Calculate probability of joint distribution pdf and marginal pdf

832 Views Asked by At

I came across a research paper where they proceed in the Appendix to calculate the probability:

$$P(R_m+R_n>\bar{R}_m+\bar{R}_n)$$

Here $R_m+R_n$ is the non-orthogonal scheme achievable rate and $\bar{R}_m+\bar{R}_n$ is the orthogonal scheme achievable rate.

The above probability can be written as:

$$P\left(\left(\frac{1+\rho a_m^2|h_m|^2}{\rho a_n^2|h_m|^2+1}\right)(1+\rho a_n^2|h_n|^2)>(1+\rho|h_m|^2)^{1/2}(1+\rho|h_n|^2)^{1/2}\right)$$

where $|h_m|^2$ and $|h_n|^2$ is the channel gain with constraint $|h_m|^2\leq|h_n|^2$, and power coefficients $a_m^2\geq a_n^2$.

First they calculate the joint PDF using order of statistics of $|h_m|^2$ and $|h_n|^2$, and after that they also calculate the marginal PDF of $|h_n|^2$. These are denoted $f_{|h_m|^2,|h_n|^2}(x,y)$ and $f_{|h_n|^2}(y)$, respectively.

They finally state that applying both joint and marginal density above, we can expressed the addressed probability as:

$P(R_m+R_n>\bar{R}_m+\bar{R}_n)={\displaystyle \int f_{|h_m|^2,|h_n|^2}(x,y)dxdy}+{\displaystyle \int f_{|h_n|^2}(y)dy} $

I don't understand why the addressed probability is calculating the sum integral of joint PDF and marginal PDF. What is the basic idea?

Can someone help me with the understanding since I'm still learning the statistics?