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"?