# How to derive the solution of $F_S(x)=P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right)$?

[EDIT]

I came across a received signal-to-interference-plus-noise-ratio (SINR), $$S$$, of a wireless communication system as \begin{align*} S = \frac{\phi|h|^2\rho_1}{1+|g|^{2} \rho _{2} }, \tag{1} \end{align*}

where $$\phi$$ is the power allocation coefficient, $$\rho_1$$ is the SNR (signal-to-noise-ratio) of channel 1, $$\rho_2$$ is the SNR of channel 2, $$h \sim\mathcal{CN}(0,\sigma^2)$$ is channel 1's gain, and $$g \sim\mathcal{CN}(0,\sigma^2)$$ is channel 2's gain. Furthermore, $$H= \mathbb{E}[|h|^2]$$ and $$G= \mathbb{E}[|g|^2]$$.

The authors state that the CDF of the received SINR is

\begin{align*} F_S(x)=&P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right), \end{align*}

where the RHS is found to be \begin{align*} P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right) = 1 - \frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}},\tag{2} \end{align*}

where $$a=\frac{G \rho_2}{\phi H\rho_1}$$.

I understand that $$z=h_r^2+h_i^2=\frac{2}{\sigma^2} |h|^2$$ is a chi squared distributed random variable with $$k=2$$ degrees of freedom. Then, the CDF for $$k=2$$ is

$$$$F(z;k=2)=1-e^{-z/2}, \tag{3}$$$$

which allows us to obtain the structure of (2).

However, I am unable to figure out how the $$\frac {1}{1 + a x}$$ in $$\frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}}$$ was derived.

Does the fact that $$g$$ is a complex random variable also affect the outcome?

Could someone help me understand how the expression in (2) was derived?

• There are some strange aspects to this question. In particular, since $\sigma^2$ determines $E[|h|^2]$ and $E[|g|^2],$ why do you introduce these latter variables? Indeed, your result (2) is incorrect, further suggesting there are serious typographical errors in this question.
– whuber
Dec 13, 2021 at 18:38
• @whuber Thank you for pointing out the issue. I have added more context in the hopes to make it clear. Regarding the correctness of result (2), I doubled checked the article, and I am afraid that it is correct in the sense that it is exactly how the authors have derived it. Dec 14, 2021 at 10:49
• Unfortunately, without strong additional assumptions, this result is false.
– whuber
Dec 14, 2021 at 18:14
• @whuber I believe that could be the reason why I am not able to figure this out. May I know what are these assumptions you are referring to? Or under what assumption would the result be correct? Dec 15, 2021 at 1:43
• I am beginning to suspect this can be solved, after all, assuming $G=H=2\sigma^2$ as implied by your assumptions. One has to do the integral to compare one exponential variable $|h|^2$ to another one $|g|^2.$
– whuber
Dec 15, 2021 at 2:25

The integral turns out to be easier than it looks.

Let's simplify notation a little. By choosing suitable units for the variables, we may make $$\sigma^2=1/2,$$ entailing $$H=G=1.$$ Consequently everything depends on the values of $$\frac{1}{\phi\rho_1} = \beta \gt 0,$$ say, and $$a = \rho_2 \beta \gt 0.$$ $$Y=|h|^2$$ and $$X=|g|^2$$ are independent Exponential variables, which means that for any number $$\lambda \ge 0,$$

$$\Pr(X \le \lambda) = \Pr(Y \le \lambda) = 1 - e^{-\lambda}.$$

The problem is for any $$\lambda\ge 0$$ to compute

\begin{aligned} \Pr(Y \le \beta \lambda(1 + \rho_2 X)) &= E\left[1 - e^{-( \beta\lambda(1 + \rho_2 X))}\right] \\ &= \int_0^\infty \left(1 - e^{-( \beta\lambda(1 + \rho_2 x))}\right) \,\mathrm{d}\left(1-e^{-x}\right) \\ &= \int_0^\infty e^{-x}\,\mathrm{d}x - e^{-\lambda}\int_0^\infty e^{- \beta\lambda \rho_2 x - x}\,\mathrm{d}x \\ &= 1 - e^{-\beta\lambda} \frac{1}{\beta\lambda\rho_1+1} \\ &= 1 - \frac{1}{1 + a\lambda}e^{-\lambda/(\phi \rho_1)}. \end{aligned}

In the question, "$$x$$" is used instead of "$$\lambda:$$" because $$H=1,$$ this result obviously agrees with the quoted formula in the question, QED.

• Thank you very much @whuber. Appreciate it. Dec 21, 2021 at 8:33