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-   -   *ANSWER* Hw 2, problem 3 (http://book.caltech.edu/bookforum/showthread.php?t=371)

useral 04-18-2012 11:13 AM

*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?

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 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.

useral 04-19-2012 02:25 PM

Re: Hw 2, problem 3
 
Quote:

Originally Posted by htlin (Post 1428)
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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