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.
Stacey Warren - Expert brainly.com
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X being 1
N being \[\phi(x-\mu/1)\]
we add to 1 on top so
and divde by conditional probability
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)
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I'll put this subject on my "to do" list. In the meantime....
I think @phi is correct. As long as X is given as 1, then the probability of x occurring doesn't affect the pdf.
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.