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:


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.


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!

Weak Learning, Boosting, and the AdaBoost algorithm

When addressing the question of what it means for an algorithm to learn, one can imagine many different models, and there are quite a few. This invariably raises the question of which models are “the same” and which are “different,” along with a precise description of how we’re comparing models. We’ve seen one learning model so far, called Probably Approximately Correct (PAC), which espouses the following answer to the learning question:

An algorithm can “solve” a classification task using labeled examples drawn from some distribution if it can achieve accuracy that is arbitrarily close to perfect on the distribution, and it can meet this goal with arbitrarily high probability, where it’s runtime and the number of examples needed scales efficiently with all the parameters (accuracy, confidence, size of an example). Moreover, the algorithm needs to succeed no matter what distribution generates the examples.

You can think of this as a game between the algorithm designer and an adversary. First, the learning problem is fixed and everyone involved knows what the task is. Then the algorithm designer has to pick an algorithm. Then the adversary, knowing the chosen algorithm, chooses a nasty distribution D over examples that are fed to the learning algorithm. The algorithm designer “wins” if the algorithm produces a hypothesis with low error on D when given samples from D. And our goal is to prove that the algorithm designer can pick a single algorithm that is extremely likely to win no matter what D the adversary picks.

We’ll momentarily restate this with a more precise definition, because in this post we will compare it to a slightly different model, which is called the weak PAC-learning model. It’s essentially the same as PAC, except it only requires the algorithm to have accuracy that is slightly better than random guessing. That is, the algorithm will output a classification function which will correctly classify a random label with probability at least \frac{1}{2} + \eta for some small, but fixed, \eta > 0. The quantity \eta (the Greek “eta”) is called the edge as in “the edge over random guessing.” We call an algorithm that produces such a hypothesis a weak learner, and in contrast we’ll call a successful algorithm in the usual PAC model a strong learner.

The amazing fact is that strong learning and weak learning are equivalent! Of course a weak learner is not the same thing as a strong learner. What we mean by “equivalent” is that:

A problem can be weak-learned if and only if it can be strong-learned.

So they are computationally the same. One direction of this equivalence is trivial: if you have a strong learner for a classification task then it’s automatically a weak learner for the same task. The reverse is much harder, and this is the crux: there is an algorithm for transforming a weak learner into a strong learner! Informally, we “boost” the weak learning algorithm by feeding it examples from carefully constructed distributions, and then take a majority vote. This “reduction” from strong to weak learning is where all the magic happens.

In this post we’ll get into the depths of this boosting technique. We’ll review the model of PAC-learning, define what it means to be a weak learner, “organically” come up with the AdaBoost algorithm from some intuitive principles, prove that AdaBoost reduces error on the training data, and then run it on data. It turns out that despite the origin of boosting being a purely theoretical question, boosting algorithms have had a wide impact on practical machine learning as well.

As usual, all of the code and data used in this post is available on this blog’s Github page.

History and multiplicative weights

Before we get into the details, here’s a bit of history and context. PAC learning was introduced by Leslie Valiant in 1984, laying the foundation for a flurry of innovation. In 1988 Michael Kearns posed the question of whether one can “boost” a weak learner to a strong learner. Two years later Rob Schapire published his landmark paper “The Strength of Weak Learnability” closing the theoretical question by providing the first “boosting” algorithm. Schapire and Yoav Freund worked together for the next few years to produce a simpler and more versatile algorithm called AdaBoost, and for this they won the Gödel Prize, one of the highest honors in theoretical computer science. AdaBoost is also the standard boosting algorithm used in practice, though there are enough variants to warrant a book on the subject.

I’m going to define and prove that AdaBoost works in this post, and implement it and test it on some data. But first I want to give some high level discussion of the technique, and afterward the goal is to make that wispy intuition rigorous.

The central technique of AdaBoost has been discovered and rediscovered in computer science, and recently it was recognized abstractly in its own right. It is called the Multiplicative Weights Update Algorithm (MWUA), and it has applications in everything from learning theory to combinatorial optimization and game theory. The idea is to

  1. Maintain a nonnegative weight for the elements of some set,
  2. Draw a random element proportionally to the weights,
  3. So something with the chosen element, and based on the outcome of the “something…”
  4. Update the weights and repeat.

The “something” is usually a black box algorithm like “solve this simple optimization problem.” The output of the “something” is interpreted as a reward or penalty, and the weights are updated according to the severity of the penalty (the details of how this is done differ depending on the goal). In this light one can interpret MWUA as minimizing regret with respect to the best alternative element one could have chosen in hindsight. In fact, this was precisely the technique we used to attack the adversarial bandit learning problem (the Exp3 algorithm is a multiplicative weight scheme). See this lengthy technical survey of Arora and Kale for a research-level discussion of the algorithm and its applications.

Now let’s remind ourselves of the formal definition of PAC. If you’ve read the previous post on the PAC model, this next section will be redundant.

Distributions, hypotheses, and targets

In PAC-learning you are trying to give labels to data from some set X. There is a distribution D producing data from X, and it’s used for everything: to provide data the algorithm uses to learn, to measure your accuracy, and every other time you might get samples from X. You as the algorithm designer don’t know what D is, and a successful learning algorithm has to work no matter what D is. There’s some unknown function c called the target concept, which assigns a \pm 1 label to each data point in X. The target is the function we’re trying to “learn.” When the algorithm draws an example from D, it’s allowed to query the label c(x) and use all of the labels it’s seen to come up with some hypothesis h that is used for new examples that the algorithm may not have seen before. The problem is “solved” if h has low error on all of D.

To give a concrete example let’s do spam emails. Say that X is the set of all emails, and D is the distribution over emails that get sent to my personal inbox. A PAC-learning algorithm would take all my emails, along with my classification of which are spam and which are not spam (plus and minus 1). The algorithm would produce a hypothesis h that can be used to label new emails, and if the algorithm is truly a PAC-learner, then our guarantee is that with high probability (over the randomness in which emails I receive) the algorithm will produce an h that has low error on the entire distribution of emails that get sent to me (relative to my personal spam labeling function).

Of course there are practical issues with this model. I don’t have a consistent function for calling things spam, the distribution of emails I get and my labeling function can change over time, and emails don’t come according to a distribution with independent random draws. But that’s the theoretical model, and we can hope that algorithms we devise for this model happen to work well in practice.

Here’s the formal definition of the error of a hypothesis h(x) produced by the learning algorithm:

\textup{err}_{c,D}(h) = P_{x \sim D}(h(x) \neq c(x))

It’s read “The error of h with respect to the concept c we’re trying to learn and the distribution D is the probability over x drawn from D that the hypothesis produces the wrong label.” We can now define PAC-learning formally, introducing the parameters \delta for “probably” and \varepsilon for “approximately.” Let me say it informally first:

An algorithm PAC-learns if, for any \varepsilon, \delta > 0 and any distribution D, with probability at least 1-\delta the hypothesis h produced by the algorithm has error at most \varepsilon.

To flush out the other things hiding, here’s the full definition.

Definition (PAC): An algorithm A(\varepsilon, \delta) is said to PAC-learn the concept class H over the set X if, for any distribution D over X and for any 0 < \varepsilon, \delta < 1/2 and for any target concept c \in H, the probability that A produces a hypothesis h of error at most \varepsilon is at least 1-\delta. In symbols, \Pr_D(\textup{err}_{c,D}(h) \leq \varepsilon) > 1 - \delta. Moreover, A must run in time polynomial in 1/\varepsilon, 1/\delta and n, where n is the size of an element x \in X.

The reason we need a class of concepts (instead of just one target concept) is that otherwise we could just have a constant algorithm that outputs the correct labeling function. Indeed, when we get a problem we ask whether there exists an algorithm that can solve it. I.e., a problem is “PAC-learnable” if there is some algorithm that learns it as described above. With just one target concept there can exist an algorithm to solve the problem by hard-coding a description of the concept in the source code. So we need to have some “class of possible answers” that the algorithm is searching through so that the algorithm actually has a job to do.

We call an algorithm that gets this guarantee a strong learner. A weak learner has the same definition, except that we replace \textup{err}_{c,D}(h) \leq \varepsilon by the weak error bound: for some fixed 0 < \eta < 1/2. the error \textup{err}_{c,D}(h) \leq 1/2 - \eta. So we don’t require the algorithm to achieve any desired accuracy, it just has to get some accuracy slightly better than random guessing, which we don’t get to choose. As we will see, the value of \eta influences the convergence of the boosting algorithm. One important thing to note is that \eta is a constant independent of n, the size of an example, and m, the number of examples. In particular, we need to avoid the “degenerate” possibility that \eta(n) = 2^{-n} so that as our learning problem scales the quality of the weak learner degrades toward 1/2. We want it to be bounded away from 1/2.

So just to clarify all the parameters floating around, \delta will always be the “probably” part of PAC, \varepsilon is the error bound (the “approximately” part) for strong learners, and \eta is the error bound for weak learners.

What could a weak learner be?

Now before we prove that you can “boost” a weak learner to a strong learner, we should have some idea of what a weak learner is. Informally, it’s just a ‘rule of thumb’ that you can somehow guarantee does a little bit better than random guessing.

In practice, however, people sort of just make things up and they work. It’s kind of funny, but until recently nobody has really studied what makes a “good weak learner.” They just use an example like the one we’re about to show, and as long as they get a good error rate they don’t care if it has any mathematical guarantees. Likewise, they don’t expect the final “boosted” algorithm to do arbitrarily well, they just want low error rates.

The weak learner we’ll use in this post produces “decision stumps.” If you know what a decision tree is, then a decision stump is trivial: it’s a decision tree where the whole tree is just one node. If you don’t know what a decision tree is, a decision stump is a classification rule of the form:

Pick some feature i and some value of that feature v, and output label +1 if the input example has value v for feature i, and output label -1 otherwise.

Concretely, a decision stump might mark an email spam if it contains the word “viagra.” Or it might deny a loan applicant a loan if their credit score is less than some number.

Our weak learner produces a decision stump by simply looking through all the features and all the values of the features until it finds a decision stump that has the best error rate. It’s brute force, baby! Actually we’ll do something a little bit different. We’ll make our data numeric and look for a threshold of the feature value to split positive labels from negative labels. Here’s the Python code we’ll use in this post for boosting. This code was part of a collaboration with my two colleagues Adam Lelkes and Ben Fish. As usual, all of the code used in this post is available on Github.

First we make a class for a decision stump. The attributes represent a feature, a threshold value for that feature, and a choice of labels for the two cases. The classify function shows how simple the hypothesis is.

class Stump:
   def __init__(self):
      self.gtLabel = None
      self.ltLabel = None
      self.splitThreshold = None
      self.splitFeature = None

   def classify(self, point):
      if point[self.splitFeature] >= self.splitThreshold:
         return self.gtLabel
         return self.ltLabel

   def __call__(self, point):
      return self.classify(point)

Then for a fixed feature index we’ll define a function that computes the best threshold value for that index.

def minLabelErrorOfHypothesisAndNegation(data, h):
   posData, negData = ([(x, y) for (x, y) in data if h(x) == 1],
                       [(x, y) for (x, y) in data if h(x) == -1])

   posError = sum(y == -1 for (x, y) in posData) + sum(y == 1 for (x, y) in negData)
   negError = sum(y == 1 for (x, y) in posData) + sum(y == -1 for (x, y) in negData)
   return min(posError, negError) / len(data)

def bestThreshold(data, index, errorFunction):
   '''Compute best threshold for a given feature. Returns (threshold, error)'''

   thresholds = [point[index] for (point, label) in data]
   def makeThreshold(t):
      return lambda x: 1 if x[index] >= t else -1
   errors = [(threshold, errorFunction(data, makeThreshold(threshold))) for threshold in thresholds]
   return min(errors, key=lambda p: p[1])

Here we allow the user to provide a generic error function that the weak learner tries to minimize, but in our case it will just be minLabelErrorOfHypothesisAndNegation. In words, our threshold function will label an example as +1 if feature i has value greater than the threshold and -1 otherwise. But we might want to do the opposite, labeling -1 above the threshold and +1 below. The bestThreshold function doesn’t care, it just wants to know which threshold value is the best. Then we compute what the right hypothesis is in the next function.

def buildDecisionStump(drawExample, errorFunction=defaultError):
   # find the index of the best feature to split on, and the best threshold for
   # that index. A labeled example is a pair (example, label) and drawExample() 
   # accepts no arguments and returns a labeled example. 

   data = [drawExample() for _ in range(500)]

   bestThresholds = [(i,) + bestThreshold(data, i, errorFunction) for i in range(len(data[0][0]))]
   feature, thresh, _ = min(bestThresholds, key = lambda p: p[2])

   stump = Stump()
   stump.splitFeature = feature
   stump.splitThreshold = thresh
   stump.gtLabel = majorityVote([x for x in data if x[0][feature] >= thresh])
   stump.ltLabel = majorityVote([x for x in data if x[0][feature] < thresh])

   return stump

It’s a little bit inefficient but no matter. To illustrate the PAC framework we emphasize that the weak learner needs nothing except the ability to draw from a distribution. It does so, and then it computes the best threshold and creates a new stump reflecting that. The majorityVote function just picks the most common label of examples in the list. Note that drawing 500 samples is arbitrary, and in general we might increase it to increase the success probability of finding a good hypothesis. In fact, when proving PAC-learning theorems the number of samples drawn often depends on the accuracy and confidence parameters \varepsilon, \delta. We omit them here for simplicity.

Strong learners from weak learners

So suppose we have a weak learner A for a concept class H, and for any concept c from H it can produce with probability at least 1 - \delta a hypothesis h with error bound 1/2 - \eta. How can we modify this algorithm to get a strong learner? Here is an idea: we can maintain a large number of separate instances of the weak learner A, run them on our dataset, and then combine their hypotheses with a majority vote. In code this might look like the following python snippet. For now examples are binary vectors and the labels are \pm 1, so the sign of a real number will be its label.

def boost(learner, data, rounds=100):
   m = len(data)
   learners = [learner(random.choice(data, m/rounds)) for _ in range(rounds)]
   def hypothesis(example):
      return sign(sum(1/rounds * h(example) for h in learners))

   return hypothesis

This is a bit too simplistic: what if the majority of the weak learners are wrong? In fact, with an overly naive mindset one might imagine a scenario in which the different instances of A have high disagreement, so is the prediction going to depend on which random subset the learner happens to get? We can do better: instead of taking a majority vote we can take a weighted majority vote. That is, give the weak learner a random subset of your data, and then test its hypothesis on the data to get a good estimate of its error. Then you can use this error to say whether the hypothesis is any good, and give good hypotheses high weight and bad hypotheses low weight (proportionally to the error). Then the “boosted” hypothesis would take a weighted majority vote of all your hypotheses on an example. This might look like the following.

# data is a list of (example, label) pairs
def error(hypothesis, data):
   return sum(1 for x,y in data if hypothesis(x) != y) / len(data)

def boost(learner, data, rounds=100):
   m = len(data)
   weights = [0] * rounds
   learners = [None] * rounds

   for t in range(rounds):
      learners[t] = learner(random.choice(data, m/rounds))
      weights[t] = 1 - error(learners[t], data)
   def hypothesis(example):
      return sign(sum(weight * h(example) for (h, weight) in zip(learners, weights)))

   return hypothesis

This might be better, but we can do something even cleverer. Rather than use the estimated error just to say something about the hypothesis, we can identify the mislabeled examples in a round and somehow encourage A to do better at classifying those examples in later rounds. This turns out to be the key insight, and it’s why the algorithm is called AdaBoost (Ada stands for “adaptive”). We’re adaptively modifying the distribution over the training data we feed to A based on which data A learns “easily” and which it does not. So as the boosting algorithm runs, the distribution given to A has more and more probability weight on the examples that A misclassified. And, this is the key, A has the guarantee that it will weak learn no matter what the distribution over the data is. Of course, it’s error is also measured relative to the adaptively chosen distribution, and the crux of the argument will be relating this error to the error on the original distribution we’re trying to strong learn.

To implement this idea in mathematics, we will start with a fixed sample X = \{x_1, \dots, x_m\} drawn from D and assign a weight 0 \leq \mu_i \leq 1 to each x_i. Call c(x) the true label of an example. Initially, set \mu_i to be 1. Since our dataset can have repetitions, normalizing the \mu_i to a probability distribution gives an estimate of D. Now we’ll pick some “update” parameter \zeta > 1 (this is intentionally vague). Then we’ll repeat the following procedure for some number of rounds t = 1, \dots, T.

  1. Renormalize the \mu_i to a probability distribution.
  2. Train the weak learner A, and provide it with a simulated distribution D' that draws examples x_i according to their weights \mu_i. The weak learner outputs a hypothesis h_t.
  3. For every example x_i mislabeled by h_t, update \mu_i by replacing it with \mu_i \zeta.
  4. For every correctly labeled example replace \mu_i with \mu_i / \zeta.

At the end our final hypothesis will be a weighted majority vote of all the h_t, where the weights depend on the amount of error in each round. Note that when the weak learner misclassifies an example we increase the weight of that example, which means we’re increasing the likelihood it will be drawn in future rounds. In particular, in order to maintain good accuracy the weak learner will eventually have to produce a hypothesis that fixes its mistakes in previous rounds. Likewise, when examples are correctly classified, we reduce their weights. So examples that are “easy” to learn are given lower emphasis. And that’s it. That’s the prize-winning idea. It’s elegant, powerful, and easy to understand. The rest is working out the values of all the parameters and proving it does what it’s supposed to.

The details and a proof

Let’s jump straight into a Python program that performs boosting.

First we pick a data representation. Examples are pairs (x,c(x)) whose type is the tuple (object, int). Our labels will be \pm 1 valued. Since our algorithm is entirely black-box, we don’t need to assume anything about how the examples X are represented. Our dataset is just a list of labeled examples, and the weights are floats. So our boosting function prototype looks like this

# boost: [(object, int)], learner, int -> (object -> int)
# boost the given weak learner into a strong learner
def boost(examples, weakLearner, rounds):

And a weak learner, as we saw for decision stumps, has the following function prototype.

# weakLearner: (() -> (list, label)) -> (list -> label)
# accept as input a function that draws labeled examples from a distribution,
# and output a hypothesis list -> label
def weakLearner(draw):
   return hypothesis

Assuming we have a weak learner, we can fill in the rest of the boosting algorithm with some mysterious details. First, a helper function to compute the weighted error of a hypothesis on some exmaples. It also returns the correctness of the hypothesis on each example which we’ll use later.

# compute the weighted error of a given hypothesis on a distribution
# return all of the hypothesis results and the error
def weightedLabelError(h, examples, weights):
   hypothesisResults = [h(x)*y for (x,y) in examples] # +1 if correct, else -1
   return hypothesisResults, sum(w for (z,w) in zip(hypothesisResults, weights) if z < 0)

Next we have the main boosting algorithm. Here draw is a function that accepts as input a list of floats that sum to 1 and picks an index proportional to the weight of the entry at that index.

def boost(examples, weakLearner, rounds):
   distr = normalize([1.] * len(examples))
   hypotheses = [None] * rounds
   alpha = [0] * rounds

   for t in range(rounds):
      def drawExample():
         return examples[draw(distr)]

      hypotheses[t] = weakLearner(drawExample)
      hypothesisResults, error = computeError(hypotheses[t], examples, distr)

      alpha[t] = 0.5 * math.log((1 - error) / (.0001 + error))
      distr = normalize([d * math.exp(-alpha[t] * h)
                         for (d,h) in zip(distr, hypothesisResults)])
      print("Round %d, error %.3f" % (t, error))

   def finalHypothesis(x):
      return sign(sum(a * h(x) for (a, h) in zip(alpha, hypotheses)))

   return finalHypothesis

The code is almost clear. For each round we run the weak learner on our hand-crafted distribution. We compute the error of the resulting hypothesis on that distribution, and then we update the distribution in this mysterious way depending on some alphas and logs and exponentials. In particular, we use the expression c(x) h(x), the product of the true label and predicted label, as computed in weightedLabelError. As the comment says, this will either be +1 or -1 depending on whether the predicted label is correct or incorrect, respectively. The choice of those strange logarithms and exponentials are the result of some optimization: they allow us to minimize training error as quickly as possible (we’ll see this in the proof to follow). The rest of this section will prove that this works when the weak learner is correct. One small caveat: in the proof we will assume the error of the hypothesis is not zero (because a weak learner is not supposed to return a perfect hypothesis!), but in practice we want to avoid dividing by zero so we add the small 0.0001 to avoid that. As a quick self-check: why wouldn’t we just stop in the middle and output that “perfect” hypothesis? (What distribution is it “perfect” over? It might not be the original distribution!)

If we wanted to define the algorithm in pseudocode (which helps for the proof) we would write it this way. Given T rounds, start with D_1 being the uniform distribution over labeled input examples X, where x has label c(x). Say there are m input examples.

  1. For each t=1, \dots T:
    1. Let h_t be the weak learning algorithm run on D_t.
    2. Let \varepsilon_t be the error of h_t on D_t.
    3. Let \alpha_t = \frac{1}{2} \log ((1- \varepsilon) / \varepsilon).
    4. Update each entry of D_{t+1} by the rule D_{t+1}(x) = \frac{D_t(x)}{Z_t} e^{- h_t(x) c(x) \alpha_t}, where Z_t is chosen to normalize D_{t+1} to a distribution.
  2. Output as the final hypothesis the sign of h(x) = \sum_{t=1}^T \alpha_t h_t(x), i.e. h'(x) = \textup{sign}(h(x)).

Now let’s prove this works. That is, we’ll prove the error on the input dataset (the training set) decreases exponentially quickly in the number of rounds. Then we’ll run it on an example and save generalization error for the next post. Over many years this algorithm and tweaked so that the proof is very straightforward.

Theorem: If AdaBoost is given a weak learner and stopped on round t, and the edge \eta_t over random choice satisfies \varepsilon_t = 1/2 - \eta_t, then the training error of the AdaBoost is at most e^{-2 \sum_t \eta_t^2}.

Proof. Let m be the number of examples given to the boosting algorithm. First, we derive a closed-form expression for D_{t} in terms of the normalization constants Z_t. Expanding the recurrence relation gives

\displaystyle D_{t}(x) = D_1(x)\frac{e^{-\alpha_1 c(x) h_1(x)}}{Z_1} \dots \frac{e^{- \alpha_t c(x) h_t(x)}}{Z_t}

Because the starting distribution is uniform, and combining the products into a sum of the exponents, this simplifies to

\displaystyle \frac{1}{m} \frac{e^{-c(x) \sum_{s=1}^t \alpha_s h_t(x)}}{\prod_{s=1}^t Z_s} = \frac{1}{m}\frac{e^{-c(x) h(x)}}{\prod_s Z_s}

Next, we show that the training error is bounded by the product of the normalization terms \prod_{s=1}^t Z_s. This part has always seemed strange to me, that the training error of boosting depends on the factors you need to normalize a distribution. But it’s just a different perspective on the multiplicative weights scheme. If we didn’t explicitly normalize the distribution at each step, we’d get nonnegative weights (which we could convert to a distribution just for the sampling step) and the training error would depend on the product of the weight updates in each step. Anyway let’s prove it.

The training error is defined to be \frac{1}{m} (\textup{\# incorrect predictions by } h). This can be written with an indicator function as follows:

\displaystyle \frac{1}{m} \sum_{x \in X} 1_{c(x) h(x) \leq 0}

Because the sign of h(x) determines its prediction, the product is negative when h is incorrect. Now we can do a strange thing, we’re going to upper bound the indicator function (which is either zero or one) by e^{-c(x)h(x)}. This works because if h predicts correctly then the indicator function is zero while the exponential is greater than zero. On the other hand if h is incorrect the exponential is greater than one because e^z \geq 1 when z \geq 0. So we get

\displaystyle \leq \sum_i \frac{1}{m} e^{-c(x)h(x)}

and rearranging the formula for D_t from the first part gives

\displaystyle \sum_{x \in X} D_T(x) \prod_{t=1}^T Z_t

Since the D_T forms a distribution, it sums to 1 and we can factor the Z_t out. So the training error is just bounded by the \prod_{t=1}^T Z_t.

The last step is to bound the product of the normalization factors. It’s enough to show that Z_t \leq e^{-2 \eta_t^2}. The normalization constant is just defined as the sum of the numerator of the terms in step D. i.e.

\displaystyle Z_t = \sum_i D_t(i) e^{-\alpha_t c(x) h_t(x)}

We can split this up into the correct and incorrect terms (that contribute to +1 or -1 in the exponent) to get

\displaystyle Z_t = e^{-\alpha_t} \sum_{\textup{correct } x} D_t(x) + e^{\alpha_t} \sum_{\textup{incorrect } x} D_t(x)

But by definition the sum of the incorrect part of D is \varepsilon_t and 1-\varepsilon_t for the correct part. So we get

\displaystyle e^{-\alpha_t}(1-\varepsilon_t) + e^{\alpha_t} \varepsilon_t

Finally, since this is an upper bound we want to pick \alpha_t so as to minimize this expression. With a little calculus you can see the \alpha_t we chose in the algorithm pseudocode achieves the minimum, and this simplifies to 2 \sqrt{\varepsilon_t (1-\varepsilon_t)}. Plug in \varepsilon_t = 1/2 - \eta_t to get \sqrt{1 - 4 \eta_t^2} and use the calculus fact that 1 - z \leq e^{-z} to get e^{-2\eta_t^2} as desired.


This is fine and dandy, it says that if you have a true weak learner then the training error of AdaBoost vanishes exponentially fast in the number of boosting rounds. But what about generalization error? What we really care about is whether the hypothesis produced by boosting has low error on the original distribution D as a whole, not just the training sample we started with.

One might expect that if you run boosting for more and more rounds, then it will eventually overfit the training data and its generalization accuracy will degrade. However, in practice this is not the case! The longer you boost, even if you get down to zero training error, the better generalization tends to be. For a long time this was sort of a mystery, and we’ll resolve the mystery in the sequel to this post. For now, we’ll close by showing a run of AdaBoost on some real world data.

The “adult” census dataset

The “adult” dataset is a standard dataset taken from the 1994 US census. It tracks a number of demographic and employment features (including gender, age, employment sector, etc.) and the goal is to predict whether an individual makes over $50k per year. Here are the first few lines from the training set.

39, State-gov, 77516, Bachelors, 13, Never-married, Adm-clerical, Not-in-family, White, Male, 2174, 0, 40, United-States, <=50K
50, Self-emp-not-inc, 83311, Bachelors, 13, Married-civ-spouse, Exec-managerial, Husband, White, Male, 0, 0, 13, United-States, <=50K
38, Private, 215646, HS-grad, 9, Divorced, Handlers-cleaners, Not-in-family, White, Male, 0, 0, 40, United-States, <=50K
53, Private, 234721, 11th, 7, Married-civ-spouse, Handlers-cleaners, Husband, Black, Male, 0, 0, 40, United-States, <=50K
28, Private, 338409, Bachelors, 13, Married-civ-spouse, Prof-specialty, Wife, Black, Female, 0, 0, 40, Cuba, <=50K
37, Private, 284582, Masters, 14, Married-civ-spouse, Exec-managerial, Wife, White, Female, 0, 0, 40, United-States, <=50K

We perform some preprocessing of the data, so that the categorical examples turn into binary features. You can see the full details in the github repository for this post; here are the first few post-processed lines (my newlines added).

>>> from data import adult
>>> train, test = adult.load()
>>> train[:3]
[((39, 1, 0, 0, 0, 0, 0, 1, 0, 0, 13, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2174, 0, 40, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), -1), 

((50, 1, 0, 1, 0, 0, 0, 0, 0, 0, 13, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 13, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), -1), 

((38, 1, 1, 0, 0, 0, 0, 0, 0, 0, 9, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 40, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), -1)]

Now we can run boosting on the training data, and compute its error on the test data.

>>> from boosting import boost
>>> from data import adult
>>> from decisionstump import buildDecisionStump
>>> train, test = adult.load()
>>> weakLearner = buildDecisionStump
>>> rounds = 20
>>> h = boost(train, weakLearner, rounds)
Round 0, error 0.199
Round 1, error 0.231
Round 2, error 0.308
Round 3, error 0.380
Round 4, error 0.392
Round 5, error 0.451
Round 6, error 0.436
Round 7, error 0.459
Round 8, error 0.452
Round 9, error 0.432
Round 10, error 0.444
Round 11, error 0.447
Round 12, error 0.450
Round 13, error 0.454
Round 14, error 0.505
Round 15, error 0.476
Round 16, error 0.484
Round 17, error 0.500
Round 18, error 0.493
Round 19, error 0.473
>>> error(h, train)
>>> error(h, test)

This isn’t too shabby. I’ve tried running boosting for more rounds (a hundred) and the error doesn’t seem to improve by much. This implies that finding the best decision stump is not a weak learner (or at least it fails for this dataset), and we can see that indeed the training errors across rounds roughly tend to 1/2.

Though we have not compared our results above to any baseline, AdaBoost seems to work pretty well. This is kind of a meta point about theoretical computer science research. One spends years trying to devise algorithms that work in theory (and finding conditions under which we can get good algorithms in theory), but when it comes to practice we can’t do anything but hope the algorithms will work well. It’s kind of amazing that something like Boosting works in practice. It’s not clear to me that weak learners should exist at all, even for a given real world problem. But the results speak for themselves.

Next time

Next time we’ll get a bit deeper into the theory of boosting. We’ll derive the notion of a “margin” that quantifies the confidence of boosting in its prediction. Then we’ll describe (and maybe prove) a theorem that says if the “minimum margin” of AdaBoost on the training data is large, then the generalization error of AdaBoost on the entire distribution is small. The notion of a margin is actually quite a deep one, and it shows up in another famous machine learning technique called the Support Vector Machine. In fact, it’s part of some recent research I’ve been working on as well. More on that in the future.

If you’re dying to learn more about Boosting, but don’t want to wait for me, check out the book Boosting: Foundations and Algorithms, by Freund and Schapire.

Until next time!

Markov Chain Monte Carlo Without all the Bullshit

I have a little secret: I don’t like the terminology, notation, and style of writing in statistics. I find it unnecessarily complicated. This shows up when trying to read about Markov Chain Monte Carlo methods. Take, for example, the abstract to the Markov Chain Monte Carlo article in the Encyclopedia of Biostatistics.

Markov chain Monte Carlo (MCMC) is a technique for estimating by simulation the expectation of a statistic in a complex model. Successive random selections form a Markov chain, the stationary distribution of which is the target distribution. It is particularly useful for the evaluation of posterior distributions in complex Bayesian models. In the Metropolis–Hastings algorithm, items are selected from an arbitrary “proposal” distribution and are retained or not according to an acceptance rule. The Gibbs sampler is a special case in which the proposal distributions are conditional distributions of single components of a vector parameter. Various special cases and applications are considered.

I can only vaguely understand what the author is saying here (and really only because I know ahead of time what MCMC is). There are certainly references to more advanced things than what I’m going to cover in this post. But it seems very difficult to find an explanation of Markov Chain Monte Carlo without all any superfluous jargon. The “bullshit” here is the implicit claim of an author that such jargon is needed. Maybe it is to explain advanced applications (like attempts to do “inference in Bayesian networks”), but it is certainly not needed to define or analyze the basic ideas.

So to counter, here’s my own explanation of Markov Chain Monte Carlo, inspired by the treatment of John Hopcroft and Ravi Kannan.

The Problem is Drawing from a Distribution

Markov Chain Monte Carlo is a technique to solve the problem of sampling from a complicated distribution. Let me explain by the following imaginary scenario. Say I have a magic box which can estimate probabilities of baby names very well. I can give it a string like “Malcolm” and it will tell me the exact probability p_{\textup{Malcolm}} that you will choose this name for your next child. So there’s a distribution D over all names, it’s very specific to your preferences, and for the sake of argument say this distribution is fixed and you don’t get to tamper with it.

Now comes the problem: I want to efficiently draw a name from this distribution D. This is the problem that Markov Chain Monte Carlo aims to solve. Why is it a problem? Because I have no idea what process you use to pick a name, so I can’t simulate that process myself. Here’s another method you could try: generate a name x uniformly at random, ask the machine for p_x, and then flip a biased coin with probability p_x and use x if the coin lands heads. The problem with this is that there are exponentially many names! The variable here is the number of bits needed to write down a name n = |x|. So either the probabilities p_x will be exponentially small and I’ll be flipping for a very long time to get a single name, or else there will only be a few names with nonzero probability and it will take me exponentially many draws to find them. Inefficiency is the death of me.

So this is a serious problem! Let’s restate it formally just to be clear.

Definition (The sampling problem):  Let D be a distribution over a finite set X. You are given black-box access to the probability distribution function p(x) which outputs the probability of drawing x \in X according to D. Design an efficient randomized algorithm A which outputs an element of X so that the probability of outputting x is approximately p(x). More generally, output a sample of elements from X drawn according to p(x).

Assume that A has access to only fair random coins, though this allows one to efficiently simulate flipping a biased coin of any desired probability.

Notice that with such an algorithm we’d be able to do things like estimate the expected value of some random variable f : X \to \mathbb{R}. We could take a large sample S \subset X via the solution to the sampling problem, and then compute the average value of f on that sample. This is what a Monte Carlo method does when sampling is easy. In fact, the Markov Chain solution to the sampling problem will allow us to do the sampling and the estimation of \mathbb{E}(f) in one fell swoop if you want.

But the core problem is really a sampling problem, and “Markov Chain Monte Carlo” would be more accurately called the “Markov Chain Sampling Method.” So let’s see why a Markov Chain could possibly help us.

Random Walks, the “Markov Chain” part of MCMC

Markov Chain is essentially a fancy term for a random walk on a graph.

You give me a directed graph G = (V,E), and for each edge e = (u,v) \in E you give me a number p_{u,v} \in [0,1]. In order to make a random walk make sense, the p_{u,v} need to satisfy the following constraint:

For any vertex x \in V, the set all values p_{x,y} on outgoing edges (x,y) must sum to 1, i.e. form a probability distribution.

If this is satisfied then we can take a random walk on G according to the probabilities as follows: start at some vertex x_0. Then pick an outgoing edge at random according to the probabilities on the outgoing edges, and follow it to x_1. Repeat if possible.

I say “if possible” because an arbitrary graph will not necessarily have any outgoing edges from a given vertex. We’ll need to impose some additional conditions on the graph in order to apply random walks to Markov Chain Monte Carlo, but in any case the idea of randomly walking is well-defined, and we call the whole object (V,E, \{ p_e \}_{e \in E})Markov chain.

Here is an example where the vertices in the graph correspond to emotional states.

An example Markov chain [image source]

An example Markov chain; image source

In statistics land, they take the “state” interpretation of a random walk very seriously. They call the edge probabilities “state-to-state transitions.”

The main theorem we need to do anything useful with Markov chains is the stationary distribution theorem (sometimes called the “Fundamental Theorem of Markov Chains,” and for good reason). What it says intuitively is that for a very long random walk, the probability that you end at some vertex v is independent of where you started! All of these probabilities taken together is called the stationary distribution of the random walk, and it is uniquely determined by the Markov chain.

However, for the reasons we stated above (“if possible”), the stationary distribution theorem is not true of every Markov chain. The main property we need is that the graph G is strongly connected. Recall that a directed graph is called connected if, when you ignore direction, there is a path from every vertex to every other vertex. It is called strongly connected if you still get paths everywhere when considering direction. If we additionally require the stupid edge-case-catcher that no edge can have zero probability, then strong connectivity (of one component of a graph) is equivalent to the following property:

For every vertex v \in V(G), an infinite random walk started at v will return to v with probability 1.

In fact it will return infinitely often. This property is called the persistence of the state v by statisticians. I dislike this term because it appears to describe a property of a vertex, when to me it describes a property of the connected component containing that vertex. In any case, since in Markov Chain Monte Carlo we’ll be picking the graph to walk on (spoiler!) we will ensure the graph is strongly connected by design.

Finally, in order to describe the stationary distribution in a more familiar manner (using linear algebra), we will write the transition probabilities as a matrix A where entry a_{j,i} = p_{(i,j)} if there is an edge (i,j) \in E and zero otherwise. Here the rows and columns correspond to vertices of G, and each column i forms the probability distribution of going from state i to some other state in one step of the random walk. Note A is the transpose of the weighted adjacency matrix of the directed weighted graph G where the weights are the transition probabilities (the reason I do it this way is because matrix-vector multiplication will have the matrix on the left instead of the right; see below).

This matrix allows me to describe things nicely using the language of linear algebra. In particular if you give me a basis vector e_i interpreted as “the random walk currently at vertex i,” then Ae_i gives a vector whose j-th coordinate is the probability that the random walk would be at vertex j after one more step in the random walk. Likewise, if you give me a probability distribution q over the vertices, then Aq gives a probability vector interpreted as follows:

If a random walk is in state i with probability q_i, then the j-th entry of Aq is the probability that after one more step in the random walk you get to vertex j.

Interpreted this way, the stationary distribution is a probability distribution \pi such that A \pi = \pi, in other words \pi is an eigenvector of A with eigenvalue 1.

A quick side note for avid readers of this blog: this analysis of a random walk is exactly what we did back in the early days of this blog when we studied the PageRank algorithm for ranking webpages. There we called the matrix A “a web matrix,” noted it was column stochastic (as it is here), and appealed to a special case of the Perron-Frobenius theorem to show that there is a unique maximal eigenvalue equal to one (with a dimension one eigenspace) whose eigenvector we used as a sort of “stationary distribution” and the final ranking of web pages. There we described an algorithm to actually find that eigenvector by iterated multiplication by A. The following theorem is essentially a variant of this algorithm but works under weaker conditions; for the web matrix we added additional “fake” edges that give the needed stronger conditions.

Theorem: Let G be a strongly connected graph with associated edge probabilities \{ p_e \}_e \in E forming a Markov chain. For a probability vector x_0, define x_{t+1} = Ax_t for all t \geq 1, and let v_t be the long-term average v_t = \frac1t \sum_{s=1}^t x_s. Then:

  1. There is a unique probability vector \pi with A \pi = \pi.
  2. For all x_0, the limit \lim_{t \to \infty} v_t = \pi.

Proof. Since v_t is a probability vector we just want to show that |Av_t - v_t| \to 0 as t \to \infty. Indeed, we can expand this quantity as

\displaystyle \begin{aligned} Av_t - v_t &=\frac1t (Ax_0 + Ax_1 + \dots + Ax_{t-1}) - \frac1t (x_0 + \dots + x_{t-1}) \\ &= \frac1t (x_t - x_0) \end{aligned}

But x_t, x_0 are unit vectors, so their difference is at most 2, meaning |Av_t - v_t| \leq \frac2t \to 0. Now it’s clear that this does not depend on v_0. For uniqueness we will cop out and appeal to the Perron-Frobenius theorem that says any matrix of this form has a unique such (normalized) eigenvector.


One additional remark is that, in addition to computing the stationary distribution by actually computing this average or using an eigensolver, one can analytically solve for it as the inverse of a particular matrix. Define B = A-I_n, where I_n is the n \times n identity matrix. Let C be B with a row of ones appended to the bottom and the topmost row removed. Then one can show (quite opaquely) that the last column of C^{-1} is \pi. We leave this as an exercise to the reader, because I’m pretty sure nobody uses this method in practice.

One final remark is about why we need to take an average over all our x_t in the theorem above. There is an extra technical condition one can add to strong connectivity, called aperiodicity, which allows one to beef up the theorem so that x_t itself converges to the stationary distribution. Rigorously, aperiodicity is the property that, regardless of where you start your random walk, after some sufficiently large number of steps n the random walk has a positive probability of being at every vertex at every subsequent step. As an example of a graph where aperiodicity fails: an undirected cycle on an even number of vertices. In that case there will only be a positive probability of being at certain vertices every other step, and averaging those two long term sequences gives the actual stationary distribution.

Screen Shot 2015-04-07 at 6.55.39 PM

Image source: Wikipedia

One way to guarantee that your Markov chain is aperiodic is to ensure there is a positive probability of staying at any vertex. I.e., that your graph has a self-loop. This is what we’ll do in the next section.

Constructing a graph to walk on

Recall that the problem we’re trying to solve is to draw from a distribution over a finite set X with probability function p(x). The MCMC method is to construct a Markov chain whose stationary distribution is exactly p, even when you just have black-box access to evaluating p. That is, you (implicitly) pick a graph G and (implicitly) choose transition probabilities for the edges to make the stationary distribution p. Then you take a long enough random walk on G and output the x corresponding to whatever state you land on.

The easy part is coming up with a graph that has the right stationary distribution (in fact, “most” graphs will work). The hard part is to come up with a graph where you can prove that the convergence of a random walk to the stationary distribution is fast in comparison to the size of X. Such a proof is beyond the scope of this post, but the “right” choice of a graph is not hard to understand.

The one we’ll pick for this post is called the Metropolis-Hastings algorithm. The input is your black-box access to p(x), and the output is a set of rules that implicitly define a random walk on a graph whose vertex set is X.

It works as follows: you pick some way to put X on a lattice, so that each state corresponds to some vector in \{ 0,1, \dots, n\}^d. Then you add (two-way directed) edges to all neighboring lattice points. For n=5, d=2 it would look like this:

And for d=3, n \in \{2,3\} it would look like this:

You have to be careful here to ensure the vertices you choose for X are not disconnected, but in many applications X is naturally already a lattice.

Now we have to describe the transition probabilities. Let r be the maximum degree of a vertex in this lattice (r=2d). Suppose we’re at vertex i and we want to know where to go next. We do the following:

  1. Pick neighbor j with probability 1/r (there is some chance to stay at i).
  2. If you picked neighbor j and p(j) \geq p(i) then deterministically go to j.
  3. Otherwise, p(j) < p(i), and you go to j with probability p(j) / p(i).

We can state the probability weight p_{i,j} on edge (i,j) more compactly as

\displaystyle p_{i,j} = \frac1r \min(1, p(j) / p(i)) \\ p_{i,i} = 1 - \sum_{(i,j) \in E(G); j \neq i} p_{i,j}

It is easy to check that this is indeed a probability distribution for each vertex i. So we just have to show that p(x) is the stationary distribution for this random walk.

Here’s a fact to do that: if a probability distribution v with entries v(x) for each x \in X has the property that v(x)p_{x,y} = v(y)p_{y,x} for all x,y \in X, the v is the stationary distribution. To prove it, fix x and take the sum of both sides of that equation over all y. The result is exactly the equation v(x) = \sum_{y} v(y)p_{y,x}, which is the same as v = Av. Since the stationary distribution is the unique vector satisfying this equation, v has to be it.

Doing this with out chosen p(i) is easy, since p(i)p_{i,j} and p(i)p_{j,i} are both equal to \frac1r \min(p(i), p(j)) by applying a tiny bit of algebra to the definition. So we’re done! One can just randomly walk according to these probabilities and get a sample.

Last words

The last thing I want to say about MCMC is to show that you can estimate the expected value of a function \mathbb{E}(f) simultaneously while random-walking through your Metropolis-Hastings graph (or any graph whose stationary distribution is p(x)). By definition the expected value of f is \sum_x f(x) p(x).

Now what we can do is compute the average value of f(x) just among those states we’ve visited during our random walk. With a little bit of extra work you can show that this quantity will converge to the true expected value of f at about the same time that the random walk converges to the stationary distribution. (Here the “about” means we’re off by a constant factor depending on f). In order to prove this you need some extra tools I’m too lazy to write about in this post, but the point is that it works.

The reason I did not start by describing MCMC in terms of estimating the expected value of a function is because the core problem is a sampling problem. Moreover, there are many applications of MCMC that need nothing more than a sample. For example, MCMC can be used to estimate the volume of an arbitrary (maybe high dimensional) convex set. See these lecture notes of Alistair Sinclair for more.

If demand is popular enough, I could implement the Metropolis-Hastings algorithm in code (it wouldn’t be industry-strength, but perhaps illuminating? I’m not so sure…).

Until next time!

Zero-One Laws for Random Graphs

Last time we saw a number of properties of graphs, such as connectivity, where the probability that an Erdős–Rényi random graph G(n,p) satisfies the property is asymptotically either zero or one. And this zero or one depends on whether the parameter p is above or below a universal threshold (that depends only on n and the property in question).

To remind the reader, the Erdős–Rényi random “graph” G(n,p) is a distribution over graphs that you draw from by including each edge independently with probability p. Last time we saw that the existence of an isolated vertex has a sharp threshold at (\log n) / n, meaning if p is asymptotically smaller than the threshold there will certainly be isolated vertices, and if p is larger there will certainly be no isolated vertices. We also gave a laundry list of other properties with such thresholds.

One might want to study this phenomenon in general. Even if we might not be able to find all the thresholds we want for a given property, can we classify which properties have thresholds and which do not?

The answer turns out to be mostly yes! For large classes of properties, there are proofs that say things like, “either this property holds with probability tending to one, or it holds with probability tending to zero.” These are called “zero-one laws,” and they’re sort of meta theorems. We’ll see one such theorem in this post relating to constant edge-probabilities in random graphs, and we’ll remark on another at the end.

Sentences about graphs in first order logic

A zero-one law generally works by defining a class of properties, and then applying a generic first/second moment-type argument to every property in the class.

So first we define what kinds of properties we’ll discuss. We’ll pick a large class: anything that can be expressed in first-order logic in the language of graphs. That is, any finite logical statement that uses existential and universal quantifiers over variables, and whose only relation (test) is whether an edge exists between two vertices. We’ll call this test e(x,y). So you write some sentence P in this language, and you take a graph G, and you can ask P(G) = 1, whether the graph satisfies the sentence.

This seems like a really large class of properties, and it is, but let’s think carefully about what kinds of properties can be expressed this way. Clearly the existence of a triangle can be written this way, it’s just the sentence

\exists x,y,z : e(x,y) \wedge e(y,z) \wedge e(x,z)

I’m using \wedge for AND, and \vee for OR, and \neg for NOT. Similarly, one can express the existence of a clique of size k, or the existence of an independent set of size k, or a path of a fixed length, or whether there is a vertex of maximal degree n-1.

Here’s a question: can we write a formula which will be true for a graph if and only if it’s connected? Well such a formula seems like it would have to know about how many vertices there are in the graph, so it could say something like “for all x,y there is a path from x to y.” It seems like you’d need a family of such formulas that grows with n to make anything work. But this isn’t a proof; the question remains whether there is some other tricky way to encode connectivity.

But as it turns out, connectivity is not a formula you can express in propositional logic. We won’t prove it here, but we will note at the end of the article that connectivity is in a different class of properties that you can prove has a similar zero-one law.

The zero-one law for first order logic

So the theorem about first-order expressible sentences is as follows.

Theorem: Let P be a property of graphs that can be expressed in the first order language of graphs (with the e(x,y) relation). Then for any constant p, the probability that P holds in G(n,p) has a limit of zero or one as n \to \infty.

Proof. We’ll prove the simpler case of p=1/2, but the general case is analogous. Given such a graph G drawn from G(n,p), what we’ll do is define a countably infinite family of propositional formulas \varphi_{k,l}, and argue that they form a sort of “basis” for all first-order sentences about graphs.

First let’s describe the \varphi_{k,l}. For any k,l \in \mathbb{N}, the sentence will assert that for every set of k vertices and every set of l vertices, there is some other vertex connected to the first k but not the last l.

\displaystyle \varphi_{k,l} : \forall x_1, \dots, x_k, y_1, \dots, y_l \exists z : \\ e(z,x_1) \wedge \dots \wedge e(z,x_k) \wedge \neg e(z,y_1) \wedge \dots \wedge \neg e(z,y_l).

In other words, these formulas encapsulate every possible incidence pattern for a single vertex. It is a strange set of formulas, but they have a very nice property we’re about to get to. So for a fixed \varphi_{k,l}, what is the probability that it’s false on n vertices? We want to give an upper bound and hence show that the formula is true with probability approaching 1. That is, we want to show that all the \varphi_{k,l} are true with probability tending to 1.

Computing the probability: we have \binom{n}{k} \binom{n-k}{l} possibilities to choose these sets, and the probability that some other fixed vertex z has the good connections is 2^{-(k+l)} so the probability z is not good is 1 - 2^{-(k+l)}, and taking a product over all choices of z gives the probability that there is some bad vertex z with an exponent of (n - (k + l)). Combining all this together gives an upper bound of \varphi_{k,l} being false of:

\displaystyle \binom{n}{k}\binom{n-k}{l} (1-2^{-k-1})^{n-k-l}

And k, l are constant, so the left two terms are polynomials while the rightmost term is an exponentially small function, and this implies that the whole expression tends to zero, as desired.

Break from proof.

A bit of model theory

So what we’ve proved so far is that the probability of every formula of the form \varphi_{k,l} being satisfied in G(n,1/2) tends to 1.

Now look at the set of all such formulas

\displaystyle \Phi = \{ \varphi_{k,l} : k,l \in \mathbb{N} \}

We ask: is there any graph which satisfies all of these formulas? Certainly it cannot be finite, because a finite graph would not be able to satisfy formulas with sufficiently large values of l, k > n. But indeed, there is a countably infinite graph that works. It’s called the Rado graph, pictured below.


The Rado graph has some really interesting properties, such as that it contains every finite and countably infinite graph as induced subgraphs. Basically this means, as far as countably infinite graphs go, it’s the big momma of all graphs. It’s the graph in a very concrete sense of the word. It satisfies all of the formulas in \Phi, and in fact it’s uniquely determined by this, meaning that if any other countably infinite graph satisfies all the formulas in \Phi, then that graph is isomorphic to the Rado graph.

But for our purposes (proving a zero-one law), there’s a better perspective than graph theory on this object. In the logic perspective, the set \Phi is called a theory, meaning a set of statements that you consider “axioms” in some logical system. And we’re asking whether there any model realizing the theory. That is, is there some logical system with a semantic interpretation (some mathematical object based on numbers, or sets, or whatever) that satisfies all the axioms?

A good analogy comes from the rational numbers, because they satisfy a similar property among all ordered sets. In fact, the rational numbers are the unique countable, ordered set with the property that it has no biggest/smallest element and is dense. That is, in the ordering there is always another element between any two elements you want. So the theorem says if you have two countable sets with these properties, then they are actually isomorphic as ordered sets, and they are isomorphic to the rational numbers.

So, while we won’t prove that the Rado graph is a model for our theory \Phi, we will use that fact to great benefit. One consequence of having a theory with a model is that the theory is consistent, meaning it can’t imply any contradictions. Another fact is that this theory \Phi is complete. Completeness means that any formula or it’s negation is logically implied by the theory. Note these are syntactical implications (using standard rules of propositional logic), and have nothing to do with the model interpreting the theory.

The proof that \Phi is complete actually follows from the uniqueness of the Rado graph as the only countable model of \Phi. Suppose the contrary, that \Phi is not consistent, then there has to be some formula \psi that is not provable, and it’s negation is also not provable, by starting from \Phi. Now extend \Phi in two ways: by adding \psi and by adding \neg \psi. Both of the new theories are still countable, and by a theorem from logic this means they both still have countable models. But both of these new models are also countable models of \Phi, so they have to both be the Rado graph. But this is very embarrassing for them, because we assumed they disagree on the truth of \psi.

So now we can go ahead and prove the zero-one law theorem.

Return to proof.

Given an arbitrary property \varphi \not \in \Psi. Now either \varphi or it’s negation can be derived from \Phi. Without loss of generality suppose it’s \varphi. Take all the formulas from the theory you need to derive \varphi, and note that since it is a proof in propositional logic you will only finitely many such \varphi_{k,l}. Now look at the probabilities of the \varphi_{k,l}: they are all true with probability tending to 1, so the implied statement of the proof of \varphi (i.e., \varphi itself) must also hold with probability tending to 1. And we’re done!


If you don’t like model theory, there is another “purely combinatorial” proof of the zero-one law using something called Ehrenfeucht–Fraïssé games. It is a bit longer, though.

Other zero-one laws

One might naturally ask two questions: what if your probability is not constant, and what other kinds of properties have zero-one laws? Both great questions.

For the first, there are some extra theorems. I’ll just describe one that has always seemed very strange to me. If your probability is of the form p = n^{-\alpha} but \alpha is irrational, then the zero-one law still holds! This is a theorem of Baldwin-Shelah-Spencer, and it really makes you wonder why irrational numbers would be so well behaved while rational numbers are not :)

For the second question, there is another theorem about monotone properties of graphs. Monotone properties come in two flavors, so called “increasing” and “decreasing.” I’ll describe increasing monotone properties and the decreasing counterpart should be obvious. A property is called monotone increasing if adding edges can never destroy the property. That is, with an empty graph you don’t have the property (or maybe you do), and as you start adding edges eventually you suddenly get the property, but then adding more edges can’t cause you to lose the property again. Good examples of this include connectivity, or the existence of a triangle.

So the theorem is that there is an identical zero-one law for monotone properties. Great!

It’s not so often that you get to see these neat applications of logic and model theory to graph theory and (by extension) computer science. But when you do get to apply them they seem very powerful and mysterious. I think it’s a good thing.

Until next time!