#1




Hw5 Q8 E_out
I am struggling to understand how to calculate in this question. I have two competing theories, which I will describe below. Any help is greatly appreciated.
Once the algorithm terminates, I have . I now generate a new set of data points . Using my original target function to generate the corresponding . Case 1. Just use the same cross entropy error calculation but on this new data set. Case 2. Directly calculate the expected output of our hypothesis function and compare to . with probability Ultimately this gives us the probability that our hypothesis aligns with Y: In the lectures/book, we would multiply these probabilities to get the "likelihood" that the data was generated by this hypothesis. However, it seems that averaging over these should give the expected error in this sample. It feels as though the first approach is the correct one, but I struggle because the second approach makes intuitive sense since that is how I historically I would have calculated . To make matters worse, the two approaches very closely approximate different answers in the question! 
#2




Re: Hw5 Q8 E_out
Quote:
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Where everyone thinks alike, no one thinks very much 
#3




Re: Hw5 Q8 E_out
I suspected as much. I'll try to figure out why my other approach is wrong tomorrow. I think I've burned out on it today and am probably not seeing something obvious. Thanks for your help!

#4




Re: Hw5 Q8 E_out
I am also a little unsure about exactly how this equation works:
Obviously the more negative is the closer E_out is to zero which is good. So is w supposed to be normalized? I presume so because otherwise I could just scale w and then E_out becomes very small. And if it is normalized then the values I'm getting for E_in and E_out are both much greater than any of the options. (Maybe it's meant to be like that, if so it's quite unnerving.) 
#5




Re: Hw5 Q8 E_out
Quote:
__________________
Where everyone thinks alike, no one thinks very much 
#6




Re: Hw5 Q8 E_out
Quote:

#7




Re: Hw5 Q8 E_out
Thank you for explaining that normalization is not required. Your explanation now makes a bit more sense why I see the weights continue to increase in value the more iterations that I run.

#8




Re: Hw5 Q8 E_out
One very minor point which may help error prone folks such as myself:
In Linear methods we always have a d+1 dimensional weight vector by adding an extra pseudocoordinate of 1 for each training and test point. I completely overlooked this and was struggling with the error never getting as small as the expected answer. Spent a long time rechecking code etc. As soon as I fixed this, things fell into place. Likely I will never forget this :) 
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