LFD Book Forum Exercise 1.13 noisy targets
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#1
10-21-2014, 05:11 PM
 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
10-21-2014, 11:20 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,474
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!
Answering your questions (i) and (ii): Yes and yes.

In Exercise 1.13(b): Independent of means that changing the value of does not affect how well predicts .
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#3
10-22-2014, 04:45 PM
 mahaitao Junior Member Join Date: Oct 2014 Posts: 6
Re: Exercise 1.13 noisy targets

Thank you very much, professor.
#4
08-06-2015, 04:36 AM
 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
08-06-2015, 04:02 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,474
Re: Exercise 1.13 noisy targets

Correct.
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#6
05-12-2016, 02:24 AM
 elyoum Junior Member Join Date: May 2016 Posts: 3
Re: Exercise 1.13 noisy targets

Quote:
 Originally Posted by yaser Correct.
can i ask you some questions please?
#7
10-09-2017, 05:25 PM
 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
11-14-2017, 02:52 PM
 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.

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