1

Proofs of the Results in [1] Gaurav Nigam and Paolo Minero

I. P ROOF

OF

L EMMA 1

By defining gik := |hik |2 for k = 1, 2 and g = (g1 , g2 ), the interference in the k-th transmission P is Ik = i>nk gik kxi k−α . Now, the joint ccdf of SIR1 and SIR2 , as φ(n1 , n2 , θ1 , θ2 ) for JT, is expressed as

P

2 \

{SIRk > θk }

k=1

=P

!

2 n X 2 o \ −α hik > θk Ik D

k=1

i≤nk

!

     θ2 I2 θ1 I1 (a) · exp − = ED,Φ,g exp − n1 D −α n2 D −α   θ D α I1 θ Dα I − 1 − 2 n 2 2 = ED EΦ,g e n1 D=d  Z∞  α θ1 d θ2 dα fD (d) dd, = L , n1 n2

(1)

0

P 2 where (a) follows due to the fact that hi1 and hi2 are mutually independent and i≤nk hik is exponentially distributed with mean nk because of the Rayleigh fading assumption; L(s1 , s2 ) is

the Laplace transform of the interference vector [I1 , I2 ] and fD (d) is the probability distribution function (pdf) of D which is given by [2] 2

fD (d) = 2(λπ)2 d3 e−λπd , d ≥ 0.

(2)

Manuscript date March 11, 2015. Gaurav Nigam {[email protected]} and Paolo Minero {[email protected]} are with the Department of Electrical Engineering, University of Notre Dame, IN 46556, USA. This work was supported by the U.S. NSF grants CCF 1117728.

2

Given D = d, the Laplace transform of the interference vector, L(s1 , s2 ), can be expressed as   EΦ,g e−s1 I1 −s2 I2 | D = d   3 3 −(s1

= EΦ,g e

P

n1 +1

P

gi1 +s2

i=n2 +1

gi2 )d−α −

P

(s1 gj1 +s2 gj2 )kxj k−α

j>3

hQ i 1 E j>3 (1+s1 kxj k−α )(1+s2 kxj k−α ) (a) Φ = (1 + s1 d−α )3−n1 (1 + s2 d−α )3−n2 Z ∞  −2πλ

(b) e

=

1−

d

1

(1+s1 r −α )(1+s2 r −α )



rdr

(1 + s1 d−α )3−n1 (1 + s2 d−α )3−n2 2 −α , s2 d−α )) (c) exp (−2πλd G (s1 d = , (1 + s1 d−α )3−n1 (1 + s2 d−α )3−n2

(3)

where (a) uses the fact that gi1 and gi2 are mutually independent and exponentially distributed with unit mean; (b) is due to the probability generating functional for a PPP [3, Theorem 4.9]; (c) follows from the transformation r = td and the definition of G(x, y). Substituting (2) and (3) in (1), then using the transformation u = λπd2 and integrating over u gives the result in Lemma 1 for JT. II. P ROOF

OF

T HEOREM 1

We first prove the result for JT without MRC. The coverage probability for JT without MRC can be re-written as P = P(SIR1 > θ) + P(SIR2 > θ, SIR1 < θ) (a)

=

2 X k=1

P(SIRk > θ) − P(SIR2 > θ, SIR1 > θ)

= P ({SIR1 > θ} ∪ {SIR2 > θ}) ,

(4)

where (a) follows from basic set theory. Now, we can apply the inclusion-exclusion formula with P(SIRk > θ) in [1, Remark 1] and P(SIR1 > θ, SIR2 > θ) as in [1, Lemma 1] to derive the result. Next, we prove the result for JT with MRC. As in the proof in Appendix I, we assume n1 ≤ n2 and define γi , gi , g and Ii . The coverage probability for JT with MRC can be expressed as PMRC = P(SIR1 > θ) + P(SIR1 + SIR2 > θ, SIR1 < θ)

3 (a)

= P(SIR1 + SIR2 > θ)   = P θ − SIR2 < SIR1 " = EΦ,Z,g

!# X 2 hi1 Φ, Z , P (θ − Z)I1 < D −α i≤n1

where (a) follows from set theory and the fact that SIRk > 0 and we define the random variable P Z = SIR2 for simplicity. Using the fact that | i≤n1 hi1 |2 is exponentially distributed with mean n1 because of the Rayleigh fading assumption, the coverage probability can be expressed as    (θ − Z)+ I1 EΦ,Z,g1 exp − n1 D −α    3 P P (θ−Z)+  gi1 kxi k−α  gi1 D −α + i>3 i=n1 +1 − n1 D −α

 = EΦ,Z,g1  e 

 = EΦ,Z   1+ "

1  + 3−n1

(θ−Z) n1

= EΦ,Z (f1 (D))3−n1

where we define f1 (x) :=

Y i>3

1 . α 1+ D (θ−Z)+ x−α n

Y i>3

1+

f1 (kxi k)

Dα (θ n1

#

  



1   − Z)+ kxi k−α (5)

Next, we derive the conditional probability distribution

1

function (pdf) of Z as follows   Pn2 −α/2 2 hi2   i=1 D ≤ z Ξ P(Z ≤ z | Ξ) = P  I2 "

= 1 − Eg2 exp − =1−  1+

1 z n2

3−n2

= 1 − (f2 (D))3−n2

where we define f2 (x) :=

z

1

α

P3

i=n2 +1

Y i>3

Y i>3

1+ D zx−α n

gi2 D −α + z n2 D −α 1

1+

Dα zkxi k−α n2

P

i>3 gi2 kxi k

−α

!#

,

f2 (kxi k),

. Differentiating the above equation with respect to z, we

2

get the pdf of Z as fZ|Ξ(z) = (f2 (D))3−n2

"

3 − n2 Y f2 (kxi k)+ z + n2 i>3

4

XD j>3

α

n2

kxj k−α (f2 (kxj k))2

Y

k>3 k6=j



 f2 (kxk k) .

(6)

Now we can compute the expectation over Z in (5) using (6) as " # Z ∞ Y (f1 (D))3−n1 EΞ f1 (kxi k) × fZ|Ξ(z) dz 0

i>3

" 3 − n2 Y (f1 (D))3−n1 f1 (kxi k)f2 (kxi k) E E = D Ξ|D (f2 (D))n2 −3 z + n2 i>3 0 X Dα kxj k−α f1 (kxj k)(f2 (kxj k))2 × + n2 j>3  Y  f1 (kxk k)f2 (kxk k) dz Z

∞

k>3 k6=j

 (f1 (D))3−n1 3 − n2 −2πλ R ∞ (1−f1 (r)f2 (r))rdr D = E e D n2 −3 (f (D)) z + n 2 2 0 Z Dα ∞ + 2πλr −α f1 (r)(f2 (r))2 n2 D # # " Y f1 (r)f2 (r) rdr dz × EΞ

(a)

Z

∞

k>3

    (θ − z)+ z (f1 (D))3−n1 2 ED exp −2πλD G , = (f2 (D))n2 −3 n1 n2 0    2 + (θ − z) z 3 − n2 2πλD dz + H , × z + n2 n2 n1 n2    Z ∞ Z ∞ exp −2tG (θ−z)+ , z n1 n2 (c) = dz   3−n2  3−n 1 (θ−z)+ z 0 0 1 + n2 1 + n1    3 − n2 2t (θ − z)+ z × te−t dt + H , z + n2 n2 n1 n2 −2   (θ−z)+ z Z ∞ 1 + 2G , n1 n2 (d) =   3−n2 ×  3−n 1 + z 0 1 + 1 + (θ−z) n1 n2     z (θ−z)+ , 4H n2 n1  3 − n2 +    dz  z + n2 n 1 + 2G z , (θ−z)+ (b)

Z

∞

2

n2

(7)

n1

where (a) is due to the probability generating functional for a PPP [3, Theorem 4.9] and the

5

Campbell-Mecke formula [3]; (b) is due to the probability generating functional for a PPP, the transformation r = tD and the definitions of G(x, y) and H(x, y) in [1] ; (c) uses the pdf of D in (2) and then uses the transformation t = λπd2 ; (d) follows by integrating over t. Hence, we get the desired result in Theorem 1 for JT. III. P ROOF

OF

L EMMA 2

We first prove the finiteness result and then prove the result for the case j1 = j2 = 1. Using Cauchy Schwartz’s inequality, we have " 2 # v u 2 Y uY α jk (Ik D ) ≤ t ED,Ik [(Ik D α )2jk ]. ED,Ik

(8)

k=1

k=1

Given D = d, we can derive the Laplace transform of Ik , LIk (s) using the result in (3) and it can be written as

=

exp (−2πλd2 G (sd−α , 0)) (1 + sd−α )3−nk R∞ R∞
α α exp −2πλd2 1 rdr 0 e−yr d /s e−y dy

(1 + sd−α )3−nk  R
 2πλs2/α ∞ −2/α −y 2 ydα exp − , dy y e Γ α α s 0 (a) = (1 + sd−α )3−nk    Z 2 2πλsd2−α ∞ −2/α −sd−α t (b) t e Γ , t dt− = exp − α α 0

(3 − nk ) log 1 + sd−α = exp(−f (s)).

(9)

where Γ(s, x) is the upper incomplete gamma function. (a) follows by using the transformation u = yr αdα /s, and integrating over u; (b) follows from the transformation y = std−α . Therefore, we can write 2jk  2jk  2jk ∂ L (s) EIk Ik | D = d =(−1) I ∂s2jk k s=0 (a)

=B2jk −f (1) (0), . . . , −f (2jk ) (0)

where f (n) (0) =



∂n f (x) x=0 ∂xn




(10)

and Bn (y1 , . . . , yn ) is the nth complete Bell polynomial. (a) is

due to Faá di Bruno’s formula. We can derive f (n) (0) as f (n) (0) = f (n) (s) s=0

6

  Z 2πλd2−α ∞ t−2/α (−td−α )n−1 (n − std−α ) 2 = Γ , t dt α estd−α α 0 (−1)n−1 (n − 1)!(3 − nk )d−nα + (1 + sd−α )n s=0   2 2πλd n (b) n−1 −nα + 3 − nk =(−1) (n − 1)!d αn − 2

(a)

where (a) follows by computing the nth derivative of f (s) and (b) follows from

(11) R∞ 0

xn−s−1 Γ(s, x)dx =

(n − 1)!/(n − s). Substituting (10) into (8), we can write the expected value inside the square root as Z∞ 0

B2jk (−f (1) (0), . . . , −f (2jk ) (0))d2jk α fD (d)dd

Z∞

   2πλd2 −α = B2jk −d + 3 − nk , . . . α−2 0   4πλd2 jk −2jk α , (2jk − 1)!d + 3 − nk d2jk α fD (d)dd 2jk α − 2  Z∞ 2πλd2 (a) + 3 − nk , . . . = B2jk α−2 0   4πλd2 jk 2 , (2jk − 1)! + 3 − nk 2(λπ)2d3 e−λπd dd 2jk α − 2 ∞  Z 2t (b) = B2jk + 3 − nk , . . . α−2 0   4tjk + 3 − nk te−t dt , (2jk − 1)! 2jk α − 2 Z∞ 2jk X (c) −t = te ai ti dt 0

=

2jk X

i=0

(i + 1)! ai

(12)

i=0

where (a) follows from the definition of fD (d) in (2) and Lemma 1 which is stated after this paragraph; (b) follows from the transformation t = λπd2 and (c) follows by expanding the bell polynomial. Here, ai s are finite. Notice that it is not possible to get a closed-form expression for
Q2 α jk (I D ) is the coefficients ai but their finiteness is enough for this proof. Hence, ED,Ik k k=1 finite.

7

Lemma 1: The n-th order complete Bell polynomial satisfies Bn (ay1 , a2 y2 , . . . , an yn ) = an Bn (y1 , y2, . . . , yn )

(13)

for some finite a and yi ’s. Proof: We can express the left side of the equation in Lemma 1 as n X Bn,l (ay1 , a2 y2 , . . . , an−l+1 yn−l+1) l=1

=

n X l=1

(a) n

=a

j ,...,jn−l+1 : P1n−l+1 ji =l, i=1 Pn−l+1 i ji =n i=1

n X l=1

=an

X

n X

X

n−l+1 Y  ai yi ji n! j1 !j2 ! . . . jn−l+1 ! i=1 i!

j ,...,jn−l+1 : P1n−l+1 ji =l, i=1 Pn−l+1 i ji =n i=1

n−l+1 Y  yi ji n! j1 !j2 ! . . . jn−l+1 ! i=1 i!

Bn,l (y1 , y2 , . . . , yn−l+1 )

l=1

=an Bn (y1, y2 , . . . , yn ), P i ji = n. where (a) follows due to the fact that n−l+1 i=1

(14)

To prove the second result in Lemma 2, E[I1 I2 D 2α ] can be written as Z ∞ E[I1 I2 | D = d] d2α fD (d)dd 0 Z ∞ ∂2 = d2α fD (d)dd L(s1 , s2 ) ∂s1 ∂s2 0 s1 =s2 =0    Z ∞ 2λπd2 2λπd2 λπd2 + + 3 − n1 + 3 − n2 = α−1 α−2 α−2 0 2

× 2(λπ)2 d3 e−λπd dd    Z ∞ 2t 2t t (a) + + 3 − n1 + 3 − n2 = α−1 α−2 α−2 0 × te−t dt

4(6 − n1 − n2 ) 2 24 + + , (15) α−2 α − 1 (α − 2)2 where L(s1 , s2 ) and fD (d) are given by (3) and (2), respectively. (a) follows due to the trans=(3 − n1 )(3 − n2 ) +

formation t = λπd2 .

8

IV. P ROOF

OF

T HEOREM 3

We first prove the result for JT. First, we write the outage events for retransmission with and without MRC in form of the events considered in [4, Lemma 2] with N = 2. Then, we use the result in [1, Lemma 2] to derive the power gain. Focusing on retransmission without MRC, we can write the outage event as 2 ( n ) 2 2 X k \ \ {SIRk < θ} = kxj k−α/2 hjk < θIk j=1 k=1 k=1 ) ( Pn 2 k D −α/2 hjk 2 2θIk (a) \ j=1 < , = −α /2 −α n n kD kD k=1 | {z } | {z }

(16)

θJk

Yk

where (a) follows due to the assumptions nk ≤ 3 and kxj k = D for j ≤ 3. Here, Yk is exponentially distributed with mean 2 or, equivalently, chi-squared distributed with m = 2 P k D −α/2 hjk |2 is exponentially distributed degrees of freedom due to the fact that given D, | nj=1 with mean nk D −α . Y1 and Y2 are independent since the fading coefficients hjk are mutually

independent. In case of retransmission with MRC, the coverage probability for JT with MRC can be further expressed as PMRC = P(SIR1 > θ) + P(SIR1 + SIR2 > θ, SIR1 < θ) (a)

= P(SIR1 + SIR2 > θ)

(17)

where (a) follows from set theory and the fact that SIRk > 0. Hence, the outage event can be written as {SIR1 + SIR2 < θ} ≡



Y1 Y1 + <θ J1 J1



(18)

where Yk and Jk are defined as for the case of retransmission without MRC above. Now, we need to prove that EJ1 ,J2 [J1 J2 ] is finite to apply the result in [4, Lemma 2]. We have " 2 # Y Ik EJ1 ,J2 [J1 J2 ] = 4 ED,Ik nk D −α k=1

(19)

9

It can be proved that the above expression is finite using the result in [1, Lemma 2]. Applying [4, Lemma 2] with N = 2, we have P

2 \

k=1

P



Yk <θ Jk

2 X Yk k=1

Jk

!

∼ θ2

  E J1 J2

!

∼ θ2

8

<θ

4   E J1 J2

.

(20)

Hence, we get a diversity gain of 2 for retransmission with and without MRC. Using (19) and [1, Lemma 2], we can derive the power gain using the definition of gn1 ,n2 . Next we prove the results for BSS. Similar to the proof for JT, we can derive the outage events for retransmission without MRC and with MRC in the form of (16) and (18), respectively with Yk = 2|h1k |2 which is chi-squared distributed with 2 degrees of freedom and Jk = 2Ik /D −α . Again, EJ1 ,J2 [J1 J2 ] is finite due to [1, Lemma 2]. Hence, we get a diversity gain of 2 for retransmission with and without MRC by applying [4, Lemma 2]. The power gain can be derived using the result in [1, Lemma 2] and the definition of gn1 ,n2 . V. P ROOF

OF

T HEOREM 4

Given the point processes, the sum in the numerator of the SIR expression with CSI is a sum of Rayleigh distributed random variables. We henceforth denote these random variables as Sik = |hik |. For retransmission without MRC, we can express the outage probability as    !2 nk 2   \ X −α  P D Sik < θIk    i=1 k=1   Z nk 2 Y  Y  gSik (xik )dxk  = ED,Ik  , k=1

An ,√θI k

kD

α (xk )

i=1

2

where gSik (x) = 2xe−x , x ≥ 0 is the probability density function of Sik and we define An,r (x) =

{x ∈ (R+ )n : kxk1 < r}. The outage probability can be further expressed by substituting √ xik = θIk D α tik as:   n k Z P n 2 k t2ik Y −θIk D α Y  i=1 ED,Ik  (2θIk D α )nk tik dtk  e k=1

Ank ,1 (tk )

i=1

10



(a)

 ∼ θn1 +n2 ED,Ik 

2 Y

(2Ik D α )nk

k=1

= θn1 +n2 ED,Ik

i=1

Ank ,1 (tk )

2 Y (2Ik D α )nk

(b)

nk Y

Z

(2nk )!

k=1

!



 tik dtk 

,

(21)

where (a) is due to the fact that exp(x) ∼ 1 as x → 0 and the fact that ED,Ik

Q2

k=1 (Ik D

α nk

)




is finite due to [1, Lemma 2]; (b) is due to [5, Equation 4.634]. Hence, we get a diversity gain of n1 + n2 for retransmission without MRC. (

For retransmission with MRC, we define Wk =

Pnk

Sik ) Ik D α

2

i=1

for k = 1, 2 and use the SIR

expression with CSI to express the outage probability as ! 2 X P Wk < θ k=1

= ED,Ik

Z

∞

P (W1 < θ − w | D, I1) fW2 |D,I2 (w)dw

0



(22)

Following similar steps as in (21), we can derive the cdf of Wk given D and Ik , FWk |D,Ik (x) for x ≥ 0 and k = 1, 2 as follows. P (Wk < x | D, Ik ) Z α nk = (2xIk D )

n k P

−xIk D α

e

i=1

t2ik

∂ (x). F ∂x Wk |D,Ik

tik dtk

(23)

i=1

Ank ,1 (tk )

and fWk |D,Ik (x) =

nk Y

Substituting these values in (22), the outage probability is

expressed as ED,Ik

Z

∞

0

(a)

= ED,Ik

Z

1

0



= ED,Ik 

Z1 0

Z

FW1 |D,I1 (θ − w)

+




fW2 |D,I2 (w)dw

 FW1 |D,I1 (θ(1 − t)) fW2 |D,I2 (θt)θdt

(2θ(1 − t)I1 D α )n1 (2θI2 D α )n2 tn2 −1 × Z

−θ(1−t)I1 D α

e

n1 P

i=1

t2i1

e

n2 P

i=1

t2i2

n1 Y i=1

An2 ,1 (t2 ) An1 ,1 (t1 ) −θtI2 D α



n2 − θtI2 D α

n2 X i=1

t2i2

ti1 dt1 ×

!

n2 Y i=1

#

ti2 dt2 dt

11

(b) n1 +n2

∼θ

ED,Ik

# 2 Y (2Ik D α )nk Q k 2 n2 B(n1 + 1, n2 ), li (2nk )! ni=1 k=1

"

(24)

where (a) is due to the transformation w = θt and the fact that FW1 |I1 (0) = 0; (b) is due to
Q2 α nk is finite due to [1, the fact that exp(x) ∼ 1 as x → 0 and the fact that ED,Ik k=1 (Ik D )

Lemma 2]. Then we use [5, Equation 4.634] for nk dimensional integrals and the definition of

the beta function for the integral in t. Hence, we get a diversity gain of n1 +n2 for retransmission with MRC. VI. P ROOF

OF

T HEOREM 5

By defining gi = |hi |2 with n1 = n2 = 2, we can express the coverage probability for Alamouti code as Pc = P (g1 + g2 > θD α I)  α  (a) = ED,I (1 + θD α I)e−θD I   ∂ (b) , = ED L(s, 0) − s L(s, 0) ∂s s=θD α

(25)

where (a) follows from the fact that g1 and g2 are exponentially distributed with unit mean and the cdf of the gamma distribution; (b) follows by the definition of the Laplace transform in (3). Using the value of L(., .) in (3) and the pdf of D in (2), the above coverage probability can be further expressed as Z∞ 0

Z∞

  exp (−2λπd2 G(θ, 0)) θ 2 1 + 2λπd θH(θ, 0) + 1+θ 1+θ × fD (d)dd

  exp (−2tG(θ, 0)) θ = 1 + 2tθH(θ, 0) + te−t dt 1+θ 1+θ 0   −2 1 + 2θ 4θH(θ, 0) (b) (1 + 2G(θ, 0)) = , + 1+θ 1+θ 1 + 2G(θ, 0)

(a)

(26)

where (a) follows from the change of variable t = λπd2 and (b) follows by integrating over t.

12

VII. P ROOF

OF

T HEOREM 6

As θ → 0, the coverage probability for Alamouti code can be written as (1 + 2G(θ, 0))−2

4θH(θ, 0) 1 + 2θ + (1 + 2G(θ, 0))−3 2 (1 + θ) 1+θ

(a)

∼(1 − 4G(θ, 0) + 12(G(θ, 0)2 ))(1 + 2θ)(1 − 2θ + 3θ2 ) + 4θH(θ, 0)(1 − θ + θ2 )(1 − 6G(θ, 0) + 6(G(θ, 0))2)

4θ 2θ2 12θ2 ∼1 − θ2 − + + α − 2 α − 1 (α − 2)2   θ 6θ θ 1 − − − + 4θ α − 2 α − 1 (α − 2)2 α − 2   2 4 12 2 ⇒1 − Pc ∼ θ 1 + + + , α − 1 α − 2 (α − 2)2 (b)

(27)

where (a) is due to the fact that (1 + x)−n ∼ 1 − nx + n(n + 1)x2 /2 as x → 0; (b) follows from the fact that G(x, 0) ∼

x α−2

−

x2 2(α−1)

and H(x, 0) ∼

1 α−2

−

x α−1

as x → 0 and by ignoring the

terms with higher powers of θ. Hence, we get the diversity gain of 2. R EFERENCES [1] G. Nigam and P. Minero, “Spatiotemporal cooperation for the worst-case user in single-tier cellular networks,” IEEE GLOBECOM Communication Theory Symposium, December 2015, submitted. [2] V. Baumstark and G. Last, “Some distributional results for Poisson-Voronoi tessellations,” in Advances in Applied Probability. Applied Probability Trust, March 2007, vol. 39, no. 1, pp. 16–40. [3] M. Haenggi, Stochastic Geometry for Wireless Networks.

Cambridge University Press, 2013.

[4] G. Nigam, P. Minero, and M. Haenggi, “Spatiotemporal cooperation in heterogeneous cellular networks,” IEEE Journal on Selected Areas in Communications: Recent Advances in Heterogeneous Cellular Networks, 2015, accepted. [5] I. Gradshteyn and I. Ryzhik, Table of Integrals, Series, and Products, 7th ed.

Academic Press, 2007.