The Boosting Margin, or Why Boosting Doesn’t Overfit

There’s a well-understood phenomenon in machine learning called overfitting. The idea is best shown by a graph:

overfitting

Let me explain. The vertical axis represents the error of a hypothesis. The horizontal axis represents the complexity of the hypothesis. The blue curve represents the error of a machine learning algorithm’s output on its training data, and the red curve represents the generalization of that hypothesis to the real world. The overfitting phenomenon is marker in the middle of the graph, before which the training error and generalization error both go down, but after which the training error continues to fall while the generalization error rises.

The explanation is a sort of numerical version of Occam’s Razor that says more complex hypotheses can model a fixed data set better and better, but at some point a simpler hypothesis better models the underlying phenomenon that generates the data. To optimize a particular learning algorithm, one wants to set parameters of their model to hit the minimum of the red curve.

This is where things get juicy. Boosting, which we covered in gruesome detail previously, has a natural measure of complexity represented by the number of rounds you run the algorithm for. Each round adds one additional “weak learner” weighted vote. So running for a thousand rounds gives a vote of a thousand weak learners. Despite this, boosting doesn’t overfit on many datasets. In fact, and this is a shocking fact, researchers observed that Boosting would hit zero training error, they kept running it for more rounds, and the generalization error kept going down! It seemed like the complexity could grow arbitrarily without penalty.

Schapire, Freund, Bartlett, and Lee proposed a theoretical explanation for this based on the notion of a margin, and the goal of this post is to go through the details of their theorem and proof. Remember that the standard AdaBoost algorithm produces a set of weak hypotheses h_i(x) and a corresponding weight \alpha_i \in [-1,1] for each round i=1, \dots, T. The classifier at the end is a weighted majority vote of all the weak learners (roughly: weak learners with high error on “hard” data points get less weight).

Definition: The signed confidence of a labeled example (x,y) is the weighted sum:

\displaystyle \textup{conf}(x) = \sum_{i=1}^T \alpha_i h_i(x)

The margin of (x,y) is the quantity \textup{margin}(x,y) = y \textup{conf}(x). The notation implicitly depends on the outputs of the AdaBoost algorithm via “conf.”

We use the product of the label and the confidence for the observation that y \cdot \textup{conf}(x) \leq 0 if and only if the classifier is incorrect. The theorem we’ll prove in this post is

Theorem: With high probability over a random choice of training data, for any 0 < \theta < 1 generalization error of boosting is bounded from above by

\displaystyle \Pr_{\textup{train}}[\textup{margin}(x) \leq \theta] + O \left ( \frac{1}{\theta} (\textup{typical error terms}) \right )

In words, the generalization error of the boosting hypothesis is bounded by the distribution of margins observed on the training data. To state and prove the theorem more generally we have to return to the details of PAC-learning. Here and in the rest of this post, \Pr_D denotes \Pr_{x \sim D}, the probability over a random example drawn from the distribution D, and \Pr_S denotes the probability over a random (training) set of examples drawn from D.

Theorem: Let S be a set of m random examples chosen from the distribution D generating the data. Assume the weak learner corresponds to a finite hypothesis space H of size |H|, and let \delta > 0. Then with probability at least 1 - \delta (over the choice of S), every weighted-majority vote function f satisfies the following generalization bound for every \theta > 0.

\displaystyle \Pr_D[y f(x) \leq 0] \leq \Pr_S[y f(x) \leq \theta] + O \left ( \frac{1}{\sqrt{m}} \sqrt{\frac{\log m \log |H|}{\theta^2} + \log(1/\delta)} \right )

In other words, this phenomenon is a fact about voting schemes, not boosting in particular. From now on, a “majority vote” function f(x) will mean to take the sign of a sum of the form \sum_{i=1}^N a_i h_i(x), where a_i \geq 0 and \sum_i a_i = 1. This is the “convex hull” of the set of weak learners H. If H is infinite (in our proof it will be finite, but we’ll state a generalization afterward), then only finitely many of the a_i in the sum may be nonzero.

To prove the theorem, we’ll start by defining a class of functions corresponding to “unweighted majority votes with duplicates:”

Definition: Let C_N be the set of functions f(x) of the form \frac{1}{N} \sum_{i=1}^N h_i(x) where h_i \in H and the h_i may contain duplicates (some of the h_i may be equal to some other of the h_j).

Now every majority vote function f can be written as a weighted sum of h_i with weights a_i (I’m using a instead of \alpha to distinguish arbitrary weights from those weights arising from Boosting). So any such f(x) defines a natural distribution over H where you draw function h_i with probability a_i. I’ll call this distribution A_f. If we draw from this distribution N times and take an unweighted sum, we’ll get a function g(x) \in C_N. Call the random process (distribution) generating functions in this way Q_f. In diagram form, the logic goes

f \to weights a_i \to distribution over H \to function in C_N by drawing N times according to H.

The main fact about the relationship between f and Q_f is that each is completely determined by the other. Obviously Q_f is determined by f because we defined it that way, but f is also completely determined by Q_f as follows:

\displaystyle f(x) = \mathbb{E}_{g \sim Q_f}[g(x)]

Proving the equality is an exercise for the reader.

Proof of Theorem. First we’ll split the probability \Pr_D[y f(x) \leq 0] into two pieces, and then bound each piece.

First a probability reminder. If we have two events A and B (in what’s below, this will be yg(x) \leq \theta/2 and yf(x) \leq 0, we can split up \Pr[A] into \Pr[A \textup{ and } B] + \Pr[A \textup{ and } \overline{B}] (where \overline{B} is the opposite of B). This is called the law of total probability. Moreover, because \Pr[A \textup{ and } B] = \Pr[A | B] \Pr[B] and because these quantities are all at most 1, it’s true that \Pr[A \textup{ and } B] \leq \Pr[A \mid B] (the conditional probability) and that \Pr[A \textup{ and } B] \leq \Pr[B].

Back to the proof. Notice that for any g(x) \in C_N and any \theta > 0, we can write \Pr_D[y f(x) \leq 0] as a sum:

\displaystyle \Pr_D[y f(x) \leq 0] =\\ \Pr_D[yg(x) \leq \theta/2 \textup{ and } y f(x) \leq 0] + \Pr_D[yg(x) > \theta/2 \textup{ and } y f(x) \leq 0]

Now I’ll loosen the first term by removing the second event (that only makes the whole probability bigger) and loosen the second term by relaxing it to a conditional:

\displaystyle \Pr_D[y f(x) \leq 0] \leq \Pr_D[y g(x) \leq \theta / 2] + \Pr_D[yg(x) > \theta/2 \mid yf(x) \leq 0]

Now because the inequality is true for every g(x) \in C_N, it’s also true if we take an expectation of the RHS over any distribution we choose. We’ll choose the distribution Q_f to get

\displaystyle \Pr_D[yf(x) \leq 0] \leq T_1 + T_2

And T_1 (term 1) is

\displaystyle T_1 = \Pr_{x \sim D, g \sim Q_f} [yg(x) \leq \theta /2] = \mathbb{E}_{g \sim Q_f}[\Pr_D[yg(x) \leq \theta/2]]

And T_2 is

\displaystyle \Pr_{x \sim D, g \sim Q_f}[yg(x) > \theta/2 \mid yf(x) \leq 0] = \mathbb{E}_D[\Pr_{g \sim Q_f}[yg(x) > \theta/2 \mid yf(x) \leq 0]]

We can rewrite the probabilities using expectations because (1) the variables being drawn in the distributions are independent, and (2) the probability of an event is the expectation of the indicator function of the event.

Now we’ll bound the terms T_1, T_2 separately. We’ll start with T_2.

Fix (x,y) and look at the quantity inside the expectation of T_2.

\displaystyle \Pr_{g \sim Q_f}[yg(x) > \theta/2 \mid yf(x) \leq 0]

This should intuitively be very small for the following reason. We’re sampling g according to a distribution whose expectation is f, and we know that yf(x) \leq 0. Of course yg(x) is unlikely to be large.

Mathematically we can prove this by transforming the thing inside the probability to a form suitable for the Chernoff bound. Saying yg(x) > \theta / 2 is the same as saying |yg(x) - \mathbb{E}[yg(x)]| > \theta /2, i.e. that some random variable which is a sum of independent random variables (the h_i) deviates from its expectation by at least \theta/2. Since the y‘s are all \pm 1 and constant inside the expectation, they can be removed from the absolute value to get

\displaystyle \leq \Pr_{g \sim Q_f}[g(x) - \mathbb{E}[g(x)] > \theta/2]

The Chernoff bound allows us to bound this by an exponential in the number of random variables in the sum, i.e. N. It turns out the bound is e^{-N \theta^2 / 8}.

Now recall T_1

\displaystyle T_1 = \Pr_{x \sim D, g \sim Q_f} [yg(x) \leq \theta /2] = \mathbb{E}_{g \sim Q_f}[\Pr_D[yg(x) \leq \theta/2]]

For T_1, we don’t want to bound it absolutely like we did for T_2, because there is nothing stopping the classifier f from being a bad classifier and having lots of error. Rather, we want to bound it in terms of the probability that yf(x) \leq \theta. We’ll do this in two steps. In step 1, we’ll go from \Pr_D of the g‘s to \Pr_S of the g‘s.

Step 1: For any fixed g, \theta, if we take a sample S of size m, then consider the event in which the sample probability deviates from the true distribution by some value \varepsilon_N, i.e. the event

\displaystyle \Pr_D[yg(x) \leq \theta /2] > \Pr_{S, x \sim S}[yg(x) \leq \theta/2] + \varepsilon_N

The claim is this happens with probability at most e^{-2m\varepsilon_N^2}. This is again the Chernoff bound in disguise, because the expected value of \Pr_S is \Pr_D, and the probability over S is an average of random variables (it’s a slightly different form of the Chernoff bound; see this post for more). From now on we’ll drop the x \sim S when writing \Pr_S.

The bound above holds true for any fixed g,\theta, but we want a bound over all g and \theta. To do that we use the union bound. Note that there are only (N+1) possible choices for a nonnegative \theta because g(x) is a sum of N values each of which is either \pm1. And there are only |C_N| \leq |H|^N possibilities for g(x). So the union bound says the above event will occur with probability at most (N+1)|H|^N e^{-2m\varepsilon_N^2}.

If we want the event to occur with probability at most \delta_N, we can judiciously pick

\displaystyle \varepsilon_N = \sqrt{(1/2m) \log ((N+1)|H|^N / \delta_N)}

And since the bound holds in general, we can take expectation with respect to Q_f and nothing changes. This means that for any \delta_N, our chosen \varepsilon_N ensures that the following is true with probability at least 1-\delta_N:

\displaystyle \Pr_{D, g \sim Q_f}[yg(x) \leq \theta/2] \leq \Pr_{S, g \sim Q_f}[yg(x) \leq \theta/2] + \varepsilon_N

Now for step 2, we bound the probability that yg(x) \leq \theta/2 on a sample to the probability that yf(x) \leq \theta on a sample.

Step 2: The first claim is that

\displaystyle \Pr_{S, g \sim Q_f}[yg(x) \leq \theta / 2] \leq \Pr_{S} [yf(x) \leq \theta] + \mathbb{E}_{S}[\Pr_{g \sim Q_f}[yg(x) \leq \theta/2 \mid yf(x) \geq \theta]]

What we did was break up the LHS into two “and”s, when yf(x) > \theta and yf(x) \leq \theta (this was still an equality). Then we loosened the first term to \Pr_{S}[yf(x) \leq \theta] since that is only more likely than both yg(x) \leq \theta/2 and yf(x) \leq \theta. Then we loosened the second term again using the fact that a probability of an “and” is bounded by the conditional probability.

Now we have the probability of yg(x) \leq \theta / 2 bounded by the probability that yf(x) \leq 0 plus some stuff. We just need to bound the “plus some stuff” absolutely and then we’ll be done. The argument is the same as our previous use of the Chernoff bound: we assume yf(x) \geq \theta, and yet yg(x) \leq \theta / 2. So the deviation of yg(x) from its expectation is large, and the probability that happens is exponentially small in the amount of deviation. The bound you get is

\displaystyle \Pr_{g \sim Q}[yg(x) \leq \theta/2 \mid yf(x) > \theta] \leq e^{-N\theta^2 / 8}.

And again we use the union bound to ensure the failure of this bound for any N will be very small. Specifically, if we want the total failure probability to be at most \delta, then we need to pick some \delta_j‘s so that \delta = \sum_{j=0}^{\infty} \delta_j. Choosing \delta_N = \frac{\delta}{N(N+1)} works.

Putting everything together, we get that with probability at least 1-\delta for every \theta and every N, this bound on the failure probability of f(x):

\displaystyle \Pr_{x \sim D}[yf(x) \leq 0] \leq \Pr_{S, x \sim S}[yf(x) \leq \theta] + 2e^{-N \theta^2 / 8} + \sqrt{\frac{1}{2m} \log \left ( \frac{N(N+1)^2 |H|^N}{\delta} \right )}.

This claim is true for every N, so we can pick N that minimizes it. Doing a little bit of behind-the-scenes calculus that is left as an exercise to the reader, a tight choice of N is (4/ \theta)^2 \log(m/ \log |H|). And this gives the statement of the theorem.

\square

We proved this for finite hypothesis classes, and if you know what VC-dimension is, you’ll know that it’s a central tool for reasoning about the complexity of infinite hypothesis classes. An analogous theorem can be proved in terms of the VC dimension. In that case, calling d the VC-dimension of the weak learner’s output hypothesis class, the bound is

\displaystyle \Pr_D[yf(x) \leq 0] \leq \Pr_S[yf(x) \leq \theta] + O \left ( \frac{1}{\sqrt{m}} \sqrt{\frac{d \log^2(m/d)}{\theta^2} + \log(1/\delta)} \right )

How can we interpret these bounds with so many parameters floating around? That’s where asymptotic notation comes in handy. If we fix \theta \leq 1/2 and \delta = 0.01, then the big-O part of the theorem simplifies to \sqrt{(\log |H| \cdot \log m) / m}, which is easier to think about since (\log m)/m goes to zero very fast.

Now the theorem we just proved was about any weighted majority function. The question still remains: why is AdaBoost good? That follows from another theorem, which we’ll state and leave as an exercise (it essentially follows by unwrapping the definition of the AdaBoost algorithm from last time).

Theorem: Suppose that during AdaBoost the weak learners produce hypotheses with training errors \varepsilon_1, \dots , \varepsilon_T. Then for any \theta,

\displaystyle \Pr_{(x,y) \sim S} [yf(x) \leq \theta] \leq 2^T \prod_{t=1}^T \sqrt{\varepsilon_t^{(1-\theta)} (1-\varepsilon_t)^{(1+\theta)}}

Let’s interpret this for some concrete numbers. Say that \theta = 0 and \varepsilon_t is any fixed value less than 1/2. In this case the term inside product becomes \sqrt{\varepsilon (1-\varepsilon)} < 1/2 and the whole bound tends exponentially quickly to zero in the number of rounds T. On the other hand, if we raise \theta to about 1/3, then in order to maintain the LHS tending to zero we would need \varepsilon < \frac{1}{4} ( 3 - \sqrt{5} ) which is about 20% error.

If you’re interested in learning more about Boosting, there is an excellent book by Freund and Schapire (the inventors of boosting) called Boosting: Foundations and Algorithms. There they include a tighter analysis based on the idea of Rademacher complexity. The bound I presented in this post is nice because the proof doesn’t require any machinery past basic probability, but if you want to reach the cutting edge of knowledge about boosting you need to invest in the technical stuff.

Until next time!

Occam’s Razor and PAC-learning

So far our discussion of learning theory has been seeing the definition of PAC-learningtinkering with it, and seeing simple examples of learnable concept classes. We’ve said that our real interest is in proving big theorems about what big classes of problems can and can’t be learned. One major tool for doing this with PAC is the concept of VC-dimension, but to set the stage we’re going to prove a simpler theorem that gives a nice picture of PAC-learning when your hypothesis class is small. In short, the theorem we’ll prove says that if you have a finite set of hypotheses to work with, and you can always find a hypothesis that’s consistent with the data you’ve seen, then you can learn efficiently. It’s obvious, but we want to quantify exactly how much data you need to ensure low error. This will also give us some concrete mathematical justification for philosophical claims about simplicity, and the theorems won’t change much when we generalize to VC-dimension in a future post.

The Chernoff bound

One tool we will need in this post, which shows up all across learning theory, is the Chernoff-Hoeffding bound. We covered this famous inequality in detail previously on this blog, but the part of that post we need is the following theorem that says, informally, that if you average a bunch of bounded random variables, then the probability this average random variable deviates from its expectation is exponentially small in the amount of deviation. Here’s the slightly simplified version we’ll use:

Theorem: Let X_1, \dots, X_m be independent random variables whose values are in the range [0,1]. Call \mu_i = \mathbf{E}[X_i], X = \sum_i X_i, and \mu = \mathbf{E}[X] = \sum_i \mu_i. Then for all t > 0,

\displaystyle \Pr(|X-\mu| > t) \leq 2e^{-2t^2 / m}

One nice thing about the Chernoff bound is that it doesn’t matter how the variables are distributed. This is important because in PAC we need guarantees that hold for any distribution generating data. Indeed, in our case the random variables above will be individual examples drawn from the distribution generating the data. We’ll be estimating the probability that our hypothesis has error deviating more than \varepsilon, and we’ll want to bound this by \delta, as in the definition of PAC-learning. Since the amount of deviation (error) and the number of samples (m) both occur in the exponent, the trick is in balancing the two values to get what we want.

Realizability and finite hypothesis classes

Let’s recall the PAC model once more. We have a distribution D generating labeled examples (x, c(x)), where c is an unknown function coming from some concept class C. Our algorithm can draw a polynomial number of these examples, and it must produce a hypothesis h from some hypothesis class H (which may or may not contain c). The guarantee we need is that, for any \delta, \varepsilon > 0, the algorithm produces a hypothesis whose error on D is at most \varepsilon, and this event happens with probability at least 1-\delta. All of these probabilities are taken over the randomness in the algorithm’s choices and the distribution D, and it has to work no matter what the distribution D is.

Let’s introduce some simplifications. First, we’ll assume that the hypothesis and concept classes H and C are finite. Second, we’ll assume that C \subset H, so that you can actually hope to find a hypothesis of zero error. This is called realizability. Later we’ll relax these first two assumptions, but they make the analysis a bit cleaner. Finally, we’ll assume that we have an algorithm which, when given labeled examples, can find in polynomial time a hypothesis h \in H that is consistent with every example.

These assumptions give a trivial learning algorithm: draw a bunch of examples and output any consistent hypothesis. The question is, how many examples do we need to guarantee that the hypothesis we find has the prescribed generalization error? It will certainly grow with 1 / \varepsilon, but we need to ensure it will only grow polynomially fast in this parameter. Indeed, realizability is such a strong assumption that we can prove a polynomial bound using even more basic probability theory than the Chernoff bound.

Theorem: A algorithm that efficiently finds a consistent hypothesis will PAC-learn any finite concept class provided it has at least m samples, where

\displaystyle m \geq \frac{1}{\varepsilon} \left ( \log |H| + \log \left ( \frac{1}{\delta} \right ) \right )

Proof. All we need to do is bound the probability that a bad hypothesis (one with error more than \varepsilon) is consistent with the given data. Now fix D, c, \delta, \varepsilon, and draw m examples and let h be any hypothesis that is consistent with the drawn examples. Suppose that the bad thing happens, that \Pr_D(h(x) \neq c(x)) > \varepsilon.

Because the examples are all drawn independently from D, the chance that all m examples are consistent with h is

\displaystyle (1 - \Pr_{x \sim D}(h(x) \neq c(x)))^m < (1 - \varepsilon)^m

What we’re saying here is, the probability that a specific bad hypothesis is actually consistent with your drawn examples is exponentially small in the error tolerance. So if we apply the union bound, the probability that some hypothesis you could produce is bad is at most (1 - \varepsilon)^m S, where S is the number of hypotheses the algorithm might produce.

A crude upper bound on the number of hypotheses you could produce is just the total number of hypotheses, |H|. Even cruder, let’s use the inequality (1 - x) < e^{-x} to give the bound

\displaystyle (1 - \varepsilon)^m |H| < e^{-\varepsilon m} |H|

Now we want to make sure that this probability, the probability of choosing a high-error (yet consistent) hypothesis, is at most \delta. So we can set the above quantity less than \delta and solve for m:

\displaystyle e^{-\varepsilon m} |H| \leq \delta

Taking logs and solving for m gives the desired bound.

\square

An obvious objection is: what if you aren’t working with a hypothesis class where you can guarantee that you’ll find a consistent hypothesis? Well, in that case we’ll need to inspect the definition of PAC again and reevaluate our measures of error. It turns out we’ll get a similar theorem as above, but with the stipulation that we’re only achieving error within epsilon of the error of the best available hypothesis.

But before we go on, this theorem has some deep philosophical interpretations. In particular, suppose that, before drawing your data, you could choose to work with one of two finite hypothesis classes H_1, H_2, with |H_1| > |H_2|. If you can find a consistent hypothesis no matter which hypothesis class you use, then this theorem says that your generalization guarantees are much stronger if you start with the smaller hypothesis class.

In other words, all else being equal, the smaller set of hypotheses is better. For this reason, the theorem is sometimes called the “Occam’s Razor” theorem. We’ll see a generalization of this theorem in the next section.

Unrealizability and an extra epsilon

Now suppose that $H$ doesn’t contain any hypotheses with error less than \varepsilon. What can we hope to do in this case? One thing is that we can hope to find a hypothesis whose error is within \varepsilon of the minimal error of any hypothesis in H. Moreover, we might not have any consistent hypotheses for some data samples! So rather than require an algorithm to produce an h \in H that is perfectly consistent with the data, we just need it to produce a hypothesis that has minimal empirical error, in the sense that it is as close to consistent as the best hypothesis of h on the data you happened to draw. It seems like such a strategy would find you a hypothesis that’s close to the best one in H, but we need to prove it and determine how many samples we need to draw to succeed.

So let’s make some definitions to codify this. For a given hypothesis, call \textup{err}(h) the true error of h on the distribution D. Our assumption is that there may be no hypotheses in H with \textup{err}(h) = 0. Next we’ll call the empirical error \hat{\textup{err}}(h).

Definition: We say a concept class C is agnostically learnable using the hypothesis class H if for all c \in C and all distributions D (and all \varepsilon, \delta > 0), there is a learning algorithm A which produces a hypothesis h that with probability at least 1 - \delta satisfies

\displaystyle \text{err}(h) \leq \min_{h' \in H} \text{err}(h') + \varepsilon

and everything runs in the same sort of polynomial time as for vanilla PAC-learning. This is called the agnostic setting or the unrealizable setting, in the sense that we may not be able to find a hypothesis with perfect empirical error.

We seek to prove that all concept classes are agnostically learnable with a finite hypothesis class, provided you have an algorithm that can minimize empirical error. But actually we’ll prove something stronger.

Theorem: Let H be a finite hypothesis class and m the number of samples drawn. Then for any \delta > 0, with probability 1-\delta the following holds:

\displaystyle \forall h \in H, \hat{\text{err}}(h) \leq \text{err}(h) + \sqrt{\frac{\log |H| + \log(2 / \delta)}{2m}}

In other words, we can precisely quantify how the empirical error converges to the true error as the number of samples grows. But this holds for all hypotheses in H, so this provides a uniform bound of the difference between true and empirical error for the entire hypothesis class.

Proving this requires the Chernoff bound. Fix a single hypothesis h \in H. If you draw an example x, call Z the random variable which is 1 when h(x) \neq c(x), and 0 otherwise. So if you draw m samples and call the i-th variable Z_i, the empirical error of the hypothesis is \frac{1}{m}\sum_i Z_i. Moreover, the actual error is the expectation of this random variable since \mathbf{E}[1/m \sum_i Z_i] = Z.

So what we’re asking is the probability that the empirical error deviates from the true error by a lot. Let’s call “a lot” some parameter \varepsilon/2 > 0 (the reason for dividing by two will become clear in the corollary to the theorem). Then plugging things into the Chernoff-Hoeffding bound gives a bound on the probability of the “bad event,” that the empirical error deviates too much.

\displaystyle \Pr[|\hat{\text{err}}(h) - \text{err}(h)| > \varepsilon / 2] < 2e^{-\frac{\varepsilon^2m}{2}}

Now to get a bound on the probability that some hypothesis is bad, we apply the union bound and use the fact that |H| is finite to get

\displaystyle \Pr[|\hat{\text{err}}(h) - \text{err}(h)| > \varepsilon / 2] < 2|H|e^{-\frac{\varepsilon^2m}{2}}

Now say we want to bound this probability by \delta. We set 2|H|e^{-\varepsilon^2m/2} \leq \delta, solve for m, and get

\displaystyle m \geq \frac{2}{\varepsilon^2}\left ( \log |H| + \log \frac{2}{\delta} \right )

This gives us a concrete quantification of the tradeoff between m, \varepsilon, \delta, and |H|. Indeed, if we pick m to be this large, then solving for \varepsilon / 2 gives the exact inequality from the theorem.

\square

Now we know that if we pick enough samples (polynomially many in all the parameters), and our algorithm can find a hypothesis h of minimal empirical error, then we get the following corollary:

Corollary: For any \varepsilon, \delta > 0, the algorithm that draws m \geq \frac{2}{\varepsilon^2}(\log |H| + \log(2/ \delta)) examples and finds any hypothesis of minimal empirical error will, with probability at least 1-\delta, produce a hypothesis that is within \varepsilon of the best hypothesis in H.

Proof. By the previous theorem, with the desired probability, for all h \in H we have |\hat{\text{err}}(h) - \text{err}(h)| < \varepsilon/2. Call g = \min_{h' \in H} \text{err}(h'). Then because the empirical error of h is also minimal, we have |\hat{\text{err}}(g) - \text{err}(h)| < \varepsilon / 2. And using the previous theorem again and the triangle inequality, we get |\text{err}(g) - \text{err}(h)| < 2 \varepsilon / 2 = \varepsilon. In words, the true error of the algorithm’s hypothesis is close to the error of the best hypothesis, as desired.

\square

Next time

Both of these theorems tell us something about the generalization guarantees for learning with hypothesis classes of a certain size. But this isn’t exactly the most reasonable measure of the “complexity” of a family of hypotheses. For example, one could have a hypothesis class with a billion intervals on \mathbb{R} (say you’re trying to learn intervals, or thresholds, or something easy), and the guarantees we proved in this post are nowhere near optimal.

So the question is: say you have a potentially infinite class of hypotheses, but the hypotheses are all “simple” in some way. First, what is the right notion of simplicity? And second, how can you get guarantees based on that analogous to these? We’ll discuss this next time when we define the VC-dimension.

Until then!