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  #1  
Old 04-18-2012, 11:13 AM
useral useral is offline
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Default *ANSWER* Hw 2, problem 3

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?
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Old 04-18-2012, 04:39 PM
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htlin htlin is offline
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Default Re: Hw 2, problem 3

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Originally Posted by useral View Post
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 h(\mathbf{x}) would be different from y is when either, but not both, of these events happen: h(\mathbf{x}) \neq f(\mathbf{x}), or f(\mathbf{x}) \neq y. Hope this helps.
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Old 04-19-2012, 02:25 PM
useral useral is offline
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Default Re: Hw 2, problem 3

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Originally Posted by htlin View Post
The cases that h(\mathbf{x}) would be different from y is when either, but not both, of these events happen: h(\mathbf{x}) \neq f(\mathbf{x}), or f(\mathbf{x}) \neq y. Hope this helps.
Thanks, that clarifies it.
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