In probability theory, Le Cam's theorem, named after Lucien Le Cam, states the following.^[1][2][3]

Suppose:

• ${\displaystyle X_{1},X_{2},X_{3},\ldots }$ are independent random variables, each with a Bernoulli distribution (i.e., equal to either 0 or 1), not necessarily identically distributed.
• ${\displaystyle \Pr(X_{i}=1)=p_{i},{\text{ for }}i=1,2,3,\ldots .}$
• ${\displaystyle \lambda _{n}=p_{1}+\cdots +p_{n}.}$
• ${\displaystyle S_{n}=X_{1}+\cdots +X_{n}.}$ (i.e. ${\displaystyle S_{n}}$ follows a Poisson binomial distribution)

Then

${\displaystyle \sum _{k=0}^{\infty }\left|\Pr(S_{n}=k)-{\lambda _{n}^{k}e^{-\lambda _{n}} \over k!}\right|<2\left(\sum _{i=1}^{n}p_{i}^{2}\right).}$

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

When ${\displaystyle \lambda _{n}}$ is large a better bound is possible: ${\displaystyle \sum _{k=0}^{\infty }\left|\Pr(S_{n}=k)-{\lambda _{n}^{k}e^{-\lambda _{n}} \over k!}\right|<2\left(1\wedge {\frac {1}{\lambda }}_{n}\right)\left(\sum _{i=1}^{n}p_{i}^{2}\right)}$,^[4] where ${\displaystyle \wedge }$ represents the ${\displaystyle \min }$ operator.

It is also possible to weaken the independence requirement.^[4]

• Weisstein, Eric W. "Le Cam's Inequality". MathWorld.