Page 125 - ITUJournal Future and evolving technologies Volume 2 (2021), Issue 1

ITU Journal on Future and Evolving Technologies, Volume 2 (2021), Issue 1

By (57), > 0. By (58), > 0 if and only if By (25), ̃ ( , ) = 0 if and only if for all the following

inequality holds:
2 ℎ ℎ ℎ
+ > 1 + 1 + 4 + . ℎ ℎ ℎ
ℎ 2 + ≤ 1 + 1 + 4 + .
ℎ
The last inequality is equivalent to (67)
Since ≥ Ω ∞ > 0, substituting = Ω into the right‑
∞
side of (67) implies that if the following inequality holds
1 ℎ ℎ ℎ
2 + > 1 + 4 + . (59) then (67) also holds:
ℎ
ℎ ℎ ℎ

A straightforward calculation shows that (59) is equiva‑ 2 + ≤ 4 ℎ . (68)
lent to
2
> . (60) Taking into account notations (27) and that ≤ Ω , (68)
0
2
ℎ holds for all if the following inequality holds:

Thus, , given by (58), is positive if and only if (60) holds. + ≤ 3/2 (69)

Finally, combining (I) and (II) implies the result.
with , and given by (27). It is clear that Equation
(26) has the unique positive root = . Then, (69) holds
9.3 Proof of Proposition 2 for ≥ , and (66) follows.
Thus, (25), (64), (65) and (66) imply (b). (c) follows
2
2
A straightforward substituting = into (14) im‑ from (25) and (61)‑(63). Also, (b), (c), (66) and Propo‑
ℎ sition 2(e) imply (d).

plies (a). Since the right‑side of Equation (23) is increasing from
Let zero for = 0 to in inity for ↑ ∞ while, by (d), the right‑
1 + √1 + 1 1 side of Equation (23) is decreasing and reaches zero for
( ) = = + + , (61) ≥ , (e) follows.
2
where is a positive parameter. It is clear that
9.5 Proof of Theorem 2
( ) is decreasing in > 0. (62) By Proposition 1, all the equilibrium strategies have to
have the form given by (14) and (15), where ( , ) is a
Let positive solution of (16) and (17). By Proposition 2, (14)
ℎ
= + and = 4 . (63) establishes a bijection relation between and given by
ℎ function = Ω( ). Substituting this function into (15)
yields into Equation (23) of one variable . By Proposi‑
Substituting (63) into (61), by (14) and (62), yields (b) and
(c). (d) follows from (b). (e) follows from (a) and (b), tion 3, this equation has the unique root. Thus, the equi‑
while (21) and (22) follow from (18), (19) and (20). librium also is unique, and the result follows.

9.6 Proof of Proposition 4
9.4 Proof of Proposition 3
By (30), (52) cannot hold for any . Thus, by (14) and (15),
(a) follows from (25). > 0 and > 0 for any . Substituting (30) into (14)

Let and (15) implies
1 + ( + ) 1
( ) = + , (64) ( , ) = , (70)

+ + + ℎ /( )

/
where = ℎ /( ), = 4 /ℎ and = . Then ( , ) = + ℎ /( ) for any . (71)

2
2
2
( ) ( + 3 + 2 + 2 ) Dividing (71) by (70) implies that
= + > 0.
( + ) 2 2( + ) √ + + 1
2

(65) ( , ) = ( / ) ( , ). (72)
Summing up (72) by and taking into account that ∈
Now we establish that and ∈ we have that
̃ ( , ) = 0 for ≥ and ∈ [Ω , Ω ] (66) = ( / ) . (73)
∞ 0
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