*ANSWER* Hw 2, problem 3

[useral]

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]

Originally Posted by useral

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]

Originally Posted by htlin

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.