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Those interested in finding an single formula for the BER of a Gray-coded 8PSK system operating on an AWGN channel can take the average of the three error probabilities found above (which is approximately $\frac 43 P(E_1)$), but I think considering the three bit error probabilities separately is more informative. Bear in mind that the three error events are not independent.

So, there you have it, folks, the exact expressions for the BER(s) sustained by the three bits transmitted in a 8PSK scheme with Gray coding operating on an AWGN channel. I know it won't satisfy @Loran but I hope the rest of you find it useful.

Those interested in finding a single formula for the BER of a Gray-coded 8PSK system operating on an AWGN channel can take the average of the three error probabilities found above to arrive at \begin{align} \overline{P(E)} &= \left.\left.\frac 13\right[P(E_1) + P(E_2) + P(E_3)\right]\\ &= \left.\left.\frac 23\right[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right.\\ &\qquad\quad - Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\cdot Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\bigg]\tag{12}\\ &= \left.\left.\frac 23\right[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\right.\\ &\qquad\quad - Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\cdot Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\bigg]\tag{13}.\\ \end{align} In the form $(13)$ above, the average error probability formula can be found (without the above details) at the bottom of page 340 of M.B. Pursley's Introduction to Digital Communications, Pearson Prentice-Hall, 2005, with a citation of "Computation of the Bit Error Rate of Coherent M-ary PSK with Gray Code Bit Mapping" by P.Lee, IEEE Transactions on Communications, May 1986 (which is behind IEEE's paywall for many people) as the source of the result. Note that since the last term in $(12)$ and $(13)$ is considerably smaller than the first two terms, $$\overline{P(E)} \approx \frac 43 P(E_1)\approx \frac 23 P(E_3),$$ but I think considering the three bit error probabilities separately is more informative than looking at just the average BER. Bear in mind that the three error events are not independent events.

So, there you have it, folks, the exact expressions for the BER(s) sustained by the three bits transmitted in a 8PSK scheme with Gray coding operating on an AWGN channel right here on dsp.SE without the need to look elsewhere. I know it won't satisfy @Loran but I hope the rest of you find it useful.

Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_3$ occurs if $(X,Y) \in R_1\cup R_3\cup R_5 \cup R_7$, which is two quadrants that touch at the origin. But note that the region of interest can be expressed as the Exclusive-OR union of the two half-planes whose boundaries are the lines through the origin of slopes $\frac{\pi}{8}$ and $-\frac{3\pi}{8}$. In symbols, $$R_1\cup R_3\cup R_5 \cup R_7 = (R_1\cup R_3\cup R_2 \cup R_6) \oplus (R_5 \cup R_7 \cup R_2 \cup R_6).$$ Now, \begin{align} P(A\oplus B) &= P(A) + P(B) -2P(A\cap B)\\ &= P(A) + P(B) - 2P(A)P(B) & \scriptstyle{\text{for independent $A$ and $B$}} \end{align} and since $000$ is at distance $\sin\left(\frac{\pi}{8}\sqrt{\mathscr E_s}\right)$ from one boundary and at distance $\sin\left(\frac{3\pi}{8}\sqrt{\mathscr E_s}\right)$ from the other boundary, we get that \begin{align} P(E_3 \mid 000) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{10} \end{align} Those who have followed the analysis thus far should have no trouble understanding why $$P(E_3 \mid 100) = P(E_3 \mid 000) = P(E_3\mid 010) = P(E_3 \mid 110)$$ and indeed why \begin{align}P(E_3) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{11} \end{align} Note that $P(E_3)$ is larger than $P(E_1) = P(E_2)$. The latter probability is the arithmetic average of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ while the former is the slightly smaller than the sum of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$.

Those interested in finding an single formula for the BER of a Gray-coded 8PSK system operating on an AWGN channel can take the average of the three error probabilities found above, but I think the three separately is more informative.

Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_3$ occurs if $(X,Y) \in R_1\cup R_3\cup R_5 \cup R_7$, which is two quadrants that touch at the origin. But note that the region of interest can be expressed as the Exclusive-OR union of the two half-planes whose boundaries are the lines through the origin of slopes $\frac{\pi}{8}$ and $-\frac{3\pi}{8}$. In symbols, $$R_1\cup R_3\cup R_5 \cup R_7 = (R_1\cup R_3\cup R_2 \cup R_6) \oplus (R_5 \cup R_7 \cup R_2 \cup R_6).$$ Now, \begin{align} P(A\oplus B) &= P(A) + P(B) -2P(A\cap B)\\ &= P(A) + P(B) - 2P(A)P(B) & \scriptstyle{\text{for independent $A$ and $B$}} \end{align} and since $000$ is at distance $\sin\left(\frac{\pi}{8}\sqrt{\mathscr E_s}\right)$ from one boundary and at distance $\sin\left(\frac{3\pi}{8}\sqrt{\mathscr E_s}\right)$ from the other boundary, we get that \begin{align} P(E_3 \mid 000) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{10} \end{align} Those who have followed the analysis thus far should have no trouble understanding why $$P(E_3 \mid 100) = P(E_3 \mid 000) = P(E_3\mid 010) = P(E_3 \mid 110)$$ and indeed why \begin{align}P(E_3) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{11} \end{align} Note that $P(E_3)$ is larger than $P(E_1) = P(E_2)$. The latter probability is the arithmetic average of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ while the former is slightly smaller than the sum of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$. Thus, we have the approximation that is worth keeping in mind: $$P(E_3)\approx 2P(E_1) = 2P(E_2).$$

Those interested in finding an single formula for the BER of a Gray-coded 8PSK system operating on an AWGN channel can take the average of the three error probabilities found above (which is approximately $\frac 43 P(E_1)$), but I think considering the three bit error probabilities separately is more informative. Bear in mind that the three error events are not independent.

Analysis for $E_2$: Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_2$ occurs if $(X,Y) \in R_2\cup R_3\cup R_6 \cup R_7$ which compound region has boundary which is the straight-line of slope $\frac{3\pi}{8}$ through the origin. Going through a similar analysis mutatis mutandis, we get that $$P(E_\2\mid 000) = P(E_2\mid 100) = Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right), \tag{7}$$ and $$P(E_2\mid 001) = P(E_2\mid 101) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{8}$$ The conditional probability of $E_2$ given the other points can be calculated in a similar fashion and thus we arrive at $$P(E_2) = P(E_1) = \frac{1}{2}\left[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right].\tag{9}$$

Analysis for $E_2$: Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_2$ occurs if $(X,Y) \in R_2\cup R_3\cup R_6 \cup R_7$ which compound region has boundary which is the straight-line of slope $\frac{3\pi}{8}$ through the origin. Going through a similar analysis mutatis mutandis, we get that $$P(E_2\mid 000) = P(E_2\mid 100) = Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right), \tag{7}$$ and $$P(E_2\mid 001) = P(E_2\mid 101) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{8}$$ The conditional probability of $E_2$ given the other points can be calculated in a similar fashion and thus we arrive at $$P(E_2) = P(E_1) = \frac{1}{2}\left[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right].\tag{9}$$