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Libniz
Group Title
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.
 one year ago
 one year ago
Libniz Group Title
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.
 one year ago
 one year ago

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Libniz Group TitleBest ResponseYou've already chosen the best response.0
so 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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
I'll put this subject on my "to do" list. In the meantime....
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
I think @phi is correct. As long as X is given as 1, then the probability of x occurring doesn't affect the pdf.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
A more interesting (read: "hard") pdf is the overall pdf that encompasses both x = 1 and x = 1. Thankfully, that wasn't the question :)
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
my "to do" list seems to grow with every day I sign on here... (sigh)
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
"then the probability of x occurring doesn't affect the pdf." can you explain why?
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
@satellite73
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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(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.
 one year ago

Jessica_Moore Group TitleBest ResponseYou've already chosen the best response.0
https://onlinecourses.science.psu.edu/stat414/book/export/html/95 This site might help you with Conditional Distributions. It might help you.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
thank you so much for not ignoring my question
 one year ago

Jessica_Moore Group TitleBest ResponseYou've already chosen the best response.0
Your welcome. anytime.
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
thanks to jessica as well
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago
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