some small changes

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Josia Pietsch 2023-07-06 00:50:07 +02:00
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@ -41,7 +41,7 @@
\end{question} \end{question}
We have We have
\begin{IEEEeqnarray*}{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]\\
@ -63,7 +63,8 @@
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}
@ -89,8 +90,10 @@
\item $\bP[X_0 = i] = \alpha(i)$ \item $\bP[X_0 = i] = \alpha(i)$
for all $i \in E$, for all $i \in E$,
\item $\bP[X_{n+1} = i_{n+1} | X_0 = i_0, X_1 = i_1, \ldots, X_{n} = i_{n}] \item \begin{IEEEeqnarray*}{rCl}
= \bP[X_{n+1} = i_{n+1} | X_n = i_n]$ &&\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 $n = 0, \ldots$, $i_0,\ldots, i_{n+1} \in E$ for all $n = 0, \ldots$, $i_0,\ldots, i_{n+1} \in E$
(provided $\bP[X_0 = i_0, X_1 = i_1, \ldots, X_n = i_n] \neq 0$ ). (provided $\bP[X_0 = i_0, X_1 = i_1, \ldots, X_n = i_n] \neq 0$ ).
\end{enumerate} \end{enumerate}