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This seems a very dummy question, but I cannot figure out how the following equation has been solved by authors:

$$P_{secure} = \text{Pr}\{\epsilon cP_sd_{su_i}^{-n}|h_{su_i}|^2> cP_sd_{se_i}^{-n}|h_{se_i}|^2\}$$

(The details of the parameters are present in the image attached below)

I want to learn the method and hope the community here does not mind my naiveness! The final solution is:

$$P_{secure} = \frac{\epsilon d_{se_i}^n}{d_{su_i}^n+\epsilon d_{se_i}^n}$$

enter image description here

SJa
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  • You might get a better answer on the [Signal Processing Stackexchange](https://dsp.stackexchange.com/). Though if you ask this question on there, I would advise you to delete this question and create a question on there so that you're not duplicating across two StackExchange websites. Also, if you do ask this question on there, be specific about where this article comes from. –  Sep 24 '18 at 00:36
  • ah....finally, I got the answer, not sure if I should post it as the answer for other people to benefit or just post it as the comment! – SJa Sep 24 '18 at 01:15
  • There's nothing preventing you from answering your own question. You can do that here :) Yeah put it in the answer box. –  Sep 24 '18 at 01:16

1 Answers1

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As I have determined the answer, I am just posting it so that it is helpful for others.

First, let $$X = |h_{su_i}|^2, Y = |h_{se_i}|^2 \text{ where } X,Y\sim \exp(1) \ $$

This detail is missing in the research paper. Then:

$$P_{secure} = \text{Pr}\{X> \frac{d_{se_i}^{-n}Y}{\epsilon d_{su_i}^{-n}}\}$$

$$P_{secure} = \text{Pr}\{X> sY\}$$ Following Complimentary CDF of exponential random variable: $$P_{secure} = \exp(-sY)$$

Since Y is a exponentially distribued random variable with mean 1, we will take its Expected value:

$$P_{secure} = \mathbb{E}[\exp(-sY)]$$

Following the Laplace Transform property for exponentially distributed Random variable, we can write as:

$$P_{secure} = \frac{1}{1+s}$$

which can be simplified as $$P_{secure} = \frac{\epsilon d_{se_i}^n}{d_{su_i}^n+\epsilon d_{se_i}^n}$$

SJa
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    And this is a perfectly fine answer :) It probably feels a little weird answering your own question because it feels like you're talking to yourself but this isn't against the rules of our website. –  Sep 24 '18 at 01:48