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I'm working on the below question and I'm a bit stuck on all of the definition stuff, I have a more specific question below, but I thought it would be best to give some context to my question. I'm pretty stuck on how to approach this so any help would be much appreciated!

Consider a binary communication channel in which the input $X$ is either $c$ or $−c$ with probabilities $1/3$ and $2/3$, respectively.

Here, $c≥0$ is a given constant. The channel output is given by $Y=X+N$,where $N$ is a standard normal random variable. For parts (b) and (c), you should provide closed-form expressions for your answers in terms of the standard normal CDF, $Φ$.

so my logic is, to find the conditional PDF, I'm trying to find the joint PDF of $x$ and $y$ and divide it by the PDF of $X$

Now I have the PDF of X down, so I was trying to get the PDF of $Y$ which is $N+X$. I know $f_X(x)$ and $f_N(n)$ separately but how do I put them together to get the PDF of $Y$? Does it make sense for me to just add them together? AKA $f_Y(y)= f_X(x)+f_N(n)?$

(a) Find the conditional PDF $f_Y|X(y|x)$ of $Y$ given $X=x$. What kind of random variable is this?

(b) Find the marginal PDF $f_Y(y)$ of $Y$.

(c) Find the conditional probability $P(X=c|Y >0)$. What happens as $c$ gets larger?

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    $\begingroup$ Please cancel every "Thank you", every exclamation mark and every use of boldfaced characters. $\endgroup$
    – Did
    Commented Apr 9, 2017 at 15:09
  • $\begingroup$ Good, you got rid of the "thank you"s, now for the exclamation marks and the uses of boldface characters... $\endgroup$
    – Did
    Commented Apr 9, 2017 at 20:33

1 Answer 1

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You are given that $X$ can be equal to $-c$ and $c$ only. If $X=c$, then $Y=N+c$ has normal distribution $N(c,1)$. And for this case $$f_{Y\mid X}(y\mid c)=\frac{1}{\sqrt{2\pi}}e^{-(y-c)^2/2}.$$

Find $f_{Y\mid X}(y\mid -c)$ and then by total probability law $$ f_Y(y)=f_{Y\mid X}(y\mid c)\mathbb P(X=c)+f_{Y\mid X}(y\mid-c)\mathbb P(X=-c). $$

For (c) use the definition of conditional probability.

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