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anonymous
 4 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.
anonymous
 4 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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so obiviously Y=X+N X being 1 Y=1+N Y=1+N N being \[\phi(x\mu/1)\] \[\phi(x0/1)\] \[\phi(10/1)\]=\[\phi(1)\] we add to 1 on top so \[\phi(1)+1\] and divde by conditional probability 1/3

phi
 4 years ago
Best ResponseYou've already chosen the best response.0I 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)

phi
 4 years ago
Best ResponseYou've already chosen the best response.0I'll put this subject on my "to do" list. In the meantime....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think @phi is correct. As long as X is given as 1, then the probability of x occurring doesn't affect the pdf.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A more interesting (read: "hard") pdf is the overall pdf that encompasses both x = 1 and x = 1. Thankfully, that wasn't the question :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my "to do" list seems to grow with every day I sign on here... (sigh)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"then the probability of x occurring doesn't affect the pdf." can you explain why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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(xy), 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0https://onlinecourses.science.psu.edu/stat414/book/export/html/95 This site might help you with Conditional Distributions. It might help you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you so much for not ignoring my question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your welcome. anytime.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks to jessica as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My 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.
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