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! 
Re: Exercise 1.13 noisy targets
Thank you very much, professor.

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 ? 
Re: Exercise 1.13 noisy targets
Correct. :)

Re: Exercise 1.13 noisy targets
Quote:

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

Re: Exercise 1.13 noisy targets
Quote:

All times are GMT 7. The time now is 02:34 AM. 
Powered by vBulletin® Version 3.8.3
Copyright ©2000  2020, Jelsoft Enterprises Ltd.
The contents of this forum are to be used ONLY by readers of the Learning From Data book by Yaser S. AbuMostafa, Malik MagdonIsmail, and HsuanTien Lin, and participants in the Learning From Data MOOC by Yaser S. AbuMostafa. No part of these contents is to be communicated or made accessible to ANY other person or entity.