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Old 05-31-2021, 12:15 PM
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htlin htlin is offline
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Default Re: Exercise 1.13 noisy targets

Quote:
Originally Posted by anon4 View Post
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?
This case is about binary classification, where all the outputs are +/- 1. So your cae (4) actually implies that y = h(x). Hope this helps.
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