I am reading an engineering book. The authors mentioned that variables $x,y$ are exponentially distributed with a mean $a^2+N_o$ and $N_o$, respectively. We can calculate the probability of error by direction integration, that is: $$p_e=\mathbb{P}\{y>x\}=[2+\frac{a^2}{N_o}]^{-1}$$ Where the probability density function of the r.v. is $f(u)=\frac{1}{\mu}\cdot\exp^{\frac{-u}{\mu}}$ with $\mu$ is the mean. My question is how to calculate the error probability since $x$ and $y$ have different mean values? To my knowledge, I think we can calculate $\mathbb{P}\{y>0\}$ with $\int_{0}^{\infty} \frac{1}{N_o}\cdot\exp^{\frac{-u}{N_o}}du$ (please correct me if I am wrong). Thus, how do I extend it to two r.v. with different mean values?
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I don't understand -- you say "We can calculate ..." and give the result, but then you seem to ask how to obtain it. Is this a result from the book? Who's the "we" who can calculate it? – joriki Jul 22 '18 at 10:41
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I think this post will answer your question – David M. Jul 22 '18 at 15:34
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Thanks for the help guys. I have found a relevant link. I apologize for the repeated question. The answer can be found here: https://math.stackexchange.com/questions/1805587/prove-that-mathbb-pxy-fracba-b-if-x-y-are-exponentially-distrib?noredirect=1&lq=1 – M.A.N Jul 23 '18 at 01:02