# The fundamental theorem of statistical learning

We state and prove the fundamental theorem of statistical learning, which says that a class of functions is learnable if and only if it has finite VC dimension.

This post uses all the notations of the previous post.

Theorem:

- If $H$ has infinite VC dimension, then it is not PAC-learnable.
- If $H$ has finite VC dimension, then it is PAC-learnable.

## First implication: no free lunch

We prove that if $H$ has infinite VC dimension, then it is not PAC-learnable, which is sometimes called the “no free lunch theorem”.

Fix $m$. Assume that $H$ has infinite VC dimension, in particular it has dimension at least $2m$, so there exists $Y$ of size $2m$ which is shattered by $H$. Recall that to be PAC-learnable, there must exist an algorithm working against any distributions of the inputs. Here we will restrict ourselves to the uniform distribution over $Y$, so in particular we can only consider hypotheses \(h : Y \to \left\{0,1\right\}\)

We will show the following: if we are given $m$ samples, then there is no way that with high probability an algorithm can correctly pick an almost correct function. The intuitive reason is that the correct function can be any function \(h : Y \to \left\{0,1\right\}\) and $Y$ has size $2m$: using $m$ samples we only get (at most) half of the inputs.

We show that for any algorithm there exists a target function for which the algorithm fails on a proportion at least $\frac{1}{8}$ of the inputs with probability at least $\frac{1}{7}$. These strange constants come from the following lemma.

Lemma:Let $X$ be a random variable taking values in $[0,1]$ such that \(E[X] \ge \frac{1}{4}\) Then \(P(X \ge \frac{1}{8}) \ge \frac{1}{7}\)

This is a simple application of Markov’s inequality: \(P(X \ge a) \le \frac{E[X]}{a}\) with double complementation:

\[P(X \ge \frac{1}{8}) = 1 - P(X \le \frac{1}{8}) = 1 - P(1 - X \ge \frac{7}{8}) \ge 1 - \frac{E[1 - X]}{\frac{7}{8}} = 1 - \frac{8}{7} \left(1 - E[X] \right) = \frac{8}{7} E[X] - \frac{1}{7} \ge \frac{8}{7} \frac{1}{4} - \frac{1}{7} = \frac{1}{7}\]We write $U$ for the uniform distribution over $Y$.

An algorithm is a function $A$ taking as input the labelled samples $S_f$ and outputting \(A(S_f) : Y \to \left\{0,1\right\}\)
Let us say that the algorithm **fails** on $f$ if

Thanks to the lemma above, it is enough to show that

\[E_{S \sim U^m} \left[ L_{U,f}(A(S_f)) \right] \ge \frac{1}{4}\]We show that any algorithm fails on at least one target function $f$. To this end we draw a function $f$ uniformly at random, and show that the expectation of failing is at least $\frac{1}{4}$, implying that there indeed exists a function $f$ on which the algorithm fails.

Lemma:In the following quantity, we draw a function \(f : Y \to \left\{0,1\right\}\) uniformly at random. \(E_f \left[ E_{S \sim U^m} \left[ L_{U,f}(A(S_f)) \right] \right] \ge \frac{1}{4}\)

**Proof:**
Recall that \(L_{U,f}(A(S_f))\) is defined by
\(E_{x \sim U} [L_f(A(S_f),x)]\)

We write the following case distinction: either $x \in S$ or $x \notin S$.

\[E_{x \sim U} [L_f(A(S_f),x)] = P_{x \sim U} (x \notin S) \cdot E_{x \sim U} \left[ A(S_f)(x) \neq f(x) \mid x \notin S \right] + P_{x \sim U} (x \in S) \cdot E_{x \sim U} \left[ A(S_f)(x) \neq f(x) \mid x \in S \right]\]Intuitively, in the case where $x \in S$ the algorithm should not err, it is reasonable to think that $A(S_f)(x) = f(x)$, because the algorithm has access to the value $f(x)$ since $x \in S$. However if $x \notin S$, then the algorithm can only guess the value of $f(x)$. Hence we focus on the first term, lower bounding the second by $0$:

\[E_{x \sim U} [L_f(A(S_f),x)] \ge P_{x \sim U} (x \notin S) \cdot E_{x \sim U} \left[ A(S_f)(x) \neq f(x) \mid x \notin S \right]\]We make two observations:

- $P_{x \sim U} (x \notin S) \ge \frac{1}{2}$: indeed $U$ is the uniform distribution over $2m$ elements, and $S$ is an iid sample of $m$ elements, hence with probability at least half $x$ does not belong to the samples
- $E_f \left[ E_{S \sim U^m} \left[ E_{x \sim U} \left[ A(S_f)(x) \neq f(x) \mid x \notin S \right] \right] \right] = \frac{1}{2}$: indeed for $x \notin S$ the value of $f(x)$ is either $1$ or $0$ each with probability $\frac{1}{2}$ since $f$ is drawn uniformly at random.

This concludes the proof of this lemma.

We now wrap up the proof of the no free lunch theorem. Fix an algorithm $A$. It follows from the second lemma that there exists a function $f$ such that \(E_{S \sim U^m} \left[ L_{U,f}(A(S_f)) \right] \ge \frac{1}{4}\) Thanks to the first lemma, we obtain \(P_{S \sim U^m} \left( L_{U,f}(A(S_f)) \ge \frac{1}{8} \right) \ge \frac{1}{7}\)

## Converse implication: finite VC dimension implies PAC-learnability

Let $d$ be the VC dimension of $H$. We combine a few ingredients.

The first ingredient is Sauer’s lemma.

Sauer’s Lemma:\(\tau_H(m) \le \sum_{i = 0}^d \binom{m}{i} = O(m^d)\)

There are different proofs of Sauer’s lemma, some of them are very beautiful. My favourite is the second one in this lecture notes. Note that it is a purely combinatorial statement.

The following theorem relates the Rademacher complexity to the growth function:

Massart’s Theorem:\(R_H(m) \le \sqrt{\frac{2 \log(\pi_H(m))}{m}}\)

The proof is analytical and relies on Hoeffding’s inequality. See p39/40 of the Foundations of Machine Learning book for a proof.

Combining these two results yields

\[R_H(m) = O\left( \sqrt{ \frac{\log(m)}{m} } \right)\]In particular it goes to $0$ as $m$ goes to infinity.

The main technical piece of work to be done is to obtain generalization bounds using Rademacher complexity. What is a generalization bound? It’s an inequality relating the empirical loss, i.e. $L_{S,f}(h)$, and the actual loss, i.e. $L_{D,f}(h)$. We say that an hypothesis $h$ generalises if \(L_{D,f}(h) - L_{S,f}(h)\) is small.

Lemma:\(E_{S \sim D^m} \left[ L_{D,f}(h) - L_{S,f}(h) \right] \le 2 R_H(2m)\)

**Proof:**
By definition \(L_{D,f}(h) = E_{x \sim D} [L_f(h,x)]\) and
\(L_{S,f}(h) = \frac{1}{m} \sum_{i = 1}^m L_f(h,x_i)\)

To compare these two quantities we first make them look similar: observe that \(L_{D,f}(h) = E_{S' \sim D^m} [L_{S,f}(h)]\) We write \(S' = (x'_i)_{i \in [1,m]}\)

We obtain \(E_{S,S' \sim D^m} \left[ L_{S,f}(h) - L_{S',f}(h) \right] = E_{S,S' \sim D^m} \left[ \frac{1}{m} \sum_{i = 1}^m L_{f}(h,x_i) - L_{f}(h,x'_i) \right]\)

Recall that the goal is to upper bound by the Rademacher complexity. We claim that the quantity above is equal to

\[E_{\sigma,\sigma' \in \left\{-1,+1\right\}^m} E_{S,S' \sim D^m} \left[ \frac{1}{m} \sum_{i = 1}^m \sigma_i L_{f}(h,x_i) - \sigma'_i L_{f}(h,x'_i)) \right]\]Indeed, fix $\sigma$ and $\sigma’$. For each $i$ there are four cases:

- either $\sigma_i = \sigma’_i = +1$, then this is the term as in the quantity above,
- or $\sigma_i = \sigma’_i = -1$, this is the term as in the quantity above once $x_i$ and $x’_i$ are swapped in $S$ and $S’$,
- the other two terms cancel out ($\sigma_i = +1$ and $\sigma’_i = -1$ with $\sigma_i = -1$ and $\sigma’_i = +1$).

This quantity is smaller than \(2 R_H(2m)\) which concludes the proof of this lemma.

#### Proof wrap up

It follows using Markov’s inequality that for $\delta > 0$,

\[P_{S \sim D^m} \left( L_{D,f}(h) - L_{S,f}(h) \le \frac{2}{\delta} R_H(2m) \right) \le 1 - \delta\]Thanks to the discussion above (using Sauer’s Lemma and Massart’s Theorem) we know that \(R_H(2m)\) converges to $0$ when $m$ goes to infinity. So there exists some $m$ (which depend on $\varepsilon$ and $\delta$) such that

\[P_{S \sim D^m} \left( L_{D,f}(h) - L_{S,f}(h) \le \varepsilon \right) \le 1 - \delta\]An Empirical Risk Minimisation (ERM) algorithm is one that outputs an hypothesis exactly matching the training set, i.e. such that $L_{S,f}(A(S_f)) = 0$. Hence what we proved is that any ERM algorithm ensures \(P_{S \sim D^m} \left( L_{D,f}(h) \le \varepsilon \right) \le 1 - \delta\) for $m$ chosen as above. Hence $H$ is PAC-learnable. Note that we did not say anything about how to construct an ERM algorithm.