LFD Book Forum Exercise 1.13 noisy targets
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#1
10-21-2014, 06:11 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
10-22-2014, 12:20 AM
 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!
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, 05:45 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, 05:36 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, 05:02 PM
 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
05-12-2016, 03:24 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, 06:25 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, 03:52 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.
#9
11-09-2018, 04:40 AM
 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.
#10
05-11-2021, 07:00 PM
 ckong41 Junior Member Join Date: Apr 2021 Posts: 2
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 (1-mu), then h models y with probability (1-mu)*0.5 + mu*0.5 = (1-mu+mu)*0.5 = 0.5.

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