# The dirichlet() function in the hyper2 package

dirichlet
function (powers, alpha)
{
if (!xor(missing(powers), missing(alpha))) {
stop("supply exactly one of powers, alpha")
}
if (missing(powers)) {
powers <- alpha - 1
}
np <- names(powers)
if (is.null(np)) {
stop("supply a named vector")
}
hyper2(as.list(np), d = powers) - hyper2(list(np), d = sum(powers))
}

To cite the hyper2 package in publications, please use Hankin (2017). 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

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

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))
## log(a^4 * (a + b + c)^-12 * 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)))
## log(a^4 * (a + b + c)^-12 * 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)))
## log( b * (b + c + d)^-10 * c * d^8)

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

L1+L2
## log(a^4 * (a + b + c)^-12 * b^6 * (b + c + d)^-10 * c^4 * d^8)
maxp(L1+L2)
##          a          b          c          d
## 0.09090991 0.10909114 0.07272658 0.72727237

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, $$p=0.94$$]. Observe further that we can include four-way observations easily:

(L3 <- dirichlet(c(a=4,b=3,c=2,d=6)))
## log(a^4 * (a + b + c + d)^-15 * b^3 * c^2 * d^6)
L1+L2+L3
## log(a^8 * (a + b + c)^-12 * (a + b + c + d)^-15 * b^9 * (b + c + d)^-10
## * c^6 * d^14)

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)))
## log(a^5 * (a + b + c + d)^-11 * b^3 * 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)
##            a            b            c            d
## 4.546526e-01 2.727224e-01 1.020077e-06 2.726240e-01

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]:

specificp.test(L4,'c',0.25)
##
##  Constrained support maximization
##
## data:  H
## null hypothesis: sum p_i=1, c = 0.25
## null estimate:
##         a         b         c         d
## 0.3122093 0.2162741 0.2500000 0.2215167
## (argmax, constrained optimization)
## Support for null:  -14.93581 + K
##
## alternative hypothesis:  sum p_i=1
## alternative estimate:
##            a            b            c            d
## 4.546526e-01 2.727224e-01 1.020077e-06 2.726240e-01
## (argmax, free optimization)
## Support for alternative:  -11.738 + K
##
## degrees of freedom: 1
## support difference = 3.197811
## p-value: 0.01144022 (two sided)

However, observe that we cannot reject the equality hypothesis, that is, $$p_a=p_b=p_c=p_d=\frac{1}{4}$$:

equalp.test(L4)
##
##  Constrained support maximization
##
## data:  L4
## null hypothesis: a = b = c = d
## null estimate:
##    a    b    c    d
## 0.25 0.25 0.25 0.25
## (argmax, constrained optimization)
## Support for null:  -15.24924 + K
##
## alternative hypothesis:  sum p_i=1
## alternative estimate:
##            a            b            c            d
## 4.546526e-01 2.727224e-01 1.020077e-06 2.726240e-01
## (argmax, free optimization)
## Support for alternative:  -11.738 + K
##
## degrees of freedom: 3
## support difference = 3.511242
## p-value: 0.07118459

## References

Hankin, R. K. S. 2017. “Partial Rank Data with the hyper2 Package: Likelihood Functions for Generalized Bradley-Terry Models.” The R Journal 9 (2): 429–39.