Markov Chain problem

**Bucephalus** · April 26th 2010, 04:04 AM

Given a markov chain with states { $e_1,e_2,e_5$ } and the transition probability matrix

$P = \begin{bmatrix}0.2 & 0.8 & 0\\1 & 0 & 0\\0.7 & 0 & 0.3\end{bmatrix}$

which on of the following is false?
a) $e_1$ communicates with $e_2$
b) $e_1$ is accessible from $e_3$
c) $e_3$ communicates with $e_1$
d) $e_2$ is accessible from $e_3$

**Soroban** · April 26th 2010, 05:14 AM

Hello, Bucephalus!

Given a Markov chain with states { $e_1,e_2,e_5$ } and the transition probability matrix

. . $P \;=\; \begin{bmatrix}0.2 & 0.8 & 0\\1 & 0 & 0\\0.7 & 0 & 0.3\end{bmatrix}$

Which one of the following is false?

a) $e_1$ communicates with $e_2$

I assume that "communicates" is a two-way relationship.
. . That is: . $e_1\to e_2$ .and . $e_2\to e_1$

Since . $\begin{Bmatrix}P(e_1\to e_2) &=& 0.8 \\ P(e_2\to e_1) &=& 1\end{Bmatrix}$ . the statement is true.

b) $e_1$ is accessible from $e_3$

Since . $P(e_3\to e_1) \:=\: 0.7,$ .the statement is true.

c) $e_3$ communicates with $e_1$

We have: . $P(e_3\to e_1) \:=\:0.7$

. . . .but: . $P(e_1\to e_3) \:=\:0$

$e_1$ and $e_3$ do not communicate.

This statement is false.

d) $e_2$ is accessible from $e_3$

Since . $\begin{Bmatrix}P(e_3\to e_1) &=& 0.7 \\<br /> P(e_1\to e_2) &=& 0.8 \end{Bmatrix}$

. . then: . $P\left([e_3\to e_1] \wedge [e_1\to e_2]\right) \;=\;(0.7)(0.8)$

Hence: . $P(e_3\to e_2) \:=\:0.56$ . . . The statement is true.

**Bucephalus** · April 26th 2010, 05:22 AM

Thanks for your help.
David.

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