Now we calculate P{=ry=y}i.e.,the probability of ing expansion of the Taylor series of g(Y,R): having r runs of b in a frame of size f given that y out of f slots are b.As tags choose the slots independently,each g6,)=ga,)+[6-0)器+(-)2】 0g occurrence with r runs having y slots of b is equally likely. Therefore,we determine the total number of ways,denoted +2(Y%-0A)(B-02) 82g by gff,y,r},in which y occurrences of b and f-y occur- OYioR rences of b can be arranged such that the number of runs of b is r.We treat this as an ordered partition problem.First, +(B-2)22g】 aR +0G-1) we separate all the y occurrences of b from the frame and make r partitions of these y occurrences.Then,we create appropriate number of partitions of f-y occurrences of 6 Taking the expectation of both sides,we get such that between consecutive partitions of b,the partitions 8g of b can be interleaved.For r partitions of b,there are 4 E[g(Ya,R6】≈ 21 Var(Y) possible partitions of b. +2Co06,)8品 8g 1.The frame starts with b and ends with b,implying that +Var(Ro) +g(01,02)】 (16) there arer-1 partitions of b,each interleaved between adjacent partitions of b. Evaluating the partial derivatives of g as required in Equa- tion (16),we get 2.The frame starts with b and ends with b,implying that there are r partitions of b. 8g(Yu,R) =0 ayp 3.The frame starts with b and ends with b,implying that there are r partitions of b. 82g(Yi,R) 1 8Y%8R6 ￥==一 Rb=82 4.The frame starts with b and ends with b,implying that 82g(Yi,Ru) there are r+1 partitions of b. a We can make r partitions of y occurrences of b in ( ways and r partitions of f-y occurrences of in ( Putting these values in Equation (16)and using 01=E[Y] and 02 E[R],we get Equation (14). ways.Similarly,we can make r+1 partitions of f-y occur- The variance can be calculated as follows: rences of B in (1)ways and r-1 partitions ofof f-y occurrences of in (ways.The equation ofy Var(g(6,R)=E[{g(%,R)-E[g(,)J}](17) in the lemma statement follows from this discussion.The to- tal number of ways in which y zeros can be arranged among Considering that E[g(Y,R)]is being squared in the expres- f slots is ()Thus,we get sion above,we use first order Taylor series expansion to get the value of E[g(Yo,R)]and substitute it in Equation (17). P{=r%==U (13) ( ,=E[s-器+风-器】 +g(01,02)+0G1) Substituting values from Equations (12)and (13)in (11) and (10)results in Equation (9). =[o鼎+o]+9a,+oU=9e.) THEOREM 2.Given tag population size t,frame size f, Substituting the value of Elg(Y,Ro)]and using the first and persistence probability p,we have: order Taylor series expansion of g(Yo,R)in (17),we get E[X]= EY Cov(Y.R)E[Yo]Var(R) ERu] ES Ru (14) E2R Var(a(6A=Ec-8器+(-器月 +00-1) dg og Var(X)= Var(Y) E2YolVar(R) 2E[Yo]Cov(Y R)+ER] ≈Var(Ya)( E2R]E3[Ru] +2Cov(Yo.Rw)aYR +Var(Ro)( 0g2 (15) (18) PROOF.Let g(R)==The Taylor series ex- Evaluating the partial derivatives of g as required in the pansion of g around (01,02)is given by: equation above,we get 9,-6-品+（风-品] 81 Og(Yi,R) 1 ∂Y6 Y6=81 h=2 ÷02 1=0 Og(Y,R) 01 g6,R)}￥=a: aR ==一房 R6=2 R=a2 According to Bienayme-Chebyshev inequality,we have Putting these values in Equation(18)and using 01=E[Y] 01 E[Yo]and 02 E[R].Therefore,we get the follow- and 02=E[R],we get Equation(15). 370Now we calculate P {Rb = r|Yb = y} i.e., the probability of having r runs of b in a frame of size f given that y out of f slots are b. As tags choose the slots independently, each occurrence with r runs having y slots of b is equally likely. Therefore, we determine the total number of ways, denoted by ξ {f, y, r}, in which y occurrences of b and f − y occur￾rences of b can be arranged such that the number of runs of b is r. We treat this as an ordered partition problem. First, we separate all the y occurrences of b from the frame and make r partitions of these y occurrences. Then, we create appropriate number of partitions of f − y occurrences of b such that between consecutive partitions of b, the partitions of b can be interleaved. For r partitions of b, there are 4 possible partitions of b. 1. The frame starts with b and ends with b, implying that there are r−1 partitions of b, each interleaved between adjacent partitions of b. 2. The frame starts with b and ends with b, implying that there are r partitions of b. 3. The frame starts with b and ends with b, implying that there are r partitions of b. 4. The frame starts with b and ends with b, implying that there are r + 1 partitions of b. We can make r partitions of y occurrences of b in y−1 r−1  ways and r partitions of f − y occurrences of b in f−y−1 r−1  ways. Similarly, we can make r + 1 partitions of f −y occur￾rences of b in f−y−1 r  ways and r − 1 partitions of of f − y occurrences of b in f−y−1 r−2  ways. The equation of ξ {f, y, r} in the lemma statement follows from this discussion. The to￾tal number of ways in which y zeros can be arranged among f slots is f y  . Thus, we get P {Rb = r|Yb = y} = ξ {f, y, r} f y  (13) Substituting values from Equations (12) and (13) in (11) and (10) results in Equation (9). Theorem 2. Given tag population size t, frame size f, and persistence probability p, we have: E[Xb] = E[Yb] E[Rb] − Cov(Yb, Rb) E2[Rb] + E[Yb] E3[Rb] Var(Rb) (14) Var(Xb) = Var(Yb) E2[Rb] − 2E[Yb] E3[Rb] Cov(Yb, Rb) + E2[Yb] E4[Rb] Var(Rb) (15) Proof. Let g(Yb, Rb) = Xb = Yb Rb . The Taylor series ex￾pansion of g around (θ1, θ2) is given by: g(Yb, Rb) = ∞ j=0  1 j! (Yb − θ1) ∂ ∂Y b + (Rb − θ2) ∂ ∂R b j × g(Y b , R b)  Y b =a1 R b=a2 According to Bienaym´e-Chebyshev inequality, we have θ1 = E[Yb] and θ2 = E[Rb]. Therefore, we get the follow￾ing expansion of the Taylor series of g(Yb, Rb): g(Yb, Rb) = g(θ1, θ2) + (Yb − θ1) ∂g ∂Yb + (Rb − θ2) ∂g ∂Rb  + 1 2 (Yb − θ1) 2 ∂2g ∂Y 2 b + 2(Yb − θ1)(Rb − θ2) ∂2g ∂Yb∂Rb + (Rb − θ2) 2 ∂2g ∂R2 b  + O(j −1 ) Taking the expectation of both sides, we get E[g(Yb, Rb)] ≈ 1 2 Var(Yb) ∂2g ∂Y 2 b + 2Cov(Yb, Rb) ∂2g ∂Yb∂Rb +Var(Rb) ∂2g ∂R2 b  + g(θ1, θ2) (16) Evaluating the partial derivatives of g as required in Equa￾tion (16), we get ∂2g(Yb, Rb) ∂Y 2 b   Yb=θ1 Rb=θ2 = 0 ∂2g(Yb, Rb) ∂Yb∂Rb   Yb=θ1 Rb=θ2 = − 1 θ2 2 ∂2g(Yb, Rb) ∂R2 b   Yb=θ1 Rb=θ2 = 2 θ1 θ3 1 Putting these values in Equation (16) and using θ1 = E[Yb] and θ2 = E[Rb], we get Equation (14). The variance can be calculated as follows: Var g(Yb, Rb)  = Eg(Yb, Rb) − E[g(Yb, Rb)]2 (17) Considering that E[g(Yb, Rb)] is being squared in the expres￾sion above, we use first order Taylor series expansion to get the value of E[g(Yb, Rb)] and substitute it in Equation (17). E[g(Yb, Rb)] = E (Yb − θ1) ∂g ∂Yb + (Rb − θ2) ∂g ∂Rb  +g(θ1, θ2) + O(j −1 ) = (0) ∂g ∂Yb + (0) ∂g ∂Rb  + g(θ1, θ2) + O(j −1 ) ≈ g(θ1, θ2) Substituting the value of E[g(Yb, Rb)] and using the first order Taylor series expansion of g(Yb, Rb) in (17), we get Var g(Yb, Rb)  = E(Yb − θ1) ∂g ∂Yb + (Rb − θ2) ∂g ∂Rb 2 +O(j −1 ) ≈ Var(Yb)( ∂g ∂Yb ) 2 + 2Cov(Yb, Rb) ∂g ∂Yb ∂g ∂Rb + Var(Rb)( ∂g ∂Rb ) 2 (18) Evaluating the partial derivatives of g as required in the equation above, we get ∂g(Yb, Rb) ∂Yb   Yb=θ1 Rb=θ2 = 1 θ2 ∂g(Yb, Rb) ∂Rb   Yb=θ1 Rb=θ2 = −θ1 θ2 2 Putting these values in Equation (18) and using θ1 = E[Yb] and θ2 = E[Rb], we get Equation (15). 370