2
$\begingroup$

I was trying to understand PAC bounds on the realizable case (i.e. when there is some perfect $h^* \in \mathcal{H}$ and its generalization error is zero).


Notation:

Training data: $$S_n$$ Training error on training data $S_n$ $$\mathcal{E_{n}(h)}$$ Generalization error $$\mathcal{E(h)}$$ Hypothesis class $$\mathcal{H}$$ Size of hypothesis class: $$|\mathcal{H}| = K$$ Probability over the choice of training set that the PAC bound fails: $$\delta$$


I would like to derive how $\epsilon$ relates to other PAC variables, n, $K$ and $\delta$.

Attempted derivation:

Let $h_{bad}$ be any classifier that generalizes poorly $\mathcal{E(h)} > \epsilon$ . What is the probability that we would choose it after seeing the training set? If we choose it then given the current $S_n$ then we would choose it, if it has zero training error. Then, what is the probability that this bad classifier makes no training error?

$$P[h_{bad} \text{makes no error on } S_n] = (1 - \mathcal{E(h)})^n < (1-\epsilon)^n$$

this number is an upper bound on the "fraction/precentage of classifiers" that are bad and are chosen with the current training data.

From this I was trying to relate that upper bound to $\delta$ and was trying to figure out if the following statement was true:

$$(1 - \mathcal{E(h)})^n = \delta$$

If its true then one can conclude that:

$$\delta < (1-\epsilon)^n$$

Which is the bound that makes sense to me, but it seems that I am missing one important variable $|\mathcal{H}| = k$. For some reason the correct bound is suppose to be $\delta \leq |\mathcal{H}|(1-\epsilon)^n|$ but I can't figure out why. I think my problem is that I can't seem to understand what the difference of $\delta$ and $1 - \mathcal{E(h)})^n$ is, or the relation between them. I believe I got the correct relation but I am not sure.

$\endgroup$
0
$\begingroup$

$P[h_{bad} \text{makes no error on } S_n]$ is the probability that a particular bad classifier would work on the training data. To get a bound on the probability that any bad classifier would work on the training data, we need to multiply by the number of potentially bad classifiers, $K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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