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 useral 04-18-2012 11:13 AM

Hello,

I must be missing the forest for the trees, but I was thinking that the answer should be the weighted sum of the probabilities of (y = f(x)) and (y =/= f(x)), and the weight mu (corresponding to an error, i.e. y =/= f(x)) should be assigned to the probability of that error, i.e. to 1-lambda.

This reasoning gives the answer [d]. Is it flawed?

 htlin 04-18-2012 04:39 PM

Re: Hw 2, problem 3

Quote:
 Originally Posted by useral (Post 1409) Hello, I must be missing the forest for the trees, but I was thinking that the answer should be the weighted sum of the probabilities of (y = f(x)) and (y =/= f(x)), and the weight mu (corresponding to an error, i.e. y =/= f(x)) should be assigned to the probability of that error, i.e. to 1-lambda. This reasoning gives the answer [d]. Is it flawed?
The cases that would be different from is when either, but not both, of these events happen: , or . Hope this helps.

 useral 04-19-2012 02:25 PM

Re: Hw 2, problem 3

Quote:
 Originally Posted by htlin (Post 1428) The cases that would be different from is when either, but not both, of these events happen: , or . Hope this helps.
Thanks, that clarifies it.

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