A state transition diagram with states $A$, $B$, and $C$, and transition probabilities $p_1, p_2, \ldots, p_7$ is shown in the figure (e.g., $p_1$ denotes the probability of transition from state $A$ to $B$). For this state diagram, select the statement(s) which is/are universally true.
Show Hint
In Markov chains, only the sum of outgoing transition probabilities from the same state must be 1. Comparing transitions from different states is not valid.
For any Markov state, the sum of outgoing transition probabilities must equal 1.
For state A:
Outgoing transitions are labeled:
\[
p_1: A \to B,\quad
p_4: A \to C,\quad
p_7: A \to A.
\]
Thus,
\[
p_1 + p_4 + p_7 = 1,
\]
which matches option (C) and is universally true.
For state B:
Outgoing transitions:
\[
p_2: B \to B,\quad
p_3: B \to A.
\]
Therefore,
\[
p_2 + p_3 = 1.
\]
For state C:
Outgoing transitions:
\[
p_5: C \to C,\quad
p_6: C \to A.
\]
Thus,
\[
p_5 + p_6 = 1.
\]
Comparing the equalities:
\[
p_2 + p_3 = 1 = p_5 + p_6,
\]
which makes option (A) universally true.
Check the other options:
- (B) $p_1 + p_3 = p_4 + p_6$
No Markov rule forces this equality. Not universally true.
- (D) $p_2 + p_5 + p_7 = 1$
These probabilities originate from different states (B, C, A).
They cannot be summed to 1. Not universally true.
Thus, the universally valid statements are (A) and (C).
Final Answer: (A), (C)
