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\title{The Two Sample Bernoulli Model}
\author{Dr David A. Stephens}
\institute{
  Department of Mathematics \& Statistics\\
 McGill University\\
 Montreal, QC, Canada.

  \vspace{0.1 in}

  \texttt{david.stephens@mcgill.ca}\\
  \texttt{\texttt{www.math.mcgill.ca/dstephens/SISCER2020/}}
}

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numberstyle=\tiny, stepnumber=1, numbersep=5pt}


\begin{document}

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<<setup, cache=FALSE, echo=FALSE>>=
library(knitr)
# global chunk options
opts_chunk$set(cache=TRUE, autodep=TRUE, size = "tiny")
options(scipen=5)
options(repos=c(CRAN="https://cloud.r-project.org/"))
@

%\frame{\titlepage}

\begin{frame}[fragile,allowframebreaks]\frametitle{A Two Sample Problem}

Suppose we have two infinite sequences
\begin{align*}
\{\bY_{1}\} & =  Y_{11},Y_{12},\ldots,Y_{1n},\ldots\\
\{\bY_{2}\} & =  Y_{21},Y_{22},\ldots,Y_{2n},\ldots
 \end{align*}
For two finite collections of sizes $n_1$ and $n_2$, we have the de Finetti representation
\[
f_{\bY_1,\bY_2}(\by_1,\by_2) = \iint \prod_{i=1}^{n_1} f_{1}(y_{1i};\theta_1) \prod_{i=1}^{n_2} f_{2}(y_{2i};\theta_2) \pi_0(d \theta_1, d \theta_2)
\]

\framebreak

We can make different assumptions about a potential joint structure.

\begin{enumerate}[label=(\alph*)]
  \item exchangeability \emph{within} each sequence and \emph{independence} across sequences
\[
\pi_0(d \theta_1, d \theta_2) = \pi_0(d \theta_1) \pi_0(d \theta_2)
\]
that is, $\theta_1$ and $\theta_2$ are presumed independent \textit{a priori}.

\framebreak



  \item exchangeability \emph{within} each sequence
\[
\pi_0(d \theta_1, d \theta_2)
\]
has a general joint form; this is an example of \emph{partial exchangeability}, based on the sequence labels 1 and 2.

\framebreak

  \item \emph{complete} exchangeability amongst all observables
\[
\pi_0(d \theta_1, d \theta_2)
\]
is singular, and concentrates on the line $\theta_1 = \theta_2$.  That is, there is in fact only one parameter
\[
\theta = \theta_1 = \theta_2
\]
and a prior $\pi_0$ on this parameter.
\end{enumerate}


\framebreak
We focus on the case (a).  In the binary observable case.
\[
p_{Y_{ji}}(y;\theta_j) = \theta_j^{y} (1-\theta_j)^{1-y} \qquad j = 1,2
\]
for each $0 \leq \theta \leq 1$. Suppose that, independently,
\[
\pi_0(\theta_j) \equiv Beta(\alpha_0,\beta_0).
\]
for $\alpha_0, \beta_0 > 0$.

\framebreak

We have for the posterior densities for $j=1,2$.
\[
\pi_n(\theta_j) \equiv Beta(s_{jn}+\alpha_0,n_j-s_{jn}+\beta_0) \equiv Beta(\alpha_{jn},\beta_{jn})
\]
where
\[
s_{1n} = \sum_{i=1}^{n_1} y_{1i} \qquad s_{2n} = \sum_{i=1}^{n_2} y_{2i}
\]



\framebreak

Simulation: $n_1=8, n_2=12$, $\alpha_0 = 2,\beta_0 = 1.5$.  Suppose that, in reality the true values of the parameters are
\[
\theta_{10} = 0.5 \qquad \theta_{20} = 0.7
\]
which are in the support of the prior.

<<Ex01,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in'>>=
set.seed(213)
al0<-2.0
be0<-1.5
n1<-8
n2<-12
nreps<-4
th10<-0.5
th20<-0.7
ymat1<-t(replicate(nreps,rbinom(n1,1,th10)))
ymat2<-t(replicate(nreps,rbinom(n2,1,th20)))
@


\framebreak

<<Ex07,comment='+', fig.width=9, fig.height=6, fig.align='center',out.width='4.2in'>>=
#Data
s1<-apply(ymat1,1,sum)
s2<-apply(ymat2,1,sum)
al.n1<-al0+s1
be.n1<-be0+n1-s1
al.n2<-al0+s2
be.n2<-be0+n2-s2
xv<-seq(0,1,length=1001)
par(mar=c(4,4,2,1),mfrow=c(2,2))
for(irep in 1:4){
    plot(xv,dbeta(xv,al.n1[irep],be.n1[irep]),type='l',col='red',
        ylim=range(0,6),xlab='',ylab=expression(pi[n]))
    lines(xv,dbeta(xv,al.n2[irep],be.n2[irep]),type='l',col='blue')
    legend(0,6,c('j=1','j=2'),col=c('red','blue'),lty=1)
}
mtext("Four data sets", side = 3, line = -1, cex=1.15, outer = TRUE)
@

\framebreak
We can report

\begin{enumerate}[label=\arabic*.]
  \item Parameter Estimates:
  \begin{itemize}
    \item Posterior Mean: $\alpha_n/(\alpha_n+\beta_n)$
    \item Posterior Median: 50\% quantile of $\pi_n(\theta)$
    \item Posterior Mode: $(\alpha_n-1)/(\alpha_n+\beta_n-2)$
  \end{itemize}

\begin{table}
  \centering
  \begin{tabular}{|l|c|c|c|c|}
     \hline
     $\theta_{1}$     & 1 & 2 & 3 & 4 \\
          \hline
     Mean & \Sexpr{round(al.n1[1]/(al.n1[1]+be.n1[1]),4)} & \Sexpr{round(al.n1[2]/(al.n1[2]+be.n1[2]),4)}  & \Sexpr{round(al.n1[3]/(al.n1[3]+be.n1[3]),4)}  & \Sexpr{round(al.n1[4]/(al.n1[4]+be.n1[4]),4)}  \\
     Median & \Sexpr{round(qbeta(0.5,al.n1[1],be.n1[1]),4)} & \Sexpr{round(qbeta(0.5,al.n1[2],be.n1[2]),4)} & \Sexpr{round(qbeta(0.5,al.n1[3],be.n1[3]),4)}  & \Sexpr{round(qbeta(0.5,al.n1[4],be.n1[4]),4)}  \\
     Mode & \Sexpr{round((al.n1[1]-1)/(al.n1[1]+be.n1[1]-2),4)} & \Sexpr{round((al.n1[2]-1)/(al.n1[2]+be.n1[2]-2),4)}  & \Sexpr{round((al.n1[3]-1)/(al.n1[3]+be.n1[3]-2),4)}  & \Sexpr{round((al.n1[4]-1)/(al.n1[4]+be.n1[4]-2),4)}  \\
     \hline
   \end{tabular}
\end{table}

\begin{table}
  \centering
  \begin{tabular}{|l|c|c|c|c|}
     \hline
     $\theta_{2}$     & 1 & 2 & 3 & 4 \\
          \hline
     Mean & \Sexpr{round(al.n2[1]/(al.n2[1]+be.n2[1]),4)} & \Sexpr{round(al.n2[2]/(al.n2[2]+be.n2[2]),4)}  & \Sexpr{round(al.n2[3]/(al.n2[3]+be.n2[3]),4)}  & \Sexpr{round(al.n2[4]/(al.n2[4]+be.n2[4]),4)}  \\
     Median & \Sexpr{round(qbeta(0.5,al.n2[1],be.n2[1]),4)} & \Sexpr{round(qbeta(0.5,al.n2[2],be.n2[2]),4)} & \Sexpr{round(qbeta(0.5,al.n2[3],be.n2[3]),4)}  & \Sexpr{round(qbeta(0.5,al.n2[4],be.n2[4]),4)}  \\
     Mode & \Sexpr{round((al.n2[1]-1)/(al.n2[1]+be.n2[1]-2),4)} & \Sexpr{round((al.n2[2]-1)/(al.n2[2]+be.n2[2]-2),4)}  & \Sexpr{round((al.n2[3]-1)/(al.n2[3]+be.n2[3]-2),4)}  & \Sexpr{round((al.n2[4]-1)/(al.n2[4]+be.n2[4]-2),4)}  \\
     \hline
   \end{tabular}
\end{table}

\framebreak

  \item Posterior Credible Intervals: For $0 < \gamma < 1$, find $\mathcal{C}_{\gamma}$ such that
  \[
  \int_{\mathcal{C}_{\gamma}} \pi_n(\theta) \ d \theta = \gamma
  \]
  which records a region where $\theta$ is inferred to lie with posterior probability $\gamma$.

  \medskip

  Typically we choose $\gamma=0.95$.  There are several ways to construct such an interval

\framebreak

\emph{Equal tail probability interval:}  find the $1-\gamma/2$ and $1-(1-\gamma/2)$ quantiles of the posterior.

<<Ex08,comment='+', fig.width=8, fig.height=7, fig.align='center',out.width='3.2in'>>=
gam<-0.95
par(mar=c(4,2,2,1))
plot(xv,dbeta(xv,al.n1[1],be.n1[1]),type='l',
        ylim=range(0,4),xlab='',ylab=expression(pi[n]))
eti.l<-qbeta((1-gam)/2,al.n1[1],be.n1[1])    #Lower tail quantile
eti.u<-qbeta(1-(1-gam)/2,al.n1[1],be.n1[1])  #Upper tail quantile
eti.x<-seq(eti.l,eti.u,length=1001)
eti.y<-dbeta(eti.x,al.n1[1],be.n1[1])
polygon(c(eti.x,eti.u,rev(eti.x),eti.l),
        c(rep(0,length(eti.x)),eti.y[length(eti.y)],rev(eti.y),0),
        col='cyan')
@

\framebreak

\emph{Highest posterior density (HPD) interval:} find $t_L$ and $t_U$ such that $\pi_n(t_L) = \pi_n(t_U)$ with
  \[
  \int_{t_L}^{t_U} \pi_n(\theta) \ d \theta = \gamma
  \]
<<Ex09,comment='+', fig.width=8, fig.height=7, fig.align='center',out.width='3.2in'>>=
library(HDInterval)
par(mar=c(4,2,2,1))
xv<-seq(0,1,length=1001)
yv<-dbeta(xv,al.n1[1],be.n1[1])

hpd.int<-hdi(qbeta, 0.95, shape1=al.n1[1], shape2=be.n1[1])
hpd.int

plot(xv,yv,type='l',ylim=range(0,4),xlab='',ylab=expression(pi[n]))
hpd.l<-hpd.int[1]    #Lower tail value
hpd.u<-hpd.int[2]    #Upper tail value
hpd.x<-seq(hpd.l,hpd.u,length=1001)
hpd.y<-dbeta(hpd.x,al.n1[1],be.n1[1])
polygon(c(hpd.x,hpd.u,rev(hpd.x),hpd.l),
        c(rep(0,length(hpd.x)),hpd.y[length(hpd.y)],rev(hpd.y),0),
        col='cyan')


@

\framebreak

\emph{Shortest interval:}

<<Ex10,comment='+', fig.width=8, fig.height=7, fig.align='center',out.width='3.2in'>>=
lowp<-seq(0.001,1-gam-0.001,by=0.001)
lowq<-qbeta(lowp,al.n1[1],be.n1[1])
upq<-qbeta(lowp+gam,al.n1[1],be.n1[1])
minval<-which.min(upq-lowq)

plot(xv,yv,type='l',ylim=range(0,4),xlab='',ylab=expression(pi[n]))
spd.l<-lowq[minval]    #Lower tail value
spd.u<-upq[minval]     #Upper tail value
spd.x<-seq(spd.l,spd.u,length=1001)
spd.y<-dbeta(spd.x,al.n1[1],be.n1[1])
polygon(c(spd.x,spd.u,rev(hpd.x),hpd.l),
        c(rep(0,length(spd.x)),spd.y[length(hpd.y)],rev(spd.y),0),
        col='cyan')
@

\framebreak

<<Ex11,comment='+', fig.width=8, fig.height=7, fig.align='center',out.width='3.2in'>>=
#Equal-tailed
c(eti.l,eti.u)                 #Ends of interval
pbeta(eti.l,al.n1[1],be.n1[1]) #Left-tail probability

#Highest posterior density
c(hpd.l,hpd.u)                 #Ends of interval
pbeta(hpd.l,al.n1[1],be.n1[1]) #Left-tail probability

#Shortest
c(spd.l,spd.u)                 #Ends of interval
pbeta(spd.l,al.n1[1],be.n1[1]) #Left-tail probability
@

\end{enumerate}

\framebreak
We may also examine parameters that \emph{compare} the two samples:

\begin{itemize}
  \item Difference
  \[
  \delta = \theta_2 - \theta_1
  \]
  \item Ratio
  \[
  \lambda = \frac{\theta_2}{\theta_1}
  \]
  \item Odds Ratio
  \[
  \phi = \frac{\theta_2/(1-\theta_2)}{\theta_1/(1-\theta_1)}
  \]
\end{itemize}

In principle the posteriors for each of these derived parameters can be computed directly from
\[
\pi_n(\theta_1) \qquad \pi_n(\theta_2)
\]
using standard transformation methods.

\framebreak
For example, for $t > 0$,
\[
\Pr[\lambda \leq t] = \displaystyle \iint_{\mathcal{A}_t} \pi_n(\theta_1) \pi_n(\theta_2) \  d \theta_2 \ d \theta_1
\]
where
\[
\mathcal{A}_t = \{ (x_1,x_2) : 0 < x_1,x_2 < 1, x_2 \leq t x_1\}.
\]
that is
\[
\Pr[\lambda \leq t] = \int_0^1 \int_0^{t \theta_1} \pi_n(\theta_1) \pi_n(\theta_2) \ d \theta_2 \ d \theta_1
\]
This can be computed numerically.

\framebreak
However, it is very straightforward to use a \emph{sampling} approach:

\begin{itemize}

  \item sample 10000 values from $\pi_n(\theta_1)$
  \item sample 10000 values from $\pi_n(\theta_2)$
  \item transform samples to obtain samples of $\lambda$

\end{itemize}

<<Ex12,comment='+', fig.width=8, fig.height=6.5, fig.align='center',out.width='3.6in'>>=
nsamp<-10000
th1.samp<-rbeta(nsamp,al.n1[1],be.n1[1])
th2.samp<-rbeta(nsamp,al.n2[1],be.n2[1])
lam.samp<-th2.samp/th1.samp
hist(lam.samp,freq=FALSE,nclass=50,xlab=expression(lambda),
        main=expression(pi[n](lambda)))
box()
@

\framebreak

We can compute a HPD credible interval from the sample using the \texttt{hdi} function:
<<Ex13,comment='+', fig.width=8, fig.height=7, fig.align='center',out.width='3.6in'>>=
hdi(lam.samp, 0.95)
@

\end{frame}



\end{document}