Bandits and Stocks

So far in this series we’ve seen two nontrivial algorithms for bandit learning in two different settings. The first was the UCB1 algorithm, which operated under the assumption that the rewards for the trials were independent and stochastic. That is, each slot machine was essentially a biased coin flip, and the algorithm was trying to find the machine with the best odds. The second was the Exp3 algorithm, which held the belief that the payoffs were arbitrary. In particular this includes the possibility that an adversary is setting the payoffs against you, and so we measured the success of an algorithm in terms of how it fares against the best single action (just as we did with UCB1, but with Exp3 it’s a nontrivial decision).

Before we move on to other bandit settings it’s natural to try to experiment with the ones we have on real world data. On one hand it’s interesting to see how they fare outside academia. And more relevantly to the design of the future bandit algorithms we’ll see on this blog, we need to know what worldly problems actually provide in terms of inputs to our learning algorithm in each round.

But another interesting issue goes like this. In the real world we can’t ever really know whether the rewards of the actions are stochastic or adversarial. Many people believe that adversarial settings are far too pathological to be realistic, while others claim that the assumptions made by stochastic models are too strict. To weigh in on this dispute, we’ll dip into a bit of experimental science and see which of the two algorithms performs better on the problem of stock trading. The result is then evidence that stocks behave stochastically or adversarially. But we don’t want to stir up too many flames, so we can always back up behind the veil of applied mathematics (“this model is too simple anyway”).

Indeed the model we use in this post is rather simplistic. I don’t know as much as I should (or as my father would have me know) about stock markets. In fact, I’m more partial to not trade stocks on principle. But I must admit that average-quality stock data is easy to come by, and the basic notions of market interactions lend themselves naturally to many machine learning problems. If the reader has any ideas about how to strengthen the model, I welcome suggestions in the comments (or a fork on github).

A fair warning to the reader, we do not solve the problem of trading stocks by any means. We use a model that’s almost entirely unrealistic, and the results aren’t even that good. I’m quite nervous to publish this at all, just because above all else it reveals my gaping ignorance on how stock markets work. But this author believes in revealing ignorance as learning, if for nothing else than that it provides extremely valuable insight into the nature of a problem and an appreciation of its complexity. So criticize away, dear readers.

As usual, all of the code and data we use in this post is available on this blog’s Github page. Our language of choice for this post is Python.

This little trader got lucky.

This little trader got lucky. Could it be because he’s got TEN MONITORS?!

Stocks for Dummies (me)

A quick primer on stocks, which is only as detailed as it needs to be for this post: a stock is essentially the sum of the value of all the assets of a company. A publicly traded company divides their stock into a number of “shares,” and owning a share represents partial ownership of the company. If you own 50% of the shares, you own 50% of the company. Companies sell shares or give them to employees as benefits (or options), and use the money gained through their sale for whatever they see fit. The increase in the price of a stock generally signifies the company is successful and growing; for example, stocks generally rise when a hotly anticipated product is announced.

The stock of a company is traded through one of a number of markets called stock exchanges. The buying and selling interactions are recorded and public, and there are many people in the world who monitor the interactions as they happen (via television, or programmatically) in the hopes of noticing opportunities before others and capitalizing on them. Each interaction induces a change on the price of a share of stock: whenever a share is bought at a certain price, that is the established and recorded price of a share (up to some fudging by brokers which is entirely mysterious to me). In any case, the prices go up and down, and they’re often bundled into “bars” which summarize the data over a certain period of time. The bars we use in this post are daily, and consist of four numbers: the open, the price at the beginning of the day, the high and low, which are self-explanatory, and the close, which is the price at the end of the day.

Bandits and Daily Stock Trading

Now let’s simplify things as much as possible. Our bandit learning algorithm will interact with the market as follows: each day it chooses whether or not to buy a single dollar’s worth of a stock, and at the end of the day it sells the stock and observes the profit. There are no brokers involved, and the price the algorithm sees is the price it gets. In bandit language: the stocks represent actions, and the amount of profit at the end of a day constitutes the payoff of an action in one round. Since small-scale stock price movement is generally very poorly understood, it makes some level of sense to assume the price movements within a given day are adversarial. On the other hand, since we understand them so poorly, we might be tempted to just call them “random” fluctuations, i.e. stochastic. So this is a nice little testbed for seeing which assumption yields a more successful algorithm.

Unlike the traditional image of stock trading where an individual owns shares of a stock over a long period of time, our program will operate on a daily time scale, and hence cannot experience the typical kinds of growth. Nevertheless, we can try to make some money over time, and if it’s a good strategy, we could scale up the single dollar to whatever we’re willing to risk. Specifically, the code we used to compute the payoff is

def payoff(stockTable, t, stock, amountToInvest=1.0):
   openPrice, closePrice = stockTable[stock][t]

   sharesBought = amountToInvest / openPrice
   amountAfterSale = sharesBought * closePrice

   return amountAfterSale - amountToInvest

The remainder of the code is interfacing with the Exp3 and UCB1 functions we gave in previous posts, and data shuffling. We got our data from Google Finance, and we provide it, along with all of the code used in the making of this post, on this blog’s Github page. Before we run our experiments, let’s give a few reasons why this model is unrealistic.

  1. We assume we can buy/sell fractional shares of a stock, which to my knowledge is not possible. Though this experiment could be redone where you buy a single share of a stock, or with mutual funds/currency exchange/whatever replacing stocks, we didn’t do it this way.
  2. Brokerage fees can drastically change the success of an algorithm which trades frequently.
  3. Open and close prices are not typical prices. People will often make decisions based on the time of day, but then again we might expect this to be just another reason that Exp3 would perform better than UCB1.
  4. We’re not actually trading in the stock market, and so we’re ignoring the effects of our own algorithm on the prices in the market.
  5. It’s impossible to guarantee you get to use the opening price and closing price in your transactions.
  6. UCB1 and Exp3 don’t use all of the information available. Indeed, they assume that they would not be able to see the outcome of an action they did not take, but with stocks you can get a good estimate of how much money you would have made had you chosen a different stock.
  7. Each trial in a bandit learning problem is identical from the learner’s perspective, but one often keeps a stock around while making other decisions.

We’ll come back to #6 after seeing the raw experiments for an unaltered UCB1 and Exp3, because there is a natural extension of the algorithm to handle additional information. I’m sure there are other glaring issues with the experimental setup, and the reader should feel free to rant about it in the comments. It won’t stop me from running the algorithm and seeing what happens just for fun.

Data Sets

We ran the experiment on two sets of stocks. The first set consisted of nine random stocks, taken from the random stocks twitter feed, with 5 years of past data. The stocks are:

lxrx, keg, cuba, tdi, brks, mux, cadx, belfb, htr

And you can view more information about these particular stocks via Google Finance. The second set was a non-random choice of nine Fortune 500 companies with 10 years of past data. The stocks were

amzn, cost, jpm, gs, wfc, msft, tgt, aapl, wmt

And again more information about these stocks is available via Google Finance. For the record, here were the cumulative payoffs of each of the nine Fortune 500 companies:


The cumulative rewards for the nine Fortune 500 companies over the last ten years of data.

Interestingly, the company which started off with the best prospects (Apple), turned out to have the worst cumulative reward by the end. The long-term winners in our little imaginary world happen to be Amazon, Costco, and Goldman Sachs. Perhaps this gives credence to the assumption that payoffs are adversarial. A learner can easily get tricked into putting too much faith in one action early on.

And for the random stocks:

The cumulative payoff for the nine randomly chosen stocks.

The cumulative payoff for the nine randomly chosen stocks for the last five years of data.

The random stocks clearly perform worse and more variably overall (although HTR surpasses most of the Fortune 500 companies, despite its otherwise relatively modest stock growth over the last five years). To my untrained eyes these movements look more like a stochastic model than an adversarial one.


Here is a typical example of a run of Exp3 on the Fortune 500 data set (using $ \gamma = 0.33$, recall $ \gamma$ measures the amount of uniform exploration performed):

(Expected payoff, variance) over 1000 trials is (1.122463919564572, 0.5518037498918705)
For a single run: 
Payoff was 1.12
Regret was 2.91
Best stock was amzn at 4.02
weights: '0.00, 0.00, 0.00, 0.46, 0.52, 0.00, 0.00, 0.00, 0.01'

And one for UCB1:

(Expected payoff, variance) over 1000 trials is (1.1529891576139333, 0.5012825847001482)
For a single run: 
Payoff was 1.73
Regret was 2.29
Best stock was amzn at 4.02
ucbs: '0.234, 0.234, 0.234, 0.234, 0.234, 0.234, 0.234, 0.234, 0.234'

The results are quite curious. Indeed, the expected payoff seems to be a whopping 110% return! The variance of these results is quite high, and so it’s not at all impossible that the algorithm could have a negative return. But just as often it would return around 200% profit. 

Before we go risking all our money on this strategy, let’s take a closer look at what’s happening in the algorithm. It appears that for UCB1 the upper confidence bounds assigned to each action are the same! In other words, even after ten years of trials, no single stock “shined” above the others in the eyes of UCB1. It may seem that Exp3 has a leg up on UCB1 in this respect, because it’s clear that it gives higher weights to some stocks over others. However, running the algorithm multiple times shows drastically different weight distributions, and if we average the resulting weights over a thousand rounds, we see that they all have roughly the same mean and variance (the mean being first in the pair):

weight stats for msft: (0.107, 0.025)
weight stats for jpm: (0.109, 0.027)
weight stats for tgt: (0.110, 0.029)
weight stats for gs: (0.112, 0.025)
weight stats for wmt: (0.110, 0.027)
weight stats for aapl: (0.111, 0.027)
weight stats for amzn: (0.120, 0.029)
weight stats for cost: (0.113, 0.026)
weight stats for wfc: (0.107, 0.023)
Indeed, the best stock, Amazon, had an average weight just barely larger (and more variable) than any of the other stocks. So this evidence points to the conclusion that neither EXP3 nor UCB1 has any clue which stock is better. Pairing this with the fact that both algorithms nevertheless perform well would suggest that a random choice of action at each step is equally likely to do well. Indeed, when we run with a “random bandit” that just chooses actions uniformly at random, we get the following results:
(Expected payoff, variance) over 1000 trials is (1.1094227056931132, 0.4403783017367529)
For a single run: 
Payoff was 3.13
Regret was 0.90
Best stock was amzn at 4.02

It’s not quite as good as either Exp3 or UCB1, but it’s close and less variable, which means a lot to an investor. In other words, it’s starting to look like Exp3 and UCB1 aren’t doing significantly better than random at all, and that a monkey would do well in this system (for these particular stocks).

Of course, Fortune 500 companies are pretty successful by definition, so let’s turn our attention to the random stocks:

For the random bandit learner:

(Expected payoff, variance) over 1000 trials is (-0.23952295977625776, 1.0787311145181104)
For a single run: 
Payoff was -2.01
Regret was 3.92
Best stock was htr at 1.91

For UCB1:

(Expected payoff, variance) over 1000 trials is (-0.3503593899029112, 1.1136234992964154)
For a single run: 
Payoff was 0.26
Regret was 1.65
Best stock was htr at 1.91
ucbs: '0.315, 0.315, 0.315, 0.316, 0.315, 0.315, 0.315, 0.315, 0.316'

And for Exp3:

(Expected payoff, variance) over 1000 trials is (-0.25827976810345593, 1.2946101887058519)
For a single run: 
Payoff was -0.34
Regret was 2.25
Best stock was htr at 1.91
weights: '0.08, 0.00, 0.14, 0.06, 0.48, 0.00, 0.00, 0.04, 0.19'

But again Exp3 has no idea what stocks are actually best, with the average, variance over 1000 trials being:

weight stats for lxrx: '0.11, 0.02'
weight stats for keg: '0.11, 0.02'
weight stats for htr: '0.12, 0.02'
weight stats for cadx: '0.10, 0.02'
weight stats for belfb: '0.11, 0.02'
weight stats for tdi: '0.11, 0.02'
weight stats for cuba: '0.11, 0.02'
weight stats for mux: '0.11, 0.02'
weight stats for brks: '0.11, 0.02'

The long and short of it is that the choice of Fortune 500 stocks was inherently so biased toward success than a monkey could have made money investing in them, while the average choice of stocks had, if anything, a bias toward loss. And unfortunately using an algorithm like UCB1 or Exp3 straight out of the box doesn’t produce anything better than a monkey.

Issues and Improvements

There are two glaring theoretical issues here that we haven’t yet addressed. One of these goes back to issue #5 in that list we gave at the beginning of the post: the bandit algorithms are assuming they have less information than they actually have! Indeed, at the end of a day of stock trading, you have a good idea what would have happened to you had you bought a different stock, and in our simplified world you can know exactly what your profit would have been. Recalling that UCB1 and Exp3 both maintained some numbers representing the strength of an action (Exp3 had a “weight” and UCB1 an upper confidence bound), the natural extension to both UCB1 and Exp3 is simply to modify the beliefs about all actions after any given round. This is a pretty simple improvement to make in our implementation, since it just changes a single weight update to a loop. For Exp3:

for choice in range(numActions):
   rewardForUpdate = reward(choice, t)
   scaledReward = (rewardForUpdate - rewardMin) / (rewardMax - rewardMin)
   estimatedReward = 1.0 * scaledReward / probabilityDistribution[choice]
   weights[choice] *= math.exp(estimatedReward * gamma / numActions)

With a similar loop for UCB1. This code should be familiar from our previous posts on bandits. We then rerun the new algorithms on the same data sets, and the results are somewhat surprising. First, UCB1 on Fortune 500:

(Expected payoff, variance) over 1000 trials is (3.530670654982728, 0.007713190816014095)
For a single run: 
Payoff was 3.56
Regret was 0.47

This is clearly outperforming the random bandit learning algorithm, with an average return of 350%! In fact, it does almost as well as the best stock, and the variance is quite low. UCB1 also outperforms Exp3, which fares comparably to its pre-improved self. That is, it’s still not much better than random:

(Expected payoff, variance) over 1000 trials is (1.1424797906901956, 0.434335471375294)
For a single run: 
Payoff was 1.24
Regret was 2.79

And also for the random stocks, UCB1 with improvements outperforms Exp3 and UCB1 without improvements. UCB1:

(Expected payoff, variance) over 1000 trials is (0.680211923900068, 0.04226672915962647)
For a single run:
Payoff was 0.82
Regret was 1.09

And Exp3:

(Expected payoff, variance) over 1000 trials is (-0.2242152508929378, 1.1312843329929194)
For a single run: 
Payoff was -0.16
Regret was 2.07

We might wonder why this is the case, and there is a plausible explanation. See, Exp3 has a difficult life: it has to assume that at any time the adversary can completely change the game. And so Exp3 must remain vigilant, continuing to try options it knows to be terrible for fear that they may spontaneously do well. Exp3 is the grandfather who, after 75 years of not winning the lotto, continues to buy tickets every week. A better analogy might be a lioness who, even after being moved to the zoo, stays up all night to protect a cub from predators. This gives us quite a new perspective on Exp3: the world really has to be that messed up for Exp3 to be useful. As we saw, UCB1 is much more eager to jump on a winning bandwagon, and it paid off in both the good (Fortune 500) and bad (random stock) scenarios. All in all, this experiment would provide some minor evidence that the stock market (or just this cheesy version of it) is more stochastic than adversarial.

The second problem is that we’re treating these stocks as if they were isolated from the rest of the world. Indeed, along with each stock comes some kind of context in the form of information about that stock. Historical prices, corporate announcements, cyclic boom and bust, what the talking heads think, all of this may be relevant to the price fluctuations of a stock on any given day. While Exp3 and UCB1 are ill-equipped to handle such a rich landscape, researchers in bandit learning have recognized the importance of context in decision making. So much so, in fact, that an entire subfield of “Contextual Bandits” was born. John Langford, perhaps the world’s leading expert on bandit learning, wrote on his blog in 2007,

I’m having difficulty finding interesting real-world k-Armed Bandit settings which aren’t better thought of as Contextual Bandits in practice. For myself, bandit algorithms are (at best) motivational because they can not be applied to real-world problems without altering them to take context into account.

I tend to agree with him. Bandit problems almost always come with some inherent additional structure in the real world, and the best algorithms will always take advantage of that structure. A “context” associated with each round is perhaps the weakest kind of structure, so it’s a natural place to look for better algorithms.

So that’s what we’ll do in the future of this series. But before then we might decide to come up with another experiment to run Exp3 and UCB1 on. It would be nice to see an instance in which Exp3 seriously outperforms UCB1, but maybe the real world is just stochastic and there’s nothing we can do about it.

Until next time!

Adversarial Bandits and the Exp3 Algorithm

In the last twenty years there has been a lot of research in a subfield of machine learning called Bandit Learning. The name comes from the problem of being faced with a large sequence of slot machines (once called one-armed bandits) each with a potentially different payout scheme. The problems in this field all focus on one central question:

If I have many available actions with uncertain outcomes, how should I act to maximize the quality of my results over many trials?

The deep question here is how to balance exploitation, the desire to choose an action which has payed off well in the past, with exploration, the desire to try options which may produce even better results. The ideas are general enough that it’s hard not to find applications: choosing which drug to test in a clinical study, choosing which companies to invest in, choosing which ads or news stories to display to users, and even (as Richard Feynman once wondered) how to maximize your dining enjoyment.

Herbet Robbins, one of the first to study bandit learning algorithms.

Herbert Robbins, one of the first to study bandit learning algorithms. Image credit

In less recent times (circa 1960’s), this problem was posed and considered in the case where the payoff mechanisms had a very simple structure: each slot machine is a coin flip with a different probability $ p$ of winning, and the player’s goal is to find the best machine as quickly as possible. We called this the “stochastic” setting, and last time we saw a modern strategy called UCB1 which maintained statistical estimates on the payoffs of the actions and chose the action with the highest estimate. The underlying philosophy was “optimism in the face of uncertainty,” and it gave us something provably close to optimal.

Unfortunately payoff structures are more complex than coin flips in the real world. Having “optimism” is arguably naive, especially when it comes to competitive scenarios like stock trading. Indeed the algorithm we’ll analyze in this post will take the polar opposite stance, that payoffs could conceivably operate in any manner. This is called the adversarial model, because even though the payoffs are fixed in advance of the game beginning, it can always be the case that the next choice you make results in the worst possible payoff.

One might wonder how we can hope to do anything in such a pessimistic model. As we’ll see, our notion of performing well is relative to the best single slot machine, and we will argue that this is the only reasonable notion of success. On the other hand, one might argue that real world payoffs are almost never entirely adversarial, and so we would hope that algorithms which do well theoretically in the adversarial model excel beyond their minimal guarantees in practice.

In this post we’ll explore and implement one algorithm for adversarial bandit learning, called Exp3, and in the next post we’ll see how it fares against UCB1 in some applications. Some prerequisites: since the main algorithm presented in this post is randomized, its analysis requires some familiarity with techniques and notation from probability theory. Specifically, we will assume that the reader is familiar with the content of this blog’s basic probability theory primers (1, 2), though the real difficulty in the analysis will be keeping up with all of the notation.

In case the reader is curious, Exp3 was invented in 2001 by Auer, Cesa-Bianchi, Freund, and Schapire. Here is their original paper, which contains lots of other mathematical goodies.

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

Model Formalization and Notions of Regret

Before we describe the algorithm and analyze its we have to set up the problem formally. The first few paragraphs of our last post give a high-level picture of general bandit learning, so we won’t repeat that here. Recall, however, that we have to describe both the structure of the payoffs and how success is measured. So let’s describe the former first.

Definition: An adversarial bandit problem is a pair $ (K, \mathbf{x})$, where $ K$ represents the number of actions (henceforth indexed by $ i$), and $ \mathbf{x}$ is an infinite sequence of payoff vectors $ \mathbf{x} = \mathbf{x}(1), \mathbf{x}(2), \dots$, where $ \mathbf{x}(t) = (x_1(t), \dots, x_K(t))$ is a vector of length $ K$ and $ x_i(t) \in [0,1]$ is the reward of action $ i$ on step $ t$.

In English, the game is played in rounds (or “time steps”) indexed by $ t = 1, 2, \dots$, and the payoffs are fixed for each action and time before the game even starts. Note that we assume the reward of an action is a number in the interval $ [0,1]$, but all of our arguments in this post can be extended to payoffs in some range $ [a,b]$ by shifting by $ a$ and dividing by $ b-a$.

Let’s specify what the player (algorithm designer) knows during the course of the game. First, the value of $ K$ is given, and total number of rounds is kept secret. In each round, the player has access to the history of rewards for the actions that were chosen by the algorithm in previous rounds, but not the rewards of unchosen actions. In other words, it will only ever know one $ x_i(t)$ for each $ t$. To set up some notation, if we call $ i_1, \dots, i_t$ the list of chosen actions over $ t$ rounds, then at step $ t+1$ the player has access to the values of $ x_{i_1}(1), \dots, x_{i_t}(t)$ and must pick $ i_{t+1}$ to continue.

So to be completely clear, the game progresses as follows:

The player is given access to $ K$.
For each time step $ t$:

The player must pick an action $ i_t$.
The player observes the reward $ x_i(t) \in [0,1]$, which he may save for future use.

The problem gives no explicit limit on the amount of computation performed during each step, but in general we want it to run in polynomial time and not depend on the round number $ t$. If runtime even logarithmically depended on $ t$, then we’d have a big problem using it for high-frequency applications. For example in ad serving, Google processes on the order of $ 10^9$ ads per day; so a logarithmic dependence wouldn’t be that bad, but at some point in the distance future Google wouldn’t be able to keep up (and we all want long-term solutions to our problems).

Note that the reward vectors $ \mathbf{x}_t$ must be fixed in advance of the algorithm running, but this still allows a lot of counterintuitive things. For example, the payoffs can depend adversarially on the algorithm the player decides to use. For example, if the player chooses the stupid strategy of always picking the first action, then the adversary can just make that the worst possible action to choose. However, the rewards cannot depend on the random choices made by the player during the game.

So now let’s talk about measuring success. For an algorithm $ A$ which chooses the sequence $ i_1, \dots, i_t$ of actions, define $ G_A(t)$ to be the sum of the observed rewards

$ \displaystyle G_A(t) = \sum_{s=1}^t x_{i_s}(s)$.

And because $ A$ will often be randomized, this value is a random variable depending on the decisions made by $ A$. As such, we will often only consider the payoff up to expectation. That is, we’ll be interested in how $ \textup{E}(G_A(t))$ relates to other possible courses of action. To be completely rigorous, the randomization is not over “choices made by an algorithm,” but rather the probability distribution over sequences of actions that the algorithm induces. It’s a fine distinction but a necessary one. In other words, we could define any sequence of actions $ \mathbf{j} = (j_1, \dots, j_t)$ and define $ G_{\mathbf{j}}(t)$ analogously as above:

$ \displaystyle G_{\mathbf{j}}(t) = \sum_{s=1}^t x_{j_s}(s)$.

Any algorithm and choice of reward vectors induces a probability distribution over sequences of actions in a natural way (if you want to draw from the distribution, just run the algorithm). So instead of conditioning our probabilities and expectations on previous choices made by the algorithm, we do it over histories of actions $ i_1, \dots, i_t$.

An obvious question we might ask is: why can’t the adversary just make all the payoffs zero? (or negative!) In this event the player won’t get any reward, but he can emotionally and psychologically accept this fate. If he never stood a chance to get any reward in the first place, why should he feel bad about the inevitable result? What a truly cruel adversary wants is, at the end of the game, to show the player what he could have won, and have it far exceed what he actually won. In this way the player feels regret for not using a more sensible strategy, and likely turns to binge eating cookie dough ice cream. Or more likely he returns to the casino to lose more money. The trick that the player has up his sleeve is precisely the randomness in his choice of actions, and he can use its objectivity to partially overcome even the nastiest of adversaries.

The adversary would love to show you this bluff after you choose to fold your hand. What a jerk.

The adversary would love to show you this bluff after you choose to fold your hand. What a jerk. Image credit

Sadism aside, this thought brings us to a few mathematical notions of regret that the player algorithm may seek to minimize. The first, most obvious, and least reasonable is the worst-case regret. Given a stopping time $ T$ and a sequence of actions $ \mathbf{j} = (j_1, \dots, j_T)$, the expected regret of algorithm $ A$ with respect to $ \mathbf{j}$ is the difference $ G_{\mathbf{j}}(T) – \mathbb{E}(G_A(T))$. This notion of regret measures the regret of a player if he knew what would have happened had he played $ \mathbf{j}$.  The expected worst-case regret of $ A$ is then the maximum over all sequences $ \mathbf{j}$ of the regret of $ A$ with respect to $ \mathbf{j}$. This notion of regret seems particularly unruly, especially considering that the payoffs are adversarial, but there are techniques to reason about it.

However, the focus of this post is on a slightly easier notion of regret, called weak regret, which instead compares the results of $ A$ to the best single action over all rounds. That is, this quantity is just 

$ \displaystyle \left ( \max_{j} \sum_{t=1}^T x_j(t) \right ) – \mathbb{E}(G_A(T))$

We call the parenthetical term $ G_{\textup{max}}(T)$. This kind of regret is a bit easier to analyze, and the main theorem of this post will given an upper bound on it for Exp3. The reader who read our last post on UCB1 will wonder why we make a big distinction here just to arrive at the same definition of regret that we had in the stochastic setting. But with UCB1 the best sequence of actions to take just happened to be to play the best action over and over again. Here, the payoff difference between the best sequence of actions and the best single action can be arbitrarily large.

Exp3 and an Upper Bound on Weak Regret

We now describe at the Exp3 algorithm.

Exp3 stands for Exponential-weight algorithm for Exploration and Exploitation. It works by maintaining a list of weights for each of the actions, using these weights to decide randomly which action to take next, and increasing (decreasing) the relevant weights when a payoff is good (bad). We further introduce an egalitarianism factor $ \gamma \in [0,1]$ which tunes the desire to pick an action uniformly at random. That is, if $ \gamma = 1$, the weights have no effect on the choices at any step.

The algorithm is readily described in Python code, but we need to set up some notation used in the proof of the theorem. The pseudocode for the algorithm is as follows.


  1. Given $ \gamma \in [0,1]$, initialize the weights $ w_i(1) = 1$ for $ i = 1, \dots, K$.
  2. In each round $ t$:
    1.  Set $ \displaystyle p_i(t) = (1-\gamma)\frac{w_i(t)}{\sum_{j=1}^K w_j(t)} + \frac{\gamma}{K}$ for each $ i$.
    2. Draw the next action $ i_t$ randomly according to the distribution of $ p_i(t)$.
    3. Observe reward $ x_{i_t}(t)$.
    4. Define the estimated reward $ \hat{x}_{i_t}(t)$ to be $ x_{i_t}(t) / p_{i_t}(t)$.
    5. Set $ \displaystyle w_{i_t}(t+1) = w_{i_t}(t) e^{\gamma \hat{x}_{i_t}(t) / K}$
    6. Set all other $ w_j(t+1) = w_j(t)$.

The choices of these particular mathematical quantities (in steps 1, 4, and 5) are a priori mysterious, but we will explain them momentarily. In the proof that follows, we will extend $ \hat{x}_{i_t}(t)$ to indices other than $ i_t$ and define those values to be zero.

The Python implementation is perhaps more legible, and implements the possibly infinite loop as a generator:

def exp3(numActions, reward, gamma):
   weights = [1.0] * numActions

   t = 0
   while True:
      probabilityDistribution = distr(weights, gamma)
      choice = draw(probabilityDistribution)
      theReward = reward(choice, t)

      estimatedReward = 1.0 * theReward / probabilityDistribution[choice]
      weights[choice] *= math.exp(estimatedReward * gamma / numActions) # important that we use estimated reward here!

      yield choice, theReward, estimatedReward, weights
      t = t + 1

Here the “rewards” variable refers to a callable which accepts as input the action chosen in round $ t$ (keeps track of $ t$, assuming we’ll play nice), and returns as output the reward for that choice. The distr and draw functions are also easily defined, with the former depending on the gamma parameter as follows:

def distr(weights, gamma=0.0):
    theSum = float(sum(weights))
    return tuple((1.0 - gamma) * (w / theSum) + (gamma / len(weights)) for w in weights)

There is one odd part of the algorithm above, and that’s the “estimated reward” $ \hat{x}_{i_t}(t) = x_{i_t}(t) / p_{i_t}(t)$. The intuitive reason to do this is to compensate for a potentially small probability of getting the observed reward. More formally, it ensures that the conditional expectation of the “estimated reward” is the actual reward. We will explore this formally during the proof of the main theorem.

As usual, the programs we write in this post are available on this blog’s Github page.

We can now state and prove the upper bound on the weak regret of Exp3. Note all logarithms are base $ e$.

Theorem: For any $ K > 0, \gamma \in (0, 1]$, and any stopping time $ T \in \mathbb{N}$

$ \displaystyle G_{\textup{max}}(T) – \mathbb{E}(G_{\textup{Exp3}}(T)) \leq (e-1)\gamma G_{\textup{max}}(T) + \frac{K \log K}{\gamma}$.

This is a purely analytical result because we don’t actually know what $ G_{\textup{max}}(T)$ is ahead of time. Also note how the factor of $ \gamma$ occurs: in the first term, having a large $ \gamma$ will result in a poor upper bound because it occurs in the numerator of that term: too much exploration means not enough exploitation. But it occurs in the denominator of the second term, meaning that not enough exploration can also produce an undesirably large regret. This theorem then provides a quantification of the tradeoff being made, although it is just an upper bound.


We present the proof in two parts. Part 1:

We made a notable mistake in part 1, claiming that $ e^x \leq 1 + x + (e-2)x^2$ when $ x \leq 1$. In fact, this does follow from the Taylor series expansion of $ e$, but it’s not as straightforward as I made it sound. In particular, note that $ e^x = 1 + x + \frac{x^2}{2!} + \dots$, and so $ e^1 = 2 + \sum_{k=2}^\infty \frac{1}{k!}$. Using $ (e-2)$ in place of $ \frac{1}{2}$ gives

$ \displaystyle 1 + x + \left ( \sum_{k=2}^{\infty} \frac{x^2}{k!} \right )$

And since $ 0 < x \leq 1$, each term in the sum will decrease when replaced by $ \frac{x^k}{k!}$, and we’ll be left with exactly $ e^x$. In other words, this is the tightest possible quadratic upper bound on $ e^x$. Pretty neat! On to part 2:

As usual, here is the entire canvas made over the course of both videos.

$ \square$

We can get a version of this theorem that is easier to analyze by picking a suitable choice of $ \gamma$.

Corollary: Assume that $ G_{\textup{max}}(T)$ is bounded by $ g$, and that Exp3 is run with

$ \displaystyle \gamma = \min \left ( 1, \sqrt{\frac{K \log K}{(e-1)g}} \right )$

Then the weak regret of Exp3 is bounded by $ 2.63 \sqrt{g K \log K}$ for any reward vector $ \mathbf{x}$.

Proof. Simply plug $ \gamma$ in the bound in the theorem above, and note that $ 2 \sqrt{e-1} < 2.63$.

A Simple Test Against Coin Flips

Now that we’ve analyzed the theoretical guarantees of the Exp3 algorithm, let’s use our implementation above and see how it fares in practice. Our first test will use 10 coin flips (Bernoulli trials) for our actions, with the probabilities of winning (and the actual payoff vectors) defined as follows:

biases = [1.0 / k for k in range(2,12)]
rewardVector = [[1 if random.random() &lt; bias else 0 for bias in biases] for _ in range(numRounds)]
rewards = lambda choice, t: rewardVector[t][choice]

If we are to analyze the regret of Exp3 against the best action, we must compute the payoffs for all actions ahead of time, and compute which is the best. Obviously it will be the one with the largest probability of winning (the first in the list generated above), but it might not be, so we have to compute it. Specifically, it’s the following argmax:

bestAction = max(range(numActions), key=lambda action: sum([rewardVector[t][action] for t in range(numRounds)]))

Where the max function is used as “argmax” would be in mathematics.

We also have to pick a good choice of $ \gamma$, and the corollary from the previous section gives us a good guide to the optimal $ \gamma$: simply find a good upper bound on the reward of the best action, and use that. We can cheat a little here: we know the best action has a probability of 1/2 of paying out, and so the expected reward if we always did the best action is half the number of rounds. If we use, say, $ g = 2T / 3$ and compute $ \gamma$ using the formula from the corollary, this will give us a reasonable (but perhaps not perfectly correct) upper bound.

Then we just run the exp3 generator for $ T = \textup{10,000}$ rounds, and compute some statistics as we go:

bestUpperBoundEstimate = 2 * numRounds / 3
gamma = math.sqrt(numActions * math.log(numActions) / ((math.e - 1) * bestUpperBoundEstimate))
gamma = 0.07

cumulativeReward = 0
bestActionCumulativeReward = 0
weakRegret = 0

t = 0
for (choice, reward, est, weights) in exp3(numActions, rewards, gamma):
   cumulativeReward += reward
   bestActionCumulativeReward += rewardVector[t][bestAction]

   weakRegret = (bestActionCumulativeReward - cumulativeReward)
   regretBound = (math.e - 1) * gamma * bestActionCumulativeReward + (numActions * math.log(numActions)) / gamma

   t += 1
   if t &gt;= numRounds:

At the end of one run of ten thousand rounds, the weights are overwhelmingly in favor of the best arm. The cumulative regret is 723, compared to the theoretical upper bound of 897. It’s not too shabby, but by tinkering with the value of $ \gamma$ we see that we can get regrets lower than 500 (when $ \gamma$ is around 0.7). Considering that the cumulative reward for the player is around 4,500 in this experiment, that means we spent only about 500 rounds out of ten thousand exploring non-optimal options (and also getting unlucky during said exploration). Not too shabby at all.

Here is a graph of a run of this experiment.


A run of Exp3 against Bernoulli rewards. The first graph represents the simple regret of the player algorithm against the best action; the blue line is the actual simple regret, and the green line is the theoretical O(sqrt(k log k)) upper bound. The second graph shows the weights of each action evolving over time. The blue line is the weight of the best action, while the green and red lines are the weights of the second and third best actions.

Note how the Exp3 algorithm never stops increasing its regret. This is in part because of the adversarial model; even if Exp3 finds the absolutely perfect action to take, it just can’t get over the fact that the world might try to screw it over. As long as the $ \gamma$ parameter is greater than zero, Exp3 will explore bad options just in case they turn out to be good. The benefits of this is that if the model changes over time Exp3 will adapt, but the downside is that the pessimism inherent in this worldview generally results in lower payoffs than other algorithms.

More Variations, and Future Plans

Right now we have two contesting models of how the world works: is it stochastic and independent, like the UCB1 algorithm would optimize for? Or does it follow Exp3’s world view that the payoffs are adversarial? Next time we’ll run some real-world tests to see how each fares.

But before that, we should note that there are still more models we haven’t discussed. One extremely significant model is that of contextual bandits. That is, the real world settings we care about often come with some “context” associated with each trial. Ads being displayed to users have probabilities that should take into account the information known about the user, and medical treatments should take into account past medical history. While we will not likely investigate any contextual bandit algorithms on this blog in the near future, the reader who hopes to apply this work to his or her own exploits (no pun intended) should be aware of the additional reading.

Until next time!

Postscript: years later, a cool post by Tim Vieira shows a neat data structure that asymptotically speeds up the update/sample step of the EXP3 algorithm from linear to logarithmic (among others). The weights are stored in a heap of partial sums (the leaves are the individual weights), and sampling is a binary search. See the original post and the accompanying gist for an implementation. Exercise: implement the data structure for use with our EXP3 implementation.

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.

Exploit away, you lucky ladies.

Exploit away, you lucky ladies.

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.


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.

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}$


$ \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!