A trick used in Rademacher complexity related Theorem
$begingroup$
I am currently working on the proof of Theorem 3.1 in the book "Foundations of Machine Learning" (page 35, First edition), and there is a key trick used in the proof (equation 3.10 and 3.11):
$E_{S,S'}[supfrac{1}{m}sum(g(z'_i)-g(z_i))]=E_{mathrm{sigma},S,S'}[supfrac{1}{m}sumsigma(g(z'_i)-g(z_i))]$
It is also shown in the lecture pdf page 8 in this link:
https://cs.nyu.edu/~mohri/mls/lecture_3.pdf
This is possible because $z_i$ and $z'_i$ can be swapped. My question is, why can we swap $z_i$ and $z'_i$? In the book, it says this is possible because "we are taking the expectation over all possible $S$ and $S'$, it will not affect the overall expectation." Unfortunately, I don't understand the meaning or intuition of this part.
Thank you very much for reading the question and for your time!
machine-learning pac-learning
asked 4 hours ago
ML studentML student
334
$endgroup$
add a comment |
$begingroup$
I am currently working on the proof of Theorem 3.1 in the book "Foundations of Machine Learning" (page 35, First edition), and there is a key trick used in the proof (equation 3.10 and 3.11):
$E_{S,S'}[supfrac{1}{m}sum(g(z'_i)-g(z_i))]=E_{mathrm{sigma},S,S'}[supfrac{1}{m}sumsigma(g(z'_i)-g(z_i))]$
It is also shown in the lecture pdf page 8 in this link:
https://cs.nyu.edu/~mohri/mls/lecture_3.pdf
This is possible because $z_i$ and $z'_i$ can be swapped. My question is, why can we swap $z_i$ and $z'_i$? In the book, it says this is possible because "we are taking the expectation over all possible $S$ and $S'$, it will not affect the overall expectation." Unfortunately, I don't understand the meaning or intuition of this part.
Thank you very much for reading the question and for your time!
machine-learning pac-learning
asked 4 hours ago
ML studentML student
334
$endgroup$
add a comment |
$begingroup$
I am currently working on the proof of Theorem 3.1 in the book "Foundations of Machine Learning" (page 35, First edition), and there is a key trick used in the proof (equation 3.10 and 3.11):
$E_{S,S'}[supfrac{1}{m}sum(g(z'_i)-g(z_i))]=E_{mathrm{sigma},S,S'}[supfrac{1}{m}sumsigma(g(z'_i)-g(z_i))]$
It is also shown in the lecture pdf page 8 in this link:
https://cs.nyu.edu/~mohri/mls/lecture_3.pdf
This is possible because $z_i$ and $z'_i$ can be swapped. My question is, why can we swap $z_i$ and $z'_i$? In the book, it says this is possible because "we are taking the expectation over all possible $S$ and $S'$, it will not affect the overall expectation." Unfortunately, I don't understand the meaning or intuition of this part.
Thank you very much for reading the question and for your time!
machine-learning pac-learning
asked 4 hours ago
ML studentML student
334
$endgroup$
I am currently working on the proof of Theorem 3.1 in the book "Foundations of Machine Learning" (page 35, First edition), and there is a key trick used in the proof (equation 3.10 and 3.11):
$E_{S,S'}[supfrac{1}{m}sum(g(z'_i)-g(z_i))]=E_{mathrm{sigma},S,S'}[supfrac{1}{m}sumsigma(g(z'_i)-g(z_i))]$
It is also shown in the lecture pdf page 8 in this link:
https://cs.nyu.edu/~mohri/mls/lecture_3.pdf
This is possible because $z_i$ and $z'_i$ can be swapped. My question is, why can we swap $z_i$ and $z'_i$? In the book, it says this is possible because "we are taking the expectation over all possible $S$ and $S'$, it will not affect the overall expectation." Unfortunately, I don't understand the meaning or intuition of this part.
Thank you very much for reading the question and for your time!
machine-learning pac-learning
machine-learning pac-learning
asked 4 hours ago
ML studentML student
334
asked 4 hours ago
ML studentML student
334
asked 4 hours ago
ML studentML student
334
asked 4 hours ago
ML studentML student
334
asked 4 hours ago
ML studentML student
334
334
add a comment |
add a comment |
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