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vikasatkin 02-13-2013 09:49 PM

Discussion of Lecture 11 "Overfitting"
Links: [Lecture 11 slides] [all slides] [Lecture 11 video]

Question: (Slide 10/23) It seems, that this situation is not typical, because the data points are clashed together. Should not we space the points evenly? Would we get a different result in that case?

Answer: To make sure, that the general result is not a fluctuation, the experiment was produced multiple times for each value of noise level and order of the polynomial. In each run points were chosen independently according to the uniform distribution. You can see this result on the slide 13/23. You may see some coincidences on the example on slide 10/23, but they were probably averaged out.

So the short answer is: you may interpret the figure on the slide 10/23 as an illustration and the slide 13/23 as the final result.

vikasatkin 02-13-2013 10:00 PM

Discussion of Lecture 11 "Overfitting"
Question: (Slide 11/23) How did you generate the polynomials? How did you choose the coefficients?

Answer: Here is the technical description of the process of generating the target function and the dataset (which may be useful, if you want to reproduce the pictures from the slide 13/23). It was actually described in the "Learning From Data" book on p.123 (section 4.1.2 "Catalysts for Overfitting").

The process of generating the target function depends on two parameters: Q_f (degree of the generated polynomials) and \sigma^2 (noise level). Of course, you also need N --- amount of points in the dataset.

1. Take Legendre polynomials P_0,\dots,P_{Q_f}. Note, that they are normalized according to their value at x=1 (i.e. P_q(1)=1), not their average square.
2. Choose coefficients a_0,\dots,a_{Q_f} independently according to the standard normal distribution.
3. Generate N points (pick them randomly from [-1,1], independently from each other).
4. For every point x_i generate the noise \epsilon(x_i).

The target is given by y=c\sum_{q=0}^{Q_f} a_q P_q(x) + \epsilon(x). Here c is a normalization constant, which depends only on Q_f. It is chosen in such a way, that the mean square value of f(x)=c\sum_{q=0}^{Q_f} a_q P_q(x) is equal to 1 (mean with respect to both x and choices we made during this process: \mathbb{E}_{a,x} ((f(x))^2) = 1). On can compute, that
c = \left( \sum_{q=0}^{Q_f} \frac1{2q+1} \right)^{-1/2}.

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