$\require{mathtools}
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% Greek
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% .. over and under
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% .. relations
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% .. miscer
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% (I needed to create my own because the MathJax version of \DeclarePairedDelimiter doesn't have \mathopen{} and that messes up the spacing)
% .. one-part
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Adapted from a talk by Guy Blanc at the teach-us-anything seminar at Stanford.
We explain the link between PAC learning and VC dimension.
Theorem
Let $X$ be a set, $F \subseteq \set{f: X \to \zo}$ be a concept class over $X$, and let $\Pi_F(m)$ be largest possible the number of different ways functions in $F$ can label $m$ given points. Then except with probability $2 \cdot \Pi_F(2m) \cdot 2^{-\eps m /2}$, a consistent learner will output a hypothesis with error at most $\eps$.
Proof sketch
Let $f \in F$ be the function being learned, and $D$ be any distribution over $X$. For any hypothesis $h$, let $L_D(f,h)\coloneqq \Pr_{x \sim D}[f(x) \neq h(x)]$ be the “true error”. For some sample $S$ from $X$, let $L_S(f,h)$ be the “training error”.
Let $S$ be a set of $m$ samples. Let $A$ be the event that some $h \in F$ gets $L_S(f,h) = 0$ (i.e. $h$ is consistent with the sample) and yet $L_D(f,h) > \eps$ ($h$’s true error is big). We want to upper bound $\Pr[A]$.
However, it is hard to reason about $A$ directly: $A$ involves the quantity $L_D(f,h)$ which depends on the behavior of $h$ over the entire domain, so if we wanted to union bound over $h$, we would need to union bound over a potentially infinite number of functions. Crucial trick number 1 is to instead consider a proxy for event $A$, which instead of looking at $L_D(f,h)$, looks at an approximation $L_T(f,h)$ over an imaginary fresh sample $T$ of $m$ samples. That is, let $B$ be the event that some $h \in F$ gets $L_S(f,h)=0$ and yet $L_T(f,h) \geq \eps/2$ (for some fresh sample $T$ of $m$ samples).
We’ll upper bound $\Pr[B]$ instead. Why is this okay? Because conditioned on $A$ happening, $B$ is likely to happen. This can be seen from any old concentration bound: given that we know $h$ has true error $> \eps$, it’s likely to display a large error on a (large enough, on the order of $\Theta(1/\eps)$?) fresh sample $T$. So at the cost of making $m$ large enough, we get $\Pr[B \mid A] \geq 1/2$, and thus $\Pr[B] \geq \Pr[A] \cdot \Pr[B \mid A] \geq \Pr[A]/2$. Therefore, it is enough to show that $\Pr[B] \leq \Pi_F(2m) \cdot 2^{-\eps m/2}$.
All that’s good, but it isn’t clear that we’ve made any progress. Sure, now technically only $2m$ points are involved, but we’re still “selecting” $h$ from a potentially infinite set of functions. Why would “testing it again once” be enough to counterbalance this potentially infinite selection bias and make this sound? Crucial trick number 2 is not to focus on the chance that some $h$ can fool us, but rather to focus on the imbalance between the number of errors that show up in $S$ and $T$, irrespective of what $h$ actually ends up being.
Here’s the thought experiment: Fix some $h$, and suppose that you first run $h$ on a sample $U$ of $2m$ points, and only then split $U$ randomly into $S$ and $T$. Let $U_\mathrm{bad} \subseteq U$ be the set of the $\geq \eps m/2$ points in $U$ that $h$ labels incorrectly. In order for $B$ to happen, all points in $U_\mathrm{bad}$ need to all end up in $S$. Each point in $U_\mathrm{bad}$ has a $1/2$ chance to end up in $S$, and these events are negatively correlated, so the chance that they all end up in $S$ is $\leq 2^{-\eps m/2}$. Now, $U_\mathrm{bad}$ only depends on the behavior of $h$ on the $2m$ points of the sample $U$, so there are only $\Pi_F(2m)$ possible values for $U_\mathrm{bad}$. Therefore,
\[\Pr[B] \leq \text{``number of $U_\mathrm{bad}$''} \times \text{``probability for each $U_\mathrm{bad}$''} \leq \Pi_F(2m) \cdot 2^{-\eps m/2}.\]