HW 2 Problem 6

[dbaksi@gmail.com]

How is this different from problem 5 other than N=1000 and the fact that these simulated 'out of sample' points (E_out) are generated fresh ? I may be missing something but it seems to boil down to running the same program as in problem 5 with N=1000 for 1000 times; can someone clarify please ? thanks

[yaser]

Originally Posted by dbaksi@gmail.com

How is this different from problem 5 other than N=1000 and the fact that these simulated 'out of sample' points (E_out) are generated fresh ? I may be missing something but it seems to boil down to running the same program as in problem 5 with N=1000 for 1000 times; can someone clarify please ? thanks

There are indeed instances in the homeworks where the same experiment covers a number of homework problems.

Problem 5 asks about while Problem 6 asks about (an estimate of) . In both problems, ( stands for the number of training examples in our notation).

[MLearning]

It is my understanding that "fresh data" refers to cross-validation data. Do we then compute Eout using the weights obtained in problem 5? When I do this, Eout < Ein. When I design the weights using the fresh data, Eout is approximately equal to Ein. Does this makes sense?

[yaser]

Originally Posted by MLearning

It is my understanding that "fresh data" refers to cross-validation data. Do we then compute Eout using the weights obtained in problem 5?

It is simpler than cross validation (a topic that will be covered in detail in a later lecture). You just generate new data points that were not involved in training and evaluate the final hypothesis on those points.

The final hypothesis is indeed the one whose weights were determined in Problem 5, where the training took place.

[dsvav]

I am confused here , I don't understand what is final hypothesis here.

There are 1000 target function and corresponding 1000 weight vectors/hypothesis in problem 5 .

So for problem 6 , 1000 times I generate 1000 out-of-sample data and then for each weight vector and target function(from problem 5) I evaluate E_out for that out-of-sample data and finally average them. This is how I have done.

I don't see final hypothesis here , what I am missing , any hint

Could it be that in problem 5 there is supposed to be only one target function and many in-sample data ? If so then the final hypothesis/weights could be that produces minimum in-sample error E_in .

Please clarify.
Thanks a lot.

[yaser]

Originally Posted by dsvav

I am confused here , I don't understand what is final hypothesis here.

There are 1000 target function and corresponding 1000 weight vectors/hypothesis in problem 5 .

So for problem 6 , 1000 times I generate 1000 out-of-sample data and then for each weight vector and target function(from problem 5) I evaluate E_out for that out-of-sample data and finally average them. This is how I have done.

I don't see final hypothesis here , what I am missing , any hint

Could it be that in problem 5 there is supposed to be only one target function and many in-sample data ? If so then the final hypothesis/weights could be that produces minimum in-sample error E_in .

Please clarify.
Thanks a lot.

There is a final hypothesis for each of the 1000 runs. The only reason we are repeating the runs is to average out statistical fluctuations, but all the notions of the learning problem, including the final hypothesis, pertain to a single run.

[dbaksi@gmail.com]

Thanks a lot. The statements about (i) N being the number of 'in-sample' training data in both problems and (ii) the freshly generated 1000 points being disjoint from the first set clarified the confusion I had.