Let no player deviate from the agreement till the stage
τ. Let, at stage
τ+1, the player
k deviate from the strategy and transmits at
\(\widetilde {s}_{k}^{(\tau +1)}=\overline {\alpha }_{k}\neq \alpha _{k}^{*}\). There are two sub-cases. (1) First, we consider the case when
\(\overline {\alpha }_{k}<\alpha _{k}^{*}\). In this case, from stage
τ+2 onwards, the node A conforms to the initial strategy again. Since there has been a deviation, conforming to the strategy implies
\(\widetilde {s}_{k}^{(t)}=0\) for all
t≥
τ+2. We calculate the discounted payoffs in two cases. Firstly, in case there is no deviation, the discounted payoff is given by
$$ v_{k}(s^{*})=\frac{(1-\delta)}{(1-\delta^{T})}\sum_{t=1}^{T}\delta^{t-1}u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})=u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*}) $$
(16)
Secondly, in case there is a deviation by player
k at stage
τ+1, the discounted payoff is given by
\(v_{k}(\widetilde {s})=\)
$$\begin{aligned} \frac{(1-\delta)}{(1-\delta^{T})}&\left(\sum_{t=1}^{\tau}\delta^{t-1}u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})\right.\\ &\left.+\delta^{\tau}u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+\sum_{t=\tau+2}^{T}\delta^{t-1}u_{k}(0,0)\right) \end{aligned} $$
where
\(\overline {\alpha }_{k}<\alpha _{k}^{*}\). We simplify the above expression as
$${} {\small{\begin{aligned} v_{k}(\widetilde{s})&=\frac{(1-\delta)}{(1-\delta^{T})}\left(u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})\left[\frac{1-\delta^{\tau}}{1-\delta}\right]\right.\\ &\left.\quad+\delta^{\tau}u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+u_{k}(0,0)\left[\sum_{t=1}^{T}\delta^{t-1}-\sum_{t=1}^{\tau+1}\delta^{t-1}\right]\right)\\ &\quad\times\ v_{k}(\widetilde{s})=\frac{(1-\delta)}{(1-\delta^{T})}\left(u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})\left[\frac{1-\delta^{\tau}}{1-\delta}\right]\right.\\ &\left.\quad+\delta^{\tau}u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+\left(u_{k}(0,0)\left[\frac{1-\delta^{T}}{1-\delta}-\frac{1-\delta^{\tau+1}}{1-\delta}\right]\right)\right) \end{aligned}}} $$
The discounted payoff of the strategy with one-step deviation should be lesser than the strategy with no deviation, for the latter to be SPE. Therefore, we identify the
δ such that
$$v_{k}(s^{*})>v_{-k}(\widetilde{s}) $$
Inserting the values from (
16) and (
17), we get
$${} {\small{\begin{aligned} u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})& > \left(u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})\delta^{\tau}+ \delta^{\tau}u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})\right.\\ &\left.-\delta^{\tau+1}u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+\delta^{\tau+1}u_{k}(0,0)\right) \end{aligned}}} $$
By (
12),
\(u_{k}(\alpha _{k}^{*},\alpha _{-k}^{*})>0\) because
u
k
(0,0)>0 for sustainable points
1, we get
$${} {{\begin{aligned} 0>\delta^{\tau}(-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})+u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-\delta u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+\delta u_{k}(0,0)) \end{aligned}}} $$
Since
δ>0, therefore,
$${} (u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})-u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})+\delta u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-\delta u_{k}(0,0))>0 $$
Consequently,
$${} [u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})-u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})]+\delta[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]>0 $$
Collecting the terms and rewriting, we get
$$\delta>\frac{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]} $$
Therefore, the player
k has no incentive to deviate to any
\(\overline {\alpha }_{k}<\alpha _{k}^{*}\) with the above discount factor condition in order to decrease its cost. Under the following sufficient condition on the discount factor
$$ 1>\delta>\max_{k\in\mathcal{P},\overline{\alpha}_{k}<\alpha_{k}^{*},\overline{\alpha}_{k}\in\mathcal{A}_{k}}\frac{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]} $$
(18)
we can see that the discounted payoff
\(v_{k}(\widetilde {s})\) is less than
v
k
(
s
∗). Additionally, it can be seen from (
12), that
\([u_{k}(\overline {\alpha }_{k},\alpha _{-k}^{*})-u_{k}(\alpha _{k}^{*},\alpha _{-k}^{*})]<[u_{k}(\overline {\alpha }_{k},\alpha _{-k}^{*})-u_{k}(0,0)]\). Therefore,
\([u_{k}(\overline {\alpha }_{k},\alpha _{-k}^{*})-u_{k}(\alpha _{k}^{*},\alpha _{-k}^{*})]/[u_{k}(\overline {\alpha }_{k},\alpha _{-k}^{*})-u_{k}(0,0)]<1\). So, condition (
18) does not imply any supplementary condition on system parameters. To further simplify the expression, we find the value of
\(\overline {\alpha }_{k}\) which maximizes the expression in (
18). Using (
9) and (
7), we know that
\(\frac {du_{k}(\alpha _{k},\alpha _{-k})}{d\alpha _{k}}<0\). In other words, the utility of player
k monotonically decreases with increasing
α
k
for a fixed
α
−k
. We will prove that the condition in (
18) is strictly decreasing with increasing
\(\overline {\alpha }_{k}\), and reaches a maximum at
\(\overline {\alpha }_{k}=0\). Taking the derivative of right hand side in (
18), we get
$${} {{\begin{aligned} \frac{d}{d\overline{\alpha}_{k}}&\left\{ \frac{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]}\right\} =\left(\frac{du_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})}{d\overline{\alpha}_{k}}\right)\\ &\quad\times\frac{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]-[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]^{2}} \end{aligned}}} $$
With
\(\frac {du_{k}(\overline {\alpha }_{k},\alpha _{-k}^{*})}{d\overline {\alpha }_{k}}<0\), and the condition (
12), the above expression is strictly negative. Hence, the lower limit of
δ in (
18) is strictly decreasing with increasing
\(\overline {\alpha }_{k}\). Therefore, it maximizes at minimum value of
\(\overline {\alpha }_{k}\) which is 0. Thus,
$$\begin{aligned} &\max_{k\in\mathcal{P},\overline{\alpha}<\alpha^{*},\overline{\alpha}\in\mathcal{A}_{k}}\left\{ \frac{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(\overline{\alpha}_{k},\alpha_{-k}^{*})-u_{k}(0,0)]}\right\} =\\ &\quad\quad\quad\quad\max_{k\in\mathcal{P}}\left\{ \frac{[u_{k}(0,\alpha_{-k}^{*})-u_{k}(\alpha_{k}^{*},\alpha_{-k}^{*})]}{[u_{k}(0,\alpha_{-k}^{*})-u_{k}(0,0)]}\right\} \end{aligned} $$
Inserting the values of
u
k
from (
9) we get,
$${} {{\begin{aligned} &\delta> \frac{-[w_{3}f_{k}^{\text{COST}}(0)-w_{3}f_{k}^{\text{COST}}(\alpha_{k}^{*})]}{[w_{1}f_{k}^{\text{QoS}}(\alpha_{-k}^{*})+w_{2}f_{k}^{\text{QoE}}(\alpha_{-k}^{*})-w_{1}f_{k}^{\text{QoS}}(0)-w_{2}f_{k}^{\text{QoE}}(0)]} \end{aligned}}} $$
We have the limits of
δ as
$$\begin{aligned} 1>\delta>\delta_{\text{asym}}^{\text{min}}(\alpha_{k}^{*},\alpha_{-k}^{*}) \end{aligned} $$
where
\(\delta _{\text {asym}}^{\text {min}}(\alpha _{k}^{*},\alpha _{-k}^{*})\) is given by
$${} {{\begin{aligned} \max_{k\in\mathcal{P}}\frac{w_{3}[f_{k}^{\text{COST}}(\alpha^{*})-f_{k}^{\text{COST}}(0)]}{w_{1}[f_{k}^{\text{QoS}}(0)-f_{k}^{\text{QoS}}(\alpha_{-k}^{*})]+w_{2}[f_{k}^{\text{QoE}}(0)-f_{k}^{\text{QoE}}(\alpha_{-k}^{*})]} \end{aligned}}} $$
Now we consider the the second sub-case.