I have attempted the following problem:
The input X to a communications channel is $+1$ volt or $-1$ volt with equal probability. The output Y of the channel is the input plus a noise voltage N that is uniformly distributed in the interval from $-2$ volts to $+2$ volts. Find the joint probability $P(X = +1, Y ≤ 0)$.
Here is my attempt at a solution:
First we note that the probability $P(X = +1)$ is given to us in the problem as $1/2$ because the problem tells us that input $X$ can either be $+1$ or $-1$ with equal probability.
We need to find the probability $P(Y≤ 0)$. We are told that output $Y$ is the input $X$ plus $N$ which is uniformly distributed from $-2 $ to $2$.
We know then that if $ Y≤0$, then $-2≤N≤ -1$
Because we are told that $N$ is uniformly distributed, we can find the probability that $N$ meets our conditions as follows:
$P(-2≤N≤ -1)=$$
\frac{-1–2 \ }{4} = \frac18
$
So:
$P(X= +1,Y≤0)= \frac12 \times \frac14 = \frac18$
Is my solution correct? If not, could you please point out the flaw(s) in my solution?
Best Answer
Yes.$$\small\begin{align}\mathsf P(X=+1, Y\leq 0)&=\mathsf P(X=+1, N\leq -X)&&N:=Y-X\\&=\mathsf P(X=+1)\,\mathsf P(N\leq -X\mid X=+1)&&\text{Conditional Probability}\\&=\mathsf P(X=+1)\,\mathsf P(N\leq -1)&&\text{Independence}\\&=\dfrac 12\cdot\dfrac{(-1)-(-2)}{4}&&\text{pmf and CDF of given distributions}\\&=\dfrac{1}{8}&&\text{Algebra}\end{align}$$