Boosting with AdaBoost

Ensemble Methods

Ensemble Methods

Boosting Setting

Boosting Algorithm Idea

Adaboost Algorithm

Boosting Theorem

Proof Steps 1-4

Ensemble Methods

• Train multiple models/hypotheses and combine them
• Ensemble methods are plug-and-play (can be used with any base algorithm for learning)
• Last lecture: bagging, which takes high variance/low bias methods and ensembles them to reduce variance (fixes overfitting) $$H_{\text{Bagged}}(\mathbf x) = \frac{1}{m} \sum_{i=1}^{m} h_i(\mathbf x)$$
  • Sample \(m\) points with replacement uniformly from training data
  • Random forest: bagged decision trees with max depth \(m\) and \(k < d\) subsampled features (often \(k=\sqrt(d)\))
• Today: boosting, which takes high bias/low variance methods and ensembles them to obtain low bias model (fixes underfitting)

Boosting Setting & Algorithm Idea

Ensemble Methods

Boosting Setting

Boosting Algorithm Idea

Adaboost Algorithm

Boosting Theorem

Proof Steps 1-4

Boosting Setting

• Training data \(D = \{(\mathbf{x}_1, y_1), \ldots, (\mathbf{x}_n, y_n)\}\) with \(y_i \in \{+1, -1\}\) (binary classification)
• Weak learner: a high bias (simple) classification algorithm that does (slightly) better than chance
  • Examples: depth 1 decision tree, nearly random classifier
• Question: how can you combine depth 1 DT to get zero training error for the following?

Boosting Algorithm Idea

• How can we ensemble weak learners to get an algorithm with 0 training error?
• Idea: sequentially weight and sample points from training data (more weight on points with errors)
• Ensemble will have the form $$H_{\text{Boosted}}(\mathbf x) = \sum_{t=1}^{T} \alpha_t h_t(\mathbf x)$$

1. Initialize equal weights \(w_1\)
2. For \(t = 1\) to \(T\):
   1. Weight datapoints in \(D\) according to \(w_t\) to get \(D_t\)
   2. Use weak learning algorithm on \(D_t\) to obtain classifier \(h_t\)
   3. Add \(h_t\) to ensemble and update \(w_{t+1}\) based on errors
3. Return ensembled classifier

Adaboost Algorithm

Ensemble Methods

Boosting Setting

Boosting Algorithm Idea

Adaboost Algorithm

Boosting Theorem

Proof Steps 1-4

Adaboost Algorithm

1. set \(w_1[i]=\frac{1}{n}\) for all \(i\)   (initialize uniformly)
2. set \(H_0=0\)   (initialize ensemble to 0)
3. For \(t = 1\) to \(T\):
   1. Weight points in \(D\) according to \(w_t\) to get \(D_t\)
   2. Obtain \(h_t\) via weak learning algorithm on \(D_t\)
   3. Compute error and ensemble weight $$\epsilon_t = \sum_{i=1}^{n} w_t[i] \mathbf{1}\{h_t(\mathbf x_i) \neq y_i\}, \quad \alpha_t = \frac{1}{2} \log\left(\frac{1 - \epsilon_t}{\epsilon_t}\right)$$
   4. Update ensemble \(H_t=H_{t-1} + \alpha_t h_t\)
   5. Update weights \(w_{t+1}[i] = \frac{w_t[i] \exp(-\alpha_t y_i h_t(\mathbf x_i) )}{\sum_{j=1}^n w_t[j] \exp(-\alpha_t y_j h_t(\mathbf x_j) ) }\)

Boosting Theorem & Proof

• Boosting Theorem: If the weak learner can achieve weighted classification error less than \(1/2 − \gamma\), the boosted classifier will have 0 training error after $$T = O\left(\frac{\log(n)}{\gamma^2}\right)$$

Ensemble Methods

Boosting Setting

Boosting Algorithm Idea

Adaboost Algorithm

Boosting Theorem

Proof Steps 1-4

• Training error (0-1 loss) $$\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}\{H_T(\mathbf x_i) \neq y_i\}$$
• Upper bounded by exponential loss: $$ \Phi_T = \frac{1}{n} \sum_i \exp(-y_i H_T(x_i)) $$

Boosting Theorem

• Weak Learner Assumption: For all distributions over points in D, the weak learning algorithm can produce a hypothesis whose weighted classification error is less than \(1/2 − \gamma\).
• Boosting Theorem: If the weak learner assumption holds with some margin \(\gamma\), the boosted classifier will have 0 training error after $$T = O\left(\frac{\log(n)}{\gamma^2}\right)$$
• Implications of this theorem:
  • Each weak learner is high bias but low variance classifier
  • We only combine \(O(\log(n))\) of these weak learners
  • So Boosted classifier will not have too high of a variance

Proof Step 1: Bounding Error by Exponential Loss

• We will bound the training error $$\frac{1}{n} \sum_{i=1}^{n} \mathbf{1}\{H_T(\mathbf x_i) \neq y_i\}$$
• Define the exponential loss: $$ \Phi_T = \frac{1}{n} \sum_i \exp(-y_i H_T(x_i)) $$
• Notice that if \(y_i H_T(\mathbf x_i) \le 0\), then \(\exp(-y_i H_T(x_i)) \ge 1\), so \(\Phi_T\) upper bounds the training error
• We will show that the exponential loss decreases by a fixed multiplicative factor at each boosting round.

Proof Step 2: Loss Shrinks Multiplicatively

• Recall the update rule $$ H_{t+1}(x) = H_t(x) + \alpha_{t+1} h_{t+1}(x) $$
• Therefore, $$\begin{align*} \Phi_{t+1} = \frac{1}{n} \sum_{i=1}^n \exp(-y_i H_{t+1}(\mathbf x_i))\ &= \frac{1}{n} \sum_{i=1}^n \exp(-y_i H_t(\mathbf x_i)) \exp(-\alpha_{t+1} y_i h_{t+1}(\mathbf x_i)) \\&= \Phi_t  \sum_{i=1}^n \frac{\exp(-y_i H_t(\mathbf x_i))}{n \Phi_t} \exp(-\alpha_{t+1} y_i h_{t+1}(\mathbf x_i))  \end{align*} $$
• Define weights: \(p_t[i] = {\exp(-y_i H_t(\mathbf x_i))}/\left({\sum_{j=1}^n \exp(-y_j H_t(\mathbf x_j))}\right)\)
  • Can show by induction that \(p_t[i]=w_t[i]\) algorithm weights
• Define multiplier \(Z_{t+1}=\sum_{i=1}^n w_t[i]\exp(-\alpha_{t+1} y_i h_{t+1}(\mathbf x_i)) \)
• Final recursive update is $$\Phi_{t+1} = \Phi_t Z_{t+1}$$

Proof Step 3: Using Weak Learner Assumption

• Now we just need to show that \(Z_t=\sum_{i=1}^n w_{t-1}[i]\exp(-\alpha_{t} y_i h_{t}(\mathbf x_i))<1\)

• Since both \(h_t(\mathbf{x}_i),y_i \in \{-1, +1\}\), the product \(y_i h_t(\mathbf{x}_i)\) is always \(\pm 1\):

  $$Z_t =\sum_{i:, h_t(\mathbf{x}i) = y_i} w_t[i] e^{-\alpha_t} + \sum_{i:, h_t(\mathbf{x}_i) \neq y_i} w_t[i] e^{\alpha_t}$$

• Recall that $$\epsilon_t = \sum_{i=1}^{n} w_t[i] \mathbf{1}\{h_t(\mathbf x_i) \neq y_i\}, \quad \alpha_t = \frac{1}{2} \log\!\left(\frac{1 - \epsilon_t}{\epsilon_t}\right)$$

• Therefore, $$ Z_t = (1 - \epsilon_t)e^{-\alpha_t} + \epsilon_t e^{\alpha_t} = 2\sqrt{\epsilon_t(1 - \epsilon_t)} $$
• Under the weak learner assumption, \(\epsilon_t \le \frac{1}{2} - \gamma\) so $$ Z_t \le \sqrt{1 - 4\gamma^2} $$

Proof Step 4: Final Bound

• Putting it all together, we show the exponential loss decreases quickly $$ \Phi_{t+1} = \Phi_t Z_{t+1} \leq \Phi_t \sqrt{1 - 4\gamma^2} \implies \Phi_T \le (\sqrt{1 - 4\gamma^2})^T\Phi_0 $$
• Simplify using \(1 - x \le e^{-x}\) and \(\Phi_0= 1\): $$ \text{Training error} \;\le\; \Phi_T \;\le\; e^{-2\gamma^2 T} $$
• Lastly, notice that by 0-1 loss, if the training error is less than \(\frac{1}{n}\), it must be exactly zero. $$ e^{-2\gamma^2 T} < \frac{1}{n} \iff T > \frac{\log(n)}{2\gamma^2}$$