# The `dirichlet()` function in the `hyper2` package

#### Robin K. S. Hankin

    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

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

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))

    ## 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.