s23-probability-theory/inputs/lecture_22.tex
2023-07-07 17:42:38 +02:00

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\pagebreak
\lecture{22}{2023-07-04}{Introduction Markov Chains II}
\section{Markov Chains}
\todo{Merge this with the end of lecture 21}
\begin{goal}
We want to start with the basics of the theory of Markov chains.
\end{goal}
\begin{example}[Markov chains with two states]
Suppose there are two states of a phone line,
$0$,``free'', or $1$, ``busy''.
We assume that the state only changes at discrete units of time
and model this as a sequence of random variables.
Assume
\begin{IEEEeqnarray*}{rCl}
\bP[X_{n+1} = 0 | X_n = 0] &=& 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 = 1] &=& (1-q)
\end{IEEEeqnarray*}
for some $p,q \in (0,1)$.
We can write this as a matrix
\begin{IEEEeqnarray*}{rCl}
P &=& \begin{pmatrix}
p & (1-p) \\
q & (1-q)
\end{pmatrix}
\end{IEEEeqnarray*}
Note that the rows of this matrix sum up to $1$.
Additionally, we make the following assmption:
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
on the way, the phone reached state $i$.
\begin{question}
Suppose $X_0 = 0$.
What is the probability, that the phone will be free at times
$1 \& 2$ and will become busy at time $3$,
i.e.~what is $\bP[X_1 = 0, X_2 = 0, X_3 = 1]$?
\end{question}
We have
\begin{IEEEeqnarray*}{rCl}
&&\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] \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]\\
&=& P_{0,1} P_{0,0} P_{0,0}
\end{IEEEeqnarray*}
\begin{question}
Assume $X_0 = 0$. What is $\bP[X_3 = 1]$?
\end{question}
For $\{X_3 = 1\}$ to happen, we need to look at the following
disjoint events:
\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 = 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 = 1\}) &=& P_{0,1} P_{1,1}^2.
\end{IEEEeqnarray*}
More generally, consider a Matrix $P \in (0,1)^{n \times n}$
whose rows sum up to $1$.
Then we get a Markov Chain with $n$ states
by defining
\[\bP[X_{n+1} = i | X_{n} = j] = P_{i,j}.\]
\end{example}
\begin{definition}
Let $E$ denote a \vocab{discrete state space},
usually $E = \{1,\ldots, N\}$
or $E = \N$ or $E = \Z$.
Let $\alpha$ be a probability measure on $E$.
We say that $(p_{i,j})_{i \in E, j \in E}$ is a
\vocab{transition probability matrix}, if
\[
\forall i,j \in E .~p_{i,j} \ge 0 \land \forall i \in E \sum_{j \in E} p_{i,j} = 1.
\]
Given a triplet $(E, \alpha, P)$, we say that a stochastic process $(X_n)_{n \ge 0}$,
i.e.~$X_n: \Omega \to E$, is a \vocab[Markov chain!discrete]%
{Markov chain taking values
on the state space $E$
with initial distribution $\alpha$
and transition probability matrix $P$},
if the following conditions hold:
\begin{enumerate}[(i)]
\item $\bP[X_0 = i] = \alpha(i)$
for all $i \in E$,
\item \begin{IEEEeqnarray*}{rCl}
&&\bP[X_{n+1} = i_{n+1} | X_0 = i_0, X_1 = i_1, \ldots, X_{n} = i_{n}]\\
&=& \bP[X_{n+1} = i_{n+1} | X_n = i_n]
\end{IEEEeqnarray*}
for all $n = 0, \ldots$, $i_0,\ldots, i_{n+1} \in E$
(provided $\bP[X_0 = i_0, X_1 = i_1, \ldots, X_n = i_n] \neq 0$ ).
\end{enumerate}
\end{definition}
\begin{fact}
For all $n \in \N_0$ and $i_0,\ldots,i_n \in E$, we have
\[
\bP[X_0 = i_0, X_1 = i_1, \ldots, X_n = i_n] =
\alpha(i_0) \cdot p_{i_0,i_1} \cdot p_{i_1,i_2} \cdot \ldots \cdot p_{i_{n-1}, i_n}.
\]
\end{fact}
\begin{fact}
For all $n \in \N$, $i_n \in E$, we have
\[
\bP[X_n = i_n] = \sum_{i_0, \ldots, i_{n-1} \in E} \alpha_{i_0} p_{i_0,i_1} \cdot \ldots \cdot p_{i_{n-1}, i_n}.
\]
\end{fact}
\begin{example}[Simple random walk on $\Z$]
Let $E \coloneqq \Z$, $(\xi_n)_n$ i.i.d.~with $\bP[\xi_i = 1] = \bP[\xi_i = -1] = \frac{1}{2}$.
Let $X_0 = 0, X_n = \xi_1 + \ldots + \xi_n$.
Let $\alpha = \delta_0 \in M_1(\Z)$.
Consider
\begin{IEEEeqnarray*}{rCl}
P &=&
\begin{pmatrix}
& \ddots & \ddots & \ddots & & & & & 0\\
\ldots & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \ldots \\
& \ldots & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \ldots \\
& & \ldots & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & \ldots \\
0 & & & & & \ddots & \ddots & \ddots & \\
\end{pmatrix}
\end{IEEEeqnarray*}
\end{example}
% \begin{example}
% Consider a game, where a player wins or loses $1 €$ per round of the game.
% Let $p$ be the probability of winning.
% The player plays until they lose all money.
% Let $X_n$ be the capital of the gambler at time $n$.
% Define a matrix $P$
% by $P_{0,0} = 1$, $P_{i,i+1} = p$, $P_{i+1,i} = (1-p)$
% and all other entries $0$.
% \end{example}
\begin{definition}
Let $E$ be a complete, separable metric space,
$\alpha \in M_1(E)$.
For every $x \in E$,
let $\mathbf{P}(x, \cdot )$ be a probability measure on $E$.%
\footnote{$\mathbf{P}(x,\cdot )$ corresponds to a row of our matrix in the discrete case}
Given the triples $(E, \alpha, \{\mathbf{P}(x, \cdot )\}_{x \in E})$,
we say that a stochastic process $(X_n)_{n \ge 0}$
is a \vocab[Markov chain]{Markov chain taking values on $E$ %
with starting distribution $\alpha$ %
and transition probability $\{\mathbf{P}(x, \cdot )\}_{x \in E}$}
if
\begin{enumerate}[(i)]
\item $\bP[X_0 \in \cdot ] = \alpha(\cdot )$,
\item For all bounded, measurable $f: E \to \R$,
\[
\bE[f(X_{n+1}) | \cF_n] = \bE[f(X_{n+1}) | X_n]
= \int_E f(y) \mathbf{P}(X_n, \dif y) \text{ a.s.}
\]
\end{enumerate}
\end{definition}
\begin{remark}
This agrees with the definition in the discrete case,
as all bounded, measurable $f: E\to \R$ can be approximated
by simple functions,
i.e.~(ii) from the discrete case implies (ii) from the general definition.
\end{remark}
\begin{notation}
If $\{\mathbf{P}(x, \cdot )\}_{x \in E}$ is a transition probability,
then for all $f: E \to \R$ bounded and measurable,
define $\mathbf{P} : \cB_{\text{bdd}}(E) \to \cB_{\text{bdd}}$
by
\[
(\mathbf{P} f)(x) \coloneqq \int_E f(y) \mathbf{P}(x, \dif y).
\]
\end{notation}
We get the following fundamental link between martingales and Markov chains:
\begin{theorem}
\label{martingalesandmarkovchains}
Suppose $(E, \alpha, \{\mathbf{P}(x, \cdot )\}_{x \in E})$
is given.
Then a stochastic process $(X_n)_{n \ge 0}$ is a Markov chain
iff for every $f: E \to \R$ bounded, measurable,
\[
M_n(f) \coloneqq f(X_n) - f(X_0) - \sum_{j=1}^{n-1} (\mathbf{I} - \mathbf{P})f(X_j)
\]
is a martingale
with respect to the canonical filtration of $(X_n)$.
\end{theorem}
\begin{proof}
$\implies$
Fix some bounded, measurable $f : E \to \R$.
Then, for all $n$, $M_n(f)$ is bounded
and hence $M_n(f) \in L^1$.
$M_n(f)$ is $\cF_n$-measurable for all $n \in \N$.
In order to prove $\bE[M_{n+1}(f) | \cF_n] = M_n(f)$,
it suffices to show $\bE[M_{n+1}(f) - M_n(f) | \cF_n] = 0$ a.s.
We have
\begin{IEEEeqnarray*}{rCl}
\bE[M_{n+1}(f) - M_n(f) | \cF_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)\\
&=& 0
\end{IEEEeqnarray*}
$\impliedby$
Suppose $(M_n(f))_n$ is a martingale for all bounded, measurable $f$.
By the martingale property, we have
\begin{IEEEeqnarray*}{rCl}
\bE[f(X_{n+1}) | X_n]
&=& (\mathbf{P}f)(X_n)\\
&=& \int f(y) \mathbf{P}(X_n, \dif y)
\end{IEEEeqnarray*}
This proves (ii).
\end{proof}
\begin{definition}
Given $\{\mathbf{P}(x, \cdot )\}_{x \in E}$,
we say that $f: E \to \R$ is \vocab{harmonic},
iff $f(x) = (\mathbf{P}f)(x)$
for all $x \in E$.
We call $f$ \vocab{super-harmonic},
if $(\mathbf{I} - \mathbf{P}) f \ge 0$
and \vocab{sub-harmonic},
if $(\mathbf{I} - \mathbf{P}) f \le 0$.
\end{definition}
\begin{corollary}
If $f$ is (sub/super) harmonic, then for every
$(E, \{\mathbf{P}(x, \cdot )\}_{x \in E}, \alpha)$
and every Markov chain $(X_n)_{n \ge 0}$,
we have that
$f(X_n)$ is a (sub/super) martingale.
\end{corollary}
\begin{question}
Given a set $A$ and a function $f$ on a superset of $A$.
Find a function $u$, such that $u$ is harmonic,
and $u = f$ on $A$.
\end{question}
Let $u(x) \coloneqq \bE_x[f(X_{T_A}]$,
where $\bE_x$ is the expectation with respect to the Markov chain
starting in $x$,
and $T_A$ is the stopping time defined by the Markov chain hitting $A$.