---
title: "Races with hyper3 objects"
author: "R. K. S. Hankin"
output: bookdown::html_document2
bibliography: hyper2.bib  # copy vignettes/hyper2.bib to inst/
---


```{r setup, include=FALSE}
set.seed(1)
knitr::opts_chunk$set(echo = TRUE)
library("hyper2")
```

```{r out.width='20%', out.extra='style="float:right; padding:10px"',echo=FALSE}
knitr::include_graphics(system.file("help/figures/hyper2.png", package = "hyper2"))
```

(takes about an hour to process without cache).  To cite the `hyper2`
package in publications, please use @hankin2017_rmd.  Consider a race in
which there are six runners 1-6 but we happen to know that three of
the runners (1,2,3) are clones of strength $p_a$, say, two of the
runners (4,5) have strength $p_b$, and the final runner (6) is of
strength $p_c$.  The runners race and the finishing order is:


$$a\succ c\succ b\succ a\succ a \succ b$$

Thus the winner was $a$, second place for $b$, third for $a$, and so
on.  Alternatively we might say that $a$ came first, fourth, and
fifth; $b$ came third and sixth, and $c$ came second.  The
Plackett-Luce likelihood function for $p_a,p_b,p_c$ would be

$$
\frac{p_a}{3p_a+2p_b+p_c}\cdot
\frac{p_c}{2p_a+2p_b+p_c}\cdot
\frac{p_b}{2p_a+2p_b    }\cdot
\frac{p_a}{2p_a+ p_b    }\cdot
\frac{p_a}{ p_a+ p_b    }\cdot
\frac{p_b}{      p_b    },\qquad p_a+p_b+p_c=1
$$

The likelihood function is easily specified in package idiom:


```{r, label=definehyper3}
H <- ordervec2supp3(c("a","c","b","a","a","b"))
H
```

```{r, label=maximizeprobs, cache=TRUE}
mH  <-  maxp(H)
mH
```

We can test various hypotheses, for example that the competitors are of equal strength:

```{r testequality, cache=TRUE}
equalp.test(H)
```

showing that this small dataset does not allow one to say that the
competitors differ.  However, it is easy to generate synthetic order
statistics where one can reject the null of equality of strength:


```{r,biggerrace,cache=TRUE}
o <- rrace3(pn=c(a=20,b=11,c=7),ps=c(a=0.7,b=0.2,c=0.1))
o
H <- ordervec2supp3(o)
H
```

Above, we specify that there are 20 $a$'s, 11 $b$'s and 7 $c$'s with
strength $70\%$, $20\%$, and $10\%$ respectively.  Recall this means
that, in a head-to-head between $a$ and $b$ we would have $a$ winning
with probability $0.7/(0.7+0.2)=\frac{7}{9}$.

```{r,bigraceestimation,cache=TRUE}
mH <- maxp(H)
mH
pTRUE <- c(a=0.7,b=0.2,c=0.1)   # true values 
mH - pTRUE # residual
```

Above we see a reasonably close agreement between the true values and the estimate.

```{r,dothetest,cache=TRUE}
equalp.test(H)
```

Above we see a $p$-value of $<10^{-6}$, strong evidence to reject the
hypothesis that $p_a=p_b=p_c=\frac{1}{3}$.  We may also test the
hypothesis that the strength vector is equal to its true value:


```{r,isittrue,cache=TRUE}
knownp.test(H,pTRUE)
```

And above we see that there is no evidence to reject the hypothesis
that the strengths are in fact the specified strengths in `pTRUE`.

# Some simple cases: likelihood functions for order statistics with clones

## Three competitors

Here I consider some order statistics with nontrivial maximum
likelihood Bradley-Terry strength.   The simplest nontrivial case is
three competitors with strengths $a,a,b$ and finishing order $a\succ
b\succ a$.  The likelihood function would be

$$
\frac{a}{2a+b}\cdot\frac{b}{a+b}
$$

In this case we know that $a+b=1$ so this is equal to
$\mathcal{L}=\mathcal{L}(a)=\frac{a(1-a)}{1+a}$.

$$
\frac{d\mathcal{L}}{da}=\frac{(1+a)(1-2a)-a(1-a)}{(1+a)^2}=
\frac{1-2a+a-2a^2 -a + a^2}{(1+a)^2}=
\frac{1-2a-a^2}{(1+a)^2}
$$

This will be zero at $\sqrt{2}\pm 1$; we also note that
$d^2\mathcal{L}/da^2=-4(1+a)^{-3}$, manifestly strictly negative for
$0\leq a\leq 1$: the root is a maximum.


```{r label=maxaba,cache=TRUE}
maxp(ordervec2supp3(c("a","b","a")))
```

above, we see close agreement with the theoretical value of
$(\sqrt{2}-1,1-\sqrt{2})\simeq (0.414,0.586)$.  Observe that the
maximum likelihood estimate for $a$ is strictly less than 0.5, even
though the finishing order is symmetric.  Using
$\mathcal{L}(a)=\frac{a(1-a)}{1+a}$, we can show that
$\log\mathcal{L}(\hat{a})=\log\left(3-2\sqrt{2}\right)\simeq -1.76$,
where $\hat{a}=\sqrt{2}-1$ is the maximum likelihood estimate for $a$.
Defining $\mathcal{S}=\log\mathcal{L}$ as the support [log-likelihood]
we have

$$\mathcal{S}=\mathcal{S}(a)=\log\left(\frac{a(1-a)}{1+a}\right)-\log\left(3-2\sqrt{2}\right)$$

as a standard support function which has a maximum value of zero when
evaluated at $\hat{a}=\sqrt{2}-1$.  For example, we can test the null
that $a=b=\frac{1}{2}$, the statement that the competitors have equal
Bradley-Terry strengths:


```{r}
a <- 1/2 # null
(S_delta <- log( a*(1-a)/(1+a))- log(3-2*sqrt(2)))
```

Thus the additional support gained in moving from $a=\frac{1}{2}$ to
the evaluate of $a=\sqrt{2}-1$ is 0.029, rather small [as might be
expected given that we have only one rather uninformative observation,
and also given that the maximum likelihood estimate ($\simeq 0.41$) is
quite close to the null ($0.5$).  Nevertheless we can apply Wilks's
theorem for a $p$-value:

```{r}
pchisq(-2*S_delta,df=1,lower.tail=FALSE)
```

The $p$-value is about 0.81, exceeding 0.05; thus we have no strong
evidence to reject the null of $a=\frac{1}{2}$.  Further, we can use
either analytical or numerical methods to find a credible interval.
With an $n$-unit of support criterion the analytical solution to
$\mathcal{S}(p)=-n$ is given by defining $X=\log(3-2\sqrt{2})-n$ and
solving $p(1-p)/(1+p)=X$, or
$p_\pm=\left(1-X\pm\sqrt{1+4X+X^2}\right)/2$, the two roots being the
lower and upper limits of the credible interval.  Graphically:

```{r}
a <- seq(from=0,by=0.005,to=1)
S <- function(a){log(a*(1-a)/((1+a)*(3-2*sqrt(2))))}
plot(a,S(a),type='b')
abline(h=c(0,-2))
abline(v=c(0.02438102,0.9524271),col='red')
abline(v=sqrt(2)-1)
```


### all results for AAB

* $A\succ A\succ B\longrightarrow\frac{a^2}{1+a}$
* $A\succ B\succ A\longrightarrow\frac{1-2a-a^2}{(1+a)^2}$
* $B\succ A\succ A\longrightarrow\frac{1-a}{1+a}$

Graphically

```{r}
f_aab <- function(a){a^2/(1+a)}
f_aba <- function(a){a*(1-a)/(1+a)^2}
f_baa <- function(a){(1-a)/(1+a)}

p <- function(f,...){points(a,f(a)/max(f(a)),...)}
plot(0:1,0:1,xlab="p(a)",ylab="Likelihood",type="n")
p(f_aab,type="l",col="black")
p(f_aba,type="l",col="red")
p(f_baa,type="l",col="blue")
legend(x=0.85,y=0.7,col=c("black","red","blue"),lty=1,pch=NA,legend=c("AAB","ABA","BAA"))
abline(h=exp(-2),lty=2)
```

## Four competitors


With four competitors we have five nontrivial order statistics which,
without loss of generality, are:

* $a\succ b\succ a\succ a$
* $a\succ a\succ b\succ a$
* $a\succ b\succ a\succ b$
* $a\succ b\succ b\succ a$
* $a\succ b\succ c\succ a$.

Taking $a\succ b\succ a\succ b$ as an example we have

$$
\mathcal{L}_{a\succ b\succ a\succ b}(a)=
\frac{a}{2a+2b}\cdot
\frac{b}{ a+2b}\cdot
\frac{a}{ a+ b}\cdot
\frac{b}{    b}=\frac{a^2-a^3}{4-2a}
$$

the evaluate would be a root of $2a^2-7a+4$, or $(7-\sqrt{17})/4\simeq
0.72$:


```{r label=maxabab,cache=TRUE}
maxp(ordervec2supp3(c("a","b","a","b")))
```

If we allow non-finishers there is another nontrivial order statistic,
viz $a\succ b\succ\left\lbrace a,b\right\rbrace$ [thus one of the two
$a$'s won, one of the $b$'s came second, and one of each of $a$ and
$b$ failed to finish].

$$
\mathcal{L}(a)=
\frac{a}{2a+2b}\cdot
\frac{b}{ a+2b}\propto\frac{a(1-a)}{2-a}
$$

(see how the likelihood function is actually simpler than for the
complete order statistic).  The evaluate would be $2-\sqrt{2}\simeq
0.586$:

```{r label=maxnonfinishers,cache=TRUE}
o <- ordervec2supp3(c("a","b"),nonfinishers=c("a","b"))
o
maxp(o)
```

Now consider

$$a\succ b\succ c\succ a$$

with likelihood function

$$
\mathcal{L} =
\frac{a}{2a+b+c}\cdot
\frac{b}{ a+b+c}\cdot
\frac{c}{ a  +c},\qquad a+b+c=1$$

It can be shown that this has a maximum if $a^3+4a^2+4a-1=0$.
Although cubic equations have analytical roots, the expression is
complicated, and equal to about `c(a=0.2056,b=0.5466,c=0.2478)`:

```{r,label=abca,cache=TRUE}
maxp(ordervec2supp3(c("a","b","c","a")))
```

## Results for AAAB

```{r,plotaaab,cache=TRUE}
Hab <- function(i,n=4){
  jj <- rep("a",n)
  jj[i] <- "b"
  ordervec2supp3(jj)
}

small <- 1e-5
a <- seq(from=small,to=1-small,len=400)

plot(0:1,c(0,1),type="n",xlab="p[a]",ylab="likelihood")
dop <- function(w, n, ...){
  Hjj <- Hab(w,n)
  y <- sapply(a,function(p){loglik(c(a=p,b=1-p),Hjj,log=FALSE)})
  points(a,y/max(y),type="l",...)
}
dop(1,4,col="black")
dop(2,4,col="red")
dop(3,4,col="blue")
dop(4,4,col="green")
```

```{r,plotaaaaaaab,cache=TRUE}
plot(0:1,c(0,1),type="n",xlab="p[a]",ylab="likelihood")
for(i in 1:10){dop(i,10,col=rainbow(10)[i])}
```

```{r,plot40ab,cache=TRUE}
plot(0:1,c(0,1),type="n",xlab="p[a]",ylab="likelihood")
for(i in 1:40){dop(i,40,col=rainbow(40)[i])}
```







# Fisher information for simple cases for aab


We have three possible finishing orders:

$b\succ a\succ a$, probability $\frac{b}{2a+b}=\frac{1-a}{1+a}$.  And
$\frac{\partial\log\mathcal{L}}{\partial a}=\frac{2}{p^2-1}$.

$a\succ b\succ a$, probability $\frac{a}{2a+b}\cdot\frac{b}{a+b}=\frac{a(1-a)}{1+a}$. And
$\frac{\partial\log\mathcal{L}}{\partial a}=\frac{a^2+2a-1}{a^3-a}$.

$a\succ a\succ b$, probability $\frac{a}{2a+b}\cdot\frac{a}{a+b}=\frac{a^2}{1+a}$.  And
$\frac{\partial\log\mathcal{L}}{\partial a}=\frac{a+2}{a^2+a}$.


Thus the Fisher information is

$$
\frac{1-a}{1+a}\cdot\left(\frac{2}{a^2-1}\right)^2
+
\frac{a(1-a)}{1+a}\cdot\left(\frac{a^2+2a-1}{a^3-a}\right)^2
+\frac{a^2}{1+a}\cdot\left(\frac{a+2}{a^2+a}\right)^2
$$

Simplifying,

$$\mathcal{I}(a)=\frac{1+16a+18a^2-4a^3-15a^4-7a^5-a^6}{a(1-a)(1+a)^3}$$


```{r}
fisher <- function(a){(1+16*a+18*a^2-4*a^3-15*a^4-7*a^5-a^6)/(a*(1-a)*(1+a)^3)}
a <- seq(from=0.05,to=0.95,by=0.01)
plot(a,fisher(a))
```

# Estimation of strength

We will try some random simulation where the strengths of the competitors is known:


```{r,label=randomsimknownstrength,cache=TRUE}

es <- replicate(100,maxp(ordervec2supp3(rrace3(pn=c(a=10,b=10),ps=c(a=1/2,b=1/2))))[1])
```

```{r,label=showestimates}
hist(es)
abline(v=1/2,lwd=5)
```


# Power test for `hyper3` objects compared with Mann-Whitney-Wilcox




First do some stuff where the null is true:

```{r,label=definepower}
wilcox_pval <- function(a,b,pa,pb){
  o <- rrace3(pn=c(a=a,b=b),ps=c(a=pa,b=pb))
  wilcox.test(which(o=='a'),which(o=='b'))$p.value
}

hyper3_pval <- function(a,b,pa,pb){
  o <- rrace3(pn=c(a=a,b=b),ps=c(a=pa,b=pb))
  equalp.test(ordervec2supp3(o))$p.value
}
```



```{r,label=dopowernulltrue,cache=TRUE}
wil_p_h0 <- replicate(100,wilcox_pval(10,10,0.5,0.5))
hyp_p_h0 <- replicate(100,hyper3_pval(10,10,0.5,0.5))
```


```{r,label=plotpvalnulltrue}
plot(ecdf(wil_p_h0),col="black")
plot(ecdf(hyp_p_h0),col="red",add=TRUE)
legend("topleft",pch=16,lty=1,col=c("black","red"),legend=c("Wilcox","hyper3"))
```

Now some stuff where the null is false:


```{r,label=dopower,cache=TRUE}
wil_p <- replicate(100,wilcox_pval(10,10,0.7,0.3))
hyp_p <- replicate(100,hyper3_pval(10,10,0.7,0.3))
```

```{r,label=plotpower}
hist(wil_p)
hist(hyp_p)
par(pty="s")
qqplot(wil_p,hyp_p,asp=1,xlim=c(0,1),ylim=c(0,1),pty="m")
abline(0,1)
qqplot(log(wil_p),log(hyp_p),asp=1,xlim=c(-6,1),ylim=c(-6,1),pty="m")
abline(0,1)
ks.test(wil_p,hyp_p)
```



# The constructors' championship

Here we define likelihood function for the constructors' championship.

```{r,dotheconst}
jj <- read.table("constructor_2021.txt",header=TRUE)
i <- seq(from=1,by=2,to=nrow(jj)) # extract every second entry for points
constructor_points_2021 <- jj[i,ncol(jj)]
names(constructor_points_2021) <- jj[i,1]

constructor_points_2021 <- sort(constructor_points_2021,dec=TRUE)
constructor_points_2021 

constructor_table_2021 <- jj[,-ncol(jj)]
constructor_table_2021[,1:17]
constructor_2021_table <- constructor_table_2021
```

Looking at the first column we see that the constructors are Mercedes
RedBull, and so on (file `inst/constructor_names.txt` summarises the
constructors' lifespans).  Each constructor fields two cars in each
race.  Looking at the next column, we see the order statistic for
Bahrain (`BHR`).  We see that Mercedes came first and third, RedBull
came second and fifth, Ferrari sixth and eighth, and so on.

```{r,maketheconst,cache=TRUE}
constructor_2021 <- ordertable2supp3(constructor_table_2021)
```

Then


```{r,label=maxconstr,cache=TRUE}
constructor_2021_maxp <-  maxp(constructor_2021)
mc <- constructor_2021_maxp  # saves typing
```

```{r,label=drawaconstructorpie}
pie(mc)
dotchart(mc,pch=16)
ordertransplot(rank(-mc),rank(-constructor_points_2021),xlab="likelihood rank",ylab="points rank")
```

We may test the null of equal strengths:

```{r,label=testequalityconstructor,cache=TRUE}
equalp.test(constructor_2021)
```

and see that there is (very) strong evidence to reject this null.
Further, we may test the hypothesis that the big four (viz Mercedes,
Ferrari, RedBull, and McLaren) are equal:

```{r,label=testsameconstructor,cache=TRUE}
samep.test(constructor_2021,c("Mercedes","Ferrari","RedBull","McLaren")) # ~10 mins
```

and fail to reject that null.  How about the mid-range constructors:


```{r,label=testsameconstructorlowscores,cache=TRUE}
samep.test(constructor_2021,c("Alpine","AlphaTauri","AstonMartin")) # ~ 10 mins
```


## Constructors championship with `cheering` or `tether` functionality


```{r,Bahrain2021_to_support3}
constructor_2021_table
(cons <- constructor_2021_table[,1])
(BHR <- as.numeric(constructor_2021_table[,2]))  # Bahrain
names(BHR) <- cons
BHR
finishers <- BHR[!is.na(BHR)]
finishers <- names(finishers[order(finishers)])  # in order
nonfinishers <- names(BHR[is.na(BHR)])  # no particular order
finishers
nonfinishers
u <- unique(c(finishers,nonfinishers))
e <- seq_along(u)
names(e) <- u
e
help <- rep(1.8,length(u))
e
help
cheering3(v=finishers,e=e,help=help,nonfinishers=nonfinishers)
```

OK so now generalize it:

```{r,definevenue3}
venue3 <- function(ven,cons,help){
   suppressWarnings(names(ven) <- cons)
   finishers <- ven[!is.na(ven)]
   finishers <- names(finishers[order(finishers)])  # in order
   nonfinishers <- names(ven[is.na(ven)])  # no particular order
   u <- unique(c(finishers,nonfinishers))
   e <- seq_along(u)
   names(e) <- u
   help <- rep(help,length(u))
   cheering3(v=finishers,e=e,help=help,nonfinishers=nonfinishers)
}  # venue3() definition closes
```

Now use `venue3()` for all the venues in the 2021 season, with a common help:

```{r,defineallvenues,cache=TRUE}
set.seed(9)
allvenues <- function(h){
  H3 <- hyper3()
  cons <- constructor_2021_table[,1]
  for(i in seq(from=2,to=ncol(constructor_2021_table),by=1)){
     H3 <- H3 + venue3(ven=as.numeric(constructor_2021_table[,i]), cons=cons,help=h)
  }
  return(H3)
}
## And maximize it for different help values:


o <- function(h){maxp(allvenues(h),give=TRUE,n=1)$likes}
l <- seq(from=0.9,to=2.5,len=11)
L <- sapply(l,o)
```

```{r,plotsupport}
plot(l,L-max(L),type='b')
```

Above, we are testing (and rejecting) the hypothesis that the various
constructors are independent of one another.  We reject the hypothesis
that $h=1$.  Noting that we control both for constructors'
Plackett-Luce strengths, and on the assumption that the two drivers of
any given constructor are of equal strengths, we still reject the
hypothesis of conditional independence.


```{r,findmaxlikeusingoptim,cache=TRUE}
jj <- optimize(o,c(1.3,1.5),maximum=TRUE)
jj
```

```{r,likelihoodnull,cache=TRUE}
(jjnull <- o(1))
```

```{r}
pchisq(2*(jj$objective-jjnull),df=1,lower.tail=FALSE)
```



```{r,label=maxwithfour,cache=TRUE}
mv_max  <- maxp(allvenues(jj$maximum),give=TRUE,n=1)
mv_null <- maxp(allvenues(1),give=TRUE,n=1)
```

```{r,analyse4}
x <- mv_max$par
y <- mv_null$par
par(pty='s')
plot(x,y,asp=1,pch=16,xlim=c(0,0.2),ylim=c(0,0.2),xlab="null",ylab="max like")
for(i in seq_along(x)){text(x[i],y[i],names(x)[i],pos=4,cex=0.7,col='gray')}
abline(0,1)
```

# Constructors' championship 2020

```{r,label=constructor2020,cache=TRUE}
jj <- read.table("constructor_2020.txt",header=TRUE)
i <- seq(from=1,by=2,to=nrow(jj)) # extract every second entry for points
constructor_points_2020 <- jj[i,ncol(jj)]
names(constructor_points_2020) <- jj[i,1]

constructor_points_2020 <- sort(constructor_points_2020,dec=TRUE)
constructor_points_2020

constructor_table_2020 <- jj[,-ncol(jj)]
constructor_table_2020[,1:17]
constructor_2020_table <- constructor_table_2020
constructor_2020 <- ordertable2supp3(constructor_table_2020)
```

```{r,label=maxpconstructor2020,cache=TRUE}
constructor_2020_maxp <- maxp(constructor_2020,n=1)
constructor_2020_maxp <- sort(constructor_2020_maxp,dec=TRUE)
constructor_2021_maxp <- sort(constructor_2021_maxp,dec=TRUE)
constructor_2020_maxp
constructor_2021_maxp
```

```{r,label=changenames,cache=FALSE}
constructor_2020_and_2021 <- psubs(constructor_2020,"Mercedes","Merc2020") + psubs(constructor_2021,"Mercedes","Merc2021")
```


```{r,doconstructor20202021test,cache=TRUE}
ans <- samep.test(constructor_2020_and_2021,c("Merc2020","Merc2021"))
```

```{r,show20201test}
ans
```


Note that the analysis in `javelin.Rmd` is appropriate here too, a
nice example.

### Package dataset {-}

Following lines create `constructor.rda`, residing in the `data/`
directory of the package.

```{r,label=saveconstructordataset}
save(
constructor_2020_maxp,constructor_2020,constructor_2020_table,
constructor_2021_maxp,constructor_2021,constructor_2021_table,
file="constructor.rda")
```
