GATE 2022 EC Q.48: solution and answer

GATE 2022 · ECQ.48 · 2 marks · MSQ

A state transition diagram with states $A$ , $B$ , and $C$ , and transition probabilities $p_{1}, p_{2}, \dots, p_{7}$ 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.

GATE 2022 EC question 48: Three circular states A, B, C. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$p_{2} + p_{3} = p_{5} + p_{6}$
(B)
$p_{1} + p_{3} = p_{4} + p_{6}$
(C)
$p_{1} + p_{4} + p_{7} = 1$
(D)
$p_{2} + p_{5} + p_{7} = 1$

Answer and solution

ANSWER

(A) and (C) are universally true. The arrows leaving each state sum to 1, which gives $p_{1} + p_{4} + p_{7} = 1$ , $p_{2} + p_{3} = 1$ and $p_{5} + p_{6} = 1$ . (B) and (D) each fail for a valid choice of probabilities.

Official answer: A, CSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

The rule. From any state the chain must go somewhere at the next step, so the probabilities on the arrows leaving a state add up to 1. (Each row of the transition matrix sums to 1.) The figure shows every allowed transition, so a move with no arrow has probability 0.
Read the arrows leaving each state from the figure. Leaving A: $p_{1}$ (to B), $p_{4}$ (to C), $p_{7}$ (back to A). Leaving B: $p_{2}$ (back to B), $p_{3}$ (to A). Leaving C: $p_{5}$ (back to C), $p_{6}$ (to A).
The three equations that always hold: $p_{1} + p_{4} + p_{7} p_{2} + p_{3} p_{5} + p_{6} = 1 (state A) = 1 (state B) = 1 (state C)$
(A) $p_{2} + p_{3} = 1 = p_{5} + p_{6}$ , so (A) is always true.
(C) $p_{1} + p_{4} + p_{7} = 1$ is exactly the equation for state A, so (C) is always true.
(B) is false in general. Take $p_{1} = 0.5, p_{4} = 0.2, p_{7} = 0.3$ and $p_{2} = p_{3} = p_{5} = p_{6} = 0.5$ . All three equations hold, but $p_{1} + p_{3} = 1.0$ while $p_{4} + p_{6} = 0.7$ . The two sides are not equal.
(D) is false in general. The three self-loops belong to three different states, so no equation ties them together. Take $p_{2} = p_{5} = p_{7} = 0.5$ , with $p_{3} = p_{6} = 0.5$ and $p_{1} = p_{4} = 0.25$ . Every equation holds ( $0.25 + 0.25 + 0.5 = 1$ for A), but $p_{2} + p_{5} + p_{7} = 1.5 \neq = 1$ .
So the statements that are always true are (A) and (C).

KEY IDEA

In a Markov chain's state diagram, the probabilities on the arrows leaving a state (its self-loop included) add up to 1. Only statements that follow from these per-state equations are universally true. To rule out any other statement, one valid set of numbers that breaks it is enough.

WATCH OUT FOR

Summing the arrows coming INTO a state instead of the ones leaving it. That leads to equations like $p_{1} + p_{2} = 1$ for B, and it makes (B) or (D) look plausible. A second trap is marking (D) because it collects "all the self-loops". The self-loops belong to different states, so nothing forces them to add up to 1.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 33 Probability & statistics questions, with notes

GATE 2022 EC Q.48: solution and answer

Answer and solution

More Probability & statistics questions