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\title{Propensity Score Methods, Models and Adjustment}
\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}

%\part{NHANES Example}

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

%\frame{\titlepage}

\begin{frame}[fragile,allowframebreaks]\frametitle{Exchangeability: Variance components model}

A \emph{variance components} model is a form of `random effects' model for observable quantities that takes the form
\[
Y_i = M + \epsilon_i
\]
for $i=1,\ldots,n$, where

\begin{itemize}

\item $M$ is a random variable

\item $\epsilon_1,\ldots,\epsilon_n$ are identically distributed random variables with
\[
\E[\epsilon_i] = 0
\]

\end{itemize}
with \emph{all variables independent}.


\framebreak

Suppose

\begin{itemize}

\item $M \sim Normal(0,\tau^2)$,

\item $\epsilon_1,\ldots,\epsilon_n \sim Normal(0,\sigma^2)$,

\end{itemize}

so that \emph{conditional} on $M=m$
\[
Y_1,\ldots,Y_n|M=m \sim Normal(m,\sigma^2)
\]
are independent.

\framebreak

Then, by a standard marginalization calculation,

{\scriptsize
\begin{align*}
f_{Y_1,\ldots,Y_n}&(y_1,\ldots,y_n) = \int_{-\infty}^\infty \prod_{i=1}^n f_{Y_i|M}(y_i|m) f_M(m) \ dm \\[6pt]
& = \int_{-\infty}^\infty \left\{ \prod_{i=1}^n \left(\dfrac{1}{2 \pi \sigma^2}\right)^{1/2} \exp\left\{-\dfrac{1}{2 \sigma^2} (y_i-m)^2 \right\} \right\} \left(\dfrac{1}{2 \pi \tau^2}\right)^{1/2} \exp\left\{-\dfrac{1}{2 \tau^2} m^2 \right\} dm \\[6pt]
& = \left(\dfrac{1}{2 \pi}\right)^{(n+1)/2} \left(\frac{1}{\sigma^2} \right)^{n/2} \left(\frac{1}{\tau^2} \right)^{1/2} \int_{-\infty}^\infty \exp\left\{-\dfrac{1}{2} \underbrace{\left[ \frac{1}{\sigma^2} \sum_{i=1}^n (y_i - m)^2 + \frac{1}{\tau^2} m^2 \right]} \right\} \ dm.
\end{align*}}
We need to integrate with respect to $m$.

\framebreak
By a standard sums-of-squares decomposition
\[
\sum_{i=1}^n (y_i - m)^2 = \sum_{i=1}^n (y_i - \ybar)^2 + n (\ybar - m)^2.
\]

Also, recall the completing-the-square formula
\[
A(t-a)^2 + B(t-b)^2 = (A+B) \left( t - \frac{Aa+Bb}{A+B} \right)^2 + \frac{AB}{A+B} (a-b)^2.
\]

\framebreak
Therefore we may rewrite
\[
\frac{1}{\sigma^2} \sum_{i=1}^n (y_i - m)^2 + \frac{1}{\tau^2} m^2
\]
as
\[
\frac{1}{\sigma^2}\sum_{i=1}^n (y_i - \ybar)^2 + \frac{n}{\sigma^2} ( m - \ybar)^2 + \frac{1}{\tau^2} m^2
\]
and then to combine the second and third terms, take
\[
A = \frac{n}{\sigma^2} \quad a = \ybar \qquad B = \frac{1}{\tau^2} \quad b = 0
\]
in the above formula.

\framebreak
We have
{\scriptsize
\[
\frac{n}{\sigma^2} ( m - \ybar)^2 + \frac{1}{\tau^2} m^2  = \left(\frac{n}{\sigma^2} + \frac{1}{\tau^2} \right) \left(m - \frac{n \ybar/\sigma^2}{(n/\sigma^2) + (1/\tau^2)} \right)^2 + \frac{n/(\sigma^2 \tau^2)}{(n/\sigma^2)+(1/\tau^2)} \ybar^2
\]}
which we may rewrite as
\[
\frac{1}{\lambda^2} \left(m - \mu \right)^2 + \frac{n}{n \tau^2 + \sigma^2} \ybar^2
\]
where
\[
\mu = \frac{n \ybar/\sigma^2}{(n/\sigma^2) + (1/\tau^2)} = \frac{n \tau^2 \ybar }{n \tau^2 + \sigma^2}  \qquad \lambda^2 = \left(\frac{\sigma^2 \tau^2}{n \tau^2 + \sigma^2} \right)
\]

\framebreak
Therefore, for the integral
{\scriptsize
\begin{align*}
\int_{-\infty}^\infty &\exp\left\{-\dfrac{1}{2} \left[ \frac{1}{\sigma^2} \sum_{i=1}^n (y_i - m)^2 + \frac{1}{\tau^2} m^2 \right] \right\} \ dm \\[6pt]
& = \exp\left\{-\frac{1}{2} \left[\frac{1}{\sigma^2} \sum_{i=1}^n (y_i - \ybar)^2 + \frac{n}{n \tau^2 + \sigma^2} \ybar^2 \right] \right\} \int_{-\infty}^\infty \exp\left\{ -\frac{1}{2 \lambda^2} (m-\mu)^2 \right\} \ dm \\[6pt]
& = \exp\left\{-\frac{1}{2} \left[\frac{1}{\sigma^2} \sum_{i=1}^n (y_i - \ybar)^2 + \frac{n}{n \tau^2 + \sigma^2} \ybar^2 \right] \right\} \left( 2 \pi \lambda^2 \right)^{1/2}
\end{align*}}
as the integrand is proportional to a Normal pdf.

\framebreak

Thus for $(y_1,\ldots,y_n)  \in \mathbb{R}^n$,
{\scriptsize
\begin{align*}
f_{Y_1,\ldots,Y_n}& (y_1,\ldots,y_n) \\[6pt]
& = \left(\dfrac{1}{2 \pi}\right)^{n/2} \left(\frac{1}{\sigma^2} \right)^{n/2}\left(\frac{1}{\tau^2} \right)^{1/2} \lambda \exp\left\{-\frac{1}{2} \left[\frac{1}{\sigma^2} \sum_{i=1}^n (y_i - \ybar)^2 + \frac{n}{n \tau^2 + \sigma^2} \ybar^2 \right] \right\}
\end{align*}}
which also relies only upon the summary statistics
\[
s_1 = \ybar \qquad s_2 = \sum\limits_{i=1}^n (y_i - \ybar)^2
\]
and so we may deduce \emph{exchangeability}, as the statistics are invariant to the indexing of the $y$s.

\framebreak

Simulation: $n=5$, $\tau^2=3$, $\sigma^2 = 1$

<<Ex01,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in'>>=
set.seed(213)
n<-5
sim.exch01<-function(nv,tauv,sigv){  #Sample the exchangeable variables.
    Mv<-rnorm(1)*tauv
    Yv<-rnorm(nv,Mv,sigv)
}
tau<-sqrt(3)
sig<-sqrt(1)

Ymat<-t(replicate(2000,sim.exch01(n,tau,sig)))  #2000 replicate draws of Y
@

<<Ex02,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=FALSE>>=
par(pty='s',mar=c(4,4,3,1))
labs<-c(expression(Y[1]),expression(Y[2]),expression(Y[3]),expression(Y[4]),expression(Y[5]))
boxplot(Ymat,names=labs,pch=19,cex=0.5)
pairs(Ymat,pch=19,cex=0.5,labels=labs)
@
We have that for $i=1,\ldots,n$,
\[
\E_{Y_i}[Y_i] = \E_{M}[M] + \E_{\epsilon_i}[\epsilon_i] = 0
\]
and by independence
\[
\Var_{Y_i}[Y_i] = \Var_{M}[M] + \Var_{\epsilon_i}[\epsilon_i] = \tau^2 + \sigma^2 =  3 + 1 = 4.
\]
<<Ex03,comment='+', fig.width=6, fig.height=6, fig.align='center'>>=
apply(Ymat,2,mean)
apply(Ymat,2,var)
@

\framebreak

For the covariances, using iterated expectation we have
\begin{align*}
\Cov_{Y_i,Y_j}[Y_i,Y_j] & \equiv \E_{Y_i,Y_j}[Y_i Y_j] \\[6pt]
& = \E_M \left[ \E_{Y_i,Y_j|M}[Y_i Y_j|M] \right] \\[6pt]
& = \E_M \left[ \E_{Y_i|M}[Y_i|M] \E_{Y_j|M}[Y_j|M] \right]
\end{align*}
as $Y_i$ and $Y_j$ have expectation zero, and are conditionally independent given $M$.

\framebreak

Thus, as $\E_{Y_i|M}[Y_i|M] = M$ for each $i$, we have
\[
\Cov_{Y_i,Y_j}[Y_i,Y_j] = \E_M [ M^2 ] = \Var_M[M] + \{ \E_M[M]\}^2 = \tau^2
\]
and hence
\[
\Corr_{Y_i,Y_j}[Y_i,Y_j] = \frac{\Cov_{Y_i,Y_j}[Y_i,Y_j]}{\sqrt{\Var_{Y_i}[Y_i]\Var_{Y_j}[Y_j]}} = \frac{\tau^2}{\tau^2 + \sigma^2} = \frac{3}{4}.
\]
<<Ex04,comment='+', fig.width=6, fig.height=6, fig.align='center'>>=
round(cor(Ymat),3)
@

\framebreak
Note: if $\tau \longrightarrow 0$,
\[
\Corr_{Y_i,Y_j}[Y_i,Y_j] \longrightarrow 0
\]
and the $Y$s are \emph{uncorrelated}.  In fact, in this case the $Y$s are \emph{independent} as
\[
M = 0
\]
with probability 1, so $Y_i \equiv \epsilon_i$.

\framebreak
In vector form, we have
\[
\bY = M \One_n + \epsilon
\]
where 

\begin{itemize}

\item $\One_n= (1,1,\ldots,1)^\top$ is the $n \times 1$ vector of 1s.

\item $\epsilon = (\epsilon_1,\epsilon_2,\ldots,\epsilon_n)^\top$.

\end{itemize}
That is
\[
\begin{bmatrix}
  Y_1 \\
  Y_2 \\
  \cdots \\
  Y_n
\end{bmatrix} = M \begin{bmatrix}
  1 \\
  1 \\
  \cdots \\
  1
\end{bmatrix} + \begin{bmatrix}
  \epsilon_1 \\
  \epsilon_2 \\
  \cdots \\
  \epsilon_n
\end{bmatrix}.
\]

\framebreak

By properties of vector random variables, we have that marginally
\[
\bY \sim Normal_n(\Zero,\Sigma)
\]
where
\[
\Sigma = \tau^2 \One_n \One_n^\top + \sigma^2 \Ident_n.
\]
with
\begin{itemize}
  \item $\One_n \One_n^\top$ an $n \times n$ matrix of 1s
  \item $\Ident_n$ the $n \times n$ identity matrix
\end{itemize}


\end{frame}


\end{document}