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lecture 22

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@ -5,66 +5,65 @@
We want to start with the basics of the theory of Markov chains. We want to start with the basics of the theory of Markov chains.
\end{goal} \end{goal}
% \begin{example}[Markov chains with two states] \begin{example}[Markov chains with two states]
% Suppose there are two states of a phone line, Suppose there are two states of a phone line,
% $0$,``free'', or $1$, ``busy''. $0$,``free'', or $1$, ``busy''.
% We assume that the state only changes at discrete units of time We assume that the state only changes at discrete units of time
% and model this as a sequence of random variables. and model this as a sequence of random variables.
% Assume Assume
% \begin{IEEEeqnarra*}{rCl} \begin{IEEEeqnarray*}{rCl}
% \bP[X_{n+1} = 0 | X_n = 0] &=& p\\ \bP[X_{n+1} = 0 | X_n = 0] &=& p\\
% \bP[X_{n+1} = 0 | X_n = 1] &=& (1-p)\\ \bP[X_{n+1} = 0 | X_n = 1] &=& (1-p)\\
% \bP[X_{n+1} = 1 | X_n = 0] &=& q\\ \bP[X_{n+1} = 1 | X_n = 0] &=& q\\
% \bP[X_{n+1} = 1 | X_n = 1] &=& (1-q) \bP[X_{n+1} = 1 | X_n = 1] &=& (1-q)
% \end{IEEEeqnarra*} \end{IEEEeqnarray*}
% for some $p,q \in (0,1)$. for some $p,q \in (0,1)$.
% We can write this as a matrix We can write this as a matrix
% \begin{IEEEeqnarra*}{rCl} \begin{IEEEeqnarray*}{rCl}
% P &=& \begin{pmatrix} P &=& \begin{pmatrix}
% p & (1-p) \\ p & (1-p) \\
% q & (1-q) q & (1-q)
% \end{pmatrix} \end{pmatrix}
% \end{IEEEeqnarra*} \end{IEEEeqnarray*}
% Note that the rows of this matrix sum up to $1$. Note that the rows of this matrix sum up to $1$.
%
% Additionally, we make the following assmption: Additionally, we make the following assmption:
% Given that at some time $n$, the phone is in state $i \in \{0,1\}$, Given that at some time $n$, the phone is in state $i \in \{0,1\}$,
% the behavior of the phone after time $n$ does not depend the behavior of the phone after time $n$ does not depend
% on the way, the phone reached state $i$. on the way, the phone reached state $i$.
%
% \begin{question} \begin{question}
% Suppose $X_0 = 0$. Suppose $X_0 = 0$.
% What is the probability, that the phone will be free at times What is the probability, that the phone will be free at times
% $1 \& 2$ and will become busy at time $3$, $1 \& 2$ and will become busy at time $3$,
% i.e.~what is $\bP[X_1 = 0, X_2 = 0, X_3 = 1]$? i.e.~what is $\bP[X_1 = 0, X_2 = 0, X_3 = 1]$?
% \end{question} \end{question}
% We have We have
% \begin{IEEEeqnarra*}{rCl} \begin{IEEEeqnarray*}{rCl}
% \bP[X_1 = 0, X_2 = 0, X_3 = 1] \bP[X_1 = 0, X_2 = 0, X_3 = 1]
% &=& \bP[X_3 = 0 | X_2 = 0, X_1 = 0] \bP[X_2 = 0, X_1 = 0]\\ &=& \bP[X_3 = 0 | X_2 = 0, X_1 = 0] \bP[X_2 = 0, X_1 = 0]\\
% &=& \bP[X_3 = 0 | X_2 = 0] \bP[X_2 = 0, X_1 = 0]\\ &=& \bP[X_3 = 0 | X_2 = 0] \bP[X_2 = 0, X_1 = 0]\\
% &=& \bP[X_3 = 0 | X_2 = 0] \bP[X_2 = 0 | X_1 = 0] \bP[X_1 = 0]\\ &=& \bP[X_3 = 0 | X_2 = 0] \bP[X_2 = 0 | X_1 = 0] \bP[X_1 = 0]\\
% &=& P_{0,1} P_{0,0} P_{0,0} &=& P_{0,1} P_{0,0} P_{0,0}
% \end{IEEEeqnarra*} \end{IEEEeqnarray*}
%
% \begin{question} \begin{question}
% Assume $X_0 = 0$. What is $\bP[X_3 = 1]$? Assume $X_0 = 0$. What is $\bP[X_3 = 1]$?
% \end{question} \end{question}
% For $\{X_3 = 1\}$ to happen, we need to look at the following For $\{X_3 = 1\}$ to happen, we need to look at the following
% disjoint events: disjoint events:
% % \begin{IEEEeqnarra*}{rCl} \begin{IEEEeqnarray*}{rCl}
% % \bP(\{X_3 = 1, X_2 = 0, X_1 = 0\}) &=& P_{0,1} P_{0,0}^2,\\ \bP(\{X_3 = 1, X_2 = 0, X_1 = 0\}) &=& P_{0,1} P_{0,0}^2,\\
% % \bP(\{X_3 = 1, X_2 = 0, X_1 = 1\}) &=& P_{0,1}^2 P_{1,0},\\ \bP(\{X_3 = 1, X_2 = 0, X_1 = 1\}) &=& P_{0,1}^2 P_{1,0},\\
% % \bP(\{X_3 = 1, X_2 = 1, X_1 = 0\}) &=& P_{0,0} P_{0,1} P_{1,1},\\ \bP(\{X_3 = 1, X_2 = 1, X_1 = 0\}) &=& P_{0,0} P_{0,1} P_{1,1},\\
% % \bP(\{X_3 = 1, X_2 = 1, X_1 = 1\}) &=& P_{0,1} P_{1,1}^2. \bP(\{X_3 = 1, X_2 = 1, X_1 = 1\}) &=& P_{0,1} P_{1,1}^2.
% % \end{IEEEeqnarr*} \end{IEEEeqnarray*}
%
% More generally, consider a Matrix $P \in (0,1)^{n \times n}$ More generally, consider a Matrix $P \in (0,1)^{n \times n}$
% whose rows sum up to $1$. whose rows sum up to $1$.
% Then we get a Markov Chain with $n$ states Then we get a Markov Chain with $n$ states
% by defining $\bP[X_{n+1} = i | X_{n} = j] = P_{i,j}$. by defining $\bP[X_{n+1} = i | X_{n} = j] = P_{i,j}$.
% \end{example}
% \end{example}
\begin{definition} \begin{definition}
Let $E$ denote a \vocab{discrete state space}, Let $E$ denote a \vocab{discrete state space},
@ -196,21 +195,16 @@ We get the following fundamental link between martingales and Markov chains:
and hence $M_n(f) \in L^1$. and hence $M_n(f) \in L^1$.
$M_n(f)$ is $\cF_n$-measurable for all $n \in \N$. $M_n(f)$ is $\cF_n$-measurable for all $n \in \N$.
\begin{claim} In order to prove $\bE[M_{n+1}(f) | \cF_n] = M_n(f)$,
$\bE[M_{n+1}(f) | \cF_n] = M_n(f)$. it suffices to show $\bE[M_{n+1}(f) - M_n(f) | \cF_n] = 0$ a.s.
\end{claim}
\begin{subproof}
It suffices to show
$\bE[M_{n+1}(f) - M_n(f) | \cF_n] = 0$ a.s.
We have We have
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
\bE[M_{n+1}(f) - M_n(f) | \cF_n] \bE[M_{n+1}(f) - M_n(f) | \cF_n]
&=& \bE[f(X_{n+1} | \cF_n] - (\mathbf{P}f)(X_n)\\ &=& \bE[f(X_{n+1} | \cF_n] - (\mathbf{P}f)(X_n)\\
&\overset{\text{Markov property}}{=}& (\mathbf{P}f)(X_n) - (\mathbf{P}f)(X_n)\\ &\overset{\text{Markov property}}{=}& (\mathbf{P}f)(X_n) - (\mathbf{P}f)(X_n)\\
&=& 0 &=& 0
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
\end{subproof}
$\impliedby$ $\impliedby$
Suppose $(M_n(f))_n$ is a martingale for all bounded, measurable $f$. Suppose $(M_n(f))_n$ is a martingale for all bounded, measurable $f$.