lecture 22

inputs/lecture_22.tex

@@ -5,66 +5,65 @@
     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{IEEEeqnarra*}{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{IEEEeqnarra*}
-%     for some $p,q \in (0,1)$.
-%     We can write this as a matrix
-%     \begin{IEEEeqnarra*}{rCl}
-%         P &=& \begin{pmatrix}
-%             p & (1-p) \\
-%             q & (1-q)
-%         \end{pmatrix} 
-%     \end{IEEEeqnarra*}
-%     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{IEEEeqnarra*}{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{IEEEeqnarra*}
-% 
-%     \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{IEEEeqnarra*}{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{IEEEeqnarr*}
-% 
-%     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{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},
@@ -196,21 +195,16 @@ We get the following fundamental link between martingales and Markov chains:
     and hence $M_n(f) \in  L^1$.
     $M_n(f)$ is $\cF_n$-measurable for all $n \in \N$.
 
-    \begin{claim}
-        $\bE[M_{n+1}(f) | \cF_n] = M_n(f)$.
-    \end{claim}
-    \begin{subproof}
-        It suffices to show
-        $\bE[M_{n+1}(f) - M_n(f) | \cF_n] = 0$ a.s.
+    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*}
-    \end{subproof}
+    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 $f$.