Re: Exercise 1.13 noisy targets
I don't understand why the case y != f(x) and h(x) != f(x) doesn't count toward the probability that y != h(x). We have four cases:
(1) y = f(x) and h(x) = f(x) imply y = h(x);
(2) y != f(x) and h(x) = f(x) imply y != h(x);
(3) y = f(x) and h(x) != f(x) imply y != h(x);
(4) y != f(x) and h(x) != f(x) imply neither y = h(x) or y != h(x).
For instance if at x = 0 we had y = 1, h(0) = 2 and f(0) = 3, then we are in case (4) and y != h(x). But if at x = 1 we had y = 4, h(1) = 4 and f(1) = 5, then we are in case (4) and y = h(x). What am I missing?
