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[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

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

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?

Thanks in advance.

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    $\begingroup$ 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. $\endgroup$
    – whuber
    Dec 13, 2021 at 18:38
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    $\begingroup$ @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. $\endgroup$
    – nashynash
    Dec 14, 2021 at 10:49
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    $\begingroup$ Unfortunately, without strong additional assumptions, this result is false. $\endgroup$
    – whuber
    Dec 14, 2021 at 18:14
  • $\begingroup$ @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? $\endgroup$
    – nashynash
    Dec 15, 2021 at 1:43
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    $\begingroup$ 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 $|g|^2.$ $\endgroup$
    – whuber
    Dec 15, 2021 at 2:25

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much @whuber. Appreciate it. $\endgroup$
    – nashynash
    Dec 21, 2021 at 8:33

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