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{\bf Question}

Consider the Markov chain having state space $\{0,1,2\}$ and
transition probability matrix $$P = \begin{array}{c} 0 \\ 1 \\ 2
\end{array} \left( \begin{array}{ccc} 0 & 1 & 2 \\ \frac{1}{4} & 0
& \frac{3}{4} \\ 0 & 1 & 0 \end{array} \right)$$
\begin{description}
\item[(a)] Find $P^2$
\item[(b)] Show that $P^4 = P^2$
\item[(c)] Find $P^n$, $n\leq 1.$
\end{description}
If the system starts (at step 0) in state 1, find the probability
that it occupies the different states at step $n.$



\vspace{.25in}

{\bf Answer}

\begin{eqnarray*}
P &=& \begin{array}{c} 0 \\ 1 \\ 2
\end{array} \left( \begin{array}{ccc} 0 & 1 & 2 \\ \frac{1}{4} & 0
& \frac{3}{4} \\ 0 & 1 & 0 \end{array} \right)\\ P^2 &=&  \left(
\begin{array}{ccc} \frac{1}{4} & 0 & \frac{3}{4}
\\  0 & 1 & 0 \\ \frac{1}{4} & 0 & \frac{3}{4} \end{array}
\right)\\ P^3 &=&  \left( \begin{array}{ccc} 0 & 1 & 2 \\
\frac{1}{4} & 0 & \frac{3}{4} \\ 0 & 1 & 0 \end{array} \right)= P
\end{eqnarray*}

So $P^2 = P^4$

Also $P^n = P$ if $n$ is odd and $P^n = P^2$ if $n$ is even.

So if $\ds {\bf p}_0 = (0,1,0), \hspace{.1in} {\bf p}^{(n)} = {\bf
p}_0P^n = \left\{ \begin{array}{ll} (\frac{1}{4}, 0 , \frac{3}{4})
& n\ {\rm odd} \\ (0,1,0) & n\ {\rm even} \end{array} \right.$

\bigskip

So after an even number of steps the system occupies state 1.
After an odd number of steps it occupies state 0 with probability
$\frac{1}{4}$ and state 2 with probability $\frac{3}{4}.$




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