I have a system that consists of two subsystems and a user who is supposed to associate with either one of them. Let $S_1$ and $ S_2$ denote subsystem $1$ and subsystem $2$, respectively. The user associates with subsystem $k$ for $k=1,2$ if:
\begin{equation} P_{S_k} \, R_{S_k}^{-\alpha_{S_k}} > P_{S_j} \, R_{S_j}^{-\alpha_{S_j}} \quad \text{for all} \, \, j=1,2 \, \,\text{and} \, \,j \ne k \end{equation}
where $P$ and $\alpha$ are both positive real numbers that are preassigned to each subsystem (indicated by the index in the subscript) based on its features. On the other hand, $R_{S_1}$ and $R_{S_2}$ are two independent random variables. The pdf of $R_{S_1}$ is $f_{R_{S_1}}(r_{s_1})$ with $r_{s_1} \geq a$, while the pdf of $R_{S_2}$ is a piece-wise function: \begin{equation} f_{R_{S_2}}(r_{s_2}) = \begin{cases} f_{R_{S_2},1}(r_{s_2}), \quad b \leq r_{s_2} \leq c \\ f_{R_{S_2}.2}(r_{s_2}), \quad c < r_{s_2} \leq d \end{cases} \end{equation} where $a$, $b$, $c$ and $d$ are all positive real numbers such that $a\leq b<c<d<\infty$. Note that the support of $R_{S_1}$ is infinite, while the support of the $R_{S_2}$ is finite.
I want to find the association probability, denoted $\mathcal{A}_k$ for $k=1,2$. It is the probability that the user is associated with $S_1$ or $S_2$. Here is my attempt for finding $\mathcal{A}_1$ (Note that $\mathcal{A}_2 = 1 -\mathcal{A}_1$):
\begin{align*} \mathcal{A}_1&=\mathbb{P}(P_{S_1} \, R_{S_1}^{-\alpha_{S_1}} > P_{S_2} \, R_{S_2}^{-\alpha_{S_2}})\\ &=\mathbb{P}\left( R_{S_1} < \left(\frac{P_{S_2}}{P_{S_1}} \, R_{S_2}^{-\alpha_{S_2}}\right)^{-\frac{1}{\alpha_{S_1}}}\right)\\ &=\mathbb{P}\left( R_{S_1} < g(R_{S_2}) \right)\\ &= \int_{a}^{\infty}(1-F_{g(R_{S_2})}(r_1))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 \\ &= \int_{a}^{\infty}(1-F_{R_{S_2}}(g^{-1}(r_1)))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 \\ &= \int_{a}^{g^{-1}(b)}(1-F_{R_{S_2}}(g^{-1}(r_1)))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 +\int_{g^{-1}(b)}^{g^{-1}(c)}(1-F_{R_{S_2}}(g^{-1}(r_1)))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 \\ &+\int_{g^{-1}(c)}^{g^{-1}(d)}(1-F_{R_{S_2}}(g^{-1}(r_1)))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 +\int_{g^{-1}(d)}^{\infty}(1-F_{R_{S_2}}(g^{-1}(r_1)))\, f_{R_{S_1}}(r_1) \,\mathrm{d}r_1 \\ \end{align*}
Now I need to substitute $F_{R_{S_2}}(g^{-1}(r_1))$ into the four integrals. In other words, I need to find $\mathbb{P}(R_{S_2} < g^{-1}(r_1))$ in each region, but I don't how. Any hints to get me started?
This question is an extension to the question posted here.
Unfortunately there is a mistake in the last expression.
Let $[x<y]$ denote the function that takes value $1$ if $x<y$ and value $0$ otherwise.
$$\begin{aligned}\mathbb{P}\left(R_{S_{1}}<g\left(R_{S_{2}}\right)\right) & =\mathbb{P}\left(R_{S_{1}}<g\left(R_{S_{2}}\right)\wedge R_{S_{1}}\in\left[a,g\left(b\right)\right]\right)+\mathbb{P}\left(R_{S_{1}}<g\left(R_{S_{2}}\right)\wedge R_{S_{1}}\in\left(g\left(b\right),g\left(d\right)\right]\right)\\ & =\mathbb{P}\left(R_{S_{1}}\in\left[a,g\left(b\right)\right]\right)+\int_{g\left(b\right)}^{g\left(d\right)}\int\left[u<g\left(v\right)\right]f_{R_{S_{1}}}\left(u\right)f_{R_{S_{2}}}\left(v\right)dvdu\\ & =F_{R_{S_{1}}}\left(g\left(b\right)\right)+\int_{g\left(b\right)}^{g\left(d\right)}f_{R_{S_{1}}}\left(u\right)\int\left[u<g\left(v\right)\right]f_{R_{S_{2}}}\left(v\right)dvdu\\ & =F_{R_{S_{1}}}\left(g\left(b\right)\right)+\int_{g\left(b\right)}^{g\left(d\right)}f_{R_{S_{1}}}\left(u\right)\left(1-F_{g\left(R_{S_{2}}\right)}\left(u\right)\right)du\\ & =F_{R_{S_{1}}}\left(g\left(b\right)\right)+\int_{g\left(b\right)}^{g\left(d\right)}f_{R_{S_{1}}}\left(u\right)\left(1-F_{R_{S_{2}}}\left(g^{-1}\left(u\right)\right)\right)du \end{aligned} $$
So in your last line $F_{R_{S_{2}},1}(g(r_1))$ and $F_{R_{S_{2}},2}(g(r_1))$ must both be interchanged with $F_{g\left(R_{S_{2}}\right)}(g^{-1}(r_1))$.
where $g^{-1}$ denotes the inverse function of the monotonically increasing function $g$.
Also observe that a random variable has only one CDF, so the notations $F_{R_{S_{2}},1}$ and $F_{R_{S_{2}},2}$ are not okay.
The split up in $\left(g\left(b\right),g\left(c\right)\right]$ and $\left(g\left(c\right),g\left(d\right)\right]$ makes no sense.