in a markov chain, if B communicates with S, and S can reach absorbing state I but not vice versa and M can reach S but not vice versa, B and S can't be in the same class right?

In a Markov chain, states are grouped into *communicating classes*, which are sets of states that communicate with each other. Two states \( i \) and \( j \) are said to communicate if:

1. You can reach state \( j \) from state \( i \) (denoted \( i \rightarrow j \)).
2. You can also reach state \( i \) from state \( j \) (denoted \( j \rightarrow i \)).

This means that for two states to be in the same class, there must be a two-way communication between them.

Now, let's examine your scenario:

- **State \( B \)** communicates with **State \( S \)**, meaning \( B \leftrightarrow S \), which implies \( B \rightarrow S \) and \( S \rightarrow B \).
- **State \( S \)** can reach the absorbing state **\( I \)**, but not vice versa, meaning \( S \rightarrow I \) but \( I \not\rightarrow S \). Since \( I \) is an absorbing state, no state can be reached from \( I \), including \( S \).
- **State \( M \)** can reach **State \( S \)** but not vice versa, meaning \( M \rightarrow S \) but \( S \not\rightarrow M \).

For \( B \) and \( S \) to be in the same class, they must be able to communicate both ways. However, if \( B \leftrightarrow S \), then both should belong to the same class. But here’s the problem:

- \( S \) can reach the absorbing state \( I \), which breaks the symmetry because you cannot come back from \( I \) to \( S \). This lack of two-way communication for some states complicates the classification.

Thus, based on this analysis, **\( B \) and \( S \)** could still be in the same class if they communicate with each other. However, the fact that \( S \) can reach an absorbing state \( I \) (from which no return is possible) means that **\( B \) and \( S \)** might not necessarily stay in a recurrent class together if the chain contains absorbing states. Their class structure might be disrupted because absorbing states affect which states can form recurrent communicating classes.

### Clarification:
- **\( B \) and \( S \)** could still belong to the same transient class (before reaching \( I \)), but once \( S \) reaches the absorbing state \( I \), \( S \)'s dynamics change, and \( B \)'s path could diverge. So, the concept of *same class* holds only as long as both remain in transient behavior and can communicate bidirectionally.

Would you like further details or clarifications?

### 5 Related Questions:
1. What is an absorbing state in a Markov chain, and how does it affect the classification of states?
2. How are transient and recurrent states defined in Markov chains?
3. What is a communication class, and how does it affect the behavior of a Markov chain?
4. Can a Markov chain have both transient and absorbing states, and how does this impact its long-term behavior?
5. How does the concept of irreducibility apply to Markov chains?

### Tip:
Always check whether all the states can communicate with each other (in both directions) to identify which states belong to the same class.

In a markov chain, if B communicates with S, and S can reach absorbing state I but not vice versa and M can reach S but not vice versa, B and S can't be in the same class right?

Explore how communication between states in a Markov chain works, with a detailed look into absorbing states, transient behavior, and the classification of states. This problem examines the relationships between states B, S, and I, with considerations on how absorbing states impact class membership in Markov Chains. Suitable for undergraduate students.

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