Security capability analysis of cognitive radio network with secondary user capable of jamming and self-powering

n!
 I (84)
  0 ∞ 0 
 Z n Z −Msex
 J−Mse  X J e 
 Kse−V −Msex
 = e e dx + 2n n dx ,
  (Kse − V ) n! x 
 I n=1 I
where I is given in (48).
 The first integral in (84) is easily solved while the second one is solved by invoking [44, eq.
(2.324.2)]. Inserting closed forms of these two integrals into (84), one reduces (84) to (50),
completing the proof of ∆ for 1 ≤ Kse < V .
Scenario 3. V < 1
 This scenario simplifies (71) as
 V  Uy  K  K 
 Z Z Z se Z sr
 ∆ =  fXsr (x) dx fXse (y) dy +  fXsr (x) dx fXse (y) dy . (85)
 y=U −1 x=1 y=1 x=1
 | {z } | {z }
 Case 2: U −1≤yV
 Since fXse (y) = 0 when y = V < 1, the first term in (85) is zero. Additionally, fXsr (x)
is non-zero for x ∈ [1,Ksr) and fXse (y) is non-zero for y ∈ [1,Kse) and thus, the second
term in (85) is one. Plugging these results into (85), one infers ∆ = 1, which matches (45)
for V < 1. This finishes the proof of ∆ for V < 1.
 By combining the three above scenarios, it is seen that ∆ is precisely solved as (45). This
completes the proof of Theorem 1.
 APPENDIX B: PROOF OF Ψ2
 Bh¯ se
 Let Xsd = 1 + Ah¯ sd and X¯se = 1 + where A¯, B¯, and C¯ are given in (60), (61),
 Ch¯ se+κe
and (62), correspondingly. Relied on the variable change, the PDF of Xsd is expressed as
 x−1
 1 − ¯
 ϑsdA
 fXsd (x) = e , x ≥ 1. (86)
 ϑsdA¯
 Also, by invoking Lemma 1, the PDF of X¯se can be obtained as
 ¯ y−1
 E ¯
 ¯ e y−D ¯
 fX¯ (y) = G , 1 ≤ y < D, (87)
 se y − D¯
2
where D¯, E¯, and G¯ are correspondingly given in (63), (64), and (65).
 Invoking (11) and (13) for the case of Γs < Γt, one rewrites Ψ2 in (56) as
    
 µPshsd µPshse 
 Ψ2 = Pr 1 + < U 1 +  Γs < Γt , (88)
 κd (1 − µ) Pshse + κe 
 SECURITY CAPABILITY ANALYSIS OF COGNITIVE RADIO NETWORK 227
which is also represented in terms of Xsd and X¯se as
  ¯  
 Ψ2 = Pr Xsd < UXse Γs < Γt . (89)
 Since Xsd and X¯se are statistically independent, (89) can be rewritten in terms of the
individual PDFs of Xsd and X¯se as
 ZZ
 Ψ2 = fXsd (x) fX¯se (y) dxdy. (90)
 x<Uy
 Because fXsd (x) is non-zero for x ≥ 1, two cases on Uy in correlation to 1 are considered
as
   D¯  Uy 
 Z Z Z Z
 Ψ =  f (x) dx f ¯ (y) dy + f (x) dx f ¯ (y) dy . (91)
 2  Xsd  Xse  Xsd  Xse
 y<U −1 x<Uy y=1 x=1
 | {z } | {z }
 Case 1: Uy1
 R
 The first term in (91) is zero since fXsd (x) dx = 0 due to fXsd (x) = 0 for x = Uy <
 x<Uy
1. Accordingly, by substituting (86) into the second term in (91) and after solving the inner
integral in (91), one has
 D¯ D¯ D¯
 Z  Uy−1  Z 1 Z
 − ¯ ¯ ¯
 ϑsdA ϑsdA −Qy
 Ψ2 = 1 − e fX¯se (y) dy = fX¯se (y) −e e fX¯se (y) dy, (92)
 1 1 1
 | {z } | {z }
 1 H
where Q¯ is given in (66).
 The first integral in the last equality of (92) is one due to the property of the PDF while
the second integral can be rewritten after invoking fX¯se (y) in (87) as
 D¯ ¯ y−1
 E ¯
 Z ¯ e y−D
 H = e−QyG¯ dy
 y − D¯
2
 1
 −∞
 1
 x= Z ¯
 y−D¯ E¯−Q¯D¯ Gx¯ − Q
 = −Ge¯ e x dx (93)
 1/(1−D¯)
 ∞
 Z ¯
 y=−x ¯ ¯ ¯ −Gy¯ + Q
 = Ge¯ E−QD e y dy.
 1/(D¯−1)
228 NGOC PHAM-THI-DAN, et al.
 ¯
 ∞ n Q
 x P x y
 Executing the series expansion e = n! for the term e in (93), one obtains
 n=0
 ∞ ∞  n!
 ¯ ¯ ¯ Z ¯ X 1 Q¯
 H = Ge¯ E−QD e−Gy dy
 n! y
 n=0
 1/(D¯−1)
 ∞
 ∞  ¯ ¯
n −x
 x=Gy¯ ¯ ¯ ¯ X QG Z e
 = eE−QD dx. (94)
 n! xn
 n=0
 E¯
 The last integral in (94) is solved by invoking [44, eq. (3.381.6)], which reduces (94) to
 ∞ Q¯G¯
n
 E¯−Q¯D¯ X ¯−n/2 −E/¯ 2  ¯

 H = e E e W− n , 1−n E . (95)
 n! 2 2
 n=0
 Substituting (95) into (92), one reduces (92) to (59), finishing the proof.
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