (cid:18)p(cid:19)
2e−t{fp+12(fp)2} +(1+k) 4t e−(2t−1){fp+21(fp)2}
f
If we closely follow the equation we see that, the value of
(cid:18) (cid:19) (cid:21)
the common factor (cid:16)1− p(cid:17)t−1ln(cid:16)1− p(cid:17) is negative . So, −4t p e−(t−1){fp+12(fp)2} >0 (38)
f f f
for the total value to be posive the part of the equation inside
Now we have a list of approximations to make. They are,
the third bracket will have to be negative. After algebraic
manipulations and approximating ln(cid:16)1− p(cid:17) as −p we get e−t{fp+12(fp)2} ≈e−tfp (39)
f f
for t< 3f+p, e−(2t−1){fp+12(fp)2} ≈e−2tfp (40)
2p
e−2(t−1){fp+12(fp)2} ≈e−2tfpl (41)
(cid:20) (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19)(cid:21)
p p p p p
2t −4 + 1− <0 e−(t−1){fp+12(fp)2} ≈e−tfp (42)
f f f f f
After all these approximations and using (34) we have,
Or, equivalently, for t < 3f+p ≈ 3f, d2 g (t) is positive, ⇒1−(2+k)(cid:0)1−e−r−re−r(cid:1)+(2−k)re−r
2p 2p dt2 f
indicating gf(t) is a convex function of t and for the rest of +(1+k)(cid:0)1+e−2r+4r2e−2r−2e−r(cid:1)
the t values the curve is concave. Substituting t= 32fp in (34) +(1+k)(cid:0)4re−2r−4re−r(cid:1)>0
gives us r = 3. Our simulation results strongly support that
2 (43)
claim.
For the equation (17) to not hold, (43) must not hold. Simple 3) 3rd condition: From equation (19) we have the follow-
algebraic manipulations give us that for (43) to not hold the ing,
value of k must be, (cid:112)
⇒µ >(cid:15) fσ
f j,f
(cid:112)
1 er ⇒(pe−p1)>(cid:15) f[pe+p1−(pe−p1)2]
k ≥ (cid:12)(cid:12)−er(1+4r)(cid:12)(cid:12)−1 =k1 = er−1 ⇒(pe−p1)2 >(cid:15)2f(cid:2)pe+p1−(pe−p1)2(cid:3)
(cid:12) (1+2r)2 (cid:12) ⇒−(cid:15)2f(p +p )+(cid:0)1+(cid:15)2f(cid:1)(p −p )2 >0
e 1 e 1
So, k1 is the minimum value of k for which (17) does not ⇒−k(cid:34)1−(cid:18)1− p(cid:19)t(cid:35)+(1+k)(cid:34)1+(cid:18)1− p(cid:19)2t+4t2
hold. f f
2) 2nd condition: From equation (18) we have the follow-
(cid:18)p(cid:19)2(cid:18) p(cid:19)2(t−1) (cid:18) p(cid:19)t (cid:18)p(cid:19)
ing, 1− −2 1− +4t
f f f f
⇒1+µ >(cid:15)(cid:112)fσ (cid:18) p(cid:19)2t−1 (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35)
f j,f 1− −4t 1− >0 (47)
(cid:112) f f f
⇒1+(p −p )>(cid:15) f[p +p −(p −p )2]
e 1 e 1 e 1
⇒1+2(p −p )+(p −p )2 >(cid:15)2f(cid:2)P +p −(p −p )2(cid:3) Likewedidintheprevioustwoconditions,(1−x)y canbe
e 1 e 1 e 1 e 1
⇒1+(cid:0)2−(cid:15)2f(cid:1)p −(cid:0)2+(cid:15)2f(cid:1)p +(cid:0)1+(cid:15)2f(cid:1)(p −p )2 >0 appriximated as eyln[1−x] and that in turn can be reduced to
e 1 e 1(44) ey[−x−12x2] applying Taylor series. Applying this along with
approximations made in (39), (40), (41), (42) and using (34)
we have,
letting (cid:15)2f be represented by k, and inserting the expression
for p , p and p we have, ⇒−k(cid:0)1−e−r(cid:1)+(1+k)(cid:0)1+e−2r+4r2e−2r−2e−r(cid:1)
0 1 e
+(1+k)(cid:0)4re−2r−4re−r(cid:1)>0 (48)
(cid:34) (cid:18) p(cid:19)t (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35) Fortheequation(19)tonothold,(48)mustnothold.Simple
⇒1+(2−k) 1− 1− −t 1−
f f f algebraic manipulations give us that for (48) to not hold the
(cid:34) (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35) (cid:34) (cid:18) p(cid:19)2tvalue of k must be,
−(2+k) t f 1− f +(1+k) 1+ 1− f k ≥(cid:12)(cid:12)(cid:12)e2r−2er(1+2r)+(1+2r)2(cid:12)(cid:12)(cid:12)=k
(cid:18)p(cid:19)2(cid:18) p(cid:19)2(t−1) (cid:18) p(cid:19)t (cid:18)p(cid:19) (cid:12)(cid:12) (1+2r)2−er(1+4r) (cid:12)(cid:12) 3
+4t2 1− −2 1− +4t So, k is the minimum value of k for which (19) does not
f f f f 3
(cid:18) p(cid:19)2t−1 (cid:18)p(cid:19)(cid:18) p(cid:19)t−1(cid:35) holFdr.om the above three conditions we see that if we select
1− −4t 1− >0 (45)
f f f the value of k such that k =max{k ,k ,k } all of (17), (18)
1 2 3
and (19) will not hold or equivalently (16) will hold. That es-
Like the first condition (1 − x)y can be appriximated as sentially means for k =max{k1,k2,k3} the GERT estimator
eyln[1−x] and that in turn can be reduced to ey[−x−12x2] ap- Zf under {0,1,e} channel model is Gaussian distributed.
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