# How to evaluate pairwise error probability and detection in presence of gaussian noise?

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 $$d_{14}/2$$.

If you read the book carefully, you will notice and understand that the probability given by the equation

$$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$

is the error probability for the event that the received signal x is closer to $$s_k$$ than $$s_i$$, when $$s_i$$ is sent. In your figure, it is the error probability that $$x$$ is closer to $$s_4$$ than $$s_1$$ when $$s_1$$ is sent, or $$x$$ is closer to $$s_1$$ than $$s_4$$ when $$s_4$$ is sent.

The situation which makes you confused is a different situation, in which $$x$$ is closer to $$s_1$$ than $$s_4$$ when $$s_1$$ is sent. This situation does not create an error decision, and makes no contribution to the error probability.

• The other thing not being taken into account is the decision boundaries. For the four points show and zero mean, symmetric noise, I believe the decision boundaries will be the lines $y=x$ and $y=-x$. The given integral formula for the error is only applicable if the measurement point $x$ is on the line between $s_1$ and $s_4$. If it's not, $x$ may be closer to $s_2$ or $s_3$.
– Peter K.
Jun 23 at 2:05
• @Peter K. Please note we are talking about pairwise error probabilities here and therefore, the decision boundary is just between two signal points and not more than that. Jun 23 at 4:02
• @Userhanu OK. Seems like a toy problem, though, that could have been done with just two signal points $s_1$ and $s_2$.
– Peter K.
Jun 23 at 18:17