Consider the hyper3 likelihood function for the observation $a\succ b\succ b$. This should be $\frac{a}{a+2b}\cdot\frac{b}{2b}\cdot\frac{b}{b}$. It is reasonable to expect the final two terms, viz $\frac{b}{2b}$ and $\frac{b}{b}$, to cancel out. Now, the final term of $\frac{b}{b}$ is cancelled out [I think num3() adds a c(b=1) with power 1, and den3() decerements the power of c(b=1) so the back end performs the cancellation]; but the penultimate factor $\frac{b}{2b}$ is not cancelled. In idiom we have c(b=1)^1 and c(b=2)^-1:
library("hyper2")
ordervec2supp3(c("a","b","b"))
#> log( (a=1)^1 * (a=1, b=2)^-1 * (b=1)^1 * (b=2)^-1)[of course, the factor of $\frac{1}{2}$ does not matter, for likelihood is defined only up to a multiplicative constant]. However, it is computationally inefficient. It would be nice to find and perform such cancellations. Pairs such as c(a=2,b=3) and c(a=4, b=6) should similarly cancel.
Consider the$a\succ b\succ b$ . This should be $\frac{a}{a+2b}\cdot\frac{b}{2b}\cdot\frac{b}{b}$ . It is reasonable to expect the final two terms, viz $\frac{b}{2b}$ and $\frac{b}{b}$ , to cancel out. Now, the final term of $\frac{b}{b}$ is cancelled out [I think $\frac{b}{2b}$ is not cancelled. In idiom we have
hyper3likelihood function for the observationnum3()adds ac(b=1)with power 1, andden3()decerements the power ofc(b=1)so the back end performs the cancellation]; but the penultimate factorc(b=1)^1andc(b=2)^-1:[of course, the factor of$\frac{1}{2}$ does not matter, for likelihood is defined only up to a multiplicative constant]. However, it is computationally inefficient. It would be nice to find and perform such cancellations. Pairs such as
c(a=2,b=3)andc(a=4, b=6)should similarly cancel.