I am reading Digital Communication Systems by Simon Haykin and I am stuck at one point.
Consider a two-dimensional signal space that has a message constellation of four points, given by $s_1,s_2,s_3,s_4$. We need to calculate pairwise error probability in the presence of additive white Gaussian noise ($w$). The signal vectors are arranged so as to be aligned with the basis vectors, as shown in the figure
The observed vector $x$ is given as $$x=s_1+w$$ where $s_1$ is the transmitted message. We can calculate the probability of error using $\int_{Z_j} f_x(x|s_i)dx$ where $j\neq i$ and $Z$ is the observation space which is divided into various small $Z_i$'s for detection purposes. Assuming variance of noise to be $N_0/2$, we can write the error as: $$\int_{Z_j}\frac{1}{\sqrt{\pi N_0}}\exp(-(x-s_1)^2/N_0) dx$$ As shown in the figure
we take the bisector and define the distance between $s_1$ and $s_4$ as $d_{14}$, which gives the pairwise probability expression as: $$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$ where $d_{14}$ is the distance between constellation points $s_1$ and $s_4$ However, my confusion is what if we exceed the limit $d_{14}/2$ in another location as shown in the figure
As Euclidean distance is always positive, does the expression still hold? Please note I know that detection is going to work as the distance between $x$ and $s_1$ is smaller than between $x$ and $s_4$, but I am more concerned about this integral expression, how does it work here? $$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$ because the distance is greater than $d_{14}/2$.
