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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.

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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? –  Jonathan Christensen Oct 29 '12 at 19:50
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1 Answer 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 $$\frac{1}{\sqrt{2\pi}}e^{-(y-1)^2/2}.$$

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do we divide by probability of x being 1 which is 1/3? –  user46261 Oct 30 '12 at 20:22
    
No, the conditional density function is precisely the one written above. –  André Nicolas Oct 30 '12 at 20:43
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