Perceptron Convergence Proof
Last Updated: 2020-01-14 14:19

Okay, so here's the setup for proving that the perceptron learning algorithm converges in the binary classification case. If this is new to you, the perceptron is a linear classifier as follows:

  • \(\hat{y} = \text{sign}(w \cdot x)\)
    • The perceptron is a linear combination of the input vector and outputs either \(-1\) or \(+1\).
  • If \(\hat{y}_t \not = y_t\) then we update \(w_{k+1} = w_{k} + y_t(x_k)^T\)
    • It's an iterative algorithm, where the first misclassified input is added to our weight vector on iteration \(k+1\) and then we iterate again.
  • \(w_1 = 0\)
    • We initialize the starting weight vector to the zero vector.

For this proof, we'll assume a few things:

  • There exists an optimal \(w = w^*\) such that for some \(\epsilon > 0\), it's the case that \(y_i(w^* \cdot x_i) > \epsilon\) for all inputs on the training set.
    • In other words, we assume that there is a margin of at least \(\epsilon\) and that a solution exists.
  • \(||w^*|| = 1\)
    • This gives us a unique weight vector.
  • For all \(x_1, x_2, ... \in X\), it's the case that \(||x_i|| \le R\).
    • We're upper bounding the norm of the data points in our training set.

To start, we show that \(w_{k+1} \cdot (w^*)^T \ge k \epsilon \) with a proof by induction.

Base case where \(k = 0\): \(w_{0+1}\cdot (w^*)^T = 0 \ge 0 \cdot \epsilon \)

Inductive step where \(k \to k+1\): \(w_{k+1}\cdot (w^*)^T = (w^k + y_t(x_t)^T)\cdot (w^*)^T\) (WLOG, we assume that \(x_t\) is the first misclassified input) \(= w^k \cdot (w^*)^T + y_t (w^* \cdot x_t)\) \( \ge (k-1)\cdot \epsilon + \epsilon\) (Using the inductive hypothesis, and the definition of \(w^*\)) \( \ge k \epsilon\)

Next, we show that \(||w_{k+1}|| \ge k \epsilon​\) \(w_{k+1} \cdot w^* = ||w_{k+1}|| \times ||w^*|| \times \text{cos}(\theta)​\) where \(\theta​\) is the angle between \(w_{k+1}​\) and \(w^*​\) Because \(cos(\theta) \le 1​\), \(||w_{k+1}|| \times ||w^*|| \ge w_{k+1} \cdot (w^*)^T​\) ​Thus \(||w_{k+1}|| \times ||w^*|| \ge k \epsilon​\) Finally because \(||w^*|| = 1​\) by definition, \(||w_{k+1}|| \ge k \epsilon​\)

Now we notice that \(||w_{k+1}||^2 = ||w_{k} + y_t (x_t)^T||^2\) \(= ||w_k||^2 + 2y_t (w_k \cdot x_t) + ||x_k||^2\) \(\le ||w_k||^2 + ||x_k||^2\) because by definition \(\text{sign}(y_t(w_k \cdot x_t)) = -1\) as it is the first misclassified input, so \(2y_t(w_k \cdot x_t) < 0\). \(\le ||w_k||^2 + R^2\) because by definition \(||x_i|| \le R\)

Next we show that \(||w_{k+1}||^2 \le kR^2\) with a proof by induction.

Base case where \(k = 0\) \(||w_{0+1}||^2 = 0 \le 0(R)\)

Inductive step where \(k \to k+1\)

By the above part, we know that \(||w_{k+1}||^2 \le ||w_k||^2 + R^2\). Now, by the inductive hypothesis, we see that \(||w_{k+1}|| \le (k-1)R^2 + R^2 = kR^2\).

Thus, we have the full inequality: \(k^2 \epsilon^2 \le ||w_{k+1}||^2 \le kR^2\)

And thus we can see that \(k \le \frac{R^2}{\epsilon^2}\), bounding the number of iterations we run our algorithm by the margin and the farthest point.