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"?

  • $\begingroup$ 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? $\endgroup$
    – MBaz
    Jan 22 at 15:02
  • $\begingroup$ @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? $\endgroup$
    – S.H.W
    Jan 22 at 15:10
  • $\begingroup$ Yes, that looks correct. Now make sure you don't count noise events more than once. $\endgroup$
    – MBaz
    Jan 22 at 16:39
  • $\begingroup$ @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. $\endgroup$
    – S.H.W
    Jan 23 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.

  • $\begingroup$ 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? $\endgroup$
    – S.H.W
    Jan 23 at 18:22
  • $\begingroup$ Not quite; you need to average the probabilities for $a_m=1$ and $a_m=-1$, not add them. $\endgroup$
    – MBaz
    Jan 23 at 20:02
  • $\begingroup$ 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 $\frac{1}{2}(\mathbb{P}(y_m<0 |a_m = 1)+ \mathbb{P}(y_m\gt 0|a_m = -1))$ $\endgroup$
    – S.H.W
    Jan 23 at 20:12
  • $\begingroup$ 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 $P(e|s_1)P(s_1)+P(e|s_2)P(s_2)$. $\endgroup$
    – MBaz
    Jan 23 at 23:01
  • 1
    $\begingroup$ 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. $\endgroup$
    – S.H.W
    Jan 24 at 0:14

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