In RobinHankin/hyper2: The Hyperdirichlet Distribution, Mark 2

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

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

Consider a race between a competitor of Bradley-Terry strength $a$ and $n$ competitors each of strength $b$; we require $a+b=1$. NB we have $n+1$ competitors in total, 1 of strength $a$ and $n$ of strength $b$. An observation is indexed by $r$, the number of $b$ clones finishing ahead of $a$, so $0\leqslant r\leqslant n$. The initial field strength is $a+nb=1+b(n-1)$.

Suppose our bro comes first (zero $b$'s ahead of him):

[ \mathcal{L}(a)=\frac{a}{a+nb}=\frac{1-b}{1+(n-1)b} ]

Comes second, one $b$ ahead of him:

[ \mathcal{L}(a)=\frac{b}{a+nb}\cdot\frac{a}{a+(n-1)b} ]

Third, two ahead of him:

[ \mathcal{L}(a)= \frac{b}{a+ n b}\cdot \frac{b}{a+(n-1)b}\cdot \frac{a}{a+(n-2)b} ]

Fourth, three ahead:

[ \mathcal{L}(a)= \frac{b}{a+ n b}\cdot \frac{b}{a+(n-1)b}\cdot \frac{b}{a+(n-2)b}\cdot \frac{a}{a+(n-3)b} ]

Placing after $r<n$ of the $n$ clones of strength $b$ finish:

[ \mathcal{L}(a)= \frac{b^ra}{ (a+nb)(a+(n-1)b)(a+(n-2)b)\cdots(a+(n-r)b)} = \frac{B-1}{(B+n-1)(B+n-2)\cdots(B+n-r-1)} ]

where $B=1/b$ [also using the fact that $a+b=1$].

This is

[ \mathcal{L}_{r,n}(a)=\frac{(B-1)(B+n-r-2)!}{(B+n-1)!},\qquad B=1/(1-a) ]

Then, using Stirling [viz $n!\sim n^ne^{-n}\sqrt{2\pi n}$] we get

[ \log(B-1) + ((B+n-r-2)\log(B+n-r-2)-(B+n-r-2)+\frac{\log(B+n-r-2)}{2}) - ((B+n -1)\log(B+n -1)-(B+n -1)+\frac{\log(B+n -1)}{2}) ]

If you give Sage this:

var('B n r')
pretty_print_default(True)
X=log(B-1) + (
+ ((B+n-r-2)*log(B+n-r-2)-(B+n-r-2)+log(B+n-r-2)/2)
- ((B+n  -1)*log(B+n  -1)-(B+n  -1)+log(B+n  -1)/2)
)
X.taylor(n,Infinity,2)

It returns this:

[ -(r+1)\log n -\frac{(2B-3)r-r^2+2B-2}{2n} -\frac{3(2B-3)r^2-6B^2-6(B^2-3B+2)r+12B-5}{12n^2} +\log(B-1)+\mathcal{O}(n^{-3}) ]

Note that if we use instead $n~\sim n^ne^{-n}\sqrt{2\pi n}\left(1+\frac{1}{12n}\right)$, then only terms of $\mathcal{O}(n^{-2})$ and above are changed; the first-order terms and indeed the zeroth order terms are unaffected. Anyway, after simplifying and taking an asymptotic expansion to order $n^{-1}$, we find, to first order,

[ \mathcal{S}=\log\mathcal{L}_{r,n}(a)=\log(B-1)-B\frac{r+1}{n} + K + \mathcal{O}\left(n^{-2}\right) ]

(where $K$ is an arbitrary constant). Observing that $\partial^2\mathcal{S}/\partial B^2=-(B-1)^{-2}<0$, the evaluate would be unique; we have $\hat{B}=\frac{n+r+1}{r+1}+\mathcal{O}\left(n^{-1}\right)$, or alternatively $\hat{a}=\frac{n}{n+r+1}$. I present some numerical results below.

Placing last

Placing last is finishing after $n=r$ clones finish:

[ n=1\longrightarrow\mathcal{L}=b]

[ n=2\longrightarrow\mathcal{L}=\frac{b}{a+2b}\cdot\frac{b}{a+b}=\frac{b^2}{1+b} =\frac{1}{B(B+1)} ]

[ n=3\longrightarrow\mathcal{L}=\frac{b}{a+3b}\cdot\frac{b}{a+2b}\cdot\frac{b}{a+b}=\frac{b^3}{(1+2b)(1+b)}=\frac{1}{B(B+1)(B+2)} ]

[ n=4\longrightarrow\mathcal{L}=\frac{b}{a+4b}\cdot\frac{b}{a+3b}\cdot\frac{b}{a+2b}\cdot\frac{b}{a+b}= \frac{b^4}{(1+3b)(1+2b)(1+b)}= \frac{1}{B(B+1)(B+2)(B+3)} ]

For $n=4$ we have equivalently

[ \frac{1}{(B+3)(B+2)(B+1)B}=\frac{(B-1)!}{(B+3)!} ]

In general we would have

[ \frac{1}{(B+n-1)(B+n-2)\cdots(B+1)B}= \frac{(B-1)!}{(B+n-1)!} ]

(denominator has $n+1$ terms). Observe that this agrees with the $r<n$ case above (substituting $r=n$). We may introduce some R idiom:

f_single <- function(a,r,n,log=FALSE){  # not vectorised
    B <- 1/(1-a)
    if(log){
        out <- log(B-1) - lgamma(B+n) + lgamma(B+n-r-1)
    } else {  # not-log
        out <- (B-1)/prod(B+(n-r-1):(n-1))
    }
   return(out)
}

f_vec_a <- function(a,r,n, log=FALSE){  # vectorised in 'a' but not in 'r'
    sapply(a,function(a){f_single(a,r=r,n=n,log=log)})
}

f_vec_arn <- function(a,r,n,log=FALSE){
    ## vectorized in 'a'; treats 'r' and 'n' as
    ## vectors of independent observations

    M <- cbind(r,n)

    if(log){
        out <- 0
        for(i in seq_len(nrow(M))){out <- out + f_vec_a(a,r=M[i,1],n=M[i,2],log=TRUE)}
    } else {
        out <- 1
        for(i in seq_len(nrow(M))){out <- out * f_vec_a(a,r=M[i,1],n=M[i,2],log=FALSE)}
    }
    return(out)
}

fapprox <- function(a,r,n,log=FALSE){
    if(log){
        return(log(B-1)-B*(r+1)/n)
    } else {
        return(B-1 - (r+1)*exp(B)/n)
    }
}

fdash <- function(a,r,n,log=TRUE){
    B <- 1/(1-a)
    out <- (1/(B-1) - psigamma(B+n) + psigamma(B+n-r-1))*B^2
    if(!log){out <- out * f_vec_arn(a,r,n,log=FALSE)}
    return(out)
}

Now verify fdash():

a <- 0.34
d <- 1e-3
r <- 4
n <- 10
c(
  numerical  = (f_single(a+d/2,r,n,log=TRUE)-f_single(a-d/2,r,n,log=TRUE))/d,
  analytical = fdash(a,r,n)
)

c(
  numerical  = (f_single(a+d/2,r,n,log=FALSE)-f_single(a-d/2,r,n,log=FALSE))/d,
  analytical = fdash(a,r,n,log=FALSE)
)

Now use the derivative to find the maximum likelihood point

MLE <- function(r,n,give=FALSE){
    d <- 1e-6
    out <- optimize(f_vec_arn,c(d,1-d),r=r,n=n,log=TRUE,maximum=TRUE)
    if(!give){out <- out$maximum}
    return(out)
}

MLEa <- function(r,n,give=FALSE,small=1e-4){
    out <- uniroot(f=function(a){fdash(a,r,n)},interval=c(small,1-small))
    if(!give){out <- out$root}
    return(out)
}

showdiff <- function(x,y){c(way1=x,way2=y,diff=x-y,logratio=log(x/y))}

rbind(
showdiff(MLE(1,9),MLEa(1,9)),
showdiff(MLE(3,9),MLEa(3,9)),
showdiff(MLE(3,7),MLEa(3,7))
)

rbind(
showdiff(MLE(10, 100), 100 / 111),
showdiff(MLE(10, 500), 500 / 511),
showdiff(MLE(10,1000),1000 /1011),
showdiff(MLE(10,5000),5000 /5011)
)

rbind(
showdiff(MLEa(10, 100), 100 / 111),
showdiff(MLEa(10, 500), 500 / 511),
showdiff(MLEa(10,1000),1000 /1011),
showdiff(MLEa(10,5000),5000 /5011)
)

Now verification

a <- 0.23423434
b <- 1-a
B <- 1/b
n <- 9
r <- 3
c(
way1 = b^3*a/((a+9*b) * (a+8*b) * (a+7*b) * (a+6*b)),
way2 = (B-1)/prod(B+(5:8)),
way3 = f_vec_arn(a,r,n,log=FALSE)
)

out <- list()
for(n in 1:10){
  out[[n]]  <- rep(NA,n+1)
  for(r in 0:n){
    if(r==0){
      jj <- 1
    } else if(r==n){n
      jj <- 0
    } else {
      jj <- MLE(r,n)
    }
    out[[n]][r+1] <- jj
  }
}
out

plotterp <- function(...){
plot(NA,xlim=c(1,10),ylim=c(0,1),type="n",xlab="n",ylab=expression(hat(a)))
for(n in 1:10){
  for(r in 0:n){   points(n,out[[n]][r+1],pch=16) }
  for(r in 0:n){   text(n+0.2,out[[n]][r+1],r,cex=0.5,col='gray') }
}
}
plotterp()
pdf(file="dotprobs.pdf")
plotterp()
dev.off()

Note on the differences between probability and likelihood

The likelihood calculations above can be confusing becuase likelihood is not the same as probability.

Consider, for example, $n=2$ so we have 3 competitors.

probability of focal competitor coming first [i.e. $r=0$]:

[\frac{a}{a+2b}=\frac{1-b}{1+b}]

Now what is the probability of coming second, that is, $r=1$?

My first thought was:

[ \mbox{prob of coming second} = \frac{b}{a+2b}\cdot\frac{a}{a+b}=\frac{b(1-b)}{1+b}\qquad\mbox{WRONG} ]

but this is incorrect: there are two clones of (strength) $b$, and the likelihood arguments above do not distinguish between their finishing order. We can name the competitors $a$, $b_1$ and $b_2$ and specify that $b_1$ and $b_2$ both have strength $b$. Then "coming second" means that the finishing order was $b_1\succ a\succ b_2$ or $b_2\succ a\succ b_1$. Noting that these events are disjoint, the probability would be

[\mbox{prob of coming second} = \operatorname{P}(b_1\succ a\succ b_2) + \operatorname{P}(b_2\succ a\succ b_1) = \frac{b}{a+2b}\cdot\frac{a}{a+b} + \frac{b}{a+2b}\cdot\frac{a}{a+b} = \frac{2(1-b)}{1+b} ]

Similarly for the focal competitor coming last we need to sum the probabilities of finishing orders $b_1\succ b_2\succ a$ and $b_2\succ b_1\succ a$:

[\mbox{prob of coming third}= \operatorname{P}(b_1\succ b_2\succ a) + \operatorname{P}(b_2\succ b_1\succ a) = \frac{b}{a+2b}\cdot\frac{b}{a+b}+ \frac{b}{a+2b}\cdot\frac{b}{a+b}= \frac{2b^2}{1+b}. ]

Above we see a factor of two difference between the probabilities and the likelihoods presented previously. Note that there is no such factor for the probability of the focal competitor finishing first:

[ \mbox{prob of coming first} = \operatorname{P}(a\succ b_1\succ b_2) + \operatorname{P}(a\succ b_2\succ b_1) = \frac{b}{a+2b}\cdot\frac{b}{b+b} + \frac{b}{a+2b}\cdot\frac{b}{b+b} = \frac{1-b}{1+b} ]

Now we see that the probabilities sum to

[ \frac{1-b}{1+b} + \frac{2b^2}{1+b} + \frac{2(1-b)}{1+b}=1 ]

Now try $r=3$ [i.e. 4 competitors]. Taking things one step at a time:

[ \mbox{prob of coming first} = 3!\cdot\frac{a}{a+3b}\cdot\frac{b}{3b}\cdot\frac{b}{2b}\cdot\frac{b}{b}= \frac{1-b}{1+2b} ]

[ \mbox{prob of coming second} = 3!\cdot\frac{b}{a+3b}\cdot\frac{a}{a+2b}\cdot\frac{b}{2b}\cdot\frac{b}{b}= 3\cdot\frac{b(1-b)}{(1+2b)(1+b)} ]

[ \mbox{prob of coming third} = 3!\cdot\frac{b}{a+3b}\cdot\frac{b}{a+2b}\cdot\frac{a}{a+b}\cdot\frac{b}{b} = 6\cdot\frac{b^2(1-b)}{(1+2b)(1+b)} ] [ \mbox{prob of coming fourth (last)} = 3!\cdot\frac{b}{a+3b}\cdot\frac{b}{a+2b}\cdot\frac{b}{a+b}\cdot\frac{a}{a} = 6\cdot\frac{b^3}{(1+2b)(1+b)} ]

The sum would be $[(1+b)(1-b) + 3b(1-b) + 6b^2(1-b) + 6b^3]/[(1+b)(1+2b)]=1$.

Now consider the general case. Again, $n$ clones of strength $b$ and one of strength $a=1-b$, $n+1$ competitors.

[ \mbox{prob of coming first} = n!\cdot\frac{b}{a+nb}\cdot\frac{b}{nb}\cdot\frac{b}{(n-1)b}\cdots\frac{b}{b} = \frac{b}{1+(n-1)b} ]

[ \mbox{prob of coming second} = n!\cdot\frac{b}{a+nb}\cdot\frac{a}{a+(n-1)b}\cdot\frac{b}{(n-1)b}\cdot\frac{b}{(n-2)b}\cdots\frac{b}{2b}\cdot\frac{b}{b} =n\cdot X ]

[ \mbox{prob of coming third} = \frac{b}{a+nb}\cdot \frac{b}{a+(n-1)b}\cdot \frac{a}{a+(n-2)b}\cdot \frac{b}{(n-2)b}\cdots \frac{b}{2b}\cdot\frac{b}{b} =\frac{n!}{(n-2)!}\cdot X ]

Now consider the case where there are $r$ clones coming ahead of the focal competitor:

[ \mbox{prob of having $r$ clones ahead} =\frac{n!}{(n-r)!}\cdot X ]

Visuals

Now some likelihood visuals. Suppose $r=4,n=8$:

plotter <- function(...){
a <- seq(from=0,to=1,by=0.01)
n <- 8
jj <- f_vec_arn(a,3,n,log=FALSE)
plot(a,jj/max(jj,na.rm=TRUE),type='n',xlab=expression(a),ylab="likelihood")
grid()
rain <- rainbow(n+1)
for(r in seq(from=0,to=n)){
  y <- f_vec_a(a,r,n,log=FALSE)
  if(r==0){y[length(y)] <- 1}
  y <- y/max(y,na.rm=TRUE)
  if(r>0){y[length(y)] <- 0}
  points(a,y,type='l',lwd=4,col=rain[r+1])
}
abline(v=MLE(r=4,n=8))
text(0.07,0.95,"r=8",col=rain[9])
text(0.20,0.95,"r=7",col=rain[8])
text(0.26,0.91,"r=6",col=rain[7])
text(0.29,0.83,"r=5",col=rain[6])
text(0.33,0.76,"r=4",col=rain[5])
text(0.35,0.64,"r=3",col=rain[4])
text(0.39,0.53,"r=2",col=rain[3])
text(0.45,0.41,"r=1",col=rain[2])
text(0.63,0.22,"r=0",col=rain[1])
}

plotter()
pdf(file="ninelikes.pdf")
plotter()
dev.off()

MLE(r=4,n=8)

n <- 7
a <- seq(from=0,to=1,by=0.01)
fhyper3 <- function(r,n){
  out <- hyper3()
  out['a'] <- 1
  out['b'] <- r
  for(i in (n-r):n){
    out[c(a=1,b=i)] %<>% dec
  }
  return(out)
}

plot(NA,xlim=c(0,1),ylim=c(0,1),type='n')
M <- cbind(a=a,b=1-a)
for(r in 0:7){
  y <- loglik(M,fhyper3(r,n),log=FALSE)
  y <- y/max(y, na.rm=TRUE)
points(M[,1],y,type="l",lwd=8)
}

Formula 1: Ocon vs Gasly

a <- read.table("formula1_2023.txt",header=TRUE)
a <- a[,seq_len(ncol(a)-1)]
d_ocon  <- as.numeric(as.matrix(a)["Ocon",])
d_gasly <- as.numeric(as.matrix(a)["Gasly",])
d_ocon [is.na(d_ocon )] <- 22
d_gasly[is.na(d_gasly)] <- 22
d_ocon
d_gasly
MLE(d_ocon ,22)
MLE(d_gasly,22)

two sample test:

both <- c(d_ocon,d_gasly)
bothmax <- MLE(both,22)

d_ocon_max  <- MLE(d_ocon ,22)
d_gasly_max <- MLE(d_gasly,22)
f_vec_arn(bothmax,d_ocon ,22,log=TRUE)
f_vec_arn(bothmax,d_gasly,22,log=TRUE)
f_vec_arn(d_ocon_max ,d_ocon ,22,log=TRUE)
f_vec_arn(d_gasly_max,d_gasly,22,log=TRUE)

Lambda <- (
   (f_vec_arn(d_gasly_max  ,d_ocon,22,log=TRUE)+f_vec_arn(d_gasly_max  ,d_gasly,22,log=TRUE)) -
   (f_vec_arn(bothmax      ,d_ocon,22,log=TRUE)+f_vec_arn(bothmax      ,d_gasly,22,log=TRUE))
)
Lambda
pchisq(-2*Lambda,df=1,lower.tail=FALSE)

Formula 1: Formula 1 drivers: career arc

readstring <- function(year){read.table(paste("formula1_",year,".txt",sep=""))}
getfoc <- function(year,comp="Perez"){  # get focal competitor
  M <- as.matrix(readstring(year))
  out <- suppressWarnings(as.numeric(M[comp,seq_len(ncol(M)-1)]))
  out[is.na(out)] <- nrow(M)
  return(out)
}
getnum <- function(year){nrow(readstring(year))} # number of competitors

perez <- lapply(2011:2023,getfoc)
print(perez)
y <- unlist(lapply(seq_along(perez),function(i){mean(perez[[i]])}))
x <- 2011:2023
plot(x,y)
summary(lm(y~x))

checo_like <- function(a){
  out <- a*0
  for(year in 2011:2023){ out <- out + f_vec_arn(a,getfoc(year)-1,getnum(year)-1,log=TRUE) }
  return(out)
}

a <- seq(from=0.45,to=0.62,by=0.01)
cL <- checo_like(a)
cL <- cL - max(cL)
plot(a,cL,type='b')
abline(h=c(0,-2))

f1_logistic <- function(vec){
  alpha <- vec[1]
  beta  <- vec[2]
  out <- 0
  for(year in 2011:2023){
     LO <- alpha + beta*(year-2011)
     strength <- exp(LO)/(1+exp(LO))
     out <- out + f_vec_arn(strength,getfoc(year)-1,getnum(year)-1,log=TRUE)
  }
  return(out)
}

a <- seq(from=-0.8,to=0.0,by=0.01)
b <- seq(from=0,to=0.2,by=0.01)
jj <- as.matrix(expand.grid(a,b))
L <- apply(jj,1,f1_logistic)
L <- L-max(L)
L <- matrix(L,length(a),length(b))

L <- pmax(L,-40)
showchec <- function(...){
contour(a,b,L,xlab=expression(alpha),ylab=expression(beta),levels=-c(2,5*(1:5)))
points(-0.27,0.0813,pch=16,cex=2)
}
showchec()
pdf(file="showchecolike.pdf")
showchec(a,b,L)
dev.off()

f1_logistic(c(0,0))
f1_logistic(c(0,0.0001))
f1_logistic(c(0.001,0))

o <- optim(c(-0.2,0.1),fn=f1_logistic, control=list(fnscale = -1),hessian=TRUE)

o
o$par
o$hessian
eigen(o$hessian)
best <- o$par
f1_logistic(best)


jj <- best
jj[1] <- jj[1]*1.01
f1_logistic(jj) - f1_logistic(best)

jj <- best
jj[1] <- jj[1]*0.99
f1_logistic(jj) - f1_logistic(best)

jj <- best
jj[2] <- jj[2]*1.00001
f1_logistic(jj) - f1_logistic(best)

jj <- best
jj[2] <- jj[2]*0.99
f1_logistic(jj) - f1_logistic(best)

o_free <- optim(c(-0.2,0.1),fn=f1_logistic, control=list(fnscale = -1),hessian=TRUE)
o_constrained <- optim(c(-0.2,0.1),fn=function(v){
v[2] <- 0
return(f1_logistic(v))}, control=list(fnscale = -1),hessian=TRUE)

o_free
o_constrained
pchisq(2*(o_free$value - o_constrained$value),df=1,lower.tail=FALSE)

in silico formula

table(replicate(100,which(rrace3(pn=c(a=1,b=10),ps=c(a=0.9,b=0.1))=='a')))

n <- 1000
M <- matrix(0,n,4)
alpha_true <- 4
beta_true <- -6
for(i in seq_len(n)){
    x <- i/n
    LO <- alpha_true + beta_true*x
    p <- exp(LO)/(1+exp(LO))
    nn <- round(runif(1,10,100))
    M[i,1] <- x  # regressor
    M[i,2] <- which(rrace3(pn=c(a=1,b=nn),ps=c(a=p,b=1-p))=='a')
    M[i,3] <- nn+1
    M[i,4] <- p
}
head(M)
tail(M)

likeinsilico <- function(vec,M){
  alpha <- vec[1]
  beta  <- vec[2]
  S <- 0
  for(i in seq_len(nrow(M))){
    x <- M[i,1]
    LO <- alpha + beta * x
    p <- exp(LO)/(1+exp(LO))
    S <- S + f_vec_arn(p,M[i,2]-1,M[i,3]-1,log=TRUE)
  }
return(S)
}

likeinsilico(c(4,-6),M)
likeinsilico(c(4,-6.5),M)
likeinsilico(c(4,-5.5),M)

optim_ab <- optim(par=c(4,-6),likeinsilico,control=list(fnscale= -1),M=M)
optim_ab

a <- seq(from=3.8,to=4.2,by=0.01)
b <- seq(from=-6.4,to=-5.6,by=0.01)
Z <- as.matrix(expand.grid(a,b))
Z <- apply(Z,1,function(x){likeinsilico(x,M=M)})
Z <- matrix(Z,length(a),length(b))

Z <- Z-max(Z)
Z <- pmax(Z,-10)
contour(a,b,Z,levels=-c(0:10,15,20,30))
points(alpha_true,beta_true,pch=16,cex=4)
jj <- optim_ab$par
points(jj[1],jj[2],pch=4,cex=4,col='red')
grid()
persp(a,b,Z)

Mathematical Olympiad

sametest <- function(d1,d2,n){
  both <- c(d1,d2)
  bothmax <- MLE(both,n)

  d1max <- MLE(d1,n)
  d2max <- MLE(d2,n)
  Lambda <- (
  (f_vec_arn(d2max  ,d1,n,log=TRUE)+f_vec_arn(d2max  ,d2,n,log=TRUE)) -
  (f_vec_arn(bothmax,d1,n,log=TRUE)+f_vec_arn(bothmax,d2,n,log=TRUE))
)
  pchisq(-2*Lambda,df=1,lower.tail=FALSE)
}

a <- as.matrix(read.table("olympiad.txt"))
nrow(a)
d1 <- a["AUS",]
d2 <- a["NZL",]
d1
d2
sametest(d1,d2,100)
a["USA",]
a["CHN",]
sametest(a["USA",],a["CHN",],100)
sametest(a["USA",],a["CHN",],5)

Athletics

O <- read.table("olympic_athletics_mens_100m.txt",header=TRUE)
head(O)

jjf <- function(x){f_vec_arn(x,O$rank,O$n,log=TRUE)}
system.time(optolym <- optimize(jjf, c(0.45,0.5), maximum=TRUE))

optolym
jjf(0.5)
jjf(optolym$maximum)
(LR <- 2*(jjf(optolym$maximum)-jjf(0.5)))
pchisq(LR,df=1,lower.tail=FALSE)

a <- seq(from=0.3,to=0.65,len=45)
L <- f_vec_arn(a,O$rank,O$n,log=TRUE)
Lmax <- jjf(optolym$maximum)
Lmax
L <- L-Lmax

(a_lower <- uniroot(function(x){jjf(x)+2-Lmax},interval=c(0.3,0.5))$root)
(a_upper <- uniroot(function(x){jjf(x)+2-Lmax},interval=c(0.5,0.6))$root)

plotolymp <- function(...){
plot(a,L,type='l',ylab="log-likelihood",lwd=2)
abline(h=c(0,-2))
ahat <- optolym$maximum
segments(x0=ahat,y0=-1.5,y1=0)
text(ahat,-0.91,expression(hat(a)==0.573),pos=2)
abline(v=0.5,lty=3)

segments(x0=a_lower,y0=-1.5,y1=-3.5)
segments(x0=a_upper,y0=-1.5,y1=-3.5)

text(x=a_lower,y=-3,"0.380",pos=4)
text(x=a_upper,y=-3,"0.563",pos=2)
}
plotolymp()
pdf(file="plotolymp.pdf")
plotolymp()
dev.off()

Parkrun

results <- read.table("parkrun_results.txt",header=TRUE)
results

Above we see some parkrun results. In parkrun number 148, I placed 229 out of a field of 318.

S_parkrun <- function(a){f_vec_arn(a,r=results$place-1,results$runners-1,log=TRUE)}
(optparkrun <- optimize(S_parkrun, c(0.2,0.8),maximum=TRUE))

a <- seq(from=0.25,to=0.65,by=0.01)
S_parkrun_a <- S_parkrun(a) - optparkrun$objective
uniroot(function(a){S_parkrun(a) - optparkrun$objective+2},c(0.1,0.5))$root
uniroot(function(a){S_parkrun(a) - optparkrun$objective+2},c(0.5,0.9))$root
optparkrun$objective - S_parkrun(0.5)

```r\simeq 0.448$ has a support of zero. Dotted line shows $\mathcal{S}=-2$; intersection of this with the curve shows that the support interval runs from about 0.315 to 0.567"} plotparkrun <- function(...){ plot(a,S_parkrun_a,xlab="generalized Bradley-Terry strength",ylab="support",type='b',ylim=c(-5,0)) abline(h=c(0,-2),lty=2) segments(x0=0.5,y0=-1.5,y1=0) segments(x0=optparkrun$maximum,y0=-1,y1=0) text(optparkrun$maximum,-0.8,expression(hat(a)==0.448),pos=2) text(0.5,-1.5,expression(H[o]:a==0.5),pos=2) } plotparkrun() pdf("plotparkrun.pdf") plotparkrun() dev.off()

Figure \@ref(fig:plotparkrunlike) shows a support curve for the
author's generalized Bradley-Terry strength, normalized so that
$\mathcal{S}(\hat{a})=0$.  We see a support interval [that is,
$\left\lbrace a\colon\mathcal{S}(a)\geqslant -2\right\rbrace$] of
about 0.315 to 0.567.  We might also observe that the support for
$H_0\colon a=\frac{1}{2}$ [that is, that the author has a 50\% chance
of running faster than a randomly chosen Parkrun competitor] is about
0.35, less than Edwards's two units of support criterion.  We thus see
some support for Hankin's assertion that he is "as fit as anyone
else there".

```r
a <- seq(from=0.005,to=0.85,by=0.01)
La <- f_vec_arn(a,r=results$place[1]-1,results$runners[1]-1,log=TRUE)
La <- La - max(La,na.rm=TRUE)
plot(a,La,type='b',ylab='fishy')
abline(h=c(0,-2))

supp2 <- function(vec,place,runners){
  a_stirling <- vec[1]
  a_auckland <- vec[2]
  out <- 0
  for(i in 1:3){
    out <- out + f_vec_a(a_stirling,r=place[i]-1,runners[i]-1,log=TRUE)
  }
  for(i in 4:10){
    out <- out + f_vec_a(a_auckland,r=place[i]-1,runners[i]-1,log=TRUE)
  }
  return(out)
}

x <- seq(from=0.1,to=0.8,by=0.01)
jj <- as.matrix(expand.grid(x,x))
L <- apply(jj,1,supp2,results$place,results$runners)
L <- L - max(L,na.rm=TRUE)
L <- matrix(L,length(x),length(x))

par(pty='s')
contour(x,x,L,asp=1,xlim=range(x),ylim=range(x),levels=-(0:5))
abline(0,1)

f <- function(a){supp2(c(a,a),results$place,results$runners)}
a <- seq(from=0.2,to=0.6,by=0.01)
alike <- sapply(a,f)
plot(a,alike-max(alike,na.rm=TRUE),type='b',ylab='blong')
abline(h=c(0,-2))
optimize(f,c(0.1,0.6),maximum=TRUE)

f2 <- function(v){supp2(v,results$place,results$runners)}
optim(par=c(0.6,0.3),f2,control=list(fnscale=-1))

7380.314-7381.496
pchisq(2*(7381.496-7380.314),df=1,lower.tail=FALSE)

Above shows that there is no evidence to suggest that the Stirling strength differs from the Auckland strength. Now let's try a logistic-style regression:

results <- read.table("parkrun_results.txt",header=TRUE)
results
supp_logistic <- function(vec){
  alpha <- vec[1]
  beta <- vec[2]
  out <- 0
  for(i in seq_len(nrow(results))){
     LO <- alpha + beta*i
     strength <- exp(LO)/(1+exp(LO))
     out <- out + f_vec_arn(strength,results$place[i]-1,results$runners[i]-1,log=TRUE)
  }
  return(out+13400)
}

c(
supp_logistic(c(0,0)),
supp_logistic(c(0,0.01)),
supp_logistic(c(0,-0.01))
)

optim(par=c(0,0),fn=supp_logistic,control=list(fnscale=-1))
optim(par=c(0,0),fn=function(v){supp_logistic(vec=c(v[1],0))},control=list(fnscale=-1))

Education

The focal competitor

rank <- c( 9, 7, 2, 3,2 ,8 ,5 ,4 ,9 )
class_size <- c(12, 17,23, 9,13,14,13,12,15)
course <- c("rings and modules","group theory","calculus","linear algebra",
"differential equations","topology","special relativity","fluid mechanics","Lie algebra")
category=c("pure","pure","applied","applied","applied","pure","applied","applied","pure")
data.frame(course,category,rank,class_size)

a <- seq(from=0.2,by=0.01,to=0.8)
wp <- category=='pure'
wa <- category=='applied'

Pure first

Lpure <- f_vec_arn(a,rank[wp]-1,class_size[wp]-1,log=TRUE)
Lpure <- Lpure - max(Lpure,na.rm=TRUE)
plot(a,Lpure,ylab='pure strength')
optimize(function(a){f_vec_arn(a,rank[wp]-1,class_size[wp]-1,log=TRUE)},c(0.3,0.6),maximum=TRUE)
optimize(function(a){f_vec_arn(a,rank[wa]-1,class_size[wa]-1,log=TRUE)},c(0.3,0.6),maximum=TRUE)

supp2_ed <- function(vec,place1,place2,runners1,runners2){
  out <- 0
  M1 <- cbind(place1,runners1)
  M2 <- cbind(place2,runners2)
  for(i in seq_len(nrow(M1))){
    out <- out + f_vec_arn(vec[1],r=M1[i,1]-1,n=M1[i,2]-1,log=TRUE)
  }
  for(i in seq_len(nrow(M2))){
    out <- out + f_vec_arn(vec[2],r=M2[i,1]-1,n=M2[i,2]-1,log=TRUE)
  }
  return(out)
}

x <- seq(from=0.1,to=0.99,by=0.01)
jj <- as.matrix(expand.grid(x,x))
L <- apply(jj,1,supp2_ed,rank[wp]-1,rank[wa]-1,class_size[wp]-1,class_size[wa]-1)
L <- L - max(L,na.rm=TRUE)
L <- matrix(L,length(x),length(x))

plotpureandapplied <- function(...){
par(pty='s')
contour(x,x,L,asp=1,xlim=range(x),ylim=range(x),levels=-(0:5),
xlab='pure strength',ylab='applied strength')
abline(0,1)
}
plotpureandapplied()
pdf(file="plotpureandapplied.pdf")
plotpureandapplied()
dev.off()

jj1 <-
optim(
   c(.4,.8),
   function(vec){
      supp2_ed(vec,rank[wp]-1,rank[wa]-1,class_size[wp]-1,class_size[wa]-1)},
   control=list(fnscale = -1)
)
jj1

jj2 <-
optimize(function(v){supp2_ed(c(v,v),rank[wp]-1,rank[wa]-1,class_size[wp]-1,class_size[wa]-1)},
c(0.1,0.9),maximum=TRUE)
jj2

pchisq(2*(jj1$value-jj2$objective),lower.tail=FALSE,df=1)

RobinHankin/hyper2 documentation built on May 6, 2024, 4:25 p.m.