# Determining the average probability of error

I'm trying to solve the following problem:

In a binary PAM system, the input to the detector is $$y_m = a_m+n_m+i_m$$ where $$a_m = \pm1$$ is the desired signal, $$n_m$$ is a zero-mean Gaussian random variable with variance $$\sigma_n^2$$ and $$i_m$$ represents the ISI due to channel distortion. The ISI term is a random variable which takes the values, $$\frac{-1}{2}, 0, \frac{1}{2}$$ with probabilities $$\frac{1}{4}, \frac{1}{2},\frac{1}{4}$$ respectively. Determine the average probability of error as a function of $$\sigma_n^2$$.

My try: I think the probability of error is $$P_e = \mathbb{P}(y_m<0 |a_m = 1) + \mathbb{P}(y_m\gt 0|a_m = -1)$$According to the definition of conditional probability, we have $$\mathbb{P}(y_m<0 |a_m = 1) = \frac{\mathbb{P}(y_m<0 \cap a_m = 1)}{\mathbb{P}(a_m = 1)} = \frac{\mathbb{P}(n_m+i_m<-1)}{\mathbb{P}(a_m = 1)}$$ By law of total probability $$\mathbb{P}(n_m+i_m<-1) = \mathbb{P}(n_m+i_m<-1\cap i_m = \frac{-1}{2}) + \mathbb{P}(n_m+i_m<-1\cap i_m = 0) + \mathbb{P}(n_m+i_m<-1\cap i_m = \frac{1}{2}) = \mathbb{P}(n_m<\frac{-1}{2}) + \mathbb{P}(n_m<-1) + \mathbb{P}(n_m<\frac{-3}{2})$$ Each term can be written as a function of $$\sigma_n^2$$ easily. So it seems we need $$\mathbb{P}(a_m = \pm1)$$ instead of $$\mathbb{P}(i_m = \pm\frac{1}{2},0)$$ which can't be true because we should certainly use values of $$\mathbb{P}(i_m = \pm\frac{1}{2},0)$$. What's my mistake here? Also is there any difference between "average probability of error" and "probability of error"?

• Be careful not to count events twice. For example, if the noise is less than -1.5, there's an error regardless of the value of $i_m$. Also, in your first equation, the two terms on the right-hand side are equal?
– MBaz
Jan 22, 2021 at 15:02
• @MBaz Thanks for your reply, I fixed it. I think the term $\mathbb{P}(y_m<0 |a_m = 1) + \mathbb{P}(y_m\gt 0|a_m = -1)$ counts every events of interest exactly once. Is this correct? Jan 22, 2021 at 15:10
• Yes, that looks correct. Now make sure you don't count noise events more than once.
– MBaz
Jan 22, 2021 at 16:39
• @MBaz My main problem is computing $\mathbb{P}(y_m<0 |a_m = 1)$. I don't know if $$\mathbb{P}(y_m<0 |a_m = 1) = \frac{\mathbb{P}(y_m<0 \cap a_m = 1)}{\mathbb{P}(a_m = 1)} = \frac{\mathbb{P}(n_m+i_m<-1)}{\mathbb{P}(a_m = 1)}$$ is correct. Jan 23, 2021 at 1:57

Note that

$$\mathbb{P}(n_m+i_m<-1\cap i_m = \frac{-1}{2})$$ is equal to $$\mathbb{P}(n_m < -0.5)\mathbb{P}(i_m = -0.5) = 0.25\mathbb{P}(n_m < -0.5)$$ since $$n_m$$ and $$i_m$$ are independent.

Another way to obtain the solution is to add the the probability of these three disjoint events:

• $$n_m < -1.5$$,
• $$i_m = 0$$ and $$-1.5 < n_m < -1$$,
• $$i_m = -0.5$$ and $$-1 < n_m < -0.5$$

In this case, we can add the three probabilities because the events are disjoint. To keep them disjoint, it is important to keep the noise ranges disjoint too; otherwise, you would be adding them multiple times.

• Thanks. By the symmetry of the problem we have $\mathbb{P}(y_m<0 |a_m = 1) = \mathbb{P}(y_m\gt 0|a_m = -1)$, so I found the answer to be $P_e = 2\mathbb{P}(y_m<0 |a_m = 1) = 2(\frac{1}{4}Q(\frac{1}{2\sigma}) + \frac{1}{2}Q(\frac{1}{\sigma}) + \frac{1}{4}Q(\frac{3}{2\sigma}))$. Is this correct? Jan 23, 2021 at 18:22
• Not quite; you need to average the probabilities for $a_m=1$ and $a_m=-1$, not add them.
– MBaz
Jan 23, 2021 at 20:02
• Would you explain more, please? An error will occur when $\{y_m<0 |a_m = 1\}$ or $\{y_m\gt 0|a_m = -1\}$ happen. So we need to add them to get the probability of error but you are saying that the answer is $\frac{1}{2}(\mathbb{P}(y_m<0 |a_m = 1)+ \mathbb{P}(y_m\gt 0|a_m = -1))$ Jan 23, 2021 at 20:12
• Let $e$ stand for "error", $s_1$ for $a_m=1$ and $s_2$ for $a_m=-1$. You have calculated $P(e|s_1)$ and $P(e|s_2)$, but you want $P(e) = P(e \cap s_1)+P(e \cap s_2)$, which equals $P(e|s_1)P(s_1)+P(e|s_2)P(s_2)$.
– MBaz
Jan 23, 2021 at 23:01
• I see. I was confusing conditional probability with joint probability. So the correct answer is $P_e = \frac{1}{4}Q(\frac{1}{2\sigma}) + \frac{1}{2}Q(\frac{1}{\sigma}) + \frac{1}{4}Q(\frac{3}{2\sigma})$. Thank you so much for your patience. Jan 24, 2021 at 0:14