PAC-Bayes bounds for generalisation error

Author

Affiliation

Cuong Nguyen

The University of Adelaide

Properly approaximately correct (PAC) learning is a part of statistical machine learning which has been a fundamental course for most of graduate programs in machine learning. Its main idea is to upper-bound the true risk (or generalisation error) by the empirical risk with certain confidence level. In other words, it is often written in the following form: \Pr (\text{true risk} \le \text{empirical risk} + r(m, \delta)) \ge 1 - \delta where \Pr(A) is the probability of event A, \delta \in (0, 1] is the confidence parameter, and r(m, \delta) – a function of sample size m and the confidence parameter \delta – is the regularization that is satisfied: \lim_{m \to +\infty} r(m, \delta) = 0. PAC-Bayes upper generalization bound is a kind of PAC learning. It was firstly proposed in 1999 McAllester (1999), and has attracted much of research interest. There has been many subsequent improvements made to tighten further this classic PAC-Bayes bound or to extend it to more general loss functions. However, the classic PAC-Bayes theorem is still the backbone. In this post, I will show how to prove this interesting theorem.

1 Auxillary lemmas

To prove the classic PAC-Bayes theorem, we need two auxilliary lemmas shown below.

1.1 Change of measure inequality for Kullback-Leibler divergence

Lemma 1 (Banerjee 2006 - Lemma 1) For any measurable function \phi(h) on a set of predictor under consideration \mathcal{H}, and any distributions P and Q on \mathcal{H}, the following inequality holds: \mathbb{E}_{Q} [\phi(h)] \le \mathrm{KL} [Q \Vert P] + \ln \mathbb{E}_{P} [\exp(\phi(h))]. Further, \sup_{\phi} \mathbb{E}_{Q} [\phi(h)] - \ln \mathbb{E}_{P} [\exp(\phi(h))] = \mathrm{KL} [Q \Vert P].

Proof. For any measurable function \phi(h), the following holds: \begin{aligned} \mathbb{E}_{Q} [\phi(h)] & = \mathbb{E}_{Q} \left[ \ln \left( \exp(\phi(h)) \frac{Q(h)}{P(h)} \frac{P(h)}{Q(h)} \right) \right] \\ & = \mathrm{KL} [Q \Vert P] + \mathbb{E}_{Q} \left[ \ln \left( \exp(\phi(h)) \frac{P(h)}{Q(h)} \right) \right] \\ & \le \mathrm{KL} [Q \Vert P] + \ln \mathbb{E}_{Q} \left[ \exp(\phi(h)) \frac{P(h)}{Q(h)} \right] \\ & \qquad \text{(Jensen's inequality)}\\ & = \mathrm{KL} [Q \Vert P] + \ln \mathbb{E}_{P} \left[ \exp(\phi(h)) \right]. \end{aligned}

For the second part of the lemma, we need to examine the equality condition of the Jensen’s inequality. Since \ln(x) is a strictly concave function for x > 0, it follows that the equality holds when: \begin{aligned} \exp \left( \phi(h) \right) & \frac{P(h)}{Q(h)} = 1 \\ \iff \phi(h) & = \ln \left[ \frac{Q(h)}{P(h)} \right]. \end{aligned} With this choice of \phi(h), we can verify that the equality does hold.

This completes the proof.

1.2 Concentration inequality

Lemma 2 (Shalev-Shwartz and Ben-David 2014 - Exercise 31.1) Let X be a random variable that satisfies: \mathrm{Pr} (X \ge \epsilon) \le e^{-2m \epsilon^{2}}. Prove that \mathbb{E} \left[ e^{2(m - 1) X^{2}} \right] \le m.

Proof. Since the assumption is expressed in term of probability, while the conclusion is written in form of an expectation, what we need to do first is to try to present the expectation in terms of probability.

For simplicity, let Y = e^{2(m - 1) X^{2}}. Since X \in [0, +\infty), then Y \in [1, +\infty) and Y can be presented as: Y = \int_{1}^{+\infty} \pmb{1}(Y \ge t) \, \mathrm{d}t + 1, where \pmb{1}(A) is the indication function of event A. Note that the integral above is the area of a rectangle with height as 1 and the width Y - 1.

One important property of the indication function is that: \mathbb{E} \left[ \pmb{1}(Y \ge t) \right] = \mathrm{Pr}(Y \ge t). This allows to express the expectation of interest as: \begin{aligned} \mathbb{E}[Y] & = \mathbb{E} \left[ \int_{1}^{+\infty} \pmb{1}(Y \ge t) \, \mathrm{d}t \right] + 1 \\ & = \int_{1}^{+\infty} \mathbb{E} [\pmb{1}(Y \ge t)] \, \mathrm{d}t + 1 \quad \text{(Fubini's theorem)} \\ & = \int_{1}^{+\infty} \mathrm{Pr}(Y \ge t) \, \mathrm{d}t + 1. \end{aligned} Or: \mathbb{E} \left[ e^{2(m - 1) X^{2}} \right] = \int_{1}^{+\infty} \mathrm{Pr}( e^{2(m - 1) X^{2}} \ge x) \, \mathrm{d}x + 1.

We then make a change of variable from x to \epsilon to utilize the given inequality in the assumption. Let’s define: x = e^{2(m - 1) \epsilon^{2}}. Since \epsilon is assumed to be non-negative, we can express it as: \epsilon = \sqrt{\frac{\ln x}{2(m - 1)}}, and: \mathrm{d}x = 4(m - 1) \epsilon \, e^{2(m - 1) \epsilon^{2}} \, \mathrm{d} \epsilon.

The expectation of interest can, therefore, be written as: \begin{aligned} \mathbb{E} \left[ e^{2(m - 1) X^{2}} \right] & = \int_{0}^{+\infty} \mathrm{Pr} \left( e^{2(m - 1) X^{2}} \ge e^{2(m - 1) \epsilon^{2}} \right) 4(m - 1) \epsilon \, e^{2(m - 1) \epsilon^{2}} \, \mathrm{d} \epsilon + 1\\ & = \int_{0}^{+\infty} \mathrm{Pr} \underbrace{\left( X \ge \epsilon \right)}_{\le e^{-2m\epsilon^{2}}} 4(m - 1) \epsilon \, e^{2(m - 1) \epsilon^{2}} \, \mathrm{d} \epsilon + 1\\ & \le 4(m - 1) \int_{0}^{+\infty} \epsilon \, e^{-2 \epsilon^{2}} \, \mathrm{d} \epsilon + 1 = m. \end{aligned}

2 PAC-Bayes bound

Theorem 1 Let D be an arbitrary distribution over an example domain Z. Let \mathcal{H} be a hypothesis class, \ell: \mathcal{H} \times Z \to [0, 1] be a loss function, \pi be a prior distribution over \mathcal{H}, and \delta \in (0, 1]. If S = \{z_j\}_{j=1}^{m} is an i.i.d. training set sampled according to D, then for any “posterior” Q over \mathcal{H}, the following holds: \mathrm{Pr} \left( \mathbb{E}_{z_{j} \sim D} \mathbb{E}_{h \sim Q} \left[ \ell(h, z_{j}) \right] \le \mathbb{E}_{z_{j} \sim S} \mathbb{E}_{h \sim Q} \left[ \ell(h, z_{j}) \right] + \sqrt{\frac{\mathrm{KL} [Q \Vert \pi] + \frac{\ln m}{\delta}}{2(m - 1)}} \right) \ge 1 - \delta.

Proof. We define some notations to ease the proving: - L = \mathbb{E}_{z_{j} \sim D} \left[ \ell(h, z_{j}) \right] - \hat{L} = \mathbb{E}_{z_{j} \sim S} \left[ \ell(h, z_{j}) \right] = \frac{1}{m} \sum_{j=1}^{m} \ell(h, z_{j}) - \Delta L = L - \hat{L}

Applying Lemma 1 with P(h) = \pi (h) and \phi(h) = 2(m - 1) (\Delta L)^{2} gives: 2(m - 1) \mathbb{E}_{Q} \left[ (\Delta L)^{2} \right] - \mathrm{KL} [Q \Vert \pi] \le \textcolor{purple}{\ln \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right]}. \tag{1}

We upper-bound the last term in the RHS (highlighted in purple color) by Lemma 2. To do that, we consider the empirical loss on each observable data point l(h, z_{j}) as a random variable in [0, 1] with true and empirical means L and \hat{L}, respectively. Following the Hoeffding’s inequality gives: \begin{aligned} \mathrm{Pr} \left( \Delta L \ge \epsilon \right) & = \mathrm{Pr} \left( L - \hat{L} \ge \epsilon \right)\\ & \le \mathrm{Pr} \left( | L - \hat{L} | \ge \epsilon \right)\\ & \le \exp(-2m \epsilon^{2}), \quad \epsilon \ge 0. \end{aligned} According to Lemma 2, this implies: \mathbb{E}_{S} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right] \le m. Taking the expectation w.r.t. h \sim \pi(h) on both sides and applying Fubini’s theorem (to interchange the 2 expectations) gives: \begin{aligned} & \mathbb{E}_{S} \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right] \le \mathbb{E}_{\pi} \left[ m \right] = m\\ & \implies \ln \mathbb{E}_{S} \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right] \le \ln m\\ & \implies \mathbb{E}_{S} \textcolor{purple}{\ln \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right]} \le \ln m. \end{aligned} Note that the last implication is due to Jensen’s inequality.

We then apply Markov’s inequality for the term highlighted in purple: \begin{aligned} \mathrm{Pr} \left( \textcolor{purple}{\ln \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right]} \ge \varepsilon \right) & \le \frac{\mathbb{E}_{S} \textcolor{purple}{\ln \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right]}}{\varepsilon} \\ & \le \frac{\ln m}{\varepsilon}. \end{aligned}

This implies: \mathrm{Pr} \left( \textcolor{purple}{\ln \mathbb{E}_{\pi} \left[\exp \left( 2(m - 1) (\Delta L)^{2} \right) \right]} \le \varepsilon \right) \ge 1 - \frac{\ln m}{\varepsilon}. \tag{2}

Combining the results in Equation 1 and Equation 2 gives: \mathrm{Pr} \left( 2(m - 1) \mathbb{E}_{Q} \left[ (\Delta L)^{2} \right] - \mathrm{KL} [Q \Vert \pi] \le \varepsilon \right) \ge 1 - \frac{\ln m}{\varepsilon}.

This is equivalent to: \mathrm{Pr} \left( \mathbb{E}_{Q} \left[ (\Delta L)^{2} \right] \le \frac{\mathrm{KL} [Q \Vert \pi] + \varepsilon}{2(m - 1)} \right) \ge 1 - \frac{\ln m}{\varepsilon}. \tag{3}

Note that squared function is a strictly concave function, resulting in: \mathbb{E}_{Q} \left[ (\Delta L)^{2} \right] \ge \left( \mathbb{E}_{Q} \left[ \Delta L \right] \right)^{2}.

Hence, Equation 3 can be written as: \mathrm{Pr} \left( \mathbb{E}_{Q} \left[ \Delta L \right] \le \sqrt{\frac{\mathrm{KL} [Q \Vert \pi] + \varepsilon}{2(m - 1)}} \right) \ge 1 - \frac{\ln m}{\varepsilon}.

Seting \delta = \frac{\ln m}{\varepsilon}, and expanding \Delta L according to its definition complete the proof.

3 Discussion

AFAIK, the result in Theorem 1 is a seminal PAC-Bayes bound in the literature of PAC learning. Readers could refer subsequent derivations of tighter PAC-Bayes bounds developed later.

4 References

Banerjee, Arindam. 2006. “On Bayesian Bounds.” In International Conference on Machine Learning, 81–88.

McAllester, David A. 1999. “PAC-Bayesian Model Averaging.” In Conference on Computational Learning Theory, 164–70.

Shalev-Shwartz, Shai, and Shai Ben-David. 2014. Understanding Machine Learning: From Theory to Algorithms. Cambridge university press.