Le Cam's theorem

In probability theory, Le Cam's theorem, named after Lucien Le Cam, states the following.

Suppose:


 * $$X_1, X_2, X_3, \ldots $$ are independent random variables, each with a Bernoulli distribution (i.e., equal to either 0 or 1), not necessarily identically distributed.
 * $$\Pr(X_i = 1) = p_i, \text{ for } i = 1, 2, 3, \ldots.$$
 * $$\lambda_n = p_1 + \cdots + p_n.$$
 * $$S_n = X_1 + \cdots + X_n.$$ (i.e. $$S_n$$ follows a Poisson binomial distribution)

Then


 * $$\sum_{k=0}^\infty \left| \Pr(S_n=k) - {\lambda_n^k e^{-\lambda_n} \over k!} \right| < 2 \left( \sum_{i=1}^n p_i^2 \right). $$

In other words, the sum has approximately a Poisson distribution and the above inequality bounds the approximation error in terms of the total variation distance.

By setting pi = λn/n, we see that this generalizes the usual Poisson limit theorem.

When $$\lambda_n$$ is large a better bound is possible: $$\sum_{k=0}^\infty \left| \Pr(S_n=k) - {\lambda_n^k e^{-\lambda_n} \over k!} \right| < 2 \left(1 \wedge \frac 1 \lambda_n\right) \left( \sum_{i=1}^n p_i^2 \right)$$, where $$\wedge$$ represents the $$\min$$ operator.

It is also possible to weaken the independence requirement.