Derivation for Two-Sided Ockham's Razor Bound

Notation follows from these lecture notes.

Let us define our finite (binary) function class as \(\mathcal{F} = \{f_1, \dots, f_M \}\) where each function in \(\mathcal{F}\) predicts either \(-1\) or \(1\). Notation \(\mathbf{Z} \sim D^n\) indicates that a probability is taken over the randomness of data draws \(Z_1, \dots, Z_n \sim D\) where each \(Z_i = (X_i, Y_i)\). Then \(\forall \epsilon > 0\) we have:

\[\mathbb{P}_{\mathbf{Z} \sim D^n}[\exists \ f \in \mathcal{F} : | R^{\text{true}}(f) - R^{\text{emp}}(f)| > \epsilon] \leq \sum_{j = 1}^M \mathbb{P}_{\mathbf{Z} \sim D^n}[| R^{\text{true}}(f_j) - R^{\text{emp}}(f_j)| > \epsilon]\]

Applying the two-sided Hoeffding’s Inequality, we know that \(\mathbb{P}_{\mathbf{Z} \sim D^n}[\mid R^{\text{true}}(f_j) - R^{\text{emp}}(f_j) \mid > \epsilon] \leq 2\exp(-2n\epsilon^2)\) and so:

\[\mathbb{P}_{\mathbf{Z} \sim D^n}[\exists \ f \in \mathcal{F} : | R^{\text{true}}(f) - R^{\text{emp}}(f)| > \epsilon] \leq 2M\exp(-2n\epsilon^2)\]

To express this more nicely, let us define \(\delta = 2M \exp(-2n\epsilon^2)\) so we can express \(\epsilon = \sqrt{\frac{\log M + \log \frac{2}{\delta}}{2n}}\).

\[\mathbb{P}_{\mathbf{Z} \sim D^n}[\exists \ f \in \mathcal{F} : | R^{\text{true}}(f) - R^{\text{emp}}(f)| > \sqrt{\frac{\log M + \log \frac{2}{\delta}}{2n}}] \leq \delta\] \[\implies \mathbb{P}_{\mathbf{Z} \sim D^n}[\forall \ f \in \mathcal{F} : | R^{\text{true}}(f) - R^{\text{emp}}(f)| \leq \sqrt{\frac{\log M + \log \frac{2}{\delta}}{2n}}] \geq 1 - \delta\]

which is exactly the two-sided Ockham’s Razor Bound.