## Tuesday, July 24, 2007

The multiplicative Chernoff bound states that if X is the average of n independent, 0-1 random variables X_i, each with mean p, then

P(|X-p| > tp) < 2exp(-npt^2 / 3).

In other words, the empirical average of independent binary random variables is exponentially concentrated about their true mean. The probability that we're off by a muliplicative factor of t is subgaussian in t. There is a catch, however: rare events are difficult to estimate. Thus, if p is tiny, the bound becomes quite loose. Indeed, if you're flipping a coin whose probability of heads is 10^-6, you can easily see a thousand flips without a single head.

But what about the following "trick" to beat Chernoff? Since X_i=1 is a rare event, let us define Y_i = 1-X_i. Now Y, the average of Y_1,...,Y_n, has expected value p'=1-p, which is large if p is small. So instead of obtaining a loose bound for X in terms of p, we obtain a good bound for Y in terms of p' -- yet X and Y are related in a very simple way. Does this trick really work or have I cheated somewhere?