I have a question about Example 10.1 in Shalev-Shwartz and Ben-David's "Understanding Machine Learning." The example means to illustrate weak learning of 3-piece classifiers $\mathcal H$ using decision stumps $\mathcal G$, where $\mathcal H=\{h_{\theta_1, \theta_2, b}:\theta_1, \theta_2\in \mathbb R, \theta_1< \theta_2,b\in \{+1,-1\}\}$ with $$ h_{\theta_1, \theta_2, b}(x) =\left\{\begin{array}{ll} -b, & \theta_1 \le x \le\theta_2\\ +b, & \textrm{otherwise.}\end{array}\right.$$

An example of $h_{\theta_1, \theta_2, b}(x)$ is as follows: enter image description here

On the other hand, $\mathcal G=\{x\mapsto \mathrm{sign}(x-\theta)\cdot b:\theta\in \mathbb R, b\in \{+1, -1\}\}.$

What I don't understand is that the book claims that

for every distribution $\mathcal D$ that is consistent with $\mathcal H$, there exists a decision stump $g$ with $L_{\mathcal D}(g)\le 1/3,$

where $L_{\mathcal D}(g)=P_{\mathcal D}\{x: g(x)\ne \hat h_{\theta_1, \theta_2, b}(x)\}$ denotes the true/population error and $\hat h_{\theta_1, \theta_2, b}$ is the ground truth. And the reasoning is that for any pair of the three regions of $\mathbb R$ partitioned by $\theta_1$ and $\theta_2$, namely $\{x:x<\theta_1\}, \{x:\theta_1\le x\le \theta_2\}$ and $\{x:x>\theta_2\}$

there exists a decision stump that agrees with the labeling of these two components.

Suppose that $P_{\mathcal D}(\{x:x<\theta_1\})=P_{\mathcal D}(\{x:x>\theta_2\})=0.4$ and $P_{\mathcal D}(\{x:\theta_1\le x \le\theta_2\})=0.2$. I don't see how we can find a decision stump $g$ which disagrees with the correct labeling $\hat h_{\theta_1, \theta_2, b}(x)$ only on the interval $\{x:\theta_1\le x\le \theta_2\}$, so that the true error $L_{\mathcal D}(g)\le1/3$? Any decision stump would at least disagree with part of $\{x:x<\theta_1\}$ or $\{x:x>\theta_2\}$, wouldn't it?

I'd appreciate it, if someone can point out where I missed.


For example, let $g = sign(x-\theta)\cdot b$ s.t.

$P_{\mathcal D}(\{x:\theta\le x \le\theta_1\})= 0.4$

That is, $\theta = -\infty$.

Then, $g$ agrees with $\hat h_{\theta_1, \theta_2, b}(x)$ on $[\theta, \theta_1]$, which is just $[-\infty, \theta_1]$, and $(\theta_2, \infty)$.

These 2 intervals, i.e. $[-\infty, \theta_1]$, and $(\theta_2, \infty)$ are so-called "these two components" in your question.

Then $L_{\mathcal D}(g)=P_{\mathcal D}(\{x: x \le\theta\}) + P_{\mathcal D}(\{x: \theta_1 \le x \le\theta_2\})\le 0.1 + 0.2 = 0.3 \le 1/3 $

Here is a figure that may be useful. enter image description here

  • $\begingroup$ Thanks for the answer. I agree that we can find $g$ s.t. $L_D(g)\le 1/3.$ However, what really confused me is the book's reasoning that "there exists a decision stump that agrees with the labeling of these two components". In this example, $g$ doesn't agree with $\hat h$ on the $\{x: x<\theta_1\}$ component entirely, does it? $\endgroup$ – syeh_106 Dec 4 at 15:08
  • $\begingroup$ Just to clarify a bit more, the key reasoning of the book is: of the three regions $\{x:x<\theta_1\}, \{x: \theta_1 \le x \le \theta_2\}, \{x: x > \theta_2\}$, one of them must have probability of at most $1/3.$ In my example, it's the second region, $\{x: \theta_1 \le x \le \theta_2\}$. And the book claims that "A decision stump that disagrees with $\hat h$ only on such a region has an error of at most $1/3$." I don't see how we can find a decision stump that disagrees with $\hat h$ only on $\{x: \theta_1 \le x \le \theta_2\}$. $\endgroup$ – syeh_106 Dec 4 at 15:14
  • 1
    $\begingroup$ Sorry, I misunderstood your question. I have updated my answer. The key is setting $\theta = -\infty$. $\endgroup$ – Ben Dec 5 at 10:18

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