Randomized weighted majority algorithm

Imagine that every morning before the stock market opens, we get a prediction from each of our "experts" about whether the stock market will go up or down. Our goal is to somehow combine this set of predictions into a single prediction that we then use to make a buy or sell decision for the day. The RWMA gives us a way to do this combination such that our prediction record will be nearly as good as that of the single best expert in hindsight.

In machine learning, the weighted majority algorithm (WMA) is a meta-learning algorithm which "predicts from expert advice". It is not a randomized algorithm:

initialize all experts to weight 1.
for each round:
    poll all the experts and predict based on a weighted majority vote of their predictions.
    cut in half the weights of all experts that make a mistake.

Suppose there are ${\displaystyle n}$ experts and the best expert makes ${\displaystyle m}$ mistakes. The weighted majority algorithm (WMA) makes at most ${\displaystyle 2.4(\log _{2}n+m)}$ mistakes, which is not a very good bound. We can do better by introducing randomization.

The nonrandomized weighted majority algorithm (WMA) only guarantees an upper bound of ${\displaystyle 2.4(\log _{2}n+m)}$, which is problematic for highly error-prone experts (e.g. the best expert still makes a mistake 20% of the time.) Suppose we do ${\displaystyle N=100}$ rounds using ${\displaystyle n=10}$ experts. If the best expert makes ${\displaystyle m=20}$ mistakes, we can only guarantee an upper bound of ${\displaystyle 2.4(\log _{2}10+20)\approx 56}$ on our number of mistakes.

As this is a known limitation of WMA, attempts to improve this shortcoming have been explored in order to improve the dependence on ${\displaystyle m}$. Instead of predicting based on majority vote, the weights are used as probabilities: hence the name randomized weighted majority. If ${\displaystyle w_{i}}$ is the weight of expert ${\displaystyle i}$, let ${\displaystyle W=\sum _{i}w_{i}}$. We will follow expert ${\displaystyle i}$ with probability ${\displaystyle {\frac {w_{i}}{W}}}$. The goal is to bound the worst-case expected number of mistakes, assuming that the adversary (the world) has to select one of the answers as correct before we make our coin toss. Why is this better in the worst case? Idea: the worst case for the deterministic algorithm (weighted majority algorithm) was when the weights split 50/50. But, now it is not so bad since we also have a 50/50 chance of getting it right. Also, to trade-off between dependence on ${\displaystyle m}$ and ${\displaystyle \log _{2}n}$, we will generalize to multiply by ${\displaystyle \beta <1}$, instead of necessarily by ${\displaystyle {\frac {1}{2}}}$.

At the ${\displaystyle \ t}$-th round, define ${\displaystyle \ F_{t}}$ to be the fraction of weight on the wrong answers. so, ${\displaystyle \ F_{t}}$ is the probability we make a mistake on the ${\displaystyle \ t}$-th round. Let ${\displaystyle \ M}$ denote the total number of mistakes we made so far. Furthermore, we define ${\displaystyle E[M]=\ \sum _{t}F_{t}}$, using the fact that expectation is additive. On the ${\displaystyle \ t}$-th round, ${\displaystyle W}$ becomes ${\displaystyle \ W(1-(1-\beta )F_{t})}$. Reason: on ${\displaystyle \ F_{t}}$ fraction, we are multiplying by ${\displaystyle \ \beta }$. So, ${\displaystyle \ W_{final}=n*(1-(1-\beta )F_{1})*(1-(1-\beta )F_{2})...}$
Let's say that ${\displaystyle \ m}$ is the number of mistakes of the best expert so far. We can use the inequality ${\displaystyle \ W\geq \beta ^{m}}$. Now we solve. First, take the natural log of both sides. We get: ${\displaystyle \ m\ln \beta \leq \ln(n)+\sum _{t}\ln(1-(1-\beta )F_{t})}$, Simplify:
${\displaystyle \ \ln(1-x)=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-...}$, So,
${\displaystyle \ \ln(1-(1-\beta )F_{t})<-(1-\beta )F_{t}}$.
${\displaystyle \ m\ln \beta \leq \ln(n)-(1-\beta )*\sum _{t}F_{t}}$
Now, use ${\displaystyle \ E[M]=\ \sum _{t}F_{t}}$, and the result is:
${\displaystyle \ E[M]\leq {\frac {m\ln(1/\beta )+\ln(n)}{1-\beta }}}$
Let's see if we made any progress:

If ${\displaystyle \ \beta ={\frac {1}{2}}}$, we get, ${\displaystyle \ 1.39m+2\ln(n).}$,
if ${\displaystyle \ \beta ={\frac {3}{4}}}$, we get, ${\displaystyle \ 1.15m+4\ln(n)}$.
so we can see we made progress. Roughly, of the form ${\displaystyle \ (1+\epsilon )*m+\epsilon ^{-1}*\ln(n)}$.

The Randomized Weighted Majority Algorithm can be used to combine multiple algorithms in which case RWMA can be expected to perform nearly as well as the best of the original algorithms in hindsight.

Furthermore, one can apply the Randomized Weighted Majority Algorithm in situations where experts are making choices that cannot be combined (or can't be combined easily). For example, RWMA can be applied to repeated game-playing or the online shortest path problem. In the online shortest path problem, each expert is telling you a different way to drive to work. You pick one path using RWMA. Later you find out how well you would have done using all of the suggested paths and penalize appropriately. To do this right, we want to generalize from "losses" of 0 or 1 to losses in [0,1]. The goal is to have an expected loss not much larger than the loss of the best expert. We can generalize the RWMA by applying a penalty of ${\displaystyle \beta ^{loss}}$ (i.e. two losses of one half result in the same weight as one loss of 1 and one loss of 0). The analysis given in the previous section does not change significantly.

• Multi-armed bandit problem.
• Efficient algorithm for some cases with many experts.
• Sleeping experts/"specialists" setting.

• Machine learning
• Weighted majority algorithm
• Game theory
• Multi-armed bandit

• Weighted Majority & Randomized Weighted Majority
• Avrim Blum (2004) machine learning theory
• Rob Schapire 2006 Foundations of Machine Learning
• Predicting From Experts Advice
• Uri Feige, Robi Krauthgamer, Moni Naor. Algorithmic Game Theory