## Libniz 2 years ago The input X to a binary communication channel assumes the values +1 or −1 with probabilities 1/3 and 2/3 respectively. The output Y of the AWGN channel is given by Y = X + N where N is zero mean Gaussian noise with variance 2 n = 1. 1. Find the conditional pdf of Y given X = +1.

1. Libniz

@phi

2. Libniz

so obiviously Y=X+N X being 1 Y=1+N Y=1+N N being $\phi(x-\mu/1)$ $\phi(x-0/1)$ $\phi(1-0/1)$=$\phi(1)$ we add to 1 on top so $\phi(1)+1$ and divde by conditional probability 1/3

3. phi

I have not looked at this stuff, but my intuition is that if x is given as 1, the pdf for y should look like N, but centered at 1 (a gaussian with 1 mean, variance 2)

4. phi

I'll put this subject on my "to do" list. In the meantime....

5. JakeV8

I think @phi is correct. As long as X is given as 1, then the probability of x occurring doesn't affect the pdf.

6. JakeV8

A more interesting (read: "hard") pdf is the overall pdf that encompasses both x = 1 and x = -1. Thankfully, that wasn't the question :)

7. JakeV8

my "to do" list seems to grow with every day I sign on here... (sigh)

8. Libniz

"then the probability of x occurring doesn't affect the pdf." can you explain why?

9. Libniz

@satellite73

10. JakeV8

I thought of it just like simple conditional probability... if P(x) = 0.5 and P(y) = 0.3, but the problem says, "Given x, what is the probability of y?", then the P(x) doesn't affect the probability of P(x|y), since you know x is given. Here, the fact that x = 1 occurs 1/3 of the time is irrelevant to the pdf if you are given that x = 1.

11. Jessica_Moore

12. Libniz

thank you so much for not ignoring my question

13. Jessica_Moore

14. Libniz

thanks to jessica as well

15. JakeV8

So, I think if you run "x = 1" through the channel, then you would expect Y to look like 1, but it would have the shape of the pdf of the AWGN due to the channel. But the signal X = 1 would move the mean to be around 1 vs. around 0 which is what the channel looks like with no signal.

16. JakeV8

My comment earlier about the overall pdf was if you were trying to build a sensor/detector, and you didn't know whether X was +1 or -1, and you picked up an output signal Y = 0.2, you would have to decide if it was really an X= -1 input or if it was an X = + 1 input... the conditional pdfs would be overlapping.

17. Libniz

ok,