In this video at 43:00, a version of the PAC bound for generalization error $\epsilon$, which I hadn't seen before, is quoted:

$$\epsilon^2 < \frac{\log{|H_\epsilon|} + \log{1/\delta}}{2m}$$

where $m$ is the number of samples, $\delta$ is the confidence parameter, and $H_\epsilon$ is the cardinality of an "$\epsilon$-cover of the hypothesis class", where he defines an $\epsilon$-cover as a set of subsets of the hypothesis class, such that the probability that two hypothesis in the same subset disagree is less than $\epsilon$.

Apart from the fact that this isn't a formal statement, I couldn't prove this myself. Has anyone heard of this version of PAC, and if so, could they point me to resources explaining it, or give some explanation here?


2 Answers 2


This follows from Massart's finite class lemma. Let $F$ is a binary function class restricted to some set $\{X_1,\ldots,X_n\}$, and let $P_n$ be the empirical (i.e., uniform) measure on this set. Then, for any $\epsilon>0$, the empirical Rademacher complexity of $F$ is bounded by $$ R_n(F;X) \le \epsilon + \sqrt{\frac{2\log N_F(\epsilon)}{n}},$$ where $N_F(\epsilon)$ is the $\ell_2$ $\epsilon$-covering number of $F$ w.r.t. $P_n$. This is proved in display (1) here: https://www.cs.bgu.ac.il/~asml162/wiki.files/dudley-pollard.pdf -- check out the course notes: https://www.cs.bgu.ac.il/~asml162/Class_Material

  • 1
    $\begingroup$ I'd say it's less difficult than, say, the standard PAC bounds, which rely on the "deep" Sauer's Lemma -- while this is really just a clever use of Hoeffding + union bound. $\endgroup$
    – Aryeh
    Commented Mar 27, 2018 at 19:54
  • 2
    $\begingroup$ On a side note, I like how in learning theory what is essentially "Chernoff + union bound" is named after a person :) $\endgroup$ Commented Mar 28, 2018 at 12:25
  • 1
    $\begingroup$ @SashoNikolov you bring up an interesting meta-point that may be worthy of its own post. Massart's finite class lemma is indeed a rather trivial result, likely folklore long before he published it. But he's a famous probabilist, well-known for highly non-trivial results (such computing the optimal constants in the DKW inequality). I've observed that famous people often have simple results named after them, and this is not unique to learning theory. Just look at the Johnson-Lindenstrauss lemma -- certainly not the deepest or most difficult, thought probably what they're best known for. $\endgroup$
    – Aryeh
    Commented Mar 28, 2018 at 12:29
  • 1
    $\begingroup$ I agree :). I think "Chernoff bound" is an even better example. Chernoff considered it a minor lemma in a paper, whose proof he had learned from Rubin. The proof technique is at least thirty years older than Chernoff's paper, and dates back to the work of Sergei Bernstein. But the name took off. $\endgroup$ Commented Mar 28, 2018 at 16:33
  • 1
    $\begingroup$ @SashoNikolov Bring back Legendre's constant! $\endgroup$
    – Clement C.
    Commented Mar 28, 2018 at 19:47

Let the risk be a random variable, by the Chernoff-Hoeffding's inequality, you can bound the approximation to the true risk: \begin{align} P \left(\Bigg | \frac{1}{n} \sum_{i=1}^{n}\xi_i - \mathbb{E}(\xi) \Bigg | \geq \epsilon \right) &\leq 2 \exp(-2n\epsilon^2)\\ P(|\hat{R}_{S}(h) - R(h)| \geq \epsilon) &\leq 2 \exp(-2n\epsilon^2) \end{align}

Unfortunately, this bound only holds for a fixed function $h$ which does not depend on the training data, but our hypothesis certainly does depend. The reason for such constraint is intuitive. If we let the hypothesis space convey all possible functions and do not restrict our hypothesis to not depend on the training data, we can always generate a function that ``memorises'' the given sample and has no empirical error. Such function will most certainly not generalise well and invalidate the bound.

Vapnik and Chervonenkis solved this conundrum by using the union bound.

==Union Bound==

If we enumerate all the functions in $H$, using the fact that it is finite (it is a $\epsilon$-cover of $H$), the bound still holds for each hypothesis: \begin{align} P(|\hat{R}_{S}(h_1) - R(h_1) | > \epsilon) &\lor \nonumber \\ P(|\hat{R}_{S}(h_2)-R(h_2)| > \epsilon) &\lor \cdots \nonumber \\ P(|\hat{R}_{S}(h_{|H|})-R(h_{|H|})| > \epsilon) &\leq \sum^{|H|} 2 \exp(-2n\epsilon^2) \nonumber\\ \therefore P(\sup_{h \in H}|\hat{R}_{S}(h) - R(h)| >\epsilon) &\leq 2 |H| \exp(-2n\epsilon^2)\\ P\left[ \exists h \in H:~|\hat{R}_{S}(h) - R(h)| >\epsilon\right] &\leq 2 |H| \exp(-2n\epsilon^2) \end{align} \begin{align} P\left[ \exists h \in H:~|\hat{R}_{S}(h) - R(h)| >\epsilon\right] &< \delta \tag{PAC}\\ P\left[ \exists h \in H:~|\hat{R}_{S}(h) - R(h)| >\epsilon\right] &\leq 2 |H| \exp(-2n\epsilon^2) \\ \therefore \delta &> 2 |H| \exp(-2n\epsilon^2) \end{align} Assuming $\delta = 2 |H| \exp(-2n\epsilon^2)$, we have: \begin{align} \exp(-2n\epsilon^2) = \frac{\delta}{2|H|} \\ -2n\epsilon^2 = \ln{\delta} - ln{2|H|}\\ \epsilon^2 = \frac{\ln{|H|}+\ln{2}-\ln{\delta}}{2n}\\ \therefore \epsilon > 0 \rightarrow \epsilon = + \sqrt{\frac{\ln{|H|}+ \ln{2/\delta}}{2n}} \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.