 LFD Book Forum Exercise 1.13 noisy targets
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 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: 1,478 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: 1,478 Re: Exercise 1.13 noisy targets

Correct. __________________
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#6
 elyoum Junior Member Join Date: May 2016 Posts: 3 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by yaser Correct. #7
 ckong41 Junior Member Join Date: Apr 2021 Posts: 2 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by prithagupta.nsit SO final Probability of error that h makes in approximating y would be: 1+2*lamda*mu -mu -lamda.
Anyone know how this user arrived at this step?
#8 htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 610 Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by ckong41 Anyone know how this user arrived at this step?
I think it can be derived by calculating (1-mu) * (1-lambda)+mu * lambda . Hope this helps.
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#9
 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.
#10
 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. Thread Tools Show Printable Version Email this Page Display Modes Switch to Linear Mode Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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