---
title: "Weakly informative observations: identifying the weakest player"
output: html_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
library(hyper2)
```

<p style="text-align: right;">
![](`r system.file("help/figures/hyper2.png", package = "hyper2")`){width=10%}
</p>


# Partial rank observations: who lost?

(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:

```{r,label=useggrl}
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:

```{r,label=executemaxp,cache=TRUE}
(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`:

```{r,label=constrainthething,cache=TRUE}
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:

```{r,label=likesingle}
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

```{r,label=useqchis}
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`.
