1. Concentration Inequalities
One of the most important tasks in analyzing randomized algorithms is to understand what random variables arise and how well they are concentrated. A variable with good concentration is one that is close to its mean with good probability. A “concentration inequality” is a theorem proving that a random variable has good concentration. Such theorems are also known as “tail bounds”.
1.1. Markov’s inequality
This is the simplest concentration inequality. The downside is that it only gives very weak bounds, but the upside is that needs almost no assumptions about the random variable. It is often useful in scenarios where not much concentration is needed, or where the random variable is too complicated to be analyzed by more powerful inequalities.
Note that if then the right-hand side of the inequality is at least , and so the statement is trivial. So Markov’s inequality is only useful if . Typically we use Markov’s inequality to prove that has only a constant probability of exceeding its mean by a constant factor, e.g., .
where the inequality comes from taking the expectation of both sides of (1).
Note that Markov’s inequality only bounds the right tail of , i.e., the probability that is much greater than its mean.
1.2. The Reverse Markov inequality
In some scenarios, we would also like to bound the probability that is much smaller than its mean. Markov’s inequality can be used for this purpose if we know an upper-bound on . The following result is an immediate corollary of Theorem 1.
Corollary 2 Let be a random variable that is never larger than . Then, for all ,
1.3. Application to Max Cut
Recall the Max Cut problem from the last lecture. We are given a graph and wish to approximately solve . Recall our algorithm simply chooses to be a uniformly random subset of .
Let’s now analyze the probability that this algorithm gives a large cut. Let and , and note that is never larger than . Fix any and set . By the Reverse Markov inequality,
where we have used Inequality 2 from the Notes on Convexity Inequalities.
This shows that, with probability at least , the algorithm outputs a set satisfying
This statement is quite unsatisfying. If we want a approximation, then we have only shown that the algorithm has probability of succeeding. Next we will show how to increase the probability of success.
2. Amplification by Independent Trials
In many cases where the probability of success is positive but small, we can “amplify” that probability by perfoming several independent trials and taking the best outcome.
For Max Cut, consider the following algorithm. First it picks several sets , independently and uniformly at random. Let be the index for which is largest. The algorithm simply outputs the set . (Note that our Max Cut algorithm in the last lecture did not even look at the edges of the graph, but this new algorithm must look at the edges to compute .)
To analyze this algorithm, we wish to argue that is large. Well, is small only if all are small, and this is rather unlikely.
Here we have used the standard trick , which is Inequality 1 from the Notes on Convexity Inequalities.
Thus, setting , we obtain that
In summary, our new Max Cut algorithm picks sets , finds the which maximizes , then outputs . With probability at least we have
In this analysis we used the following general fact, which is elementary but often very useful.
2.1. Example: Flipping Coins
Consider flipping a fair coin. If the coin turns up heads, call this a “success”. So the probability of success is . By Claim 1,
This is a true statement, but not all that interesting: it is quite obvious that after coin flips, the probability of seeing at least one head is very close to . Note that the expected number of heads is . Can’t we say that there are probably close to heads? This takes us beyond the subject of amplification and into a discussion of the binomial distribution.
3. The binomial distribution
Again, let us flip a fair coin times and let be the number of heads seen. Then has the binomial distribution. So
Here the meaning of “small” depends on the choice of . For what values of is this sum small? Unfortunately the sum is a bit too complicated to get a feel for its magnitude. Can we simplify the expression so it’s easier to see what’s going on?
As far as I know, this sum does not have a simple closed-form expression. Instead, can we usefully approximate that sum? It would be great to have a provable upper bound for the sum that is simple, useful and asymptotically nearly tight. The Chernoff bound, which we discuss in Section 4, is a very general and powerful tool for analyzing tails of probability distributions, and it gives a fantastic bound on (2). But first, to motivate the bound that we will obtain, let us describe some interesting behavior of binomial distributions.
3.1. Different behavior in different regimes
In most introductory courses on probability theory, one often encounters statements that describe the limiting behavior of the binomial distribution. For example:
Fact 1 Let be a binomially distributed random variable with parameters and . (That is, gives the number of successes in independent trials where each trial succeeds with probability .) As gets large while remains fixed, the distribution of is well approximated by the normal distribution with mean and variance .
The following plot shows the binomial distribution for and in red and the corresponding normal approximation in blue.
However the binomial distribution has qualitatively different behavior when is very small.
The following plot shows the binomial distribution for and in red and the corresponding Poisson approximation in blue. Note that the red plot is quite asymmetric, so we would not expect a normal distribution (which is symmetric) to approximate it well.
Ideally would like an upper bound on (2) which works well for all ranges of parameters, and captures the phenomena described in Fact 1 and Fact 2. Remarkably, the Chernoff bound is able to capture both of these phenomena.
4. The Chernoff Bound
The Chernoff bound is used to bound the tails of the distribution for a sum of independent random variables, under a few mild assumptions. Since binomial random variables are sums of independent Bernoulli random variables, it can be used to bound (2). Not only is the Chernoff bound itself very useful, but its proof techniques are very general and can be applied in a wide variety of settings.
The Chernoff bound is by far the most useful tool in randomized algorithms. Numerous papers on randomized algorithms have only three ingredients: Chernoff bounds, union bounds and cleverness. Of course, there is an art in being clever and finding the right way to assemble these ingredients!
4.1. Formal Statement
Let be independent random variables such that always lies in the interval . Define and . Let and note that .
Remarks. All of these bounds are useful in different scenarios. The statement in (3) is the most general bound and it implies the statements in (4) and (5). The difference between (4) and (5) is due to the different phenomena illustrated in Facts 1 and 2.
Useful fact. This theorem does not actually require that . It suffices to have .
Theorem 4 (Chernoff Lower Tail) For any ,
Useful fact. This theorem does not actually require that . It suffices to have .
Historical remark. Chernoff actually only considered the case in which the random variables are identically distributed. The more general form stated here is due to Hoeffding.
But the Chernoff bound has nothing to do with products of random variables, it is about sums of random variables. So one trick we could try is to convert sums into products using the exponential function. Fix some parameter whose value we will choose later. Define
It is easy to check that, since the s are independent, the s are also independent. Therefore, by (6),
Proof: We would like to relate and . If were a linear function of we could just use linearity of expectation, but unfortunately it is not. However, since we know that , we can instead use a linear approximation of on that interval.
By convexity (see Inequality 3 in the Notes on Convexity Inequalities), we have
where again we have used our favorite inequality .
Now we are ready to prove the Chernoff bound. In fact, we will only prove (3); all the other inequalities follow using the same ideas and additional calculations.
Finally, substituting and using proves the desired inequality.