#1




Exercise 1.13 noisy targets
Exercise 1.13(a): what is the probability of error that h makes in approximating y if we use a noisy version of f. That means we want to compute Pr(h(x)~=y), and I consider two cases:
(1) h(x)=f(x) and f(x) != y; [(1\mu)*\(1\lambda)] (2) h(x)!=f(x) and f(x) = y. [\mu*\lambda] I am not sure the solution is right. My questions are follows: (i) Does "h makes an error with \mu in approximating a deterministic target function f" mean Pr(h(x) != f(x)) = \mu? (ii) Does the probability of Pr(h(x)~=y)=Pr(1)+Pr(2)? Exercise 1.13(b) : I am not clear what does "performance of h be independent of \mu" mean? Should I consider Pr(h(x)~=y)? thanks! 
#2




Re: Exercise 1.13 noisy targets
Quote:
In Exercise 1.13(b): Independent of means that changing the value of does not affect how well predicts .
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#3




Re: Exercise 1.13 noisy targets
Thank you very much, professor.

#4




Re: Exercise 1.13 noisy targets
SO final Probability of error that h makes in approximating y would be:
1+2*lamda*mu mu lamda. if it should be independent of mu then lamda should be 1/2 1+2*1/2*mu mu lamda =1lamda =1/2 It think this should be correct answer. Is my understanding correct for second part of the question ? 
#5




Re: Exercise 1.13 noisy targets
Correct.
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#6




Re: Exercise 1.13 noisy targets

#7




Re: Exercise 1.13 noisy targets
Dear Professor,
What about the case h(x)!=f(x) and f(x) != y? Does it count on the probability of Pr(h(x)~=y)? Thanks. 
#8




Re: Exercise 1.13 noisy targets
The case you mention would lead to h(x) = y.

#9




Re: Exercise 1.13 noisy targets
Quote:

#10




Re: Exercise 1.13 noisy targets
Was wondering if the following intuitive approach works for part (b). It's as far as I'm getting thus far.
The question is asking, "What accuracy setting of f (in its prediction of noisy target y) would render h's accuracy (in its prediction of target distribution f) inconsequential?" If f got everything wrong, that is, lamb=0: h's ability to model f matters. The worse it does on f, the better it does on y. If f got everything right, that is, lamb=1: h's ability to model f matters. The better it does by f, the better it does on y. If f got 50%, that is, lamb=0.5: h's ability to model f would not matter. For this, considering the two remaining quartiles: If h modeled f 75% of the time, that's 75% of 50% and 25% of 50% = (.75*.5)+(.25*.5) = 0.5 If h modeled f 25% of the time, that's 25% of 50% and 75% of 50% = (.75*.5)+(.25*.5) = 0.5 General case: if h models f with probability (1mu), then h models y with probability (1mu)*0.5 + mu*0.5 = (1mu+mu)*0.5 = 0.5. 
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