lecture 22
This commit is contained in:
parent
4cd05f37a0
commit
023e865eb4
1 changed files with 82 additions and 88 deletions
|
@ -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},
|
||||
|
@ -196,12 +195,8 @@ 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}
|
||||
|
@ -210,7 +205,6 @@ We get the following fundamental link between martingales and Markov chains:
|
|||
&\overset{\text{Markov property}}{=}& (\mathbf{P}f)(X_n) - (\mathbf{P}f)(X_n)\\
|
||||
&=& 0
|
||||
\end{IEEEeqnarray*}
|
||||
\end{subproof}
|
||||
|
||||
$\impliedby$
|
||||
Suppose $(M_n(f))_n$ is a martingale for all bounded, measurable $f$.
|
||||
|
|
Loading…
Reference in a new issue