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arapmv 08-06-2013 09:14 AM

calculation 4.8 on p.139
 
Hello,

I have a question regarding the calculation (4.8) on p.139 in the book. The final hypothesis g^{-} depends, albeit indirectly, on the choice of the validation set \mathcal{D}_{val}. Indeed, by construction we train on the complement of \mathcal{D}_{val} in \mathcal{D}, which makes g^{-} dependent on the choice of \mathcal{D}_{val}. The derivation (4.8) appears to rely on the assumption that g^{-} is independent of \mathcal{D}_{val}.

Does anyone have any comments on this?

magdon 08-07-2013 10:48 AM

Re: calculation 4.8 on p.139
 
I will try to reformulate the construction of Dval in such a way that the independence is patent, and then try to suggest where your confusion may be coming from.

Construction 1 of Dval: Randomly generate D. Randomly partition it into Dtrain (N-K points) and Dval (K points). Learn on Dtrain to obtain g^- and compute Eval of g^- using Dval. (This is the standard validation setting in the book.)

Construction 2 of Dval: Randomly generate N-K points to form Dtrain. Learn on Dtrain to obtain g^-. Now, randomly generate another K points to form Dval. Compute Eval of g^- using Dval.

It is patently clear in Construction 2 that we are computing Eout of g^-, essentially by definition of Eout. There is no difference between constructions 1 and 2 in terms of the Dtrain and Dval they produce (statistically). Randomly generating N points and splitting randomly into N-K and K points is statistically equivalent to first randomly generating N-K points and then another random K points. In construction 1 you generate both Dtrain and Dval at the begining, process Dtrain and then test on Dval. In construction 2, you only generate Dval after you processed Dtrain. But Dval still has the same statistical properties in both cases.

Now for where you may be getting subtly confused. It is true that the value of Eval will change based on what specific partition was selected, in part because g^- changes and in part because Dval also changes. This means that g^- depends on the partition. This is equivalently saying (in the construction 2 setting) that g^- depends on the particular Dtrain generated (no surprise there). However, g^- does not depend on the contents of Dval - if you change the data points in Dval, g^- will not change. The expectation in (4.8) is an expectation over the data points in Dval, and g^- does not depend on that (the partition is fixed and now we are looking at what points are in Dval).

Quote:

Originally Posted by arapmv (Post 11311)
Hello,

I have a question regarding the calculation (4.8) on p.139 in the book. The final hypothesis g^{-} depends, albeit indirectly, on the choice of the validation set \mathcal{D}_{val}. Indeed, by construction we train on the complement of \mathcal{D}_{val} in \mathcal{D}, which makes g^{-} dependent on the choice of \mathcal{D}_{val}. The derivation (4.8) appears to rely on the assumption that g^{-} is independent of \mathcal{D}_{val}.

Does anyone have any comments on this?


arapmv 08-09-2013 05:29 PM

Re: calculation 4.8 on p.139
 
Thank you very much for the lucid explanation!

arapmv 08-09-2013 06:02 PM

Re: calculation 4.8 on p.139
 
On the second thought, I am still not seeing the statistical equivalence of the two constructions. The first construction will always produce two disjoint sets \mathcal{D}_{val} and \mathcal{D}_{train} of sizes K and N-K, respectively. The second construction may easily produce two non-disjoint sets. Is this a problem?

Admin Edit: Replaced $ with [math] tag in math expressions.

magdon 08-10-2013 08:57 AM

Re: calculation 4.8 on p.139
 
Not a problem. The two constructions generate identically distributed Dtrain and Dval.

The partitioning approach in construction 1 generates data points with disjoint indices. That does not mean that the data points in Dtrain cannot also appear in Dval. (remember that when you generate D, the data points are iid so there can be repetitions.)

arapmv 08-11-2013 11:14 AM

Re: calculation 4.8 on p.139
 
Thanks for the explanation! I did not realize that we are doing sampling with replacement to construct the training set and the validation set. Now it is clear why the two constructions you gave are equivalent!

The first step of the Calculation 4.8 on p.139 is still not sinking into my brain somehow. It feels like in the first step we pulled out a variable (namely, $\mathcal{D}_{val}$) from under the integral sign and placed it as an indexing set for the summation. What is the correct mathematical interpretation of this step?

Thank you for your clear explanations and patience!


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