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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.
 2 years ago
 2 years 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.
 2 years ago
 2 years 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
 2 years 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)
 2 years 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....
 2 years 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.
 2 years 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 :)
 2 years 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)
 2 years 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?
 2 years ago

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

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

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

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