I have a 2-dimensional space. The receiver is at the center. The locations of transmitters are given by the PPP $\Phi$ with density $\lambda$. The serving transmitter is the one that is the closest to the receiver, and the others are interferers. I denote $r$ as the distance between the receiver and the serving transmitter, while $l_i\ge r$ are the distances of the interferers. Therefore, the index $j$ in the first equation refers to the serving transmitter. $g$ is the gain of the serving link, and $h_i$ are the gains of the interfering links. Here $f_R(r)$ is the distribution of $r$, the closest distance distribution.

Now, I need to evaluate the following expression.

$\qquad \mathbb{E}_r\left[\mathbb{E}_I\left[\mathbb{P}\left(g>\mu r^{\alpha}\epsilon+\mu r^{\alpha}I\right)\right]\right]$

where $I=\sum_{i\in\Phi\setminus j} h_il_i^{-\alpha}$

Let $g$ is distributed as ExponentialDistribution[1]., then we have

$\qquad \mathbb{E}_r\left[\exp(-\mu r^{\alpha}\epsilon)\int_0^\infty\exp(-sI)f_I(i)\text{d}i\right]$ with $s=\mu r^\alpha$

which is equal to

$\qquad \mathbb{E}_r\left[\exp(-\mu r^{\alpha}\epsilon)\mathcal{L_I(s)}\right]$ with $s=\mu r^\alpha$

Now, my problem is: If $g$ has other distribution than ExponentialDistribution[1], can I reach to similar expression along with $\mathcal{L}_I(s)$? This is because I know how to obtain $\mathcal{L}_I(s)$.

Let $f_R(r)=2\pi\lambda r\exp(-\pi\lambda r^2)$ and $h$ is distributed as ExponentialDistribution[1] or any other distribution, where $\lambda$ is the density of the Poisson process $\Phi$.

Here, $\mathbb{E}$ stand for expectation and $\mathbb{P}$ stands for probability. Also, $\mu=10$, $\alpha=4$ and $\epsilon=500$.

Is there a better way to do this in Mathematica?


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