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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)
# global chunk options
opts_chunk$set(cache=TRUE, autodep=TRUE, size = "scriptsize")
options(scipen=5)
options(repos=c(CRAN="https://cloud.r-project.org/"))
@

%\frame{\titlepage}

\begin{frame}[fragile,allowframebreaks]\frametitle{Constructing a Markov chain}

A Markov chain is a sequence of random variables
\[
X_0,X_1,X_2,\ldots,X_t,\ldots
\]
defined so that the distribution of $X_t$ conditional on the preceding variables $X_0,X_1,\ldots,X_{t-1}$ depends \emph{only} on the value of $X_{t-1}$.

\medskip

We think about generating the sequence in order
\[
X_0 \longrightarrow X_1 \longrightarrow X_2 \longrightarrow \cdots
\]

\framebreak

The simplest type of chain is defined so that each variable takes values only on a \emph{finite} set
\[
\mathcal{S}_X = \{1,2,\ldots,d\}
\]
say: each \emph{marginal} distribution
\[
\Pr[X_t = x_t]
\]
and each \emph{conditional} distribution
\[
\Pr[X_t = x_t | X_{t-1} = x_{t-1}]
\]
is simply a discrete distribution $\mathcal{S}_X$, that is, a collection of $d$ probabilities that sum to 1.

\framebreak

Suppose that
\[
\Pr[X_t = j | X_{t-1} = i] = P_{ij}  \qquad i,j=1,\ldots,d.
\]
then we need that
\[
\sum_{j=1}^d P_{ij} = 1 \qquad \textrm{for each }i
\]
The \emph{transition matrix} $P$ is the matrix of the $P_{ij}$ values.

\medskip

Note that the probabilities do not depend on $t$.

\framebreak

Suppose also that the \emph{initial} ($t=0$) probabilities are
\[
\Pr[X_0 = i] = p_i^{(0)} \qquad i=1,\ldots,d
\]
that is, the distribution of $X_0$ is the discrete distribution
\[
p^{(0)} = (p_1^{(0)},\ldots,p_d^{(0)}).
\]

\framebreak
Note that we can use the Theorem of Total Probability to conclude that
\[
\Pr[X_t = j] = \sum_{i=1}^d \Pr[X_{t-1} = i] \Pr[X_t = j|X_{t-1} = i]
\]
that is
\[
p_j^{(t)} = \sum_{i=1}^d p_i^{(t-1)} P_{ij}
\]
so in vector form
\[
p^{(t)} = p^{(t-1)} P 
\]

\framebreak

Suppose $d=4$, and
\[
P = \begin{bmatrix}
      0.1 & 0.3 & 0.1 & 0.5 \\
      0.3 & 0.4 & 0.0 & 0.3 \\
      0.3 & 0.1 & 0.2 & 0.4 \\
      0.3 & 0.3 & 0.3 & 0.1
    \end{bmatrix}
\qquad
p^{(0)} = \begin{bmatrix}
            0.2 & 0.1 & 0.2 & 0.5
          \end{bmatrix}
\]
That is
\begin{align*}
  \Pr[X_t = 2 | X_{t-1} = 1] & = 0.3 \\
  \Pr[X_t = 3 | X_{t-1} = 2] & = 0.0 \\
  \Pr[X_t = 4 | X_{t-1} = 4] & = 0.1
\end{align*}
etc.

\framebreak
<<MC1,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
par(mar=c(4,4,3,1))
n<-8
d<-4
p.t<-matrix(0,nrow=n+1,ncol=d)
p.t[1,]<-c(0.2,0.1,0.2,0.5)
P<-matrix(0,d,d)
P[1,]<-c(0.1,0.3,0.1,0.5)
P[2,]<-c(0.3,0.4,0.0,0.3)
P[3,]<-c(0.3,0.1,0.2,0.4)
P[4,]<-c(0.3,0.3,0.3,0.1)
P
@

\framebreak

<<MC2,comment='+', fig.width=8, fig.height=5.5, fig.align='center',echo=TRUE>>=
#One sample run
set.seed(234)
X.t<-rep(0,n+1)
X.t[1]<-sample(1:d,size=1,prob=p.t[1,])
for(i in 2:(n+1)){
    X.t[i]<-sample(1:d,size=1,prob=P[X.t[i-1],])
    p.t[i,]<-p.t[i-1,] %*% P
}
X.t

plot(0:n,X.t,pch=19,
    xlab='t',ylab=expression(X[t]),ylim=range(0.5,4.5))
arrows(x0=0:(n-1),y0=X.t[1:n],x1=1:n,y1=X.t[2:(n+1)])
@

\framebreak

For the state distribution $p^{(t)}$ for $t=0,1,\ldots,8$
<<MC3,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
rownames(p.t)<-paste('t=',0:8,sep='')
p.t
@
We can see that this distribution is becoming a \emph{constant} vector; it takes a few steps to become approximately constant.

\framebreak
The \emph{stationary} or \emph{invariant} distribution, $\pi$, is a discrete distribution that solves
\[
\pi = \pi P.
\]
This is an \emph{eigenvalue} problem:
<<MC4,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
eigP<-eigen(P)                           #Eigen decomposition
eigvalsP<-eigP$values                    #Eigenvalues
eigvP<-eigP$vectors                      #Right eigenvectors
eigvlP<-solve(eigvP)                     #Left eigenvectors
P - eigvP %*% diag(eigvalsP) %*% eigvlP  #Check
@

\framebreak

To find the stationary distribution we select the eigenvector corresponding to the eigenvalue equal to 1 (which here is the \emph{first} eigenvalue in \texttt{R}), suitably standardized

<<MC5,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
eigvalsP
pi.vec<-eigvlP[1,]/sum(eigvlP[1,])
pi.vec                             #Stationary distribution
pi.vec %*% P                       #Verifying the solution
@

\framebreak

Note also that by recursion
\[
p^{(t)} = p^{(t-1)} P = p^{(t-2)} P^2 = \cdots = p^{(1)} P^{t-1} = p^{(0)} P^t
\]
so if we observe
\[
p^{(t)} \longrightarrow \pi
\]
as $t \longrightarrow \infty$, whatever the value of $p^{(0)}$, then this implies that we can compute $\pi$ by inspecting $P^t$ as $t \longrightarrow \infty$, that is
\[
P^t \longrightarrow \One \pi
\]
where $\One$ is the $d \times 1$ vector of 1s.

\framebreak
<<MC6,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
P.t<-array(0,c(1000,d,d))
P.t[1,,]<-P
for(k in 2:1000){
    P.t[k,,]<-P %*% P.t[k-1,,]
}
P.t[1000,,]
pi.vec
@

\framebreak

We can use the Markov chain to simulate values from $\pi$ by collecting the $X_t$ values over a long run.  Here $n=10000$.

<<MC7,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
n<-10000
X.t<-rep(0,n+1)
X.t[1]<-sample(1:d,size=1,prob=p.t[1,])
for(i in 2:(n+1)){
    X.t[i]<-sample(1:d,size=1,prob=P[X.t[i-1],])
}
table(X.t)/n
pi.vec
@

\framebreak

The $X_t$ values are \emph{correlated}, for example
<<MC8,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
cor(X.t[-1],X.t[-n])
@
so the collected samplea are \emph{not independent}; however, this does not prevent us from using them for Monte Carlo calculations.

\framebreak

<<MC9,comment='+', fig.width=6, fig.height=6, fig.align='center',out.width='3in',echo=TRUE>>=
#Mean of pi
mu<-sum(c(1:d)*pi.vec)
mu

N<-1000
set.seed(34)
#Estimating using direct Monte Carlo
direct.MC<-replicate(N,mean(sample(1:d,size=n,rep=T,prob=pi.vec)))

#Estimating using Markov chain Monte Carlo
mcMC.run<-function(nr,p0v,Pm){
    Xt<-rep(0,nr+1)
    dv<-ncol(Pm)
    Xt[1]<-sample(1:dv,size=1,prob=p0v)
    for(i in 2:(nr+1)){
        Xt[i]<-sample(1:dv,size=1,prob=Pm[Xt[i-1],])
    }
    return(Xt)
}
MC.MC<-replicate(N,mean(mcMC.run(N,pi.vec,P)))
@

\framebreak

We can compare the estimated values over 1000 replicated runs using direct Monte Carlo and Markov chain Monte Carlo.

<<MC10,comment='+', fig.width=6, fig.height=4.2, fig.align='center',echo=TRUE>>=
par(mar=c(4,4,3,0))
boxplot(cbind(direct.MC,MC.MC),names=c('Direct MC','MCMC'))
abline(h=mu,col='red',lty=2)
@
\framebreak

<<MC11,comment='+', fig.width=6, fig.height=5, fig.align='center',echo=TRUE>>=
n*c(var(direct.MC),var(MC.MC))
var(direct.MC)/var(MC.MC)
@

Here the Markov chain method is more variable than the direct method.


\end{frame}


\end{document}