Martingales and the Optional Stopping Theorem

This is a guest post by my colleague Adam Lelkes.

The goal of this primer is to introduce an important and beautiful tool from probability theory, a model of fair betting games called martingales. In this post I will assume that the reader is familiar with the basics of probability theory. For those that need to refresh their knowledge, Jeremy’s excellent primers (1, 2) are a good place to start.

The Geometric Distribution and the ABRACADABRA Problem

Before we start playing with martingales, let’s start with an easy exercise. Consider the following experiment: we throw an ordinary die repeatedly until the first time a six appears. How many throws will this take in expectation? The reader might recognize immediately that this exercise can be easily solved using the basic properties of the geometric distribution, which models this experiment exactly. We have independent trials, every trial succeeding with some fixed probability $p$. If $X$ denotes the number of trials needed to get the first success, then clearly $\Pr(X = k) = (1-p)^{k-1} p$ (since first we need $k-1$ failures which occur independently with probability $1-p$, then we need one success which happens with probability $p$). Thus the expected value of $X$ is

$\displaystyle E(X) = \sum_{k=1}^\infty k P(X = k) = \sum_{k=1}^\infty k (1-p)^{k-1} p = \frac1p$

by basic calculus. In particular, if success is defined as getting a six, then $p=1/6$ thus the expected time is $1/p=6$.

Now let us move on to a somewhat similar, but more interesting and difficult problem, the ABRACADABRA problem. Here we need two things for our experiment, a monkey and a typewriter. The monkey is asked to start bashing random keys on a typewriter. For simplicity’s sake, we assume that the typewriter has exactly 26 keys corresponding to the 26 letters of the English alphabet and the monkey hits each key with equal probability. There is a famous theorem in probability, the infinite monkey theorem, that states that given infinite time, our monkey will almost surely type the complete works of William Shakespeare. Unfortunately, according to astronomists the sun will begin to die in a few billion years, and the expected time we need to wait until a monkey types the complete works of William Shakespeare is orders of magnitude longer, so it is not feasible to use monkeys to produce works of literature.

So let’s scale down our goals, and let’s just wait until our monkey types the word ABRACADABRA. What is the expected time we need to wait until this happens? The reader’s first idea might be to use the geometric distribution again. ABRACADABRA is eleven letters long, the probability of getting one letter right is $\frac{1}{26}$, thus the probability of a random eleven-letter word being ABRACADABRA is exactly $\left(\frac{1}{26}\right)^{11}$. So if typing 11 letters is one trial, the expected number of trials is

$\displaystyle \frac1{\left(\frac{1}{26}\right)^{11}}=26^{11}$

which means $11\cdot 26^{11}$ keystrokes, right?

Well, not exactly. The problem is that we broke up our random string into eleven-letter blocks and waited until one block was ABRACADABRA. However, this word can start in the middle of a block. In other words, we considered a string a success only if the starting position of the word ABRACADABRA was divisible by 11. For example, FRZUNWRQXKLABRACADABRA would be recognized as success by this model but the same would not be true for AABRACADABRA. However, it is at least clear from this observation that $11\cdot 26^{11}$ is a strict upper bound for the expected waiting time. To find the exact solution, we need one very clever idea, which is the following:

Let’s Open a Casino!

Do I mean that abandoning our monkey and typewriter and investing our time and money in a casino is a better idea, at least in financial terms? This might indeed be the case, but here we will use a casino to determine the expected wait time for the ABRACADABRA problem. Unfortunately we won’t make any money along the way (in expectation) since our casino will be a fair one.

Let’s do the following thought experiment: let’s open a casino next to our typewriter. Before each keystroke, a new gambler comes to our casino and bets $1 that the next letter will be A. If he loses, he goes home disappointed. If he wins, he bets all the money he won on the event that the next letter will be B. Again, if he loses, he goes home disappointed. (This won’t wreak havoc on his financial situation, though, as he only loses$1 of his own money.) If he wins again, he bets all the money on the event that the next letter will be R, and so on.

If a gambler wins, how much does he win? We said that the casino would be fair, i.e. the expected outcome should be zero. That means that it the gambler bets $1, he should receive$26 if he wins, since the probability of getting the next letter right is exactly $\frac{1}{26}$ (thus the expected value of the change in the gambler’s fortune is $\frac{25}{26}\cdot (-1) + \frac{1}{26}\cdot (+25) = 0$.

Let’s keep playing this game until the word ABRACADABRA first appears and let’s denote the number of keystrokes up to this time as $T$. As soon as we see this word, we close our casino. How much was the revenue of our casino then? Remember that before each keystroke, a new gambler comes in and bets $1, and if he wins, he will only bet the money he has received so far, so our revenue will be exactly $T$ dollars. How much will we have to pay for the winners? Note that the only winners in the last round are the players who bet on A. How many of them are there? There is one that just came in before the last keystroke and this was his first bet. He wins$26. There was one who came three keystrokes earlier and he made four successful bets (ABRA). He wins $\26^4$. Finally there is the luckiest gambler who went through the whole ABRACADABRA sequence, his prize will be $\26^{11}$. Thus our casino will have to give out $26^{11}+26^4+26$ dollars in total, which is just under the price of 200,000 WhatsApp acquisitions.

Now we will make one crucial observation: even at the time when we close the casino, the casino is fair! Thus in expectation our expenses will be equal to our income. Our income is $T$ dollars, the expected value of our expenses is $26^{11}+26^4+26$ dollars, thus $E(T)=26^{11}+26^4+26$. A beautiful solution, isn’t it? So if our monkey types at 150 characters per minute on average, we will have to wait around 47 million years until we see ABRACADABRA. Oh well.

Time to be More Formal

After giving an intuitive outline of the solution, it is time to formalize the concepts that we used, to translate our fairy tales into mathematics. The mathematical model of the fair casino is called a martingale, named after a class of betting strategies that enjoyed popularity in 18th century France. The gambler’s fortune (or the casino’s, depending on our viewpoint) can be modeled with a sequence of random variables. $X_0$ will denote the gambler’s fortune before the game starts, $X_1$ the fortune after one round and so on. Such a sequence of random variables is called a stochastic process. We will require the expected value of the gambler’s fortune to be always finite.

How can we formalize the fairness of the game? Fairness means that the gambler’s fortune does not change in expectation, i.e. the expected value of $X_n$, given $X_1, X_2, \ldots, X_{n-1}$ is the same as $X_{n-1}$. This can be written as $E(X_n | X_1, X_2, \ldots, X_{n-1}) = X_{n-1}$ or, equivalently, $E(X_n - X_{n-1} | X_1, X_2, \ldots, X_{n-1}) = 0$.

The reader might be less comfortable with the first formulation. What does it mean, after all, that the conditional expected value of a random variable is another random variable? Shouldn’t the expected value be a number? The answer is that in order to have solid theoretical foundations for the definition of a martingale, we need a more sophisticated notion of conditional expectations. Such sophistication involves measure theory, which is outside the scope of this post. We will instead naively accept the definition above, and the reader can look up all the formal details in any serious probability text (such as [1]).

Clearly the fair casino we constructed for the ABRACADABRA exercise is an example of a martingale. Another example is the simple symmetric random walk on the number line: we start at 0, toss a coin in each step, and move one step in the positive or negative direction based on the outcome of our coin toss.

The Optional Stopping Theorem

Remember that we closed our casino as soon as the word ABRACADABRA appeared and we claimed that our casino was also fair at that time. In mathematical language, the closed casino is called a stopped martingale. The stopped martingale is constructed as follows: we wait until our martingale X exhibits a certain behaviour (e.g. the word ABRACADABRA is typed by the monkey), and we define a new martingale X’ as follows: let $X'_n = X_n$ if $n < T$ and $X'_n = X_T$ if $n \ge T$ where $T$ denotes the stopping time, i.e. the time at which the desired event occurs. Notice that $T$ itself is a random variable.

We require our stopping time $T$ to depend only on the past, i.e. that at any time we should be able to decide whether the event that we are waiting for has already happened or not (without looking into the future). This is a very reasonable requirement. If we could look into the future, we could obviously cheat by closing our casino just before some gambler would win a huge prize.

We said that the expected wealth of the casino at the stopping time is the same as the initial wealth. This is guaranteed by Doob’s optional stopping theorem, which states that under certain conditions, the expected value of a martingale at the stopping time is equal to its expected initial value.

Theorem: (Doob’s optional stopping theorem) Let $X_n$ be a martingale stopped at step $T$, and suppose one of the following three conditions hold:

1. The stopping time $T$ is almost surely bounded by some constant;
2. The stopping time $T$ is almost surely finite and every step of the stopped martingale $X_n$ is almost surely bounded by some constant; or
3. The expected stopping time $E(T)$ is finite and the absolute value of the martingale increments $|X_n-X_{n-1}|$ are almost surely bounded by a constant.

Then $E(X_T) = E(X_0).$

We omit the proof because it requires measure theory, but the interested reader can see it in these notes.

For applications, (1) and (2) are the trivial cases. In the ABRACADABRA problem, the third condition holds: the expected stopping time is finite (in fact, we showed using the geometric distribution that it is less than $26^{12}$) and the absolute value of a martingale increment is either 1 or a net payoff which is bounded by $26^{11}+26^4+26$. This shows that our solution is indeed correct.

Gambler’s Ruin

Another famous application of martingales is the gambler’s ruin problem. This problem models the following game: there are two players, the first player has $a$ dollars, the second player has $b$ dollars. In each round they toss a coin and the loser gives one dollar to the winner. The game ends when one of the players runs out of money. There are two obvious questions: (1) what is the probability that the first player wins and (2) how long will the game take in expectation?

Let $X_n$ denote the change in the second player’s fortune, and set $X_0 = 0$. Let $T_k$ denote the first time $s$ when $X_s = k$. Then our first question can be formalized as trying to determine $\Pr(T_{-b} < T_a)$. Let $t = \min \{ T_{-b}, T_a\}$. Clearly $t$ is a stopping time. By the optional stopping theorem we have that

$\displaystyle 0=E(X_0)=E(X_t)=-b\Pr(T_{-b} < T_a)+a(1-\Pr(T_{-b} < T_a))$

thus $\Pr(T_{-b} < T_a)=\frac{a}{a+b}$.

I would like to ask the reader to try to answer the second question. It is a little bit trickier than the first one, though, so here is a hint: $X_n^2-n$ is also a martingale (prove it), and applying the optional stopping theorem to it leads to the answer.

A Randomized Algorithm for 2-SAT

The reader is probably familiar with 3-SAT, the first problem shown to be NP-complete. Recall that 3-SAT is the following problem: given a boolean formula in conjunctive normal form with at most three literals in each clause, decide whether there is a satisfying truth assignment. It is natural to ask if or why 3 is special, i.e. why don’t we work with $k$-SAT for some $k \ne 3$ instead? Clearly the hardness of the problem is monotone increasing in $k$ since $k$-SAT is a special case of $(k+1)$-SAT. On the other hand, SAT (without any bound on the number of literals per clause) is clearly in NP, thus 3-SAT is just as hard as $k$-SAT for any $k>3$. So the only question is: what can we say about 2-SAT?

It turns out that 2-SAT is easier than satisfiability in general: 2-SAT is in P. There are many algorithms for solving 2-SAT. Here is one deterministic algorithm: associate a graph to the 2-SAT instance such that there is one vertex for each variable and each negated variable and the literals $x$ and $y$ are connected by a directed edge if there is a clause $(\bar x \lor y)$. Recall that $\bar x \lor y$ is equivalent to $x \implies y$, so the edges show the implications between the variables. Clearly the 2-SAT instance is not satisfiable if there is a variable x such that there are directed paths $x \to \bar x$ and $\bar x \to x$ (since $x \Leftrightarrow \bar x$ is always false). It can be shown that this is not only a sufficient but also a necessary condition for unsatisfiability, hence the 2-SAT instance is satisfiable if and only if there is are no such path. If there are directed paths from one vertex of a graph to another and vice versa then they are said to belong to the same strongly connected component. There are several graph algorithms for finding strongly connected components of directed graphs, the most well-known algorithms are all based on depth-first search.

Now we give a very simple randomized algorithm for 2-SAT (due to Christos Papadimitriou in a ’91 paper): start with an arbitrary truth assignment and while there are unsatisfied clauses, pick one and flip the truth value of a random literal in it. Stop after $O(n^2)$ rounds where $n$ denotes the number of variables. Clearly if the formula is not satisfiable then nothing can go wrong, we will never find a satisfying truth assignment. If the formula is satisfiable, we want to argue that with high probability we will find a satisfying truth assignment in $O(n^2)$ steps.

The idea of the proof is the following: fix an arbitrary satisfying truth assignment and consider the Hamming distance of our current assignment from it. The Hamming distance of two truth assignments (or in general, of two binary vectors) is the number of coordinates in which they differ. Since we flip one bit in every step, this Hamming distance changes by $\pm 1$ in every round. It also easy to see that in every step the distance is at least as likely to be decreased as to be increased (since we pick an unsatisfied clause, which means at least one of the two literals in the clause differs in value from the satisfying assignment).

Thus this is an unfair “gambler’s ruin” problem where the gambler’s fortune is the Hamming distance from the solution, and it decreases with probability at least $\frac{1}{2}$. Such a stochastic process is called a supermartingale — and this is arguably a better model for real-life casinos. (If we flip the inequality, the stochastic process we get is called a submartingale.) Also, in this case the gambler’s fortune (the Hamming distance) cannot increase beyond $n$. We can also think of this process as a random walk on the set of integers: we start at some number and in each round we make one step to the left or to the right with some probability. If we use random walk terminology, 0 is called an absorbing barrier since we stop the process when we reach 0. The number $n$, on the other hand, is called a reflecting barrier: we cannot reach $n+1$, and whenever we get close we always bounce back.

There is an equivalent version of the optimal stopping theorem for supermartingales and submartingales, where the conditions are the same but the consequence holds with an inequality instead of equality. It follows from the optional stopping theorem that the gambler will be ruined (i.e. a satisfying truth assignment will be found) in $O(n^2)$ steps with high probability.

[1] For a reference on stochastic processes and martingales, see the text of Durrett .

Optimism in the Face of Uncertainty: the UCB1 Algorithm

The software world is always atwitter with predictions on the next big piece of technology. And a lot of chatter focuses on what venture capitalists express interest in. As an investor, how do you pick a good company to invest in? Do you notice quirky names like “Kaggle” and “Meebo,” require deep technical abilities, or value a charismatic sales pitch?

This author personally believes we’re not thinking as big as we should be when it comes to innovation in software engineering and computer science, and that as a society we should value big pushes forward much more than we do. But making safe investments is almost always at odds with innovation. And so every venture capitalist faces the following question. When do you focus investment in those companies that have proven to succeed, and when do you explore new options for growth? A successful venture capitalist must strike a fine balance between this kind of exploration and exploitation. Explore too much and you won’t make enough profit to sustain yourself. Narrow your view too much and you will miss out on opportunities whose return surpasses any of your current prospects.

In life and in business there is no correct answer on what to do, partly because we just don’t have a good understanding of how the world works (or markets, or people, or the weather). In mathematics, however, we can meticulously craft settings that have solid answers. In this post we’ll describe one such scenario, the so-called multi-armed bandit problem, and a simple algorithm called UCB1 which performs close to optimally. Then, in a future post, we’ll analyze the algorithm on some real world data.

As usual, all of the code used in the making of this post are available for download on this blog’s Github page.

Multi-Armed Bandits

The multi-armed bandit scenario is simple to describe, and it boils the exploration-exploitation tradeoff down to its purest form.

Suppose you have a set of $K$ actions labeled by the integers $\left \{ 1, 2, \dots, K \right \}$. We call these actions in the abstract, but in our minds they’re slot machines. We can then play a game where, in each round, we choose an action (a slot machine to play), and we observe the resulting payout. Over many rounds, we might explore the machines by trying some at random. Assuming the machines are not identical, we naturally play machines that seem to pay off well more frequently to try to maximize our total winnings.

This is the most general description of the game we could possibly give, and every bandit learning problem has these two components: actions and rewards. But in order to get to a concrete problem that we can reason about, we need to specify more details. Bandit learning is a large tree of variations and this is the point at which the field ramifies. We presently care about two of the main branches.

How are the rewards produced? There are many ways that the rewards could work. One nice option is to have the rewards for action $i$ be drawn from a fixed distribution $D_i$ (a different reward distribution for each action), and have the draws be independent across rounds and across actions. This is called the stochastic setting and it’s what we’ll use in this post. Just to pique the reader’s interest, here’s the alternative: instead of having the rewards be chosen randomly, have them be adversarial. That is, imagine a casino owner knows your algorithm and your internal beliefs about which machines are best at any given time. He then fixes the payoffs of the slot machines in advance of each round to screw you up! This sounds dismal, because the casino owner could just make all the machines pay nothing every round. But actually we can design good algorithms for this case, but “good” will mean something different than absolute winnings. And so we must ask:

How do we measure success? In both the stochastic and the adversarial setting, we’re going to have a hard time coming up with any theorems about the performance of an algorithm if we care about how much absolute reward is produced. There’s nothing to stop the distributions from having terrible expected payouts, and nothing to stop the casino owner from intentionally giving us no payout. Indeed, the problem lies in our measurement of success. A better measurement, which we can apply to both the stochastic and adversarial settings, is the notion of regret. We’ll give the definition for the stochastic case, and investigate the adversarial case in a future post.

Definition: Given a player algorithm $A$ and a set of actions $\left \{1, 2, \dots, K \right \}$, the cumulative regret of $A$ in rounds $1, \dots, T$ is the difference between the expected reward of the best action (the action with the highest expected payout) and the expected reward of $A$ for the first $T$ rounds.

We’ll add some more notation shortly to rephrase this definition in symbols, but the idea is clear: we’re competing against the best action. Had we known it ahead of time, we would have just played it every single round. Our notion of success is not in how well we do absolutely, but in how well we do relative to what is feasible.

Notation

Let’s go ahead and draw up some notation. As before the actions are labeled by integers $\left \{ 1, \dots, K \right \}$. The reward of action $i$ is a $[0,1]$-valued random variable $X_i$ distributed according to an unknown distribution and possessing an unknown expected value $\mu_i$. The game progresses in rounds $t = 1, 2, \dots$ so that in each round we have different random variables $X_{i,t}$ for the reward of action $i$ in round $t$ (in particular, $X_{i,t}$ and $X_{i,s}$ are identically distributed). The $X_{i,t}$ are independent as both $t$ and $i$ vary, although when $i$ varies the distribution changes.

So if we were to play action 2 over and over for $T$ rounds, then the total payoff would be the random variable $G_2(T) = \sum_{t=1}^T X_{2,t}$. But by independence across rounds and the linearity of expectation, the expected payoff is just $\mu_2 T$. So we can describe the best action as the action with the highest expected payoff. Define

$\displaystyle \mu^* = \max_{1 \leq i \leq K} \mu_i$

We call the action which achieves the maximum $i^*$.

A policy is a randomized algorithm $A$ which picks an action in each round based on the history of chosen actions and observed rewards so far. Define $I_t$ to be the action played by $A$ in round $t$ and $P_i(n)$ to be the number of times we’ve played action $i$ in rounds $1 \leq t \leq n$. These are both random variables. Then the cumulative payoff for the algorithm $A$ over the first $T$ rounds, denoted $G_A(T)$, is just

$\displaystyle G_A(T) = \sum_{t=1}^T X_{I_t, t}$

and its expected value is simply

$\displaystyle \mathbb{E}(G_A(T)) = \mu_1 \mathbb{E}(P_1(T)) + \dots + \mu_K \mathbb{E}(P_K(T))$.

Here the expectation is taken over all random choices made by the policy and over the distributions of rewards, and indeed both of these can affect how many times a machine is played.

Now the cumulative regret of a policy $A$ after the first $T$ steps, denoted $R_A(T)$ can be written as

$\displaystyle R_A(T) = G_{i^*}(T) - G_A(T)$

And the goal of the policy designer for this bandit problem is to minimize the expected cumulative regret, which by linearity of expectation is

$\mathbb{E}(R_A(T)) = \mu^*T - \mathbb{E}(G_A(T))$.

Before we continue, we should note that there are theorems concerning lower bounds for expected cumulative regret. Specifically, for this problem it is known that no algorithm can guarantee an expected cumulative regret better than $\Omega(\sqrt{KT})$. It is also known that there are algorithms that guarantee no worse than $O(\sqrt{KT})$ expected regret. The algorithm we’ll see in the next section, however, only guarantees $O(\sqrt{KT \log T})$. We present it on this blog because of its simplicity and ubiquity in the field.

The UCB1 Algorithm

The policy we examine is called UCB1, and it can be summed up by the principle of optimism in the face of uncertainty. That is, despite our lack of knowledge in what actions are best we will construct an optimistic guess as to how good the expected payoff of each action is, and pick the action with the highest guess. If our guess is wrong, then our optimistic guess will quickly decrease and we’ll be compelled to switch to a different action. But if we pick well, we’ll be able to exploit that action and incur little regret. In this way we balance exploration and exploitation.

The formalism is a bit more detailed than this, because we’ll need to ensure that we don’t rule out good actions that fare poorly early on. Our “optimism” comes in the form of an upper confidence bound (hence the acronym UCB). Specifically, we want to know with high probability that the true expected payoff of an action $\mu_i$ is less than our prescribed upper bound. One general (distribution independent) way to do that is to use the Chernoff-Hoeffding inequality.

As a reminder, suppose $Y_1, \dots, Y_n$ are independent random variables whose values lie in $[0,1]$ and whose expected values are $\mu_i$. Call $Y = \frac{1}{n}\sum_{i}Y_i$ and $\mu = \mathbb{E}(Y) = \frac{1}{n} \sum_{i} \mu_i$. Then the Chernoff-Hoeffding inequality gives an exponential upper bound on the probability that the value of $Y$ deviates from its mean. Specifically,

$\displaystyle \textup{P}(Y + a < \mu) \leq e^{-2na^2}$

For us, the $Y_i$ will be the payoff variables for a single action $j$ in the rounds for which we choose action $j$. Then the variable $Y$ is just the empirical average payoff for action $j$ over all the times we’ve tried it. Moreover, $a$ is our one-sided upper bound (and as a lower bound, sometimes). We can then solve this equation for $a$ to find an upper bound big enough to be confident that we’re within $a$ of the true mean.

Indeed, if we call $n_j$ the number of times we played action $j$ thus far, then $n = n_j$ in the equation above, and using $a = a(j,T) = \sqrt{2 \log(T) / n_j}$ we get that $\textup{P}(Y > \mu + a) \leq T^{-4}$, which converges to zero very quickly as the number of rounds played grows. We’ll see this pop up again in the algorithm’s analysis below. But before that note two things. First, assuming we don’t play an action $j$, its upper bound $a$ grows in the number of rounds. This means that we never permanently rule out an action no matter how poorly it performs. If we get extremely unlucky with the optimal action, we will eventually be convinced to try it again. Second, the probability that our upper bound is wrong decreases in the number of rounds independently of how many times we’ve played the action. That is because our upper bound $a(j, T)$ is getting bigger for actions we haven’t played; any round in which we play an action $j$, it must be that $a(j, T+1) = a(j,T)$, although the empirical mean will likely change.

With these two facts in mind, we can formally state the algorithm and intuitively understand why it should work.

UCB1:
Play each of the $K$ actions once, giving initial values for empirical mean payoffs $\overline{x}_i$ of each action $i$.
For each round $t = K, K+1, \dots$:

Let $n_j$ represent the number of times action $j$ was played so far.
Play the action $j$ maximizing $\overline{x}_j + \sqrt{2 \log t / n_j}$.
Observe the reward $X_{j,t}$ and update the empirical mean for the chosen action.

And that’s it. Note that we’re being super stateful here: the empirical means $x_j$ change over time, and we’ll leave this update implicit throughout the rest of our discussion (sorry, functional programmers, but the notation is horrendous otherwise).

Before we implement and test this algorithm, let’s go ahead and prove that it achieves nearly optimal regret. The reader uninterested in mathematical details should skip the proof, but the discussion of the theorem itself is important. If one wants to use this algorithm in real life, one needs to understand the guarantees it provides in order to adequately quantify the risk involved in using it.

Theorem: Suppose that UCB1 is run on the bandit game with $K$ actions, each of whose reward distribution $X_{i,t}$ has values in [0,1]. Then its expected cumulative regret after $T$ rounds is at most $O(\sqrt{KT \log T})$.

Actually, we’ll prove a more specific theorem. Let $\Delta_i$ be the difference $\mu^* - \mu_i$, where $\mu^*$ is the expected payoff of the best action, and let $\Delta$ be the minimal nonzero $\Delta_i$. That is, $\Delta_i$ represents how suboptimal an action is and $\Delta$ is the suboptimality of the second best action. These constants are called problem-dependent constants. The theorem we’ll actually prove is:

Theorem: Suppose UCB1 is run as above. Then its expected cumulative regret $\mathbb{E}(R_{\textup{UCB1}}(T))$ is at most

$\displaystyle 8 \sum_{i : \mu_i < \mu^*} \frac{\log T}{\Delta_i} + \left ( 1 + \frac{\pi^2}{3} \right ) \left ( \sum_{j=1}^K \Delta_j \right )$

Okay, this looks like one nasty puppy, but it’s actually not that bad. The first term of the sum signifies that we expect to play any suboptimal machine about a logarithmic number of times, roughly scaled by how hard it is to distinguish from the optimal machine. That is, if $\Delta_i$ is small we will require more tries to know that action $i$ is suboptimal, and hence we will incur more regret. The second term represents a small constant number (the $1 + \pi^2 / 3$ part) that caps the number of times we’ll play suboptimal machines in excess of the first term due to unlikely events occurring. So the first term is like our expected losses, and the second is our risk.

But note that this is a worst-case bound on the regret. We’re not saying we will achieve this much regret, or anywhere near it, but that UCB1 simply cannot do worse than this. Our hope is that in practice UCB1 performs much better.

Before we prove the theorem, let’s see how derive the $O(\sqrt{KT \log T})$ bound mentioned above. This will require familiarity with multivariable calculus, but such things must be endured like ripping off a band-aid. First consider the regret as a function $R(\Delta_1, \dots, \Delta_K)$ (excluding of course $\Delta^*$), and let’s look at the worst case bound by maximizing it. In particular, we’re just finding the problem with the parameters which screw our bound as badly as possible, The gradient of the regret function is given by

$\displaystyle \frac{\partial R}{\partial \Delta_i} = - \frac{8 \log T}{\Delta_i^2} + 1 + \frac{\pi^2}{3}$

and it’s zero if and only if for each $i$, $\Delta_i = \sqrt{\frac{8 \log T}{1 + \pi^2/3}} = O(\sqrt{\log T})$. However this is a minimum of the regret bound (the Hessian is diagonal and all its eigenvalues are positive). Plugging in the $\Delta_i = O(\sqrt{\log T})$ (which are all the same) gives a total bound of $O(K \sqrt{\log T})$. If we look at the only possible endpoint (the $\Delta_i = 1$), then we get a local maximum of $O(K \sqrt{\log T})$. But this isn’t the $O(\sqrt{KT \log T})$ we promised, what gives? Well, this upper bound grows arbitrarily large as the $\Delta_i$ go to zero. But at the same time, if all the $\Delta_i$ are small, then we shouldn’t be incurring much regret because we’ll be picking actions that are close to optimal!

Indeed, if we assume for simplicity that all the $\Delta_i = \Delta$ are the same, then another trivial regret bound is $\Delta T$ (why?). The true regret is hence the minimum of this regret bound and the UCB1 regret bound: as the UCB1 bound degrades we will eventually switch to the simpler bound. That will be a non-differentiable switch (and hence a critical point) and it occurs at $\Delta = O(\sqrt{(K \log T) / T})$. Hence the regret bound at the switch is $\Delta T = O(\sqrt{KT \log T})$, as desired.

Proving the Worst-Case Regret Bound

Proof. The proof works by finding a bound on $P_i(T)$, the expected number of times UCB chooses an action up to round $T$. Using the $\Delta$ notation, the regret is then just $\sum_i \Delta_i \mathbb{E}(P_i(T))$, and bounding the $P_i$‘s will bound the regret.

Recall the notation for our upper bound $a(j, T) = \sqrt{2 \log T / P_j(T)}$ and let’s loosen it a bit to $a(y, T) = \sqrt{2 \log T / y}$ so that we’re allowed to “pretend” a action has been played $y$ times. Recall further that the random variable $I_t$ has as its value the index of the machine chosen. We denote by $\chi(E)$ the indicator random variable for the event $E$. And remember that we use an asterisk to denote a quantity associated with the optimal action (e.g., $\overline{x}^*$ is the empirical mean of the optimal action).

Indeed for any action $i$, the only way we know how to write down $P_i(T)$ is as

$\displaystyle P_i(T) = 1 + \sum_{t=K}^T \chi(I_t = i)$

The 1 is from the initialization where we play each action once, and the sum is the trivial thing where just count the number of rounds in which we pick action $i$. Now we’re just going to pull some number $m-1$ of plays out of that summation, keep it variable, and try to optimize over it. Since we might play the action fewer than $m$ times overall, this requires an inequality.

$P_i(T) \leq m + \sum_{t=K}^T \chi(I_t = i \textup{ and } P_i(t-1) \geq m)$

These indicator functions should be read as sentences: we’re just saying that we’re picking action $i$ in round $t$ and we’ve already played $i$ at least $m$ times. Now we’re going to focus on the inside of the summation, and come up with an event that happens at least as frequently as this one to get an upper bound. Specifically, saying that we’ve picked action $i$ in round $t$ means that the upper bound for action $i$ exceeds the upper bound for every other action. In particular, this means its upper bound exceeds the upper bound of the best action (and $i$ might coincide with the best action, but that’s fine). In notation this event is

$\displaystyle \overline{x}_i + a(P_i(t), t-1) \geq \overline{x}^* + a(P^*(T), t-1)$

Denote the upper bound $\overline{x}_i + a(i,t)$ for action $i$ in round $t$ by $U_i(t)$. Since this event must occur every time we pick action $i$ (though not necessarily vice versa), we have

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \chi(U_i(t-1) \geq U^*(t-1) \textup{ and } P_i(t-1) \geq m)$

We’ll do this process again but with a slightly more complicated event. If the upper bound of action $i$ exceeds that of the optimal machine, it is also the case that the maximum upper bound for action $i$ we’ve seen after the first $m$ trials exceeds the minimum upper bound we’ve seen on the optimal machine (ever). But on round $t$ we don’t know how many times we’ve played the optimal machine, nor do we even know how many times we’ve played machine $i$ (except that it’s more than $m$). So we try all possibilities and look at minima and maxima. This is a pretty crude approximation, but it will allow us to write things in a nicer form.

Denote by $\overline{x}_{i,s}$ the random variable for the empirical mean after playing action $i$ a total of $s$ times, and $\overline{x}^*_s$ the corresponding quantity for the optimal machine. Realizing everything in notation, the above argument proves that

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \chi \left ( \max_{m \leq s < t} \overline{x}_{i,s} + a(s, t-1) \geq \min_{0 < s' < t} \overline{x}^*_{s'} + a(s', t-1) \right )$

Indeed, at each $t$ for which the max is greater than the min, there will be at least one pair $s,s'$ for which the values of the quantities inside the max/min will satisfy the inequality. And so, even worse, we can just count the number of pairs $s, s'$ for which it happens. That is, we can expand the event above into the double sum which is at least as large:

$\displaystyle P_i(T) \leq m + \sum_{t=K}^T \sum_{s = m}^{t-1} \sum_{s' = 1}^{t-1} \chi \left ( \overline{x}_{i,s} + a(s, t-1) \geq \overline{x}^*_{s'} + a(s', t-1) \right )$

We can make one other odd inequality by increasing the sum to go from $t=1$ to $\infty$. This will become clear later, but it means we can replace $t-1$ with $t$ and thus have

$\displaystyle P_i(T) \leq m + \sum_{t=1}^\infty \sum_{s = m}^{t-1} \sum_{s' = 1}^{t-1} \chi \left ( \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) \right )$

Now that we’ve slogged through this mess of inequalities, we can actually get to the heart of the argument. Suppose that this event actually happens, that $\overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t)$. Then what can we say? Well, consider the following three events:

(1) $\displaystyle \overline{x}^*_{s'} \leq \mu^* - a(s', t)$
(2) $\displaystyle \overline{x}_{i,s} \geq \mu_i + a(s, t)$
(3) $\displaystyle \mu^* < \mu_i + 2a(s, t)$

In words, (1) is the event that the empirical mean of the optimal action is less than the lower confidence bound. By our Chernoff bound argument earlier, this happens with probability $t^{-4}$. Likewise, (2) is the event that the empirical mean payoff of action $i$ is larger than the upper confidence bound, which also occurs with probability $t^{-4}$. We will see momentarily that (3) is impossible for a well-chosen $m$ (which is why we left it variable), but in any case the claim is that one of these three events must occur. For if they are all false, we have

$\displaystyle \begin{matrix} \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) & > & \mu^* - a(s',t) + a(s',t) = \mu^* \\ \textup{assumed} & (1) \textup{ is false} & \\ \end{matrix}$

and

$\begin{matrix} \mu_i + 2a(s,t) & > & \overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t) \\ & (2) \textup{ is false} & \textup{assumed} \\ \end{matrix}$

But putting these two inequalities together gives us precisely that (3) is true:

$\mu^* < \mu_i + 2a(s,t)$

This proves the claim.

By the union bound, the probability that at least one of these events happens is $2t^{-4}$ plus whatever the probability of (3) being true is. But as we said, we’ll pick $m$ to make (3) always false. Indeed $m$ depends on which action $i$ is being played, and if $s \geq m > 8 \log T / \Delta_i^2$ then $2a(s,t) \leq \Delta_i$, and by the definition of $\Delta_i$ we have

$\mu^* - \mu_i - 2a(s,t) \geq \mu^* - \mu_i - \Delta_i = 0$.

Now we can finally piece everything together. The expected value of an event is just its probability of occurring, and so

\displaystyle \begin{aligned} \mathbb{E}(P_i(T)) & \leq m + \sum_{t=1}^\infty \sum_{s=m}^t \sum_{s' = 1}^t \textup{P}(\overline{x}_{i,s} + a(s, t) \geq \overline{x}^*_{s'} + a(s', t)) \\ & \leq \left \lceil \frac{8 \log T}{\Delta_i^2} \right \rceil + \sum_{t=1}^\infty \sum_{s=m}^t \sum_{s' = 1}^t 2t^{-4} \\ & \leq \frac{8 \log T}{\Delta_i^2} + 1 + \sum_{t=1}^\infty \sum_{s=1}^t \sum_{s' = 1}^t 2t^{-4} \\ & = \frac{8 \log T}{\Delta_i^2} + 1 + 2 \sum_{t=1}^\infty t^{-2} \\ & = \frac{8 \log T}{\Delta_i^2} + 1 + \frac{\pi^2}{3} \\ \end{aligned}

The second line is the Chernoff bound we argued above, the third and fourth lines are relatively obvious algebraic manipulations, and the last equality uses the classic solution to Basel’s problem. Plugging this upper bound in to the regret formula we gave in the first paragraph of the proof establishes the bound and proves the theorem.

$\square$

Implementation and an Experiment

The algorithm is about as simple to write in code as it is in pseudocode. The confidence bound is trivial to implement (though note we index from zero):

def upperBound(step, numPlays):
return math.sqrt(2 * math.log(step + 1) / numPlays)


And the full algorithm is quite short as well. We define a function ub1, which accepts as input the number of actions and a function reward which accepts as input the index of the action and the time step, and draws from the appropriate reward distribution. Then implementing ub1 is simply a matter of keeping track of empirical averages and an argmax. We implement the function as a Python generator, so one can observe the steps of the algorithm and keep track of the confidence bounds and the cumulative regret.

def ucb1(numActions, reward):
payoffSums = [0] * numActions
numPlays = [1] * numActions
ucbs = [0] * numActions

# initialize empirical sums
for t in range(numActions):
payoffSums[t] = reward(t,t)
yield t, payoffSums[t], ucbs

t = numActions

while True:
ucbs = [payoffSums[i] / numPlays[i] + upperBound(t, numPlays[i]) for i in range(numActions)]
action = max(range(numActions), key=lambda i: ucbs[i])
theReward = reward(action, t)
numPlays[action] += 1
payoffSums[action] += theReward

yield action, theReward, ucbs
t = t + 1


The heart of the algorithm is the second part, where we compute the upper confidence bounds and pick the action maximizing its bound.

We tested this algorithm on synthetic data. There were ten actions and a million rounds, and the reward distributions for each action were uniform from $[0,1]$, biased by $1/k$ for some $5 \leq k \leq 15$. The regret and theoretical regret bound are given in the graph below.

The regret of ucb1 run on a simple example. The blue curve is the cumulative regret of the algorithm after a given number of steps. The green curve is the theoretical upper bound on the regret.

Note that both curves are logarithmic, and that the actual regret is quite a lot smaller than the theoretical regret. The code used to produce the example and image are available on this blog’s Github page.

Next Time

One interesting assumption that UCB1 makes in order to do its magic is that the payoffs are stochastic and independent across rounds. Next time we’ll look at an algorithm that assumes the payoffs are instead adversarial, as we described earlier. Surprisingly, in the adversarial case we can do about as well as the stochastic case. Then, we’ll experiment with the two algorithms on a real-world application.

Until then!

Probabilistic Bounds — A Primer

Probabilistic arguments are a key tool for the analysis of algorithms in machine learning theory and probability theory. They also assume a prominent role in the analysis of randomized and streaming algorithms, where one imposes a restriction on the amount of storage space an algorithm is allowed to use for its computations (usually sublinear in the size of the input).

While a whole host of probabilistic arguments are used, one theorem in particular (or family of theorems) is ubiquitous: the Chernoff bound. In its simplest form, the Chernoff bound gives an exponential bound on the deviation of sums of random variables from their expected value.

This is perhaps most important to algorithm analysis in the following mindset. Say we have a program whose output is a random variable $X$. Moreover suppose that the expected value of $X$ is the correct output of the algorithm. Then we can run the algorithm multiple times and take a median (or some sort of average) across all runs. The probability that the algorithm gives a wildly incorrect answer is the probability that more than half of the runs give values which are wildly far from their expected value. Chernoff’s bound ensures this will happen with small probability.

So this post is dedicated to presenting the main versions of the Chernoff bound that are used in learning theory and randomized algorithms. Unfortunately the proof of the Chernoff bound in its full glory is beyond the scope of this blog. However, we will give short proofs of weaker, simpler bounds as a straightforward application of this blog’s previous work laying down the theory.

If the reader has not yet intuited it, this post will rely heavily on the mathematical formalisms of probability theory. We will assume our reader is familiar with the material from our first probability theory primer, and it certainly wouldn’t hurt to have read our conditional probability theory primer, though we won’t use conditional probability directly. We will refrain from using measure-theoretic probability theory entirely (some day my colleagues in analysis will like me, but not today).

Two Easy Bounds of Markov and Chebyshev

The first bound we’ll investigate is almost trivial in nature, but comes in handy. Suppose we have a random variable $X$ which is non-negative (as a function). Markov’s inequality is the statement that, for any constant $a > 0$,

$\displaystyle \textup{P}(X \geq a) \leq \frac{\textup{E}(X)}{a}$

In words, the probability that $X$ grows larger than some fixed constant is bounded by a quantity that is inversely proportional to the constant.

The proof is quite simple. Let $\chi_a$ be the indicator random variable for the event that $X \geq a$ ($\chi_a = 1$ when $X \geq a$ and zero otherwise). As with all indicator random variables, the expected value of $\chi_a$ is the probability that the event happens (if this is mysterious, use the definition of expected value). So $\textup{E}(\chi_a) = \textup{P}(X \geq a)$, and linearity of expectation allows us to include a factor of $a$:

$\textup{E}(a \chi_a) = a \textup{P}(X \geq a)$

The rest of the proof is simply the observation that $\textup{E}(a \chi_a) \leq \textup{E}(X)$. Indeed, as random variables we have the inequality $a \chi_a \leq X$. Whenever $a < X$, the former is equal to zero while the latter is nonnegative. And whenever $a \geq X$, the former is precisely $a$ while the latter is by assumption at least $a$. It follows that $\textup{E}(a \chi_a) \leq \textup{E}(X)$.

This last point is a simple property of expectation we omitted from our first primer. It usually goes by monotonicity of expectation, and we prove it here. First, if $X \geq 0$ then $\textup{E}(X) \geq 0$ (this is trivial). Second, if $0 \leq X \leq Y$, then define a new random variable $Z = Y-X$. Since $Z \geq 0$ and using linearity of expectation, it must be that $\textup{E}(Z) = \textup{E}(Y) - \textup{E}(X) \geq 0$. Hence $\textup{E}(X) \leq \textup{E}(Y)$. Note that we do require that $X$ has a finite expected value for this argument to work, but if this is not the case then Markov’s inequality is nonsensical anyway.

Markov’s inequality by itself is not particularly impressive or useful. For example, if $X$ is the number of heads in a hundred coin flips, Markov’s inequality ensures us that the probability of getting at least 99 heads is at most 50/99, which is about 1/2. Shocking. We know that the true probability is much closer to $2^{-100}$, so Markov’s inequality is a bust.

However, it does give us a more useful bound as a corollary. This bound is known as Chebyshev’s inequality, and its use is sometimes referred to as the second moment method because it gives a bound based on the variance of a random variable (instead of the expected value, the “first moment”).

The statement is as follows.

Chebyshev’s Inequality: Let $X$ be a random variable with finite expected value and positive variance. Then we can bound the probability that $X$ deviates from its expected value by a quantity that is proportional to the variance of $X$. In particular, for any $\lambda > 0$,

$\displaystyle \textup{P}(|X - \textup{E}(X)| \geq \lambda) \leq \frac{\textup{Var}(X)}{\lambda^2}$

And without any additional assumptions on $X$, this bound is sharp.

Proof. The proof is a simple application of Markov’s inequality. Let $Y = (X - \textup{E}(X))^2$, so that $\textup{E}(Y) = \textup{Var}(X)$. Then by Markov’s inequality

$\textup{P}(Y \geq \lambda^2) \leq \frac{\textup{E}(Y)}{\lambda^2}$

Since $Y$ is nonnegative $|X - \textup{E}(X)| = \sqrt(Y)$, and $\textup{P}(Y \geq \lambda^2) = \textup{P}(|X - \textup{E}(X)| \geq \lambda)$. The theorem is proved. $\square$

Chebyshev’s inequality shows up in so many different places (and usually in rather dry, technical bits), that it’s difficult to give a good example application.  Here is one that shows up somewhat often.

Say $X$ is a nonnegative integer-valued random variable, and we want to argue about when $X = 0$ versus when $X > 0$, given that we know $\textup{E}(X)$. No matter how large $\textup{E}(X)$ is, it can still be possible that $\textup{P}(X = 0)$ is arbitrarily close to 1. As a colorful example, let $X$ is the number of alien lifeforms discovered in the next ten years. We might debate that $\textup{E}(X)$ can arbitrarily large: if some unexpected scientific and technological breakthroughs occur tomorrow, we could discover an unbounded number of lifeforms. On the other hand, we are very likely not to discover any, and probability theory allows for such a random variable to exist.

If we know everything about $\textup{Var}(X)$, however, we can get more informed bounds.

Theorem: If $\textup{E}(X) \neq 0$, then $\displaystyle \textup{P}(X = 0) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$.

Proof. Simply choose $\lambda = \textup{E}(X)$ and apply Chebyshev’s inequality.

$\displaystyle \textup{P}(X = 0) \leq \textup{P}(|X - \textup{E}(X)| \geq \textup{E}(X)) \leq \frac{\textup{Var}(X)}{\textup{E}(X)^2}$

The first inequality follows from the fact that the only time $X$ can ever be zero is when $|X - \textup{E}(X)| = \textup{E}(X)$, and $X=0$ only accounts for one such possibility. $\square$

This theorem says more. If we know that $\textup{Var}(X)$ is significantly smaller than $\textup{E}(X)^2$, then $X > 0$ is more certain to occur. More precisely, and more computationally minded, suppose we have a sequence of random variables $X_n$ so that $\textup{E}(X_n) \to \infty$ as $n \to \infty$. Then the theorem says that if $\textup{Var}(X_n) = o(\textup{E}(X_n)^2)$, then $\textup{P}(X_n > 0) \to 1$. Remembering one of our very early primers on asymptotic notation, $f = o(g)$ means that $f$ grows asymptotically slower than $g$, and in terms of this fraction $\textup{Var}(X) / \textup{E}(X)^2$, this means that the denominator dominates the fraction so that the whole thing tends to zero.

The Chernoff Bound

The Chernoff bound takes advantage of an additional hypothesis: our random variable is a sum of independent coin flips. We can use this to get exponential bounds on the deviation of the sum. More rigorously,

Theorem: Let $X_1 , \dots, X_n$ be independent random $\left \{ 0,1 \right \}$-valued variables, and let $X = \sum X_i$. Suppose that $\mu = \textup{E}(X)$. Then the probability that $X$ deviates from $\mu$ by more than a factor of $\lambda > 0$ is bounded from above:

$\displaystyle \textup{P}(X > (1+\lambda)\mu) \leq \frac{e^{\lambda \mu}}{(1+\lambda)^{(1+\lambda)\mu}}$

The proof is beyond the scope of this post, but we point the interested reader to these lecture notes.

We can apply the Chernoff bound in an easy example. Say all $X_i$ are fair coin flips, and we’re interested in the probability of getting more than 3/4 of the coins heads. Here $\mu = n/2$ and $\lambda = 1/2$, so the probability is bounded from above by

$\displaystyle \left ( \frac{e}{(3/2)^3} \right )^{n/4} \approx \frac{1}{5^n}$

So as the number of coin flips grows, the probability of seeing such an occurrence diminishes extremely quickly to zero. This is important because if we want to test to see if, say, the coins are biased toward flipping heads, we can simply run an experiment with $n$ sufficiently large. If we observe that more than 3/4 of the flips give heads, then we proclaim the coins are biased and we can be assured we are correct with high probability. Of course, after seeing 3/4 of more heads we’d be really confident that the coin is biased. A more realistic approach is to define some $\varepsilon$ that is small enough so as to say, “if some event occurs whose probability is smaller than $\varepsilon$, then I call shenanigans.” Then decide how many coins and what bound one would need to make the bad event have probability approximately $\varepsilon$. Finding this balance is one of the more difficult aspects of probabilistic algorithms, and as we’ll see later all of these quantities are left as variables and the correct values are discovered in the course of the proof.

Chernoff-Hoeffding Inequality

The Hoeffding inequality (named after the Finnish statistician, Wassily Høffding) is a variant of the Chernoff bound, but often the bounds are collectively known as Chernoff-Hoeffding inequalities. The form that Hoeffding is known for can be thought of as a simplification and a slight generalization of Chernoff’s bound above.

Theorem: Let $X_1, \dots, X_n$ be independent random variables whose values are within some range $[a,b]$. Call $\mu_i = \textup{E}(X_i)$, $X = \sum_i X_i$, and $\mu = \textup{E}(X) = \sum_i \mu_i$. Then for all $t > 0$,

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{-2t^2 / n(b-a)^2}$

For example, if we are interested in the sum of $n$ rolls of a fair six-sided die, then the probability that we deviate from $(7/2)n$ by more than $5 \sqrt{n \log n}$ is bounded by $2e^{(-2 \log n)} = 2/n^2$. Supposing we want to know how many rolls we need to guarantee with probability 0.01 that we don’t deviate too much, we just do the algebra:

$2n^{-2} < 0.01$
$n^2 > 200$
$n > \sqrt{200} \approx 14$

So with 15 rolls we can be confident that the sum of the rolls will lie between 20 and 85. It’s not the best possible bound we could come up with, because we’re completely ignoring the known structure on dice rolls (that they follow a uniform distribution!). The benefit is that it’s a quick and easy bound that works for any kind of random variable with that expected value.

Another version of this theorem concerns the average of the $X_i$, and is only a minor modification of the above.

Theorem: If $X_1, \dots, X_n$ are as above, and $X = \frac{1}{n} \sum_i X_i$, with $\mu = \frac{1}{n}(\sum_i \mu_i)$, then for all $t > 0$, we get the following bound

$\displaystyle \textup{P}(|X - \mu| > t) \leq 2e^{-2nt^2/(b-a)^2}$

The only difference here is the extra factor of $n$ in the exponent. So the deviation is exponential both in the amount of deviation ($t^2$), and in the number of trials.

This theorem comes up very often in learning theory, in particular to prove Boosting works. Mathematicians will joke about how all theorems in learning theory are just applications of Chernoff-Hoeffding-type bounds. We’ll of course be seeing it again as we investigate boosting and the PAC-learning model in future posts, so we’ll see the theorems applied to their fullest extent then.

Until next time!

Probability Theory — A Primer

It is a wonder that we have yet to officially write about probability theory on this blog. Probability theory underlies a huge portion of artificial intelligence, machine learning, and statistics, and a number of our future posts will rely on the ideas and terminology we lay out in this post. Our first formal theory of machine learning will be deeply ingrained in probability theory, we will derive and analyze probabilistic learning algorithms, and our entire treatment of mathematical finance will be framed in terms of random variables.

And so it’s about time we got to the bottom of probability theory. In this post, we will begin with a naive version of probability theory. That is, everything will be finite and framed in terms of naive set theory without the aid of measure theory. This has the benefit of making the analysis and definitions simple. The downside is that we are restricted in what kinds of probability we are allowed to speak of. For instance, we aren’t allowed to work with probabilities defined on all real numbers. But for the majority of our purposes on this blog, this treatment will be enough. Indeed, most programming applications restrict infinite problems to finite subproblems or approximations (although in their analysis we often appeal to the infinite).

We should make a quick disclaimer before we get into the thick of things: this primer is not meant to connect probability theory to the real world. Indeed, to do so would be decidedly unmathematical. We are primarily concerned with the mathematical formalisms involved in the theory of probability, and we will leave the philosophical concerns and applications to  future posts. The point of this primer is simply to lay down the terminology and basic results needed to discuss such topics to begin with.

So let us begin with probability spaces and random variables.

Finite Probability Spaces

We begin by defining probability as a set with an associated function. The intuitive idea is that the set consists of the outcomes of some experiment, and the function gives the probability of each event happening. For example, a set $\left \{ 0,1 \right \}$ might represent heads and tails outcomes of a coin flip, while the function assigns a probability of one half (or some other numbers) to the outcomes. As usual, this is just intuition and not rigorous mathematics. And so the following definition will lay out the necessary condition for this probability to make sense.

Definition: A finite set $\Omega$ equipped with a function $f: \Omega \to [0,1]$ is a probability space if the function $f$ satisfies the property

$\displaystyle \sum_{\omega \in \Omega} f(\omega) = 1$

That is, the sum of all the values of $f$ must be 1.

Sometimes the set $\Omega$ is called the sample space, and the act of choosing an element of $\Omega$ according to the probabilities given by $f$ is called drawing an example. The function $f$ is usually called the probability mass function. Despite being part of our first definition, the probability mass function is relatively useless except to build what follows. Because we don’t really care about the probability of a single outcome as much as we do the probability of an event.

Definition: An event $E \subset \Omega$ is a subset of a sample space.

For instance, suppose our probability space is $\Omega = \left \{ 1, 2, 3, 4, 5, 6 \right \}$ and $f$ is defined by setting $f(x) = 1/6$ for all $x \in \Omega$ (here the “experiment” is rolling a single die). Then we are likely interested in more exquisite kinds of outcomes; instead of asking the probability that the outcome is 4, we might ask what is the probability that the outcome is even? This event would be the subset $\left \{ 2, 4, 6 \right \}$, and if any of these are the outcome of the experiment, the event is said to occur. In this case we would expect the probability of the die roll being even to be 1/2 (but we have not yet formalized why this is the case).

As a quick exercise, the reader should formulate a two-dice experiment in terms of sets. What would the probability space consist of as a set? What would the probability mass function look like? What are some interesting events one might consider (if playing a game of craps)?

Of course, we want to extend the probability mass function $f$ (which is only defined on single outcomes) to all possible events of our probability space. That is, we want to define a probability measure $\textup{P}: 2^\Omega \to \mathbb{R}$, where $2^\Omega$ denotes the set of all subsets of $\Omega$. The example of a die roll guides our intuition: the probability of any event should be the sum of the probabilities of the outcomes contained in it. i.e. we define

$\displaystyle \textup{P}(E) = \sum_{e \in E} f(e)$

where by convention the empty sum has value zero. Note that the function $\textup{P}$ is often denoted $\textup{Pr}$.

So for example, the coin flip experiment can’t have zero probability for both of the two outcomes 0 and 1; the sum of the probabilities of all outcomes must sum to 1. More coherently: $\textup{P}(\Omega) = \sum_{\omega \in \Omega} f(\omega) = 1$ by the defining property of a probability space. And so if there are only two outcomes of the experiment, then they must have probabilities $p$ and $1-p$ for some $p$. Such a probability space is often called a Bernoulli trial.

Now that the function $\textup{P}$ is defined on all events, we can simplify our notation considerably. Because the probability mass function $f$ uniquely determines $\textup{P}$ and because $\textup{P}$ contains all information about $f$ in it ($\textup{P}(\left \{ \omega \right \}) = f(\omega)$), we may speak of $\textup{P}$ as the probability measure of $\Omega$, and leave $f$ out of the picture. Of course, when we define a probability measure, we will allow ourselves to just define the probability mass function and the definition of $\textup{P}$ is understood as above.

There are some other quick properties we can state or prove about probability measures: $\textup{P}(\left \{ \right \}) = 0$ by convention, if $E, F$ are disjoint then $\textup{P}(E \cup F) = \textup{P}(E) + \textup{P}(F)$, and if $E \subset F \subset \Omega$ then $\textup{P}(E) \leq \textup{P}(F)$. The proofs of these facts are trivial, but a good exercise for the uncomfortable reader to work out.

Random Variables

The next definition is crucial to the entire theory. In general, we want to investigate many different kinds of random quantities on the same probability space. For instance, suppose we have the experiment of rolling two dice. The probability space would be

$\displaystyle \Omega = \left \{ (1,1), (1,2), (1,3), \dots, (6,4), (6,5), (6,6) \right \}$

Where the probability measure is defined uniformly by setting all single outcomes to have probability 1/36. Now this probability space is very general, but rarely are we interested only in its events. If this probability space were interpreted as part of a game of craps, we would likely be more interested in the sum of the two dice than the actual numbers on the dice. In fact, we are really more interested in the payoff determined by our roll.

Sums of numbers on dice are certainly predictable, but a payoff can conceivably be any function of the outcomes. In particular, it should be a function of $\Omega$ because all of the randomness inherent in the game comes from the generation of an output in $\Omega$ (otherwise we would define a different probability space to begin with).

And of course, we can compare these two different quantities (the amount of money and the sum of the two dice) within the framework of the same probability space. This “quantity” we speak of goes by the name of a random variable.

Definition: random variable $X$ is a real-valued function on the sample space $\Omega \to \mathbb{R}$.

So for example the random variable for the sum of the two dice would be $X(a,b) = a+b$. We will slowly phase out the function notation as we go, reverting to it when we need to avoid ambiguity.

We can further define the set of all random variables $\textup{RV}(\Omega)$. It is important to note that this forms a vector space. For those readers unfamiliar with linear algebra, the salient fact is that we can add two random variables together and multiply them by arbitrary constants, and the result is another random variable. That is, if $X, Y$ are two random variables, so is $aX + bY$ for real numbers $a, b$. This function operates linearly, in the sense that its value is $(aX + bY)(\omega) = aX(\omega) + bY(\omega)$. We will use this property quite heavily, because in most applications the analysis of a random variable begins by decomposing it into a combination of simpler random variables.

Of course, there are plenty of other things one can do to functions. For example, $XY$ is the product of two random variables (defined by $XY(\omega) = X(\omega)Y(\omega)$) and one can imagine such awkward constructions as $X/Y$ or $X^Y$. We will see in a bit why it these last two aren’t often used (it is difficult to say anything about them).

The simplest possible kind of random variable is one which identifies events as either occurring or not. That is, for an event $E$, we can define a random variable which is 0 or 1 depending on whether the input is a member of $E$. That is,

Definition: An indicator random variable $1_E$ is defined by setting $1_E(\omega) = 1$ when $\omega \in E$ and 0 otherwise. A common abuse of notation for singleton sets is to denote $1_{\left \{ \omega \right \} }$ by $1_\omega$.

This is what we intuitively do when we compute probabilities: to get a ten when rolling two dice, one can either get a six, a five, or a four on the first die, and then the second die must match it to add to ten.

The most important thing about breaking up random variables into simpler random variables will make itself clear when we see that expected value is a linear functional. That is, probabilistic computations of linear combinations of random variables can be computed by finding the values of the simpler pieces. We can’t yet make that rigorous though, because we don’t yet know what it means to speak of the probability of a random variable’s outcome.

Definition: Denote by $\left \{ X = k \right \}$ the set of outcomes $\omega \in \Omega$ for which $X(\omega) = k$. With the function notation, $\left \{ X = k \right \} = X^{-1}(k)$.

This definition extends to constructing ranges of outcomes of a random variable. i.e., we can define $\left \{ X < 5 \right \}$ or $\left \{ X \textup{ is even} \right \}$ just as we would naively construct sets. It works in general for any subset of $S \subset \mathbb{R}$. The notation is $\left \{ X \in S \right \} = X^{-1}(S)$, and we will also call these sets events. The notation becomes useful and elegant when we combine it with the probability measure $\textup{P}$. That is, we want to write things like $\textup{P}(X \textup{ is even})$ and read it in our head “the probability that $X$ is even”.

This is made rigorous by simply setting

$\displaystyle \textup{P}(X \in S) = \sum_{\omega \in X^{-1}(S)} \textup{P}(\omega)$

In words, it is just the sum of the probabilities that individual outcomes will have a value under $X$ that lands in $S$. We will also use for $\textup{P}(\left \{ X \in S \right \} \cap \left \{ Y \in T \right \})$ the shorthand notation $\textup{P}(X \in S, Y \in T)$ or $\textup{P}(X \in S \textup{ and } Y \in T)$.

Often times $\left \{ X \in S \right \}$ will be smaller than $\Omega$ itself, even if $S$ is large. For instance, let the probability space be the set of possible lottery numbers for one week’s draw of the lottery (with uniform probabilities), let $X$ be the profit function. Then $\textup{P}(X > 0)$ is very small indeed.

We should also note that because our probability spaces are finite, the image of the random variable $\textup{im}(X)$ is a finite subset of real numbers. In other words, the set of all events of the form $\left \{ X = x_i \right \}$ where $x_i \in \textup{im}(X)$ form a partition of $\Omega$. As such, we get the following immediate identity:

$\displaystyle 1 = \sum_{x_i \in \textup{im} (X)} P(X = x_i)$

The set of such events is called the probability distribution of the random variable $X$.

The final definition we will give in this section is that of independence. There are two separate but nearly identical notions of independence here. The first is that of two events. We say that two events $E,F \subset \Omega$ are independent if the probability of both $E, F$ occurring is the product of the probabilities of each event occurring. That is, $\textup{P}(E \cap F) = \textup{P}(E)\textup{P}(F)$. There are multiple ways to realize this formally, but without the aid of conditional probability (more on that next time) this is the easiest way. One should note that this is distinct from $E,F$ being disjoint as sets, because there may be a zero-probability outcome in both sets.

The second notion of independence is that of random variables. The definition is the same idea, but implemented using events of random variables instead of regular events. In particular, $X,Y$ are independent random variables if

$\displaystyle \textup{P}(X = x, Y = y) = \textup{P}(X=x)\textup{P}(Y=y)$

for all $x,y \in \mathbb{R}$.

Expectation

We now turn to notions of expected value and variation, which form the cornerstone of the applications of probability theory.

Definition: Let $X$ be a random variable on a finite probability space $\Omega$. The expected value of $X$, denoted $\textup{E}(X)$, is the quantity

$\displaystyle \textup{E}(X) = \sum_{\omega \in \Omega} X(\omega) \textup{P}(\omega)$

Note that if we label the image of $X$ by $x_1, \dots, x_n$ then this is equivalent to

$\displaystyle \textup{E}(X) = \sum_{i=1}^n x_i \textup{P}(X = x_i)$

The most important fact about expectation is that it is a linear functional on random variables. That is,

Theorem: If $X,Y$ are random variables on a finite probability space and $a,b \in \mathbb{R}$, then

$\displaystyle \textup{E}(aX + bY) = a\textup{E}(X) + b\textup{E}(Y)$

Proof. The only real step in the proof is to note that for each possible pair of values $x, y$ in the images of $X,Y$ resp., the events $E_{x,y} = \left \{ X = x, Y=y \right \}$ form a partition of the sample space $\Omega$. That is, because $aX + bY$ has a constant value on $E_{x,y}$, the second definition of expected value gives

$\displaystyle \textup{E}(aX + bY) = \sum_{x \in \textup{im} (X)} \sum_{y \in \textup{im} (Y)} (ax + by) \textup{P}(X = x, Y = y)$

and a little bit of algebraic elbow grease reduces this expression to $a\textup{E}(X) + b\textup{E}(Y)$. We leave this as an exercise to the reader, with the additional note that the sum $\sum_{y \in \textup{im}(Y)} \textup{P}(X = x, Y = y)$ is identical to $\textup{P}(X = x)$. $\square$

If we additionally know that $X,Y$ are independent random variables, then the same technique used above allows one to say something about the expectation of the product $\textup{E}(XY)$ (again by definition, $XY(\omega) = X(\omega)Y(\omega)$). In this case $\textup{E}(XY) = \textup{E}(X)\textup{E}(Y)$. We leave the proof as an exercise to the reader.

Now intuitively the expected value of a random variable is the “center” of the values assumed by the random variable. It is important, however, to note that the expected value need not be a value assumed by the random variable itself; that is, it might not be true that $\textup{E}(X) \in \textup{im}(X)$. For instance, in an experiment where we pick a number uniformly at random between 1 and 4 (the random variable is the identity function), the expected value would be:

$\displaystyle 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{4} = \frac{5}{2}$

But the random variable never achieves this value. Nevertheless, it would not make intuitive sense to call either 2 or 3 the “center” of the random variable (for both 2 and 3, there are two outcomes on one side and one on the other).

Let’s see a nice application of the linearity of expectation to a purely mathematical problem. The power of this example lies in the method: after a shrewd decomposition of a random variable $X$ into simpler (usually indicator) random variables, the computation of $\textup{E}(X)$ becomes trivial.

tournament  $T$ is a directed graph in which every pair of distinct vertices has exactly one edge between them (going one direction or the other). We can ask whether such a graph has a Hamiltonian path, that is, a path through the graph which visits each vertex exactly once. The datum of such a path is a list of numbers $(v_1, \dots, v_n)$, where we visit vertex $v_i$ at stage $i$ of the traversal. The condition for this to be a valid Hamiltonian path is that $(v_i, v_{i+1})$ is an edge in $T$ for all $i$.

Now if we construct a tournament on $n$ vertices by choosing the direction of each edges independently with equal probability 1/2, then we have a very nice probability space and we can ask what is the expected number of Hamiltonian paths. That is, $X$ is the random variable giving the number of Hamiltonian paths in such a randomly generated tournament, and we are interested in $\textup{E}(X)$.

To compute this, simply note that we can break $X = \sum_p X_p$, where $p$ ranges over all possible lists of the vertices. Then $\textup{E}(X) = \sum_p \textup{E}(X_p)$, and it suffices to compute the number of possible paths and the expected value of any given path. It isn’t hard to see the number of paths is $n!$ as this is the number of possible lists of $n$ items. Because each edge direction is chosen with probability 1/2 and they are all chosen independently of one another, the probability that any given path forms a Hamiltonian path depends on whether each edge was chosen with the correct orientation. That’s just

$\textup{P}(\textup{first edge and second edge and } \dots \textup{ and last edge})$

which by independence is

$\displaystyle \prod_{i = 1}^n \textup{P}(i^\textup{th} \textup{ edge is chosen}) = \frac{1}{2^{n-1}}$

That is, the expected number of Hamiltonian paths is $n!2^{-(n-1)}$.

Variance and Covariance

Just as expectation is a measure of center, variance is a measure of spread. That is, variance measures how thinly distributed the values of a random variable $X$ are throughout the real line.

Definition: The variance of a random variable $X$ is the quantity $\textup{E}((X - \textup{E}(X))^2)$.

That is, $\textup{E}(X)$ is a number, and so $X - \textup{E}(X)$ is the random variable defined by $(X - \textup{E}(X))(\omega) = X(\omega) - \textup{E}(X)$. It is the expectation of the square of the deviation of $X$ from its expected value.

One often denotes the variance by $\textup{Var}(X)$ or $\sigma^2$. The square is for silly reasons: the standard deviation, denoted $\sigma$ and equivalent to $\sqrt{\textup{Var}(X)}$ has the same “units” as the outcomes of the experiment and so it’s preferred as the “base” frame of reference by some. We won’t bother with such physical nonsense here, but we will have to deal with the notation.

The variance operator has a few properties that make it quite different from expectation, but nonetheless fall our directly from the definition. We encourage the reader to prove a few:

• $\textup{Var}(X) = \textup{E}(X^2) - \textup{E}(X)^2$.
• $\textup{Var}(aX) = a^2\textup{Var}(X)$.
• When $X,Y$ are independent then variance is additive: $\textup{Var}(X+Y) = \textup{Var}(X) + \textup{Var}(Y)$.
• Variance is invariant under constant additives: $\textup{Var}(X+c) = \textup{Var}(X)$.

In addition, the quantity $\textup{Var}(aX + bY)$ is more complicated than one might first expect. In fact, to fully understand this quantity one must create a notion of correlation between two random variables. The formal name for this is covariance.

Definition: Let $X,Y$ be random variables. The covariance of $X$ and $Y$, denoted $\textup{Cov}(X,Y)$, is the quantity $\textup{E}((X - \textup{E}(X))(Y - \textup{E}(Y)))$.

Note the similarities between the variance definition and this one: if $X=Y$ then the two quantities coincide. That is, $\textup{Cov}(X,X) = \textup{Var}(X)$.

There is a nice interpretation to covariance that should accompany every treatment of probability: it measures the extent to which one random variable “follows” another. To make this rigorous, we need to derive a special property of the covariance.

Theorem: Let $X,Y$ be random variables with variances $\sigma_X^2, \sigma_Y^2$. Then their covariance is at most the product of the standard deviations in magnitude:

$|\textup{Cov}(X,Y)| \leq \sigma_X \sigma_Y$

Proof. Take any two non-constant random variables $X$ and $Y$ (we will replace these later with $X - \textup{E}(X), Y - \textup{E}(Y)$). Construct a new random variable $(tX + Y)^2$ where $t$ is a real variable and inspect its expected value. Because the function is squared, its values are all nonnegative, and hence its expected value is nonnegative. That is, $\textup{E}((tX + Y)^2)$. Expanding this and using linearity gives

$\displaystyle f(t) = t^2 \textup{E}(X^2) + 2t \textup{E}(XY) + \textup{E}(Y^2) \geq 0$

This is a quadratic function of a single variable $t$ which is nonnegative. From elementary algebra this means the discriminant is at most zero. i.e.

$\displaystyle 4 \textup{E}(XY)^2 - 4 \textup{E}(X^2) \textup{E}(Y^2) \leq 0$

and so dividing by 4 and replacing $X,Y$ with $X - \textup{E}(X), Y - \textup{E}(Y)$, resp., gives

$\textup{Cov}(X,Y)^2 \leq \sigma_X^2 \sigma_Y^2$

and the result follows. $\square$

Note that equality holds in the discriminant formula precisely when $Y = -tX$ (the discriminant is zero), and after the replacement this translates to $Y - \textup{E}(Y) = -t(X - \textup{E}(X))$ for some fixed value of $t$. In other words, for some real numbers $a,b$ we have $Y = aX + b$.

This has important consequences even in English: the covariance is maximized when $Y$ is a linear function of $X$, and otherwise is bounded from above and below. By dividing both sides of the inequality by $\sigma_X \sigma_Y$ we get the following definition:

Definition: The Pearson correlation coefficient of two random variables $X,Y$ is defined by

$\displaystyle r= \frac{\textup{Cov}(X,Y)}{\sigma_X \sigma_Y}$

If $r$ is close to 1, we call $X$ and $Y$ positively correlated. If $r$ is close to -1 we call them negatively correlated, and if $r$ is close to zero we call them uncorrelated.

The idea is that if two random variables are positively correlated, then a higher value for one variable (with respect to its expected value) corresponds to a higher value for the other. Likewise, negatively correlated variables have an inverse correspondence: a higher value for one correlates to a lower value for the other. The picture is as follows:

The  horizontal axis plots a sample of values of the random variable $X$ and the vertical plots a sample of $Y$. The linear correspondence is clear. Of course, all of this must be taken with a grain of salt: this correlation coefficient is only appropriate for analyzing random variables which have a linear correlation. There are plenty of interesting examples of random variables with non-linear correlation, and the Pearson correlation coefficient fails miserably at detecting them.

Here are some more examples of Pearson correlation coefficients applied to samples drawn from the sample spaces of various (continuous, but the issue still applies to the finite case) probability distributions:

Various examples of the Pearson correlation coefficient, credit Wikipedia.

Though we will not discuss it here, there is still a nice precedent for using the Pearson correlation coefficient. In one sense, the closer that the correlation coefficient is to 1, the better a linear predictor will perform in “guessing” values of $Y$ given values of $X$ (same goes for -1, but the predictor has negative slope).

But this strays a bit far from our original point: we still want to find a formula for $\textup{Var}(aX + bY)$. Expanding the definition, it is not hard to see that this amounts to the following proposition:

Proposition: The variance operator satisfies

$\displaystyle \textup{Var}(aX+bY) = a^2\textup{Var}(X) + b^2\textup{Var}(Y) + 2ab \textup{Cov}(X,Y)$

And using induction we get a general formula:

$\displaystyle \textup{Var} \left ( \sum_{i=1}^n a_i X_i \right ) = \sum_{i=1}^n \sum_{j = 1}^n a_i a_j \textup{Cov}(X_i,X_j)$

Note that in the general sum, we get a bunch of terms $\textup{Cov}(X_i,X_i) = \textup{Var}(X_i)$.

Another way to look at the linear relationships between a collection of random variables is via a covariance matrix.

Definition: The covariance matrix of a collection of random variables $X_1, \dots, X_n$ is the matrix whose $(i,j)$ entry is $\textup{Cov}(X_i,X_j)$.

As we have already seen on this blog in our post on eigenfaces, one can manipulate this matrix in interesting ways. In particular (and we may be busting out an unhealthy dose of new terminology here), the covariance matrix is symmetric and nonnegative, and so by the spectral theorem it has an orthonormal basis of eigenvectors, which allows us to diagonalize it. In more direct words: we can form a new collection of random variables $Y_j$ (which are linear combinations of the original variables $X_i$) such that the covariance of distinct pairs $Y_j, Y_k$ are all zero. In one sense, this is the “best perspective” with which to analyze the random variables. We gave a general algorithm to do this in our program gallery, and the technique is called principal component analysis.

Next Up

So far in this primer we’ve seen a good chunk of the kinds of theorems one can prove in probability theory. Fortunately, much of what we’ve said for finite probability spaces holds for infinite (discrete) probability spaces and has natural analogues for continuous probability spaces.

Next time, we’ll investigate how things change for discrete probability spaces, and should we need it, we’ll follow that up with a primer on continuous probability. This will get our toes wet with some basic measure theory, but as every mathematician knows: analysis builds character.

Until then!