Thread: Exercise 1.11
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Old 06-04-2016, 08:41 AM
henry2015 henry2015 is offline
Join Date: Aug 2015
Posts: 31
Default Re: Exercise 1.11

Originally Posted by henry2015 View Post
For part c, I thought:

Given p = 0.9, h1 is a better hypothesis than h2.

Hence, the probability that S produces a better hypothesis than C is the probability that S picks h1 essentially as C will pick the other hypothesis that S doesn't pick.

In other words, P[S produces a better hypothesis than C] = P[S picks h1 based on the 25 training examples].

S will pick h1 if 13 out of 25 training examples give +1, so we will have:
P[S picks h1]
= P[13 or more out of 25 training examples give +1]
= \sum_{k = 13}^{25}\binom{25}{k}(.9)^{k}(.1)^{25-k}
= 0.9999998379165839813935344

It is quite different from tatung2112's explanation for c.

Could you comment further?

I just noticed that the formula in my post actually is one form of the formula in Problem 1.7...

Now, I am even more confused.
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