In probability theory, Le Cam's theorem, named after Lucien Le Cam, states the following. Suppose:
X 1 , X 2 , X 3 , … {\displaystyle 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 , for i = 1 , 2 , 3 , … . {\displaystyle \Pr(X_{i}=1)=p_{i},{\text{ for }}i=1,2,3,\ldots .}
λ n = p 1 + ⋯ + p n . {\displaystyle \lambda _{n}=p_{1}+\cdots +p_{n}.}
S n = X 1 + ⋯ + X n . {\displaystyle S_{n}=X_{1}+\cdots +X_{n}.} (i.e. S n {\displaystyle S_{n}} follows a Poisson binomial distribution) Then
∑ k = 0 ∞ | Pr ( S n = k ) − λ n k e − λ n k ! | < 2 ( ∑ i = 1 n p i 2 ) . {\displaystyle \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 λ n {\displaystyle \lambda _{n}} is large a better bound is possible: ∑ k = 0 ∞ | Pr ( S n = k ) − λ n k e − λ n k ! | < 2 ( 1 ∧ 1 λ n ) ( ∑ i = 1 n p i 2 ) {\displaystyle \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 ∧ {\displaystyle \wedge } represents the min {\displaystyle \min } operator. It is also possible to weaken the independence requirement.
References
External links Weisstein, Eric W. "Le Cam's Inequality". MathWorld.