function $h(\cdot)$ (see Proposition 2.16 in [@avriel2010generalized] and Section 3.2.4 in [@IFACboyd2004convex]). Thus, by (\[eq:firstln\]), $$\begin{aligned}
\ln\prod_{i=t_{1}}^{t_{2}-1}h(v(i)) & \leq(t_{2}-t_{1})\ln h(\frac{1}{t_{2}-t_{1}}\sum_{i=t_{1}}^{t_{2}-1}v(i)),\label{eq:lnineq2}\end{aligned}$$ which implies (\[eq:powerbound\]). The interference power process $v(\cdot)$ satisfies (\[eq:attack\_assumption\_two\]) in Assumption \[SecondAssumption\], and hence, $\frac{1}{t_{2}-t_{1}}\sum_{i=t_{1}}^{t_{2}-1}v(i)\leq\frac{\hat{\kappa}}{t_{2}-t_{1}}+\hat{v}$, almost surely. Thus, noting that $h(\cdot)$ is a nondecreasing function, by (\[eq:powerbound\]), we obtain $\prod_{i=t_{1}}^{t_{2}-1}h(v(i))\leq h^{t_{2}-t_{1}}(\frac{\hat{\kappa}}{t_{2}-t_{1}}+\hat{v})$, almost surely. Consequently, we have $$\begin{aligned}
\mathbb{E}\big[\prod_{i=t_{1}}^{t_{2}-1}h(v(i))\big] & \leq h^{t_{2}-t_{1}}(\frac{\hat{\kappa}}{t_{2}-t_{1}}+\hat{v}).\label{eq:hfineq-1-1}\end{aligned}$$ Now, it follows from (\[eq:alpha\_ineq\]) that $h(\hat{v})<1$. Therefore, by the continuity of $h(\cdot)$, there exists $\delta>0$ such that $h(\delta+\hat{v})<1$. As a result, for sufficiently large values of $t_{2}-t_{1}$, we have $h(\frac{\hat{\kappa}}{t_{2}-t_{1}}+\hat{v})<1$.

Let $T^{*}$ be a positive integer such that $h(\frac{\hat{\kappa}}{T^{*}}+\hat{v})<1$ and let $$\begin{aligned}
\theta & \triangleq h(\frac{\hat{\kappa}}{T^{*}}+\hat{v}).\label{eq:ThetaDef}\end{aligned}$$ It follows from (\[eq:hfineq-1-1\]) that $$\begin{aligned}
\mathbb{E}\big[\prod_{i=t_{1}}^{t_{2}-1}h(v(i))\big] & \leq\theta^{t_{2}-t_{1}},\label{eq:hjt_part1}\end{aligned}$$ for all $t_{1},t_{2}\in\mathbb{N}_{0}$ such that $t_{2}-t_{1}\geq T^{*}$. If $T^{*}=1$, then (\[eq:geometric-lemma-result\]) holds, by (\[eq:hjt\_part1\]). If, on the other hand, $T^{*}>1$, then by using $h(v(t))\leq\alpha_{1}+\alpha_{0}$, $t\in\mathbb{N}_{0}$, we obtain $$\begin{aligned}
\mathbb{E}\big[\prod_{i=t_{1}}^{t_{2}-1}h(v(i))\big] & \leq(\alpha_{1}+\alpha_{0})^{t_{2}-t_{1}}\leq(\alpha_{1}+\alpha_{0})^{T^{*}-1}\label{eq:hjt_part2}\end{aligned}$$ for all $t_{1},t_{2}\in\mathbb{N}_{0}$ such that $0<t_{2}-t_{1}<T^{*}$. Letting $$\begin{aligned}
\mu & \triangleq(\alpha_{1}+\alpha_{0})^{T^{*}-1}\theta^{-(T^{*}-1)},\label{eq:MuDef}\end{aligned}$$ we obtain (\[eq:geometric-lemma-result\]), by (\[eq:hjt\_part1\]) and (\[eq:hjt\_part2\]).

Lemma \[LemmaDOne\] shows that under Assumption \[SecondAssumption\], the expectation term $\mathbb{E}[\prod_{i=t_{1}}^{t_{2}-1}(\alpha_{1}l(i)+\alpha_{0})]$ with $\alpha_{1}\geq0$, $\alpha_{0}\geq0$ satisfying (\[eq:alpha\_ineq\]), converges to zero at a geometric rate. By using this lemma, we obtain the following result.

\[LemmaSum\] 