= Poisson characterization by independent binomial splitting
{c}
Let $N$ be a nonnegative integer-valued <random variable> with finite <expected value> $\mu$, and independently label each of its items with either of two equally likely labels. If the two resulting counts are <independent random variables>, then $N$ has a <Poisson distribution>. Indeed, its <probability generating function> obeys $G(s)=G((1+s)/2)^2$. With $H(t)=G(1-t)$, iteration gives $H(t)=H(t/2^k)^{2^k}$. Finite <expected value> gives $H(h)=1-\mu h+o(h)$, hence $H(t)=e^{-\mu t}$ and $G(s)=e^{\mu(s-1)}$. Conversely, <Poisson thinning> produces independent <Poisson distributions> for the two counts.
Back to article page
