Le Cam's theorem

In probability theory, Le Cam's theorem, named after Lucien Le Cam (1924 – 2000), states the following. Suppose:

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

Then

[math]\displaystyle{ \sum_{k=0}^\infty \left| \Pr(S_n=k) - {\lambda_n^k e^{-\lambda_n} \over k!} \right| \lt 2 \left( \sum_{i=1}^n p_i^2 \right). }[/math]

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 p_i = λ_n/n, we see that this generalizes the usual Poisson limit theorem.

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

It is also possible to weaken the independence requirement.

External links

• Weisstein, Eric W.. "Le Cam's Inequality". http://mathworld.wolfram.com/LeCamsInequality.html.