$\require{mathtools}
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% .. over and under
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Adapted from the talk “Realizable learning is all you need” by Max Hopkins at Stanford’s theory lunch.
See also: PAC learning and VC dimension.
Motivation
There are two kinds of PAC learning. In both cases, you fix a hypothesis class $\CH$.
- Realizable: the adversary picks a distribution $D_x$ over inputs and a hypothesis $h \in \CH$, and you need to find a hypothesis $h^*$ that has error $\leq \eps$ over $D_x$.
- Agnostic: the adversary picks a distribution $D$ over inputs and outputs and you need to find a hypothesis $h^*$ that does at most $\eps$ worse than the best hypothesis $\CH$.
It is obvious that agnostic learning implies realizable learning. The converse turns out to be true, though computational efficiency suffers. Unfortunately, the traditional proof is a bit complicated, goes through VC-dimension, and doesn’t give a nice direct way to translate the realizable learner into an agnostic one.
Here’s a neat proof.
The algorithm
Let $n(\eps,\delta)$ be the number of samples needed for the realizable learner.
- Sample $n(\eps/2, \delta/2)$ points from $D$, discarding the labels (this effectively samples from the marginal distribution $D_x$).
- Run the realizable learner on all possible $2^{n(\eps/2, \delta/2)}$ labelings of this set, and let $\CH^*$ be the set of hypotheses it returned. Crucially, $\CH^*$ is finite and not too big.
- Use a fresh sample of $O\left(\frac{\log|\CH^*| + \log(1/\delta)}{\eps^2}\right)$ points bounds to select the best hypothesis among $\CH^*$ (up to $\eps/2$, w.p. $\geq 1-\delta/2$, using Chernoff bounds).
The total sample complexity is $O\left(\frac{n(\eps/2,\delta/2) + \log(1/\delta)}{\eps^2}\right)$, which is not a big loss.
Claim
For any distribution $D$ and any $h \in H$, w.p. $\geq 1-\delta/2$, at least one of the hypotheses in $\CH^*$ is $(\eps/2)$-close to $h$ under the marginal distribution $D_x$.
If true, we’re done
Just apply it to whatever $D$ the adversary chose and to the $h$ that gets the least error for $D$. This means that $\CH^*$ would contain a hypothesis $h'$ within $\eps/2$ of optimum. Then step 3 would find a hypothesis $h''$ within $\eps/2$ of that one.
It’s true
This is basically the definition of realizable learning: during the step where the realizable learner is run on the labeling that happens to match $h$, the scenario is exactly what the realizable learner was trained to do, so it will find a hypothesis close to $h$ under $D_x$.
Note
The claim doesn’t say that every hypothesis in $\CH$ has a close equivalent in $\CH^*$ (something that’s apparently called uniform convergence). But we only need a specific one here: we only need non-uniform covering!