#1




Query regarding Noise Targets lecture 4
As per lecture 4 : Slide 16/22
Noise target = Diterministic Target + Noise (This is clear) However it then says : Diterministic target f(x) = E(YX). Hence y ( Noisy target) = E(YX) ( Diterministic target) + noise ( yf(x) ) I do not understand the above statement. Did we not say that to accomodate for noisy targets we replace y = f(x) with a conditional distribution of y given x. Should then not the noisy target be defined by E(YX) ? 
#2




Re: Query regarding Noise Targets lecture 4
Quote:
__________________
Where everyone thinks alike, no one thinks very much 
#3




Re: Query regarding Noise Targets lecture 4
Is E(YX ) the noiseless component ? If yes, how ?

#4




Re: Query regarding Noise Targets lecture 4
Yes, it is. It is a deterministic function of . The noisy aspect has been integrated out (averaged out) by the expected value.
__________________
Where everyone thinks alike, no one thinks very much 
#5




Re: Query regarding Noise Targets lecture 4
Understood. Thanks prof Yaser

#6




Re: Query regarding Noise Targets lecture 4
One question:
If is a random sample, it is clearly a random variable (or rather, a sequence of random variables). If is also a random variable (the target), then wouldn't be a random variable and not a deterministic quantity? (unless and are independent variables in which case the conditional expectation collapses to a constant?) 
#7




Re: Query regarding Noise Targets lecture 4
Quote:
The conditional expectation, , is a deterministic quantity (more presisely, a deterministic function of ). As you point out, if and are statistically independent, that function is a constant (independent of ). However, even if and are not statistically independent, that function is still a deterministic, albeit nonconstant, function of .
__________________
Where everyone thinks alike, no one thinks very much 
#8




Re: Query regarding Noise Targets lecture 4

Thread Tools  
Display Modes  

