I am reading a paper that calculates outage probability. I had reached upto following: Step 1: The outage probability is defined as $P_o = Pr\left(\gamma \geq \gamma_{th}\right)$-----(1), where $\gamma$ is SNR and $\gamma_{th}$ is SNR threshold. Then i have the expression of SNR that i had substituted into definition of SNR which is given in (1). Thus the step 2 is obtained as follows.
Step 2: $P_o = Pr\left(\frac{P_t|h|^2}{\alpha|f|^{2}|g|^{2}P_{t}+\sigma^2}\geq\gamma_{th}\right)$-----(2), where $h,f,g $ are rayleigh random variable, $\sigma^2$ is variance of additive white gaussian noise (AWGN) and $P_t$ is transmit power. After some simple manupulation, i reached upto the step 3.
Step3: $P_o = Pr\left(|h|^{2}\geq\alpha|f|^{2}|g|^{2}\gamma_{th}+\frac{\sigma^2\gamma_{th}}{P_t}\right)$------(3). After step 3, the step 4 is given as
Step4: $P_o = \int_0^{\infty}\int_{\frac{\gamma_{th}(\alpha P_tx+\sigma^2)}{P_t}}^{\infty}f_{|h|^{2}}(y)f_{|f|^2|g|^2}(x)dydx$------(4) I am not getting how Eq. (4) is obtained. Any help please. Note: I apologize for the very long question that I had asked. However I had written so that every one understood my query. And I just want how Eq. (4) is obtained from Eq. (3)
I think that what’s going on is as follows:
First (and check the paper for notation): I think $f_{|h|^2}(y)$ designates the probability density function for a Rayleigh distribution of scale parameter $|h|$.
So the inner integral with respect to $dx$ is accumulating the portion of $P_0$ for a given value of $h$. The lower bound of this integral is the smallest valid value of $h$.
And then the outer integral with respect to $dy$ is accumulating over all values of $h$.
In more detail: $$\begin{align*} P_0 &= Pr\!\left( |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t} \right) \\&= \lim_{\delta y \to 0} \sum_{k=0}^{\infty} Pr\!\left( |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t} \,\Bigg{|}\, h = y_k \right) Pr\Bigl(h \in [y_k,y_{k+1})\Bigr) &\text{where $y_k = k\delta y$} \end{align*}$$ by the law of total probability. Intuitively, we tile the positive number line into intervals of width $\delta y$. Those are the possible values for $h$. Now $Pr\Bigl(h \in [y_k,y_{k+1})\Bigr) \approx f_{|h|^2}(y) \delta y$ and the approximation becomes perfect as $\delta y \to 0$ so the sum becomes an integral $$ P_0 = \int_{0}^{\infty} Pr\!\left( |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t} \,\Bigg{|}\, h = y \right) f_{|h|^2}(y) \, dy $$
I think that something similar is going on with the inner integral with respect to $x$, so that $$\begin{align*} Pr\!\left( |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t} \,\Bigg{|}\, h = y \right) &= \lim_{\delta x \to 0} \sum_{\ell=0}^{\infty} Pr\!\left( |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t} \,\Bigg{|}\, h = y, |f||g| = x_{\ell} \right) Pr\Bigl( |f||g| \in [x_{\ell}, x_{\ell+1}) \Bigr) & \text{where $x_{\ell} = \ell \delta x$} \\&= \int_{\alpha x \gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}}^{\infty} f_{|f|^2|g|^2}(x) \, dx \end{align*}$$
Hence $$\begin{align*} P_0 &= \int_{0}^{\infty} \left( \int_{\frac{\gamma_{th}(\alpha P_t x + \sigma^2)}{P_t}}^{\infty} f_{|f|^2|g|^2}(x) \, dx \right) f_{|h|^2}(y) \, dy \\&= \int_{0}^{\infty} \int_{\frac{\gamma_{th}(\alpha P_t x + \sigma^2)}{P_t}}^{\infty} f_{|h|^2}(y) f_{|f|^2|g|^2}(x) \, dy dx \end{align*}$$ where the last step is from rearranging the order of the integration and bringing the factor $f_{|h|^2}(y)$ inside the inner integral.