
#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
Anyone know how this user arrived at this step?

#8




Re: Exercise 1.13 noisy targets
I think it can be derived by calculating (1mu) * (1lambda)+mu * lambda . Hope this helps.
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#9




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. 
#10




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

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