11.2.2 State Transition Matrix and Diagram
State Transition Diagram:
A Markov chain is usually shown by a state transition diagram. Consider a Markov chain with three possible states $1$, $2$, and $3$ and the following transition probabilities \begin{equation} \nonumber P = \begin{bmatrix} \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\[5pt] \frac{1}{3} & 0 & \frac{2}{3} \\[5pt] \frac{1}{2} & 0 & \frac{1}{2} \end{bmatrix}. \end{equation} Figure 11.7 shows the state transition diagram for the above Markov chain. In this diagram, there are three possible states $1$, $2$, and $3$, and the arrows from each state to other states show the transition probabilities $p_{ij}$. When there is no arrow from state $i$ to state $j$, it means that $p_{ij}=0$.Example
Consider the Markov chain shown in Figure 11.7.
- Find $P(X_4=3|X_3=2)$.
- Find $P(X_3=1|X_2=1)$.
- If we know $P(X_0=1)=\frac{1}{3}$, find $P(X_0=1,X_1=2)$.
- If we know $P(X_0=1)=\frac{1}{3}$, find $P(X_0=1,X_1=2,X_2=3)$.
- Solution
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- By definition $$P(X_4=3|X_3=2)=p_{23}=\frac{2}{3}.$$
- By definition $$P(X_3=1|X_2=1)=p_{11}=\frac{1}{4}.$$
- We can write \begin{align*} P(X_0=1,X_1=2) &=P(X_0=1) P(X_1=2|X_0=1)\\ &= \frac{1}{3} \cdot\ p_{12} \\ &=\frac{1}{3} \cdot \frac{1}{2}= \frac{1}{6}. \end{align*}
- We can write \begin{align*} &P(X_0=1,X_1=2,X_2=3) \\ &\quad=P(X_0=1) P(X_1=2|X_0=1) P(X_2=3|X_1=2, X_0=1)\\ &\quad=P(X_0=1) P(X_1=2|X_0=1)P(X_2=3|X_1=2) \quad (\textrm{by Markov property}) \\ &\quad=\frac{1}{3} \cdot\ p_{12} \cdot p_{23} \\ &\quad=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{3}\\ &\quad= \frac{1}{9}. \end{align*}
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