In most bandit problems there is likely to be some additional information available at the beginning of rounds and often this information can potentially help with the action choices. For example, in a web article recommendation system, where the goal is to keep the visitors engaged with the website, *contextual information* about the visitor of the website, the time of day, information on what is trendy, etc., can likely improve the choice of the article to be put on the “front-page”. For example, a science-oriented article is more likely to grab the attention of a science geek, and a baseball fan may care little about European soccer.

If we used a standard bandit algorithm (like Exp3, Exp3-IX, or UCB), the one-size fits all approach implicitly taken by these algorithms which aim just finding the single most catching article is likely to disappoint an unnecessarily large portion of the site’s visitors. In situations like this, since the benchmark that the bandit algorithms aim to approach performs poorly by omitting available information, it is better to change the problem and redefine the benchmark! It is important to realize though that there is an inherent difficulty here:

Competing with a poor benchmark does not make sense since even an algorithm that perfectly matches the benchmark will perform poorly. At the same time, competing with a better benchmark can be harder from a learning point of view and in a specific scenario the gain from a better benchmark may very well be offset by the fact that algorithms that compete with stronger benchmark have to search in a larger space of possibilities.

The tradeoff just described is fundamental to all machine learning problems. In statistical estimation, the analogue tradeoff is known as the *bias-variance tradeoff*.

We will not attempt to answer the question of how to resolve this tradeoff in this post because first we need to see ways ways effectively competing with improved benchmarks. First, let’s talk about possible improvements to the benchmarks.

# Contextual bandits: One bandit per context

In a contextual bandit problem everything works the same as in a bandit problem, except that the learner receives a context at the beginning of the round, before it needs to select its action. The promise, as discussed, is that perhaps specializing the action taken to the context can help to collect more reward.

Assuming that the set $\cC$ of all possible contexts is finite, one simple approach then is to set up a separate bandit algorithm, such as Exp3, for each context. Indeed, if we do this then the collection of bandit algorithms should be able to compete with the best context-dependent action. In particular, the best total reward that we can achieve in $n$ rounds if we are allowed to adjust the action to the context is

\begin{align*}

S_n = \sum_{c\in \cC} \max_{k\in [K]} \sum_{t: c_t=c} x_{t,k}\,,

\end{align*}

where $c_t\in \cC$ is the context received at the beginning of round $t$. For future reference note that we can also write

\begin{align}

S_n = \max_{\phi: \cC \to [K]} \sum_{t=1}^n x_{t,\phi(c_t)}\,.

\label{eq:maxrewardunrestricted}

\end{align}

Then, the regret of a learner who incurs a reward $X_t$ in round $t$ is $R_n=S_n – \sum_t X_t$, which satisfies

\begin{align*}

R_n = \sum_{c\in \cC} \EE{\max_{k\in [K]} \sum_{t: c_t=c} (x_{t,k}-X_t)}\,.

\end{align*}

That is, $R_n$ is just the sum of the regrets suffered by the bandits assigned to the individual contexts. Let $T^c(n)$ be the number of times context $c\in \cC$ is seen during the first $n$ rounds and let $R^c(s)$ be the regret of the instance of Exp3 associated with $c$ at the end of the round when this instance is used $s$ times. With this notation we can thus write

\begin{align*}

R_n = \sum_{c\in \cC} \EE{ R^c(T^c(n)) }\,.

\end{align*}

Since $T^c(n)$ may vary from context to context and the distribution of $T^c(n)$ may be uneven, it would be wasteful to use a version of Exp3 that is tuned to achieve a small regret after a fixed number of rounds as such a version of Exp3 may suffer a larger regret when the actual number of rounds $T^c(n)$ is vastly different from the anticipated horizon. Luckily, the single parameter $\eta$ of Exp3 can be chosen in a way that does not depend on the anticipated horizon without losing much on the regret. In particular, as we hinted on this before, such an *anytime* version of Exp3 can be created if we let the $\eta$ parameter of Exp3 depend on the round index. In particular, when an Exp3 instance is used the $s$th time, if we set $\eta$ to $\eta_s = \sqrt{\log(K)/(sK)}$, then one can show that $R^c(s) \le 2\sqrt{sK \log(K)}$ will hold for any $s\le 1$.

Plugging this into the previous display, we get

\begin{align}

R_n \le 2 \sqrt{K \log(K)} \, \sum_{c\in \cC} \sqrt{ T_c(n) }\,.

\label{eq:exp3perc}

\end{align}

How big the right-hand side is depends on how skewed the context distribution $(T_c(n)/n)_{c\in \cC}$ is.

The best case is when only one context is seen, in which case $\sum_{c\in \cC} \sqrt{ T_c(n) }=\sqrt{n}$. Note that in this case the benchmark won’t improve either, but nevertheless the algorithm that keeps one Exp3 instance for each context does not lose anything compared to the algorithm that would run a single Exp3 instance, trying to compete with the single best action. This is good. The worst-case is when all context are seen equally often, in which case case $T_c(n) = n/|\cC|$ (assume for simplicity that this is an integer). In this case, $\eqref{eq:exp3perc}$ becomes

\begin{align}

R_n \le 2 \sqrt{ K \log(K) |\cC| n }\,.

\label{eq:exp3perc2}

\end{align}

It is instructive to consider what this means for the total reward:

\begin{align*}

\EE{\sum_{t=1}^n X_t} \ge S_n – 2 \sqrt{ K \log(K) |\cC| n }\,.

\end{align*}

While the benchmark $S_n$ may have improved, the second term will always exceed the first one for the first $4 K \log(K) |\cC|$ rounds! Thus, the guarantee on the total reward will be vacuous for all earlier time steps! When $|\cC|$ is large, we conclude that for a long time, the per-instance Exp3 algorithm may have a much worse total reward than if we just ran a single instance of Exp3, demonstrating the tradeoff mentioned above.

# Bandits with expert advice

For large context sets, using one bandit algorithm per context will almost always be a poor choice because the additional precision is wasted unless the amount of data is enormous. Very often the contexts themselves have some kind of internal structure that we may hope to exploit. There are many different kinds of structure. For example, we expect that a person who is deeply into computer science may share common interests with people who are deeply into math, and people who are into sport acrobatics may enjoy gymnastics and vice versa. This gives the idea to group the contexts in some way, to reduce their number, and then assign a bandit to the individual groups. Assume that we chose a suitable partitioning of contexts, which we denote by $\cP \subset 2^{\cC}$. Thus, the elements of $\cP$ are disjoint subsets of $\cC$ such that jointly they cover $\cC$: $\cup \cP = \cC$. Then, the maximum total reward that can be achieved if for every element $P$ of the partitioning $\cP$ we can select a single action is

\begin{align*}

S_n = \sum_{P\in \cP} \max_{k\in [K]} \sum_{t: c_t\in P} x_{t,k}\,.

\end{align*}

It may be worthwhile to put this into an alternate form. Defining $\Phi(\cP) = \{\phi:\cC\to[K]\,;\, \forall c,c’\in \cC \text{ s.t. } c,c’\in P \text{ for some } P\in \cP, \phi(c) = \phi(c’)\}$ to be the set of functions the map contexts in the same partition to the same action (the astute reader may recognize $\Phi(\cP)$ as the set of all $\sigma(\cP)$-measurable functions from $\cC$ to $[K]$), we can rewrite $S_n$ as

\begin{align*}

S_n = \max_{\phi\in \Phi(\cP)} \sum_{t=1}^n x_{t,\phi(c_t)}\,.

\end{align*}

Compared to $\eqref{eq:maxrewardunrestricted}$ we see that what changed is that the set of functions that we are maximizing over became smaller. The reader may readily verify that the regret of the composite algorithm where an Exp3 with a varying parameter sequence as described above is used for each element of $\cP$ is exactly of the form $\eqref{eq:exp3perc}$ except that $c\in \cC$ has to be changed $P\in \cP$ and the definition of $T_c$ also needs to be adjusted accordingly.

Of course, $\Phi(\cP)$ is not the only set of functions that come to mind. We may consider of course different partitions. Or we may bring in extra structure of $\cC$. Why not, for example, use a similarity function of $\cC$ and then consider all functions which tend to assign identical actions to contexts that are more similar. For example, if $s:\cC\times \cC \to [0,1]$ is a *similarity function*, we may consider all functions $\phi: \cC\to[K]$ such that the average dissimilarity,

\begin{align*}

\frac{1}{|\cC|^2} \sum_{c,d\in \cC} (1-s(c,d)) \one{ \phi(c)\ne \phi(d) },

\end{align*}

is below a user-tuned threshold $\theta\in (0,1)$.

Another options is to run your favorite supervised learning method, training on some batch data to find a few predictors $\phi_1,\dots,\phi_M: \cC \to [K]$ (in machine learning terms, these would be called classifiers since the range space is finite). Then we could use a bandit algorithm to compete with the “best” of these in an online fashion. This has the advantage that the offline training procedure can bring in the power of batch data and the whole army of supervised learning, without relying on potentially inaccurate evaluation methods that aim to pick the best of the pack. And why pick if one does not need to?

The possibilities are endless, but in any case, we would end up with a set of functions $\Phi$ with the goal of competing with the best of them. This gives the idea that perhaps we should think more generally about some subset $\Phi$ of functions without necessarily considering the internal structure of $\Phi$. This is the viewpoint that we will take. In fact, we will bring this one or two steps further, leading to what is called *bandits with expert advice*.

In this model, there are $M$ experts that we wish to compete with. At the beginning of each round, the experts announce their predictions of which actions are the most promising. For the sake of generality, we allow the experts to report not only a single prediction, but a probability distribution over the actions. The interpretation of this probability distribution is that the expert, if the decision was left to it, would choose the action for the round at random from the probability distribution that it reported. As discussed before, in an adversarial setting, it is natural to consider randomized algorithms, hence one should not be too surprised that the experts are also allowed to randomize. In any case, this can only increase generality. For reasons that will become clear later, it will be useful to collect the advice of the $M$ experts for round $t$ into an $M\times K$ matrix $E^{(t)}$ such that the $m$th row of $E^{(t)}$, $E_m^{(t)}$ is the probability distribution that expert $m$ recommends for round $t$. Denote by $E_{mi}^{(t)}$ the $i$th entry of the row vector $E_m^{(t)}$, we thus have $\sum_i E_{mi}^{(t)}=1$ and $E_{mi}^{(t)}\ge 0$ for every $(m,t,i)\in [M]\times \N \times [K]$. After receiving $E^{(t)}$, the learner then chooses $A_t\in [K]$ and as before observes the reward $X_t = x_{t,A_t}$, where $x_t=(x_{ti})_i$ is the $K$-dimensional vector of rewards of the individual actions. For a real-world application, see the figure below.

The regret of the learner is with respect to the total expected reward of the best *expert*:

\begin{align}

R_n = \EE{ \max_m \sum_{t=1}^n E_{m}^{(t)} x_{t} – \sum_{t=1}^n X_t }\,.

\label{eq:expertregret}

\end{align}

Below, we may allow the expert advice $E^{(t)}$ of round $t$ to depend on all the information up the beginning of round $t$. While this does allow “learning” experts, the regret definition above is not really meaningful if the experts would learn from the feedback $(A_t,X_t)$. For dealing with learning experts, it is more appropriate to measure regret as done in *reinforcement learning* where an agent controls the state of the environment, but the agent’s reward is compared to the best total reward that any other (simple) policy would incur, regardless of the “trajectory” of the agent. We will discuss reinforcement learning in some later post.

# Can it go higher? Exp4

Exp4 is actually not just an increased version number, but it stands for **E**xponential weighting for **E**xploration and **E**xplotation with **E**xperts. The idea of the algorithm is very simple: Since exponential weighting worked so well in the standard bandit problem, we should adopt it to the problem at hand. However, since now the goal is to compete with the best expert in hindsight, it is not the actions that we should score, but the experts. Hence, the algorithm will keep a probability distribution, which we will denote by $Q_t$, over the experts and use this to come up with the next action. Once the action is chosen, we can use our favorite reward estimation procedure to estimate the rewards for all the actions, which can then be used to estimate how much total reward the individual experts would have made so far, which in the end can be used to update $Q_t$.

Formally, the procedure is as follows: First, the distribution $Q_1$ is initialized to the uniform distribution $(1/M,\dots,1/M)$ (the $Q_t$ are treated as row vectors). Then, some values of $\eta,\gamma\ge 0$ are selected.

In round $t=1,2,\dots$, the following things happen:

- The advice $E^{(t)}$ is received
- Choose the action $A_t\sim P_{t,\cdot}$ at random, where $P_t = Q_t E^{(t)}$
- The reward $X_t = x_{t,A_t}$ is received
- The rewards of all the actions are estimated; say: $\hat X_{ti} = 1- \frac{\one{A_t=i}}{P_{ti}+\gamma} (1-X_t)$
- Propagate the rewards to the experts: $\tilde X_t = E^{(t)} \hat X_t$
- The distribution $Q_t$ is updated using exponential weighting:

$Q_{t+1,i} = \frac{\exp( \eta \tilde X_{ti} ) Q_{ti}}{\sum_j \exp( \eta \tilde X_{tj}) Q_{tj} }$, $i\in [M]$

Note that $A_t$ can be chosen in two steps, first sampling $M_t$ from $Q_{t}$ and then choosing $A_t$ from $E^{(t)}_{M_t,\cdot}$. The reader can verify that (given the past) the probability distribution of the so-selected action is also $P_{t}$.

# A bound on the expected regret of Exp4

The algorithm when we set $\gamma=0$ is actually what is most commonly known as Exp4 and the algorithm when $\gamma>0$ is the “IX” version of Exp4. As in the case of Exp3, setting $\gamma>0$ helps concentrating the regret. Here, for the sake of brevity we only consider the expected regret of Exp4; the analysis of the tail properties of Exp4-IX with appropriately tuned $\eta,\gamma$ is left as an exercise for the reader.

To bound the expected regret of Exp4, we apply the analysis of Exp3. In particular, the following lemma can be extracted from our earlier analysis of Exp3:

Lemma (regret of exponential weighting):Let $(\hat X_{ti})_{ti}$ and $(P_{ti})_{ti}$ satisfy for all $t\in [n]$ and $i\in [K]$ the relations $\hat X_{ti}\le 1$ and

\begin{align*}

P_{ti} = \frac{\exp( \eta \sum_{s=1}^t \hat X_{ti} )}{\sum_j\exp( \eta \sum_{s=1}^t \hat X_{tj} )}\,.

\end{align*}

Then, for any $i\in [K]$,

\begin{align*}

\sum_{t=1}^n \hat X_{ti} – \sum_{t=1}^n \sum_{j=1}^K P_{tj} \hat X_{tj} \le \frac{\log(M)}{\eta} + \frac{\eta}{2} \sum_{t,j} P_{tj} (1-\hat X_{tj})^2\,.

\end{align*}

Based on this lemma, we immediately get that for any $m\in [M]$,

\begin{align*}

\sum_{t=1}^n \tilde X_{tm} – \sum_{t=1}^n \sum_{m’} Q_{t,m’} \tilde X_{tm’} \le \frac{\log(M)}{\eta}

+ \frac{\eta}{2}\, \sum_{t,m’} Q_{t,m’} (1-\tilde X_{tm’})^2\,.

\end{align*}

Note that by our earlier argument $\EE{\hat X_{ti}} = x_{ti}$ and hence the expected value of the left-hand of the above display is the regret $R_n$ defined in $\eqref{eq:expertregret}$. Hence, to derive a regret bound it remains to bound the expectation of the right-hand side. For this, as before, we will find it useful to introduce the loss estimates $\hat Y_{ti} = 1-\hat X_{ti}$, the losses, $y_{ti} = 1-x_{ti}$, and also the loss estimates of experts: $\tilde Y_{tm} = 1-\tilde X_{tm}$, $m\in [M]$. Note that $\tilde Y_t = E^{(t)} \hat Y_t$, thanks to $E^{(t)} \mathbf{1} = \mathbf{1}$, where $\mathbf{1}$ is the vector whose components are all equal to one. Defining $A_{ti} = \one{A_t=i}$, we have $\hat Y_{ti} = \frac{A_{ti}}{P_{ti}} y_{ti}$.

Hence, using $\E_{t-1}[\cdot]$ to denote expectation conditioned on the past $(E^{(1)},A_1,X_1,\dots,E^{(t-1)},A_{t-1},X_{t-1},E^{(t)})$, we have

\begin{align*}

\E_{t-1}[ \tilde Y_{tm}^2 ] =\E_{t-1}\left[ \left(\frac{E^{(t)}_{m,A_t} y_{t,A_t}}{P_{t,A_t}}\right)^2\right]

= \sum_i \frac{(E^{(t)}_{m,i})^2 y_{t,i}^2}{P_{ti}} \le \sum_i \frac{(E^{(t)}_{m,i})^2 }{P_{ti}}\,.

\end{align*}

Therefore, \(\require{cancel}\)

\begin{align*}

\E_{t-1}\left[ \sum_m Q_{tm} \tilde Y_{tm}^2 \right]

&\le \sum_m Q_{tm} \sum_i \frac{(E^{(t)}_{m,i})^2 }{P_{ti}}

\le \sum_{i} \left(\max_{m’} E_{m’,i}^{(t)}\right) \frac{ \cancel{\sum_m Q_{tm} E^{(t)}_{m,i}} }{\cancel{P_{ti}}} \,.

\end{align*}

Defining

\begin{align*}

E^*_n = \sum_{t,i} \left(\max_{m’} E_{m’,i}^{(t)}\right)\,,

\end{align*}

we get the following result:

Theorem (Regret of Exp4):If $\eta$ is chosen appropriately, the regret of Exp4 defined in $\eqref{eq:expertregret}$ satisfies

\begin{align}

R_n \le \EE{\sqrt{ 2 \log(M) E^*_n }}\,.

\label{eq:exp4bound}

\end{align}

Note that to get this result, one should set $\eta = \sqrt{(2\log M)/E^*_n}$, which is an infeasible choice when $E^{(1)},\dots,E^{(n)}$ are not available in advance, as is typically the case. Hence, in a practical implementation, one sets $\eta_t = \sqrt{\log M/E^*_t}$, which is a decreasing sequence (the factor of $2$ inside the square root will be lost due to some technical reasons). Then, as discussed before, a regret bound of the above form with a slightly larger constant will hold.

To assess the quality of the bound in the above theorem we consider a few upper bounds on $\eqref{eq:exp4bound}$. First, the expression in \eqref{eq:exp4bound} is the smallest when all experts agree: $E_m^{(t)} = E_{m’}^{(t)}$ for all $t,m,m’$. In this case $E_n^*= n$ and the right-hand side of $\eqref{eq:exp4bound}$ becomes $\sqrt{ 2 \log(M) n }$. We see that the price we pay for not knowing in advance that the experts will agree is $O(\sqrt{\log(M)})$. This case also highlights that in a way $E_n^*$ measures the amount of agreement between the experts.

In general, we cannot know whether the experts will agree. In any case, from $E_{m,i}^{(t)}\le 1$, we get $E_n^*\le Kn$, leading to

\begin{align}

R_n \leq \sqrt{ 2n K \log(M) }\,.

\label{eq:fewactions}

\end{align}

This shows that the price paid for adding experts is relatively low, as long as we have few actions.

How about the opposite case when the number of experts is low? Using $\max_m E_{mi}^{(t)} \le \sum_m E_{mi}^{(t)}\le M$, we get $E_n^* \le M n$ and thus

\begin{align*}

R_n \leq \sqrt{ 2n M \log(M) }\,.

\end{align*}

This shows that the number of actions can be very high, yet the regret will be low if we have only a few experts. This should make intuitive sense.

The above two bounds can also be summarized as (as they hold simultaneously):

\begin{align*}

R_n \leq \sqrt{ 2n (M\wedge K) \log(M) }\,.

\end{align*}

In a way, Exp4 adapts to whether the number of experts, or the number of actions is small (and in fact adapts to the alignment between the experts).

How does this bound compare to our earlier bounds, e.g., $\eqref{eq:exp3perc2}$? The number $M$ of all possible maps of $\cC$ to $[K]$ is $|[K]^{\cC}| = K^{|\cC|}$. Taking $\log$, we see that $\log(M) =|\cC| \log(K)$. From $\eqref{eq:fewactions}$, we conclude that $R_n \le \sqrt{ 2n K |\cC| \log(K) }$, which is the same as $\eqref{eq:exp3perc2}$, up to a constant factor, which would be lost anyways if we properly bounded the regret of Exp4 that uses changing parameters. While we get back the earlier bound, we won’t be able to implement the Exp4 algorithm for even moderate context and action sizes. In particular, the memory requirement of Exp4 is $O(K^{|\cC|})$, while the memory requirement of the specialized method where an Exp3 algorithm is used with every instance is $O(|\cC| K)$. The setting when Exp4 is practically useful is when $M$ is small and $\cC$ is large.

# Conclusions

In this post we have introduced the contextual bandit setting and the Exp4 category of strategies. Perhaps the most important point of this post beyond the algorithms is to understand that there are trade-offs between having a larger comparison class and a more meaningful definition of the regret that this entails.

There are many points we have not developed in detail. One is high probability bounds, which we saw in the previous post for Exp-IX and can also be derived here. We also have not mentioned lower bounds. As it turns out, the results we have given are essentially tight, at least in some instances and we hope to return to this topic in future posts. Finally there is the “hot-topic” of adaptation. For example, what is gained in terms of the regret if the experts are often in agreement.

# References

- For a history of contextual bandits see: Tewari and Murphy. From Ads to Interventions: Contextual Bandits in Mobile Health, 2016
- The Exp4 algorithm was introduced by Auer et al. The Non-stochastic Multiarmed Bandit Problem, 2002.
- For tighter bounds when the experts agree to some extent: McMahan and Streeter. Tighter Bounds for Multi-Armed Bandits with Expert Advice, 2009.
- For the Exp4-IX: Neu. Explore no more: Improved high-probability regret bounds for non-stochastic bandits, 2015.
- Another approach for high-probability bounds generalizing Exp3.P: Beygelzimer et al. Contextual Bandit Algorithms with Supervised Learning Guarantees, 2011

The mathjax for your one-hot vector in the EXP4 reward estimate calculate is not displaying correctly.

Where is this exactly? Maybe this depends on the browser and the browser setting?

I am using Google Chrome on Mac. Here is a screenshot: https://www.dropbox.com/s/9jor149lcb9b0tu/Screenshot%202017-01-03%2011.30.41.png?dl=0

This looks like the way I think it should look. What exactly is the problem here? Note that this display determines a scalar, not a vector. One value for every $i\in [n]$ (which could have been added).

Sorry I misinterpreted the empty box as a the standard missing unicode character box. I was expected a 1 with a subscript of At = i instead of the empty box followed by {At = i}.

In images, I have always used this as my indicator function:

https://www.dropbox.com/s/v632ic0tgxhewbh/Screenshot%202017-01-09%2011.56.36.png?dl=0

So when I saw the follow I thought the box and the squiggly braces indicated a formatting error.: https://www.dropbox.com/s/35edd15q6wrb5hl/Screenshot%202017-01-09%2011.53.02.png?dl=0

But it does make it harder to misinterpret. A one with subscripts is easier to misinterpret as a one instead of an indicator function.

Had I read all the posts first I would have seen you defined your indicator function in this post: https://banditalgs.com/2016/09/04/stochastic-bandits-warm-up/

Starting from the top now and reading through. Thanks for the continued work and I look forward to the book.

Yes, our use of the syntax of the indicator is unusual, but I find it easier to read.

I think the whole Exp4 reward estimate calculation is off. It is supposed to be $\hat X_{ti} = \frac{X_ti}{P_ti}$ and $P_ti$ is supposed to already have the $\gamma$ incorporated into it to smooth the probabilities

That is also an idea. However, here we decided to use the “implicit exploration” (IX) version, which in general has better properties. This is explained near the end of this post: https://banditalgs.com/2016/10/01/adversarial-bandits/

Thanks for the clarification