1
$\begingroup$

The input $X$ to a binary communication channel assumes the value $+1$ or $-1$ with probability $\frac{1}{3}$ and $\frac{2}{3}$ respectively. The output of $Y$ of the AWGN channel is given by $Y=X+N$ where $N$ is zero mean Gaussian noise with variance $=1.$

Find the conditional pdf of $Y$ given $X=+1$.

So I started working on it:
$P_X(\{1\})=\frac{1}{3}$
$P_X(\{-1\})=\frac{2}{3}$
I am not sure how to deal with N.

  • 0
    Since you want the conditional distribution of $Y$ given $X=1$, you don't need to worry about the probability that $X=1$. What is the mean of $Y$ if $X = 1$? What is its variance? What family of distribution does it have?2012-10-29

1 Answers 1

1

We are conditioning on $X=1$. So $X$ is $1$. And therefore $Y=1+N$ is just a shifted standard normal, so a normal of mean $1$, variance $1$. The density function, if that's what you want, is therefore $\frac{1}{\sqrt{2\pi}}e^{-(y-1)^2/2}.$

  • 0
    No, the *conditional* density function is precisely the one written above.2012-10-30