us to conduct stability analysis without relying on transmission failure probabilities for each time step. In the derivation of these bounds, an essential step is to exploit the concavity of the upper-bounding function $\hat{p}$ given in (\[eq:p\_phat\_ineq\]).

\[LemmaDOne\] Suppose that the attacker’s interference power process $\{v(t)\in[0,\infty)\}_{t\in\mathbb{N}_{0}}$ satisfies Assumption \[SecondAssumption\]. Then for every $\alpha_{1}\geq0$, $\alpha_{0}\geq0$ that satisfy $$\begin{aligned}
\alpha_{1}\hat{p}(\hat{v})+\alpha_{0} & <1,\label{eq:alpha_ineq}\end{aligned}$$ there exist scalars $\mu\geq0$ and $\theta\in(0,1)$ such that $$\begin{aligned}
\mathbb{E}[\prod_{i=t_{1}}^{t_{2}-1}(\alpha_{1}l(i)+\alpha_{0})] & \leq\mu\theta^{(t_{2}-t_{1})},\label{eq:geometric-lemma-result}\end{aligned}$$ for $t_{1},t_{2}\in\mathbb{N}_{0}$ with $t_{1}<t_{2}$.

For the case where $\alpha_{1}+\alpha_{0}=0$, (\[eq:geometric-lemma-result\]) holds for any $\mu\geq0$ and $\theta\in(0,1)$. In the following, we consider the case where $\alpha_{1}+\alpha_{0}>0$. First, by Lemma 2.1 of [@ahmetifacwc2017], $$\mathbb{E}[\prod_{i=t_{1}}^{t_{2}-1}(\alpha_{1}l(i)+\alpha_{0})]=\mathbb{E}[\prod_{i=t_{1}}^{t_{2}-1}(\alpha_{1}p(v(i))+\alpha_{0})].\label{eq:first-eq}$$ Next, by (\[eq:first-eq\]), $\alpha_{1}\geq0$, and $p(v)\leq\hat{p}(v)$, $v\in[0,\infty)$, we get $$\begin{aligned}
\mathbb{E}\big[\prod_{i=t_{1}}^{t_{2}-1}(\alpha_{1}l(i)+\alpha_{0})\big] & \leq\mathbb{E}\big[\prod_{i=t_{1}}^{t_{2}-1}h(v(i))\big],\label{eq:expxhineq-1-1}\end{aligned}$$ where $h(v)\triangleq\alpha_{1}\hat{p}(v)+\alpha_{0}$. We note that $h(\cdot)$ is nondecreasing, concave, and continuous, as $\hat{p}(\cdot)$ also has such properties and $\alpha_{1}\geq0$. To obtain an upper-bound for $\mathbb{E}[\prod_{i=t_{1}}^{t_{2}-1}h(v(i))]$ in (\[eq:expxhineq-1-1\]), we first show $$\begin{aligned}
\prod_{i=t_{1}}^{t_{2}-1}h(v(i)) & \leq h^{(t_{2}-t_{1})}(\frac{1}{t_{2}-t_{1}}\sum_{i=t_{1}}^{t_{2}-1}v(i)).\label{eq:powerbound}\end{aligned}$$ We note that (\[eq:powerbound\]) holds if $h(v(i))=0$ for some $i\in\{t_{1},\ldots,t_{2}-1\}$. Now, consider the case where $h(v(i))>0$ for all $i\in\{t_{1},\ldots,t_{2}-1\}$. For this case, we have $$\begin{aligned}
\ln\prod_{i=t_{1}}^{t_{2}-1}h(v(i)) & =(t_{2}-t_{1})\Big(\frac{1}{t_{2}-t_{1}}\sum_{i=t_{1}}^{t_{2}-1}\ln h(v(i))\Big).\label{eq:firstln}\end{aligned}$$ Here, $\ln h(\cdot)$ is concave, since it is the composition of a nondecreasing concave function $\ln(\cdot)$ and a concave 