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I'm trying to derive the equation of error probability in orthogonal signaling. Here what I've tried:

Let $\vec{s_1} = (\sqrt{E_b} , 0)$ and $\vec{s_2} = (0, \sqrt{E_b} )$ be vector representation of $s_1$ and $s_2$. We know that $$f(\vec{y}|\vec{s_m}) = (\frac{1}{\sqrt{\pi N_0}})^2e^{-\frac{|\vec{y} - \vec{s_m}|^2}{N_0}} , \ m = 1,2$$ According to the Bayes' law, we can write $$P(\vec{s_m}|\vec{y}) = \frac{f(\vec{y}|\vec{s_m})P(\vec{s_m})}{f(\vec{y})}, m = 1,2$$Let $P(\vec{s_1}) = p_1$ and $P(\vec{s_1}) = p_2$. Using MAP rule, we want to find $m$ such that $P(\vec{s_m}|\vec{y})$ be maximized. So $\vec{s_1}$ is chosen if $$P(\vec{s_1}|\vec{y}) \gt P(\vec{s_2}|\vec{y})$$ and otherwise $\vec{s_2}$. After simplification we get $$\frac{N_0}{\sqrt{E_b}} \ln(\frac{p_1}{p_2}) \gt -2(y_1 - y_2)$$ Let $a = \frac{-N_0 \ln(\frac{p_1}{p_2})}{2\sqrt{E_b}}$ and note that $y_1 - y_2$ is a Gaussian random variable with mean $\sqrt{E_b}$ and variance $\sigma^2 = N_0$. So the probability of error is $$P_e = p_1\int_{-\infty}^a \frac{1}{\sqrt{2\pi N_0}}e^{-\frac{(z-\sqrt{E_b})^2}{2N_0}}dz + p_2\int_{a}^{+\infty} \frac{1}{\sqrt{2\pi N_0}}e^{-\frac{(z-\sqrt{E_b})^2}{2N_0}}dz$$ which simplifies to $$P_e = p_1\int_{-\infty}^{\frac{a - \sqrt{E_b}}{\sqrt{N_0}}} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du + p_2\int_{\frac{a - \sqrt{E_b}}{\sqrt{N_0}}}^{+\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}du = p_1Q(\frac{\sqrt{E_b} - a}{\sqrt{N_0}}) + p_2Q(\frac{a - \sqrt{E_b} }{\sqrt{N_0}})$$ Is this result correct? I couldn't find a source which includes error probability for the orthogonal signaling when prior probabilities are unequal.

Edit: Thanks to Dilip Sarwate's comment, the second exponent should be $$-\frac{(z+\sqrt{E_b})^2}{2N_0}$$ So the final answer is $$P_e = p_1Q(\frac{\sqrt{E_b} - a}{\sqrt{N_0}}) + p_2Q(\frac{\sqrt{E_b} + a }{\sqrt{N_0}})$$ Where $a = \frac{-N_0 \ln(\frac{p_1}{p_2})}{2\sqrt{E_b}}$. We can simplify this formula and get $$P_e = p_1Q(\sqrt{\frac{E_b}{N_0}} - \frac{\sqrt{N_0}}{2\sqrt{E_b}}\ln(\frac{p_2}{p_1})) + p_2Q(\sqrt{\frac{E_b}{N_0}} +\frac{\sqrt{N_0}}{2\sqrt{E_b}}\ln(\frac{p_2}{p_1}))$$ which agrees with Dilip Sarwate's answer.

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  • $\begingroup$ This is not quite right: the devil is in the details! Note that $y_1-y_2 \sim \mathcal N(E_b, N0)$ has when $x_1$ is the transmitted signal, but $y_1-y_2 \sim \mathcal N(-E_b, N0)$ has when $x_2$ is the transmitted signal, and you should use the appropriate densities when calculating $P_{e,1}$ and $P_{e,2}$. Also, I think that $a$ should be $\dfrac{-1}{2}\sqrt{\dfrac{N_0}{E_b}}\ln\left(\dfrac{p_1}{p_2}\right)$ $\endgroup$ May 12 at 20:10
  • $\begingroup$ @DilipSarwate Thanks, you're right. That was the main mistake in my answer but I think the formula for $a$ is correct and now our answers are identical. $\endgroup$
    – S.H.W
    May 12 at 20:40
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I haven't checked the details fully but the OP's revised solution is correct. Once upon a time, I had posted a to a class webpage from 1998 where I had some lecture notes giving the answer enter image description here

but the link is now defunct because my University seems to have deleted that webpage from its servers. Anyway, the answer that I came with to the OP's question is

$$P_e = \pi_0 Q\left(\mathsf{SNR} -(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) + \pi_1 Q\left(\mathsf{SNR} +(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) \tag{1}$$ where $\mathsf{SNR} = \dfrac{\Vert s_0-s_1\Vert}{\sqrt{2N_0}} = \sqrt{\dfrac{E_b}{N_0}}$ in this case. Note that when $\pi_1 > \pi_0$, $\ln\left(\frac{\pi_1}{\pi_0}\right) > 0$, and so $$Q\left(\mathsf{SNR} +(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right) < Q\left(\mathsf{SNR} -(2\cdot\mathsf{SNR})^{-1}\ln\left(\frac{\pi_1}{\pi_0}\right)\right),$$ that is, in the weighted sum of two error probabilities in $(1)$, the larger weight $\pi_1$ weights the smaller error probability, which is exactly how it should be.

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  • $\begingroup$ Thanks for the answer. You are right about the threshold and I've corrected it. I think in your answer SNR should be $\sqrt{\frac{E_b}{N_0}}$. Also I think it should be $(2(SNR))^{-1}$ instead of $2(SNR)^{-1}$. $\endgroup$
    – S.H.W
    May 12 at 8:33
  • $\begingroup$ I'm not sure whether both of probabilities should have exponent $-\frac{(z-\sqrt{E_b})^2}{2N_0}$ or one of them $-\frac{(z-\sqrt{E_b})^2}{2N_0}$ and the other one $-\frac{(z+\sqrt{E_b})^2}{2N_0}$ $\endgroup$
    – S.H.W
    May 12 at 8:37
  • $\begingroup$ @S.H.W You are correct about the typos in my answer, which I have corrected. It was a transcription error from my (nonLaTexed) lecture notes mentioned above into my answer here $\endgroup$ May 12 at 19:56
  • $\begingroup$ @S.H.W With regard to your second comment, note that $y_1-y_2 \sim \mathcal N(E_b, N_0)$ when $x_1$ is the transmitted signal, but $y_1-y_2 \sim \mathcal N(-E_b, N_0)$ when $x_2$ is the transmitted signal, and you should use the appropriate densities when calculating $P_{e,1}$ and $P_{e,2}$. $\endgroup$ May 12 at 20:14
  • $\begingroup$ It's really unfortunate that your lecture notes have been deleted. They are fantastic sources for the undergraduate students like me. $\endgroup$
    – S.H.W
    May 12 at 20:49

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