 LFD Book Forum Exercise 1.13 noisy targets

#1
 mahaitao Junior Member Join Date: Oct 2014 Posts: 6 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 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by mahaitao 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!

In Exercise 1.13(b): Independent of means that changing the value of does not affect how well predicts .
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#3
 mahaitao Junior Member Join Date: Oct 2014 Posts: 6 Re: Exercise 1.13 noisy targets

Thank you very much, professor.
#4
 prithagupta.nsit Junior Member Join Date: Jun 2015 Posts: 7 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 =1-lamda =1/2

It think this should be correct answer.

Is my understanding correct for second part of the question ?
#5 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477 Re: Exercise 1.13 noisy targets

Correct. __________________
Where everyone thinks alike, no one thinks very much
#6
 elyoum Junior Member Join Date: May 2016 Posts: 3 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by yaser Correct. #7
 Vladimir Junior Member Join Date: Oct 2017 Posts: 1 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
 don slowik Member Join Date: Nov 2017 Posts: 11 Re: Exercise 1.13 noisy targets

The case you mention would lead to h(x) = y.
#9
 Ulyssesyang Junior Member Join Date: Nov 2018 Posts: 3 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by mahaitao 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!
So why don’t you consider h(x)!=f(x) and f(x) != y? Even if there is some case here h(x) may equal to y, but we still have case here h(x)!=y.

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