#11




Re: Exercise 1.13 noisy targets
Anyone know how this user arrived at this step?

#12




Re: Exercise 1.13 noisy targets
I think it can be derived by calculating (1mu) * (1lambda)+mu * lambda . Hope this helps.
__________________
When one teaches, two learn. 
#13




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? 
#14




Re: Exercise 1.13 noisy targets
Quote:
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When one teaches, two learn. 
#15




Re: Exercise 1.13 noisy targets
Yes, it makes much more sense now hahaha

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