= Likelihood-ratio test of area-proportional Poisson means
{title2=$\Lambda=(3S_1/T)^{S_1}(3S_2/(2T))^{S_2}$}
Two equally numerous groups of independent counts with <Poisson distributions> have areas $a$ and $2a$. Under common density their group means are in ratio $1:2$. With group sums $S_1,S_2$ and $T=S_1+S_2$, the maximized alternative-to-null likelihood ratio is
$$
\Lambda=(3S_1/T)^{S_1}(3S_2/(2T))^{S_2}.
$$
Use zero-count limits and $\Lambda=1$ when $T=0$. The regular <Wilks theorem> calibration has one degree of freedom. Exactly conditioning on $T$ gives $S_1\sim\operatorname{Binomial}(T,1/3)$ under the null; this supports small-count testing without the large-sample approximation.
Back to article page
