In RobinHankin/hyper2: The Hyperdirichlet Distribution, Mark 2

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dirichlet

To cite the hyper2 package in publications, please use @hankin2017_rmd. We say that non-negative $p_1,\ldots, p_n$ satisfying $\sum p_i=1$ follow a Dirichlet distribution, and write $\mathcal{D}(p_1,\ldots,p_n)$, if their density function is

\begin{equation} \frac{\Gamma(\alpha_1+\cdots +\alpha_n)}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_n)} \prod_{i=1}^n p_i^{\alpha_i-1} \end{equation}

for some parameters $\alpha_i>0$. It is interesting in the current context because it is conjugate to the multinomial distribution; specifically, given a Dirichlet prior $\mathcal{D}(\alpha_1,\ldots,\alpha_n)$ and likelihood function $\mathcal{L}(p_1,\ldots,p_n)\propto\prod_{i=1}^n p_i^{\alpha_i}$ then the posterior is $\mathcal{D}(\alpha_1+a_1,\ldots,\alpha_n+a_n)$.

In the context of the hyper2 package, function dirichlet() presents some peculiarities, discussed here.

A simple example

Consider a three-way trial between players a, b, and c with eponymous Bradley-Terry strengths. Suppose that, out of $4+5+3=12$ trials, a wins 4, b wins 5, and c wins 3. Then an appropriate likelihood function would be:

dirichlet(c(a=4,b=5,c=3))

Note the (a+b+c)^-11 term, which is not strictly necessary according the Dirichlet density function above which has no such term. However, we see that the returned likelihood function is ${\propto} p_{a\vphantom{abc}}^4p_{b\vphantom{abc}}^5p_{c\vphantom{abc}}^3$ (because $p_a+p_b+p_c=1$).

(L1 <- dirichlet(c(a=4,b=5,c=3)))

Now suppose we observe three-way trials between b, c, and d:

(L2 <- dirichlet(c(b=1,c=1,d=8)))

The overall likelihood function would be $L_1+L_2$:

L1+L2
maxp(L1+L2)

Observe the dominance of competitor d, reasonable on the grounds that d won 8 of the 10 trials against b and c; and a, b, c are more or less evenly matched [chi square test, $p=r round(chisq.test(c(4,5,3),sim=TRUE)$p.value,2)$]. Observe further that we can include four-way observations easily:

(L3 <- dirichlet(c(a=4,b=3,c=2,d=6)))
L1+L2+L3

Package idiom L1+L2+L3 operates as expected because dirichlet() includes the denominator. Sometimes we see situations in which a competitor does not win any trials. Consider the following:

(L4 <- dirichlet(c(a=5,b=3,c=0,d=3)))

Above, we see that c won no trials and is not present in the numerator of the expression. However, L4 is informative about competitor c:

maxp(L4)

We see that the maximum likelihood estimate for c is zero (to within numerical tolerance). Further, we may reject the hypothesis that $p_c=\frac{1}{4}$ [which might be a reasonable consequence of the assumption that all four competitors have the same strength]: