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.