In RobinHankin/hyper2: The Hyperdirichlet Distribution, Mark 2

knitr::opts_chunk$set(echo = TRUE)
library(hyper2)

knitr::include_graphics(system.file("help/figures/hyper2.png", package = "hyper2"))

To cite the hyper2 package in publications, please use @hankin2017_rmd; see also inst/notthewinner.Rmd. Here we consider a situation in which we make only a rather uninformative observation: namely, the identity of the loser (which is here taken to be me, player a). We have six competitors a,b,c,d,e,f with nonnegative strengths $p_1,p_2,p_3,p_4,p_5,p_6, \sum p_i=1$. Competitor a has strength $p_1$, and loses. We are interested in how informative this observation is with regard to various assertions about the strength of a. We have three hypotheses:

$H_1\colon p_1=0$
$H_2\colon\exists i\neq 1 \mbox{ with } p_1\geq p_i>0$
$H_3\colon p_1=p_2=\ldots=p_n=\frac{1}{n}$

(observe that $H_2$ implies that someone is worse than me). All three hypotheses correspond to a restriction on allowable parameter space. The first step is to define a suitable likelihood function for our observation:

n <- 6 # number of competitors
H <- hyper2(pnames=letters[1:6])
L <- ggrl(H,letters[2:n],letters[1]) 
length(L)
L[[1]]
L[[2]]

In the above, L is a log-likelihood function for the observation that competitor a came last. In R idiom, it is a list of hyper2 objects corresponding to all $5!=120$ possible observed order statistics (only the first two, L[[1]] and L[[2]], are printed, which correspond to orders bcdefa and bcdfea respectively). Any order is permissible, so long as player a is bottom.

Hypothesis 1

Now we maximize the likelihood under $H_1$, that is, perform an unconstrained maximization:

(ans_unconstrained <- maxp(L,give=TRUE))

The constraint embodied in $H_1$ does not need to be enforced: the evaluate $\hat{p}$ has $p_1=0$, consistent with competitor a coming last; actually we have $\hat{p}=\left(0,\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5}\right)$ analytically, and the numerics are not too far from this. The likelihood at the evaluate is 1 (technically, this is wrong because likelihood is defined only up to a multiplicative constant. A correct statement is: the likelihood function evaluated by like_single_list() returns 1 at $\hat{p}$ returns 1).

Hypothesis 2

The next step is to perform a constrained maximization, the constraint being $H_2$. This says that I am better than at least one other person, here chosen to be, without loss of generality, competitor b:

small <- 1e-2  # numerical necessity
ans_constrained <-
maxp(L,startp=c(small*2,rep(small,n-2)),fcm=rbind(c(1,-1,rep(0,n-3))),fcv=0,give=TRUE)
ans_constrained

Observe that the constraint is active: $p_1=p_2$ in the numerical result. Analytically, the evaluate is indeterminate in $p_3,p_4,p_5,p_6$, but has $p_1=p_2=0$ (numerically this means that $p_1=p_2=\epsilon$, where $\epsilon$ is some small number chosen for smallness and numerical expediency).

Hypothesis 3

Recall that $H_3\colon p_1=p_2=\ldots=p_n=\frac{1}{n}$; this is equivalent to a random chance. Because this is a point hypothesis, we do not need to use any maximization techniques; it is sufficient to evaluate the likelihood at this point:

like_single_list(indep(equalp(L[[1]])),L)

The likelihood can be shown to be $\frac{1}{n}$ (which is clear on intuitive grounds: $H_3$ says that all players are equivalent and the order statistic is totally random. If this is correct, then one would identify player a as the worst with a probability $\frac{1}{n}$). Note again that this is loose language as likelihood is only defined up to a multiplicative constant.

Discussion: repeated observations

We are now in a position to define a likelihood function on the three hypotheses $H_1,H_2,H_3$:

$$ {\mathcal L}(H_1)=1\qquad {\mathcal L}(H_2)=\frac{1}{2}\qquad {\mathcal L}(H_3)=\frac{1}{n} $$

(recall that $n$ is the number of competitors). We can see that the likelihood ratio ${\mathcal L}(H_2)/{\mathcal L}(H_1)$ is 0.5 for an observation that competitor a comes last. This is exactly the same as the likelihood ratio for the following situation: we toss a coin which is either fair ($H_\mbox{fair}$) or two-headed ($H_\mbox{2heads}$). We observe a head. Then ${\mathcal L}(H_\mbox{2heads})/{\mathcal L}(H_\mbox{fair}) = 0.5$ as well (this is Edwards's valet and black and white ball thought experiment).

Suppose we make $r$ independent observations, all of which are that competitor a comes last. Then the likelihood function becomes

$$ {\mathcal L}(H_1)=1\qquad {\mathcal L}(H_2)=\frac{1}{2^r}\qquad {\mathcal L}(H_3)=\frac{1}{n^r} $$

The support for $H_1$ against $H_2$ would be $\log(2^r)=r\log 2$. To exceed Edwards's criterion of 2 units of support per degree of freedom would require $2/\log(2)\simeq 2.9$ observations, or 3 rounded up; alternatively, we might require the log-likelihood to lie in the tail area of its asymptotic $\chi^2_1$ distribution, that is

qchisq(0.95,1)

that is $2r\log 2\geq 3.84$ or $r\geq 2.77$.

The support for $H_1$ against $H_3$ would be $r\log n$ but this time we need $n-1$ degrees of freedom. The likelihood criterion would be $r\log n\geq 2(n-1)$, or $r\geq 2(n-1)/\log n$. Alternatively we might adopt a frequentist view and require $2\log(n^r)\geq\chi^2_{n-1}(0.95)$, or $r\geq Q/(2\log n)$, where Q is the 95th percentile of the chi-square distribution with $n$ degrees of freedom, qchisq(0.95,df=n). It might be interesting to compare these two approaches.

Note carefully that the observation considered here is "I came last". This is different from the observation that "I did not win" which is considered separately in inst/notthewinner.Rmd.