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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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\begin{document}

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<<setup, cache=FALSE, echo=FALSE>>=
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options(scipen=5)
options(repos=c(CRAN="https://cloud.r-project.org/"))
@

<<funcs,echo=FALSE>>=
myrnd <- function(x,d) {formatC(x, format="f", digits=d)}
library(shape)
library(diagram)
set.seed(10101)
@
%\frame{\titlepage}

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

Suppose we observe two data sets
\[
y_{11},\ldots,y_{1 n_1}
\]
and
\[
y_{21},\ldots,y_{2 n_2}
\]
and we want to assess whether the characteristics of the populations from which we believe the data are drawn are different.

\framebreak

The classical testing route proceeds as follows:

\begin{itemize}
  \item Assume $Y_{1i} \sim Normal(\mu_1,\sigma_1^2)$, $Y_{2i} \sim Normal(\mu_2,\sigma_2^2)$
  \item Carry out the hypothesis test 
  \[
  H_0: \mu_1 = \mu_2
  \]
  against some stated alternative, say
  \[
  H_a: \mu_1 \neq \mu_2
  \]
  
  \item Might assume $\sigma_1 = \sigma_2$
  
  \item Complete the test using the Rejection Region or p-value approach.
\end{itemize}

\framebreak

A common test statistic is
\[
T_{obs} = \dfrac{\overline{y}_1 - \overline{y}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}
\]
where
\[
\overline{y}_1 = \frac{1}{n_1} \sum_{i=1}^{n_1} y_{1i} \qquad \overline{y}_2 = \frac{1}{n_2} \sum_{i=1}^{n_2} y_{2i}
\]
and
\[
s_1^2 = \frac{1}{n_1-1} \sum_{i=1}^{n_1} (y_{1i} - \overline{y}_1)^2 \qquad s_2^2 = \frac{1}{n_2-1} \sum_{i=1}^{n_2} (y_{2i} - \overline{y}_2)^2
\]
and a large value of $|T_{obs}|$ supports the conclusion that $\mu_1 \neq \mu_2$.
\end{frame}


\begin{frame}[fragile,allowframebreaks]\frametitle{The Bayesian analogue}

We may reasonably assume that the data arise from two \emph{independent} infinitely exchangeable systems.

\medskip
That is, we assume that for any finite collections
\[
\mathbf{Y_1} = (Y_{11},\ldots,Y_{1 n_1})
\]
and
\[
\mathbf{Y_2} = (Y_{21},\ldots,Y_{1 n_2})
\]
we assume that
\[
f_{\bY_1,\bY_2}(\by_1,\by_2) = f_{\bY_1} (\by_1) f_{\bY_2} (\by_2)
\]

\framebreak


First assume that $\sigma_1^2$ and $\sigma_2^2$ are \emph{known}; we may reasonably assume independent Normal \emph{priors} for $\mu_1$ and $\mu_2$
\[
\pi_0(\mu_1) \equiv Normal(\eta_{10},\sigma_{10}^2/\lambda_{10}) \qquad \pi_0(\mu_2) \equiv Normal(\eta_{20},\sigma_{20}^2/\lambda_{20})
\]
so that by previous results the \emph{posteriors} are also Normal 
\[
\pi_n(\mu_1) \equiv Normal(\eta_{1n},\sigma_{10}^2/\lambda_{1n}) \qquad \pi_n(\mu_2) \equiv Normal(\eta_{2n},\sigma_{20}^2/\lambda_{2n})
\]
with the updated parameters given by the previously established formulae, and with $\mu_1$ and $\mu_2$ being \emph{independent}.

\framebreak
Also, note that the two parameters are \emph{independent}, that is
\[
\pi_n(\mu_1,\mu_2) = \pi_n(\mu_1) \pi_n(\mu_2)
\]
which follows as
\begin{itemize}
  \item the parameters are presumed to have independent priors,
  \item the two observable sequences are presumed independent of each other.
\end{itemize}

\framebreak
Consider the difference parameter
\[
\theta = \mu_1 - \mu_2.
\]
In light of the model specification and the data, and properties of the Normal distribution, we may deduce that the posterior distribution on $\theta$ is also Normal
\[
\pi_n(\theta) \equiv Normal(\eta_{1n} - \eta_{2n}, \sigma_{10}^2/\lambda_{1n}+\sigma_{20}^2/\lambda_{2n})
\]

\framebreak

The posterior parameters are given by
\[
\eta_{1n} = \frac{n_1 \overline{y}_{1} + \lambda_{10} \eta_{10}}{n_1 + \lambda_{10}} \qquad \qquad \lambda_{1n} = n_1 + \lambda_{10}
\]
and
\[
\eta_{2n} = \frac{n_2 \overline{y}_{2} + \lambda_{20} \eta_{20}}{n_2 + \lambda_{20}} \qquad \qquad \lambda_{2n} = n_2 + \lambda_{20}.
\]
The posterior uncertainty in $\theta$ is reflected by $\pi_n(\theta)$.

\framebreak

Now assume that $\sigma_1^2$ and $\sigma_2^2$ are \emph{unknown}.  We might assume
\[
\sigma_1^2 = \sigma_2^2
\]
for simplicity, but more generally would have to assume that $\sigma_1^2 \neq \sigma_2^2$.  

\medskip

In the latter case, we might specify an \emph{Inverse Gamma} prior

\begin{itemize}
  \item If $Z \sim Gamma(\alpha,\beta)$, then $X=1/Z \sim InvGamma(\alpha,\beta)$: by standard transformation results
  \[
  f_X(x) = \frac{1}{x^2} f_Z(1/x)
  \]
  \item We make this choice for analytical convenience.
  \item Here we assume that
  \[
  \pi_0(\sigma_j^2) \equiv InvGamma(\alpha_{j0},\beta_{j0}) \qquad j = 1, 2
  \]
\end{itemize}

\framebreak

After some calculations, we discover that the marginal posterior distribution for $\mu_j$, $j=1,2$ can be computed analytically as
\begin{align*}
\pi_n(\mu_j) = \frac{\Gamma((a_{jn}+1)/2)}{\Gamma(a_{jn}/2) \sqrt{\pi}} \left(\frac{\lambda_{jn}}{b_{jn}} \right)^{1/2} \left\{ 1+ \frac{\lambda_{jn}}{b_{jn}} \left(\mu_j - \eta_{jn} \right)^2 \right\}^{-(a_{jn} + 1)/2}
\end{align*}
where
\begin{align*}
a_{jn} & = n_j + \alpha_{j0}\\[6pt]
b_{jn} & = \frac{n_j \lambda_{j0}}{n_j + \lambda_{j0}} (\overline{y}_j- \eta_{j0})^2 + \sum_{i=1}^{n_j} (y_{ji} - \overline{y}_j)^2 + \beta_{j0}
\end{align*}


Often, this density is rewritten by using a rescaled version of $b_{jn}$, $j=1,2$:
\[
\phi_{jn} = \frac{ b_{jn}}{a_{jn} \lambda_{jn}}.
\]

\framebreak
Then we have
\[
\pi_n(\mu_j) = \frac{\Gamma((a_{jn}+1)/2)}{\Gamma(a_{jn}/2) \sqrt{\pi}} \frac{1}{a_{jn}^{1/2}} \left(\frac{1}{\phi_{jn}} \right)^{1/2} \left\{ 1+ \frac{1}{a_{jn}} \frac{\left(\mu_j - \eta_{jn} \right)^2}{\phi_{jn}} \right\}^{-(a_{jn} + 1)/2}
\]

This is the general \emph{Student-t} distribution with $a_{jn}$ degrees of freedom, centered at $\eta_{jn}$, with scale
parameter $\phi_{jn}^{1/2}$.

\framebreak

In this case, however, the posterior distribution of
\[
\theta = \mu_1 - \mu_2
\]
is hard to compute, as the difference of two Student-t random variables does not have a standard distribution.
\end{frame}

\begin{frame}[fragile,allowframebreaks]\frametitle{Numerical example}

Suppose we have data:
\begin{align*}
\textrm{Sample 1:} & \quad \Sexpr{n1<-6;y1<-round(rnorm(n1,2),2);myrnd(y1,2)}\\[6pt]
\textrm{Sample 2:} & \quad \Sexpr{n2<-4;y2<-round(rnorm(n2,5,2),2);myrnd(y2,2)}
\end{align*}
These data yield the following statistics:
<<echo=TRUE>>=
y1<-c(1.12, 1.25, 3.38, 2.24, 2.11, 3.20)
y2<-c(4.08, 6.28, 5.83, 6.55)
n1<-length(y1)
n2<-length(y2)
c(n1,n2)
y1bar<-mean(y1);y2bar<-mean(y2)
c(y1bar,y2bar)   #sample variances
c(var(y1),var(y2)) #sample vaeiances
@

\framebreak
\textbf{Prior specification:}  for simplicity, we might choose identical priors for the unknown parameters:
<<echo=TRUE,fig.height=5.25,fig.width=8>>=
eta10<-eta20<-0
lambda10<-lambda20<-0.1
al10<-2.0;be10<-2
al20<-2.0;be20<-2
yv<-seq(0.01,10,by=0.01)
pi0.sig<-dgamma(1/yv,al10,be10)/yv^2
par(mar=c(4,4,2,2))
plot(yv,pi0.sig,type='l',ylim=range(0,1),xlab=expression(sigma^2),
        ylab='Prior density')
title(expression(pi[0](sigma^2)))
@

\framebreak

Posterior parameters:
<<echo=TRUE>>=
eta1n<-(n1*y1bar+lambda10*eta10)/(n1+lambda10)
lambda1n<-lambda10+n1
eta2n<-(n2*y2bar+lambda20*eta20)/(n2+lambda20)
lambda2n<-lambda20+n2

c(eta1n, eta2n)

a1n<-al10+n1
b1n<-((n1*lambda10)/(n1+lambda10))*(y1bar-eta10)^2 + sum(y1-y1bar)^2 + be10
a2n<-al20+n2
b2n<-((n2*lambda20)/(n2+lambda20))*(y2bar-eta20)^2 + sum(y2-y2bar)^2 + be20

phi1n<-b1n/(a1n*lambda1n)
phi2n<-b2n/(a2n*lambda2n)
@

\framebreak
\textbf{Known variances:} Suppose we assume $\sigma_1^2 = 4$, $\sigma_2^2 = 2$.   Using the calculations above, we may compute and plot the posteriors for $\mu_1$ and $\mu_2$, and for $\theta = \mu_1 - \mu_2$:

<<echo=TRUE,fig.height=5.25,fig.width=8>>=
sig10<-2
sig20<-sqrt(2)
yv<-seq(0,10,by=0.01)
post1<-dnorm(yv,eta1n,sig10/sqrt(lambda1n))
post2<-dnorm(yv,eta2n,sig20/sqrt(lambda2n))
par(mar=c(4,4,2,2))
plot(yv,post1,type='l',ylim=range(0,1),xlab=expression(mu),
        ylab='Posterior density')
lines(yv,post2,type='l',col='red')
legend(0,1,c(expression(pi[n](mu[1])),expression(pi[n](mu[2]))),
        col=c('black','red'),lty=1)
@

\framebreak
Posterior for $\theta$:
<<echo=TRUE,fig.height=5.5,fig.width=8>>=
yv<-seq(-8,2,by=0.01)
post.th<-dnorm(yv,eta1n-eta2n,sqrt(sig10^2/lambda1n+sig20^2/lambda2n))
par(mar=c(4,4,2,2))
plot(yv,post.th,type='l',ylim=range(0,1),xlab=expression(theta),
        ylab='Posterior density')
legend(-8,1,c(expression(pi[n](theta))),lty=1)
@

\textbf{Unknown variances:}  Need to use the Student-t distributions.  

\medskip

The \texttt{dt} function computes the standard Student-t density, and we need the \emph{location-scale} transformed version: if
\[
X = c Z + m
\]
for constants $c > 0$ and $m$, then
\[
f_X(x) = \frac{1}{c} f_Z((x-m)/c)
\]

<<echo=TRUE,fig.height=5.25,fig.width=8>>=
yv<-seq(0,10,by=0.01)
post1<-dt((yv-eta1n)/sqrt(phi1n),df=a1n)/sqrt(phi1n)
post2<-dt((yv-eta2n)/sqrt(phi2n),df=a2n)/sqrt(phi2n)
par(mar=c(4,4,2,2))
plot(yv,post1,type='l',ylim=range(0,2),xlab=expression(mu),ylab='Posterior density')
lines(yv,post2,type='l',col='red')
legend(0,2,c(expression(pi[n](mu[1])),expression(pi[n](mu[2]))),
        col=c('black','red'),lty=1)
@

For the posteriors for $\sigma_1^2$ and $\sigma_2^2$:

<<echo=TRUE,fig.height=5.25,fig.width=8>>=
yv<-seq(0,5,by=0.01)
post1.sig<-dgamma(1/yv,a1n,b1n)/yv^2
post2.sig<-dgamma(1/yv,a2n,b2n)/yv^2
par(mar=c(4,4,2,2))
plot(yv,post1.sig,type='l',ylim=range(0,5), 
        xlab=expression(sigma^2),ylab='Posterior density')
lines(yv,post2.sig,type='l',col='red')
lines(yv,dgamma(1/yv,al10,be10)/yv^2,lty=3)
legend(0,5,c(expression(pi[n](sigma[1]^2)),
             expression(pi[n](sigma[2]^2)), 
             expression(pi[0])),col=c('black','red'),lty=c(1,1,3))
@

For $\theta$, it is not easy to compute the posterior distribution as the difference of two Student-t random variables does not have a standard distribution.

\medskip
Instead of using analytical calculations, we use \emph{sampling:}

\begin{itemize}
  \item draw large samples from each of the two posteriors
  
   \item compute the differences
   
   \item plot a histogram/density estimate of the differenced samples.
   
\end{itemize}

\framebreak

<<echo=TRUE,fig.height=5.25,fig.width=8>>=
N<-10000
post1.samp<-rt(N,df=a1n)*sqrt(phi1n)+eta1n
post2.samp<-rt(N,df=a2n)*sqrt(phi2n)+eta2n
theta.samp<-post1.samp-post2.samp
dest<-density(theta.samp,bw=0.1)
par(mar=c(4,4,2,2))
hist(theta.samp,breaks=seq(-8,2,by=0.1),ylim=range(0,1),
    main=expression(pi[n](theta)),xlab=expression(theta),freq=FALSE)
lines(dest,col='red')
box()
@

\framebreak

To complete the analysis, we examine $\pi_n(\theta)$:

\begin{itemize}
  \item does the value 0 lie in the \emph{high probability} region ?
  \item what is the \emph{variance} of the posterior distribution ?
\end{itemize}

All reporting and decisions are performed via the posterior distribution.

\end{frame}
\end{document}