The input $X$ to a communication channel takes on the values $±1$ with equal probabilities. The output of the channel is the random variable $Y = X + N$, where $N$ is a random variable representing noise, that is independent of $X$, and has probability density function: $f_N(z) = \frac{1}{2}e^{-|z|}$, $−∞ < z < ∞$.
(a) Find $P(X=k,Y≤y)$ for $k=±1$.
(b) Suppose you observe a negative output $Y$ and have to decide whether the input is $1$ or $−1$. What would you do?
For part (a), I think I have to do: $\frac{P(X=k, Y≤y)}{P(Y≤y)} = P(X=k|Y≤y)$. Then I was thinking about breaking that up into two cases for when $k=1$ and $k=-1$. However, I am not sure how to find the conditional probability $P(Y≤y|X=±1)$. Would I integrate $\frac{1}{2}e^{-|z|}+\frac{1}{2}$ from -∞ to y and multiply that by 2?
For part (b), I think I am interested in finding the quantity $P(X=1|Y)$ and $P(X=-1|Y)$. I think bayes rule is applicable here, but not really sure where to go with it.
Part $(a)$, It shall be easier to find $\mathsf P(Y\leq y\mid X=k)$ than $\mathsf P(X=k\mid Y\leq y)$
$X$ and $N$ are independent. $Y=X+N$. So for $k\in\{-1,+1\}$
$$\begin{split}\mathsf P(X=k, Y\leq y) &= \mathsf P(X=k)\;\mathsf P(Y\leq y\mid X=k)\\ &=\mathsf P(X=k)\,\mathsf P(X+N\leq y\mid X=k)\\ &= \mathsf P(X=k)\,\mathsf P( N\leq y-k) \\ &~~\vdots\end{split}$$
For (b) indeed use Bayes' Rule, and the Law of Total Probability.
$$\mathsf P(X=1\mid Y\leq 0) = \dfrac{\mathsf P(X=1)\;\mathsf P(Y\leq 0\mid X=1)}{\mathsf P(X=-1)\;\mathsf P(Y\leq 0\mid X=-1)+\mathsf P(X=1)\;\mathsf P(Y\leq 0\mid X=1)}$$