Iterative Weak Learnability and Multi-Class AdaBoost

Given set $X$, let $\mathinner{\!\left\lvert X\right\rvert}$ be the number of elements in $X$. Let $A$ be the set of labels, and ${\mathcal{P}}$ be the set of observations, which is endowed with a probability distribution. A classifier is

which labels each observation. We often call $h$ a weak hypothesis (or simply, a hypothesis) in place of a classifier. Let ${\mathcal{H}}$ be the set of all feasible classifiers.

Let

be the correct classifier. We say that $h$ perfectly classifies $y$ if

We assume that a decision maker is endowed with ${\mathcal{H}}$, but $y$ is often significantly more complex than any element in ${\mathcal{H}}$. As a result, no $h\in{\mathcal{H}}$ can perfectly classify $y$. An important question is whether a decision maker who is endowed with primitive ${\mathcal{H}}$ can “learn” $y$.

To incorporate learning, we need to allow a decision maker to observe data over time. We consider a dynamic interaction, where the algorithm can update its output in response to new data. Suppose that time is discrete: $t=1,2,\ldots$ and $D_{t}$ is the data available at the beginning of time $t$. $\tau(D_{t})$ is the classifier which algorithm $\tau$ generates from data $D_{t}$. Let $\tau(D_{t})(p)$ be the label assigned to observation $p$ by $\tau(D_{t})$.

Let $T\geq 1$ be the training period, when the decision maker updates $\tau(D_{t})$ in response to $D_{t}$. The final hypothesis is the output $\tau(D_{T})$ at the end of the training period.

We focus on recursive ensemble algorithm for its simplicity.

We require that an algorithm produce an accurate classifier as the final hypothesis and achieve the desired accuracy level with a reasonable amount of data.

Let us illustrate the Adaptive Boosting (AdaBoost) algorithm by \citeasnounSchapireandFreund12. Instead of the original formula, we follow the framework of SAMME (\citeasnounZhuZouRossetandHastie09). Let us denote the algorithm as $\tau_{A}$.

Initialize $d_{1}(p)=1/\mathinner{\!\left\lvert{\mathcal{P}}\right\rvert}$ as the uniform distribution over ${\mathcal{P}}$. For $k\geq 1$, $D_{k}$ is the history at the beginning of time $k$. Iterate the following steps.

1. (1)

   (calculation of forecasting error) Choose $h_{k}\in{\mathcal{H}}$.

   $$\epsilon_{k}=\sum_{p}d_{k}(p){\mathbb{I}}\left(h_{k}(p)\neq y(p)\right)$$

   If $\epsilon_{k}=0$, then stop the iteration and set $\tau(D_{k})=h_{k}$. If not, continue.

2. (2)

   (calculating the weight)³³3The following formula is from equation (11) on page 353 of \citeasnounZhuZouRossetandHastie09, which becomes the coefficient of AdaBoost if $\mathinner{\!\left\lvert A\right\rvert}=2$. This formula differs slightly from (c) on page 351.

   $$\alpha_{k}=\frac{(\mathinner{\!\left\lvert A\right\rvert}-1)^{2}}{\mathinner{\!\left\lvert A\right\rvert}}\log\left(\frac{(\mathinner{\!\left\lvert A\right\rvert}-1)(1-\epsilon_{k})}{\epsilon_{k}}\right)$$

   (2.1)

3. (3)

   (preparing for the next round)

   $$w_{k+1}(p)=\begin{cases}d_{k}(p)\exp\left(-(\mathinner{\!\left\lvert A\right\rvert}-1)\alpha_{k}\right)&\text{if }\ h_{k}(p)=y(p)\\ d_{k}(p)\exp\left(\alpha_{k}\right)&\text{if }\ h_{k}(p)\neq y(p).\end{cases}$$

   $$d_{k+1}(p)=\frac{w_{k+1}(p)}{\sum_{p^{\prime}}w_{k+1}(p^{\prime})}.$$

4. (4)

   Reset $k+1$ as $k$ and start the first step of selecting an optimal weak hypothesis.

5. (5)

   (final hypothesis)

   $$\tau_{A}(D_{t})(p)=\arg\max_{a\in A}\sum_{s=1}^{t}\alpha_{s}{\mathbb{I}}(h_{s}(p)=a)\qquad\forall p\in{\mathcal{P}}$$

If $\mathinner{\!\left\lvert A\right\rvert}=2$, $\tau_{A}$ is the Adaptive Boosting (AdaBoost) algorithm of \citeasnounSchapireandFreund12.

The description of the algorithm is incomplete because if $\alpha_{k}<0$, the algorithm is not well defined.

$\tau_{A}$ is meaningful only if $h_{k}\in{\mathcal{H}}$ performs better than a random guess.

which implies that $\alpha_{k}>0$. Note that $\frac{2-\mathinner{\!\left\lvert A\right\rvert}}{\mathinner{\!\left\lvert A\right\rvert}}$ is the score from the random guess.

The weak learnability implies the existence of such $h_{k}$. We state the condition according to \citeasnounZhuZouRossetandHastie09.

The weak learnability is sufficient for the efficiency of $\tau_{A}$ if $\mathinner{\!\left\lvert A\right\rvert}=2$.

MukherjeeSchapire2013 pointed out that the notion of weak learnability of \citeasnounZhuZouRossetandHastie09 is too weak to guarantee the convergence of the algorithm to the correct label if $\mathinner{\!\left\lvert A\right\rvert}>2$.

Let us examine an example in \citeasnounMukherjeeSchapire2013.

Since $\mathinner{\!\left\lvert A\right\rvert}=3$, the weak learnability is satisfied if the performance of a weak hypothesis is strictly better than

Let $d=(d(a),d(b))$ be a probability distribution over ${\mathcal{P}}$. The performance score of ${\mathcal{H}}$ is

where the equality holds if $d(a)=0.5$. Thus, ${\mathcal{H}}$ is weakly learnable. Any convex combination of $h_{1}$ and $h_{2}$ in ${\mathcal{H}}$ assigns the same level to two observations. Since $y$ assigns a different label on a different observation, SAMME fails to produce a consistent forecast, although ${\mathcal{H}}$ satisfies the weak learnability.

Example 2.7 reveals two issues we must address to extend AdaBoost from binary to multiple labels. First, to construct a boosting algorithm, it is necessary to strengthen the notion of weak learnability. Second, the set of weak hypotheses ${\mathcal{H}}$ cannot be too restrictive. The issue identified in the previous example is that the requirement of outperforming a random guess by $\rho$ becomes less restrictive when the number of labels gets larger. Hence some distributions can satisfy 2.3 for many labels but can fail for a small number of labels. We propose to consider the minimal departure necessary to rule this out.

We strengthen the weak learnability by requiring that (2.3) holds for any subset of $A$ with more than a single element. Let $A_{i}\subset A$ be a non-empty subset of $A$ with $\mathinner{\!\left\lvert A_{i}\right\rvert}\geq 2$. Define

as the collection of all weak hypotheses with labels from $A_{i}$. Clearly, ${\mathcal{H}}_{i}\subset{\mathcal{H}}$.

Note that $\frac{2-\mathinner{\!\left\lvert A_{i}\right\rvert}}{\mathinner{\!\left\lvert A_{i}\right\rvert}}$ is a decreasing function of $\mathinner{\!\left\lvert A_{i}\right\rvert}$. As $\mathinner{\!\left\lvert A_{i}\right\rvert}$ decreases, the requirement for weak learnability of ${\mathcal{H}}_{i}$ becomes more stringent, as the minimum performance criterion (the right hand side of (2.4) increases.

In Example 2.7, ${\mathcal{H}}$ is weakly learnable, but not iteratively weakly learnable. To see this, note that if $A_{i}=\{1,2\}\subset A=\{1,2,3\}$, then ${\mathcal{H}}={\mathcal{H}}_{i}$.

However, the weak learnability over $A_{i}$ requires that the minimum performance be strictly larger than 0.

A fundamental question is whether we can find a class of simple hypothesis which is weakly learnable. With the set of the simplest classifiers, we can do better than the random guess over any subset of $A$ with at least two elements.

Suppose that $A$ and ${\mathcal{P}}$ are subsets of a vector space or a finite dimensional Euclidean space. Consider $h$, which classifies ${\mathcal{P}}$ according to the value of a linear function, thus called a linear classifier. Let ${\sf H}$ be a hyperplane in ${\mathbb{R}}^{n}$: $\exists\lambda\in{\mathbb{R}}^{n}$ and $\omega\in{\mathbb{R}}$ such that

Define ${\sf H}_{+}$ as the close half space above ${\sf H}$:

If $h$ is a single threshold classifier, $\mathinner{\!\left\lvert h(\mathcal{P})\right\rvert}=2$, which may be strictly smaller than $\mathinner{\!\left\lvert A\right\rvert}$. On the other hand, the set of single threshold classifiers is closed under permutations: if $\pi\mathrel{\mathop{\mathchar 58\relax}}A\rightarrow A$ is a bijection, then $\pi(h(p))\in\mathcal{H}$ for all $\pi$. Being closed under permutations turns out to be the key property to obtain weak learnability.

We use this result to show that the set of single-threshold classifiers is iteratively weakly learnable.

In Example 2.7, using single-threshold classifiers ensures that an extreme point of the convex hull of $\mathcal{P}$ can always be labeled correctly, and the rest can at least do as well as a random guess. The analogous result is known to hold for the case of $\mathinner{\!\left\lvert A\right\rvert}=2$. Interestingly, the critical property in generalizing to the multi-class case is not changing the set of classifiers but instead allowing the labels to be sufficiently rich.

The algorithm is run in discrete period $k=1,2,\ldots$ within epoch $i=1,2,\ldots$. Each epoch lasts at least $K$ periods, but the terminal round of an epoch is random. At the start of an epoch, we set the period to $k=1$ and re-start the procedure.

It is convenient to enumerate the elements in $A=\{1,\ldots,\mathinner{\!\left\lvert A\right\rvert}\}$ as the first $\mathinner{\!\left\lvert A\right\rvert}$ positive integers, highlighting the fact that the weak learnability has little to do with the name of the labels but relies solely on the ordinal properties of $A$. Let $\pi\mathrel{\mathop{\mathchar 58\relax}}A\rightarrow A$ be a permutation over $A$:

Let $A_{i}=\{1,\ldots,\mathinner{\!\left\lvert A_{i}\right\rvert}\}\subset A$ be the collection of the first $\mathinner{\!\left\lvert A_{i}\right\rvert}$ positive integers which is the set of feasible labels at the beginning of epoch $i$ with $A_{1}=A$. Let $\pi_{i}$ be a permutation over $A_{i}$.

At the beginning of epoch $i$, $A_{i}\subset A$ and $y_{i}\mathrel{\mathop{\mathchar 58\relax}}{\mathcal{P}}\rightarrow A_{i}$ are given. Set $k=1$. $d_{1}(p)$ as the uniform distribution over $\mathcal{P}$. The output of each stage consists three key components: an artificial probability distribution $d_{k}(p)$ over ${\mathcal{P}}$, a threshold rule $h_{k}$ in step $k$, and a positive weight $\alpha_{k}$.

Suppose that $d_{k}(p)$ is defined $\forall p\in{\mathcal{P}}$. Choose $h_{k}$ by solving

We call $h_{k}$ an optimal classifier. Define

as the probability that the optimal classifier $h_{k}$ at $k$ round misclassifies $p$ under $d_{k}$. If $\epsilon_{k}=0$, then we stop the training and output $h_{k}$ as the final hypothesis, which perfectly forecasts $y_{i}(p)$.

Suppose that $\epsilon_{k}>0$. Define

Note that the only difference between (3.7) and (2.1) is the coefficient $\frac{1}{2(\mathinner{\!\left\lvert A_{i}\right\rvert}-1)}$. If $\mathinner{\!\left\lvert A_{i}\right\rvert}=2$, our algorithm is reduced to AdaBoost.

Define for each $p$, and each pair $(p,y(p))$,

where

Given $d_{k+1}$, we can recursively define $h_{k+1}$ and $\epsilon_{k+1}$.

Define $\forall a\in A_{i}=\{1,\ldots,\mathinner{\!\left\lvert A_{i}\right\rvert}\}$,

and

Note that

Therefore, $\forall p$,

if and only if

We use $\Psi^{k}_{a}(p)$ to predict $y_{i}(p)$, instead of $F^{k}_{a}(p)$. Since $\Psi^{k}_{a}(p)$ is a linear combination of ${\mathbb{I}}(h_{s}(p)=a)$, our algorithm is an ensemble algorithm. Since $\Psi^{k}_{a}(p)$ is calculated recursively within an epoch, and the change between two epochs depends only on the final outcome from the previous epoch, we regard our algorithm a recursive algorithm.

Except for the case of $\mathinner{\!\left\lvert A_{i}\right\rvert}=2$, (3.9) does not guarantee that $a=y_{i}(p)$ with probability close to 1, as $k\rightarrow\infty$. \citeasnounMukherjeeSchapire2013 argues that the failure is due to the fact that the weak learnability (2.3) is too weak, and proposed a stronger criterion of EOR (Edge over Random) criterion.

We take a different tack. We strengthen (2.3) to (2.4) by requiring the same condition for any subset of $A$ with at least two elements. Instead of solving (3.10), we eliminate $a\in A_{i}$ which is not an element of $\arg\max_{A_{i}}\Psi^{k}_{a}(p)$. A critical step is to show that $a=y_{i}(p)\in A_{i}$ survives the elimination process with a probability close to 1. Since $\mathinner{\!\left\lvert A\right\rvert}<\infty$, we can “solve” (3.10) in a finite number of iterations. Roughly speaking, (2.3) may be too weak to solve (3.9) directly, but is sufficiently strong to solve the equivalent problem (3.10) indirectly, if (2.3) holds for any subset with at least two elements.

Recall that $K$ is the minimum number of periods in epoch $i$, which is a parameter of the algorithm. Define

as the first period in epoch $i$ when each $p$ has a label $a\in A_{i}$ where $\Psi^{k}_{a}(p)<0$. Epoch $i$ terminates at $K_{i}$ period.

Before we start epoch $i+1$, we need to update $A_{i}$ and $y_{i}$. Roughly speaking, $A_{i+1}$ is obtained by eliminating $a\in A_{i}$ with $\Psi^{K_{i}}_{a}(p)<0$. Because the label with $\Psi^{K_{i}}_{a}(p)<0$ can vary across different $p$, we need additional work. First, we need to eliminate the equal number of labels from $A_{i}$ for each $p$, and then, permute the remaining elements so that we can “homogenize” the set of labels for each $p$.

Define

as the set of labels in $A_{i}$ with a negative value of $\Psi^{K_{i}}_{a}(p)$, which are the candidates of elimination. Because $\mathinner{\!\left\lvert B_{i}(p)\right\rvert}$ can vary across $p$, we cannot eliminate all elements in $B_{i}(p)$ from $A_{i}$, if we want to keep the number of surviving elements in $A_{i}$ the same across $p$. Define

which is at least 1, by the definition of $K_{i}$. Recall that each element in $B_{i}(p)$ is a positive integer. Instead of eliminating all elements in $B_{i}(p)$, we eliminate $N_{i}$ elements from $B_{i}(p)$ $\forall p\in{\mathcal{P}}$. Recall that

We can write

For convenience, we select the last $N_{i}$ elements in $B_{i}(p)$ to construct ${\underline{B}}_{i}(p)$:

By the construction of ${\underline{B}}_{i}(p)$,

The remaining step is to re-label each element in $A_{i}\setminus{\underline{B}}_{i}(p)$ so that each $p$ has the same set of labels.

Define

as the collection of the first $\mathinner{\!\left\lvert A_{i}\setminus{\underline{B}}_{i}(p)\right\rvert}$ positive integers. Note

by (3.11).

We arrange the elements in $A_{i}\setminus{\underline{B}}_{i}(p)$ according to the index’s order. For each $p\in{\mathcal{P}}$, define a permutation

where

and let $\pi^{p}_{i}(1)=a_{1}$. Given $a_{1},\ldots,a_{j-1}\in A_{i}\setminus{\underline{B}}_{i}(p)$,

Since $\pi^{p}_{i}$ is a bijection over $A_{i+1}$, $\left(\pi^{p}_{i}\right)^{-1}$ exists. The correct label in epoch $i$ is

Replace $A_{i}$ and $y_{i}$ by $A_{i+1}$ and $y_{i+1}$. Start the entire process described in Section 3.2. The iteration over epoch stops at the first $i$ where

Define

as the number of epoch when the algorithm stops. Since epoch $i$ last $K_{i}$ periods, the total number of periods to train the algorithm is

which is a random variable.

Let us call the constructed algorithm $\tau^{K}_{S}$ because the algorithm is parameterized by the minimum length of each epoch $K$. The element in $A_{I}\setminus{\underline{B}}_{I}(p)$ is the source of the final hypothesis of $\tau^{K}_{S}$. Since we know $\pi^{p}_{i}$ $\forall p\in{\mathcal{P}}$ and $\forall i\in\{1,\ldots,I\}$, the final hypothesis of $\tau^{K}_{S}$ is

We are interested in $\mathbf{P}\left(\tau^{K}_{S}(D_{t})(p)=y(p)\right)$.