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
 

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 .

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 =1-lamda =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
 

can i ask you some questions please?

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
 

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.