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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
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)
I'll put this subject on my "to do" list. In the meantime....
A more interesting (read: "hard") pdf is the overall pdf that encompasses both x = 1 and x = -1. Thankfully, that wasn't the question :)
my "to do" list seems to grow with every day I sign on here... (sigh)
"then the probability of x occurring doesn't affect the pdf." can you explain why?
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.
https://onlinecourses.science.psu.edu/stat414/book/export/html/95 This site might help you with Conditional Distributions. It might help you.
thank you so much for not ignoring my question
Your welcome. anytime.
thanks to jessica as well
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.
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.