VC Dimension of Linear Classifiers

Let \(d\) be the dimension of the input data and \(\mathcal{H} = \{f ~|~ f(\boldsymbol{x}) = \text{sign}(\boldsymbol{w}^\top \boldsymbol{x} + b), b \in \mathbb{R} \}\) be the set of all linear classifiers.

\(d_{VC}(\mathcal{H}) \geq d + 1\).

Proof We need to prove that \(\mathcal{H}\) can shatter at least \(d+1\) points. That is, it suffices to show that there exists a set of \(d+1\) points such that \(\mathcal{H}\) can produce any pre-specified \(\{-1, +1\}\) assignment \(\boldsymbol{y} = (y_1, \ldots, y_{d+1})\). Let \(\begin{align} \boldsymbol{X} &= \left[ \begin{array}{c} \boldsymbol{x}_1^\top \\ \vdots \\ \boldsymbol{x}_{d+1}^\top \end{array} \right] \\ &= \left[ \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & 0 & \cdots & 0 \\ 1 & 0 & 1 & \cdots & 0 \\ & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & 1 \end{array} \right] \in \mathbb{R}^{d+1 \times d+1} \end{align}\)

The first coordinate of each \(\boldsymbol{x}_i\) is 1 to allow the bias term \(b\) to be produced. Note that \(\boldsymbol{X}\) is invertible. With this definition, for any \(\boldsymbol{y}\) we can always have \(\text{sign}(\boldsymbol{X} \boldsymbol{w}) = \boldsymbol{y}\) by choosing \(\boldsymbol{w} = \boldsymbol{X}^{-1} \boldsymbol{y}\) since \(\boldsymbol{X}\boldsymbol{w} = \boldsymbol{y}\) implies \(\text{sign}(\boldsymbol{X} \boldsymbol{w}) = \boldsymbol{y}\).

\(d_{VC}(\mathcal{H}) \leq d + 1\).

Proof We need to show that no \(d+2\) points can be shattered by \(\mathcal{H}\). That is to show that there exists a certain assignment \(\boldsymbol{y} \in \mathbb{R}^{d+2}\) not achievable by \(\mathcal{H}\). We will construct one such assignment.

Assume we have \(d+2\) points \(\boldsymbol{x}_1, \ldots, \boldsymbol{x}_{d+2}\) in \((d+1)\)-dimensional space. Then, the set is linearly dependent. So, there exists \(\boldsymbol{x}_j = \sum_{i\neq j} a_i \boldsymbol{x}_i\) such that not all \(a_i\) are 0. Set \(y_j = -1\) and \(y_i = \text{sign}(a_i) = \text{sign}(\boldsymbol{w}^\top \boldsymbol{x}_i)\) . If \(a_i = 0\), \(y_i\) can be arbitrary.

\(\text{sign}(\boldsymbol{w}^\top \boldsymbol{x}_j) = \text{sign}(\sum_{i \neq j} a_i \boldsymbol{w}^\top \boldsymbol{x}_i) > 0 \neq y_j\)

Since there exists an assignment of \(d+2\) points not achievable by \(\mathcal{H}\), we can say that \(d_{VC}(\mathcal{H}) \leq d + 1 \).

\(d_{VC}(\mathcal{H}) = d + 1\).

Proof Combining the two lemmas.

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