Finite Mathematics Quiz: Markov Chains Steady State
15 questions · exam conditions
0:00
Markov Chains Steady StateQuestion 1 of 15

A Markov chain has transition matrix P=(0.70.30.40.6)P = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix} . If the system starts with initial distribution (0.2,0.8)(0.2, 0.8), what will be the probability of being in state 1 after the system reaches steady state?

47\frac{4}{7}
37\frac{3}{7}
0.20.2 (the initial probability remains unchanged)
0.70.7 (the diagonal entry for state 1)
← Back to quizzes

Finite Mathematics Quiz

Finite Mathematics Quiz: Markov Chains Steady State

Practice Markov Chains Steady State in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Markov Chains Steady State, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A Markov chain has transition matrix P=(0.70.30.40.6)P = \begin{pmatrix} 0.7 & 0.3 \\ 0.4 & 0.6 \end{pmatrix} . If the system starts with initial distribution (0.2,0.8)(0.2, 0.8), what will be the probability of being in state 1 after the system reaches steady state?

  1. 47\frac{4}{7} (correct answer)
  2. 37\frac{3}{7}
  3. 0.20.2 (the initial probability remains unchanged)
  4. 0.70.7 (the diagonal entry for state 1)
Explanation: To find the steady-state distribution, we solve πP=π\pi P = \pi where π=(π1,π2)\pi = (\pi_1, \pi_2). This gives us 0.7π1+0.4π2=π10.7\pi_1 + 0.4\pi_2 = \pi_1 and 0.3π1+0.6π2=π20.3\pi_1 + 0.6\pi_2 = \pi_2, along with π1+π2=1\pi_1 + \pi_2 = 1. From the first equation: −0.3π1+0.4π2=0-0.3\pi_1 + 0.4\pi_2 = 0, so π2=34π1\pi_2 = \frac{3}{4}\pi_1. Substituting into π1+π2=1\pi_1 + \pi_2 = 1: π1+34π1=1\pi_1 + \frac{3}{4}\pi_1 = 1, giving π1=47\pi_1 = \frac{4}{7}. The initial distribution is irrelevant for steady state. Choice B uses the wrong ratio. Choice C incorrectly assumes initial conditions matter. Choice D confuses transition probabilities with steady-state probabilities.

Question 2

A weather model uses a Markov chain where each day is classified as Sunny (S) or Rainy (R). The transition probabilities are: P(tomorrow sunny | today sunny) = 0.8, P(tomorrow rainy | today rainy) = 0.6. After many days, what is the long-run probability that any randomly selected day will be rainy?

  1. 12\frac{1}{2} because there are two equally likely states in the long run
  2. 0.60.6 because this equals the probability of staying rainy given a rainy day
  3. 25\frac{2}{5} because the limiting distribution balances inflow and outflow from each state
  4. 13\frac{1}{3} because rainy days are less persistent than sunny days in this model (correct answer)
Explanation: When you encounter Markov chain problems asking for long-run probabilities, you're looking for the steady-state distribution where the probability of being in each state remains constant over time. This occurs when the rate of flow into each state equals the rate of flow out. To find the steady-state probabilities, let πS\pi_S and πR\pi_R represent the long-run probabilities of sunny and rainy days. Since probabilities must sum to 1, we have πS+πR=1\pi_S + \pi_R = 1. At equilibrium, the flow into each state equals the flow out: For the rainy state: πR=0.2πS+0.6πR\pi_R = 0.2\pi_S + 0.6\pi_R This equation says the long-run probability of rain equals the probability of transitioning from sunny to rainy (0.2) times the probability of being sunny, plus the probability of staying rainy (0.6) times the probability of being rainy. Solving: πR=0.2(1−πR)+0.6πR\pi_R = 0.2(1-\pi_R) + 0.6\pi_R πR=0.2−0.2πR+0.6πR\pi_R = 0.2 - 0.2\pi_R + 0.6\pi_R πR=0.2+0.4πR\pi_R = 0.2 + 0.4\pi_R 0.6πR=0.20.6\pi_R = 0.2 πR=13\pi_R = \frac{1}{3} Answer A incorrectly assumes equal probabilities despite unequal transition rates. Answer B confuses the conditional probability of staying rainy with the unconditional long-run probability. Answer C gives 25\frac{2}{5}, which would result from incorrect arithmetic in the balance equations. Remember: steady-state problems require setting up flow equations, not just reading transition probabilities directly. The long-run distribution depends on the entire transition matrix structure.

Question 3

Two different Markov chains both have the same steady-state distribution (14,34)\left(\frac{1}{4}, \frac{3}{4}\right). Chain X has transition matrix PX=(0.70.30.10.9)P_X = \begin{pmatrix} 0.7 & 0.3 \\ 0.1 & 0.9 \end{pmatrix} and Chain Y has transition matrix $$P_Y = \begin{pmatrix} 0.1 & 0.9 \ 0.3 & 0.7 \end{pmatrix}

  1. Chain X converges to steady state faster because it has higher diagonal entries, providing more stability
  2. Both chains have identical long-run behavior since they share the same steady-state distribution vector
  3. Chain Y converges to steady state faster because its second eigenvalue has a smaller absolute value (correct answer)
  4. The chains are fundamentally different because PYP_Y is not a valid transition matrix for this steady state
Explanation: Both matrices are valid transition matrices (rows sum to 1) and both have steady state (14,34)\left(\frac{1}{4}, \frac{3}{4}\right). For convergence rate, we examine the second eigenvalue. For PXP_X: eigenvalues are 1 and 0.7+0.9−1=0.60.7 + 0.9 - 1 = 0.6. For PYP_Y: eigenvalues are 1 and 0.1+0.7−1=−0.20.1 + 0.7 - 1 = -0.2. Since ∣−0.2∣=0.2<0.6|{-0.2}| = 0.2 < 0.6, Chain Y converges faster. Choice A incorrectly assumes high diagonal entries mean faster convergence. Choice B ignores that convergence rates can differ. Choice D is wrong - we can verify that PYP_Y does have the given steady state by checking (14,34)PY=(14,34)\left(\frac{1}{4}, \frac{3}{4}\right) P_Y = \left(\frac{1}{4}, \frac{3}{4}\right).

Question 4

A simplified model of customer loyalty uses three states: Loyal (L), Neutral (N), and Competitor (C). Market research shows that the long-run proportions are 40% Loyal, 35% Neutral, and 25% Competitor. If the probability of moving from Loyal to Neutral in one period is 0.1, and from Loyal to Competitor is 0.05, what is the probability of remaining Loyal given that a customer is currently Loyal?

  1. 0.750.75 because the probability of not switching to competitors determines loyalty retention in competitive markets
  2. 0.400.40 because this equals the steady-state probability of being in the Loyal state
  3. 0.900.90 because 1−0.1=0.901 - 0.1 = 0.90 represents staying loyal when only Neutral transition is considered
  4. 0.850.85 because the remaining probability after accounting for transitions to other states must be self-retention (correct answer)
Explanation: This is a Markov chain problem involving transition probabilities between states. When you encounter customer loyalty or market share questions with multiple states and transition probabilities, you're working with a system where all probabilities from any given state must sum to 1. From the Loyal state, you know two transition probabilities: 0.1 to move to Neutral and 0.05 to move to Competitor. Since a customer must either stay Loyal or move to one of the other two states, these three probabilities must sum to 1. Therefore: P(stay Loyal) = 1 - P(move to Neutral) - P(move to Competitor) = 1 - 0.1 - 0.05 = 0.85. Choice A (0.75) incorrectly focuses only on competitor transitions, ignoring the movement to Neutral customers. This represents a common error of considering only some transitions rather than all possible outcomes. Choice B (0.40) confuses the steady-state proportion (long-run percentage of loyal customers) with the transition probability (likelihood of staying loyal in one period). These are completely different concepts in Markov chain analysis. Choice C (0.90) makes the mistake of only accounting for the transition to Neutral (1 - 0.1 = 0.90) while completely ignoring the transition to Competitor. This incomplete calculation misses a key piece of given information. Choice D (0.85) correctly recognizes that after accounting for all outgoing transitions, the remaining probability represents self-retention. Study tip: In transition probability problems, always verify that probabilities from each state sum to 1. Set up the equation: P(stay) + P(transition 1) + P(transition 2) + ... = 1, then solve for the unknown.

Question 5

A Markov chain has transition matrix $$P = \begin{pmatrix} 0.2 & 0.8 & 0 \ 0.3 & 0.4 & 0.3 \ 0 & 0.5 & 0.5 \end{pmatrix}

  1. State 3 is absorbing, so the steady-state probability vector is (0,0,1)(0, 0, 1) regardless of initial conditions
  2. The chain is irreducible and aperiodic, so a unique steady-state distribution exists and can be reached from any initial state (correct answer)
  3. States 1 and 3 cannot communicate directly, so the steady-state distribution depends on whether the system starts in state 1 or state 3
  4. The zero entries create absorbing barriers, preventing the establishment of any meaningful long-run equilibrium distribution
Explanation: Examining the transition matrix: from state 1 we can reach states 1 and 2; from state 2 we can reach all states; from state 3 we can reach states 2 and 3. This means all states can eventually communicate through state 2 (1→2→3 and 3→2→1), making the chain irreducible. Since there are self-loops with positive probability, the chain is aperiodic. Therefore, a unique steady state exists. Choice A is wrong - state 3 isn't absorbing (p33=0.5≠1p_{33} = 0.5 \neq 1). Choice C misses that indirect communication through state 2 makes all states accessible. Choice D incorrectly assumes zero entries prevent equilibrium.

Question 6

Consider a Markov chain with three states and transition matrix $$P = \begin{pmatrix} 0.5 & 0.3 & 0.2 \ 0.1 & 0.8 & 0.1 \ 0.4 & 0.2 & 0.4 \end{pmatrix}

  1. The steady-state probability for state 2 is approximately 0.80.8 since it has the highest self-transition probability
  2. The steady-state probability for state 2 is 1223\frac{12}{23} and represents the long-run proportion of time spent in state 2 (correct answer)
  3. The steady-state probabilities depend on the initial distribution, so no unique steady state exists for this chain
  4. State 2 will eventually absorb all probability since 0.8>0.50.8 > 0.5 and 0.8>0.40.8 > 0.4 along the main diagonal
Explanation: To find steady state, solve πP=π\pi P = \pi with π1+π2+π3=1\pi_1 + \pi_2 + \pi_3 = 1. This yields the system: 0.5π1+0.1π2+0.4π3=π10.5\pi_1 + 0.1\pi_2 + 0.4\pi_3 = \pi_1, 0.3π1+0.8π2+0.2π3=π20.3\pi_1 + 0.8\pi_2 + 0.2\pi_3 = \pi_2, 0.2π1+0.1π2+0.4π3=π30.2\pi_1 + 0.1\pi_2 + 0.4\pi_3 = \pi_3. Solving gives π1=623\pi_1 = \frac{6}{23}, π2=1223\pi_2 = \frac{12}{23}, π3=523\pi_3 = \frac{5}{23}. Choice A confuses self-transition rates with steady-state probabilities. Choice C incorrectly suggests the steady state isn't unique (it is for irreducible chains). Choice D misunderstands absorption - this isn't an absorbing chain.

Question 7

Consider the Markov chain with the transition matrix P=(0.50.20.30.10.60.30.10.10.8)P = \begin{pmatrix} 0.5 & 0.2 & 0.3 \\ 0.1 & 0.6 & 0.3 \\ 0.1 & 0.1 & 0.8 \end{pmatrix}. Let its steady-state vector be W=(w1,w2,w3)W = (w_1, w_2, w_3). What is the ratio w1/w3w_1/w_3?

  1. 5/18 (correct answer)
  2. 1/3
  3. 3/1
  4. 18/5
Explanation: The steady-state vector WW satisfies WP=WWP = W. This gives a system of linear equations. The first and third equations are: (1) 0.5w1+0.1w2+0.1w3=w1  ⟹  −0.5w1+0.1w2+0.1w3=0  ⟹  5w1=w2+w30.5w_1 + 0.1w_2 + 0.1w_3 = w_1 \implies -0.5w_1 + 0.1w_2 + 0.1w_3 = 0 \implies 5w_1 = w_2 + w_3. (2) 0.3w1+0.3w2+0.8w3=w3  ⟹  0.3w1+0.3w2−0.2w3=0  ⟹  3w1+3w2=2w30.3w_1 + 0.3w_2 + 0.8w_3 = w_3 \implies 0.3w_1 + 0.3w_2 - 0.2w_3 = 0 \implies 3w_1 + 3w_2 = 2w_3. We want to find the ratio w1/w3w_1/w_3, so we can eliminate w2w_2. From equation (1), w2=5w1−w3w_2 = 5w_1 - w_3. Substitute this into equation (2): 3w1+3(5w1−w3)=2w33w_1 + 3(5w_1 - w_3) = 2w_3. This simplifies to 3w1+15w1−3w3=2w33w_1 + 15w_1 - 3w_3 = 2w_3, which gives 18w1=5w318w_1 = 5w_3. Therefore, the ratio w1/w3=5/18w_1/w_3 = 5/18.

Question 8

Let PP be the transition matrix for a regular Markov chain, and let WW be its unique steady-state vector. Which of the following statements best describes the matrix PnP^n as the number of steps nn approaches infinity?

  1. PnP^n approaches the identity matrix II.
  2. PnP^n approaches the zero matrix.
  3. PnP^n approaches a matrix where each row is the steady-state vector WW. (correct answer)
  4. PnP^n approaches a matrix where each column is the transpose of the steady-state vector WW.
Explanation: For a regular Markov chain, the powers of the transition matrix, PnP^n, converge to a limiting matrix as n→∞n \to \infty. This limiting matrix, often denoted as Pˉ\bar{P}, has the property that all of its rows are identical and equal to the unique steady-state vector WW. This signifies that the long-run probability distribution of being in any state is independent of the initial state. The other options are incorrect: PnP^n does not generally approach the identity or zero matrix, and the convention is for rows, not columns, to be the steady-state vector.

Question 9

A lab mouse is placed in a three-chambered maze. From Chamber 1, it always moves to Chamber 2. From Chamber 2, it is equally likely to move to Chamber 1 or Chamber 3. From Chamber 3, it moves to Chamber 1 with probability 0.75 and stays in Chamber 3 with probability 0.25.

If the mouse is left to wander the maze for a very long time, what is the probability of finding it in Chamber 3?

  1. 1/4 (correct answer)
  2. 1/3
  3. 3/8
  4. 1/2
Explanation: First, construct the transition matrix PP from the description, where rows represent the current chamber and columns represent the next chamber. The states are (1, 2, 3). P=(0100.500.50.7500.25)P = \begin{pmatrix} 0 & 1 & 0 \\ 0.5 & 0 & 0.5 \\ 0.75 & 0 & 0.25 \end{pmatrix}. Let the steady-state vector be W=(w1,w2,w3)W = (w_1, w_2, w_3). We solve the system WP=WWP = W with w1+w2+w3=1w_1 + w_2 + w_3 = 1. The equations are: w1=0.5w2+0.75w3w_1 = 0.5w_2 + 0.75w_3; w2=w1w_2 = w_1; w3=0.5w2+0.25w3w_3 = 0.5w_2 + 0.25w_3. From the second equation, w1=w2w_1 = w_2. The third equation simplifies to 0.75w3=0.5w20.75w_3 = 0.5w_2, which means 3w3=2w23w_3 = 2w_2, or w3=23w2w_3 = \frac{2}{3}w_2. Now, substitute these into the sum equation: w2+w2+23w2=1w_2 + w_2 + \frac{2}{3}w_2 = 1. This gives 83w2=1\frac{8}{3}w_2 = 1, so w2=3/8w_2 = 3/8. Since w1=w2w_1=w_2, w1=3/8w_1=3/8. Finally, w3=23w2=23(38)=28=1/4w_3 = \frac{2}{3}w_2 = \frac{2}{3} (\frac{3}{8}) = \frac{2}{8} = 1/4. The long-run probability of finding the mouse in Chamber 3 is w3w_3, which is 1/4.

Question 10

A simple weather model transitions between Sunny (S) and Cloudy (C) days. The transition matrix is P=(1−ppq1−q)P = \begin{pmatrix} 1-p & p \\ q & 1-q \end{pmatrix}, where pp and qq are the probabilities of the weather changing state. In the long run, a Sunny day is observed to be twice as likely as a Cloudy day. Which of the following describes the relationship between pp and qq?

  1. q=2pq = 2p (correct answer)
  2. p=2qp = 2q
  3. p=qp = q
  4. p+q=1p + q = 1
Explanation: The long-run probabilities are given by the steady-state vector W=(wS,wC)W = (w_S, w_C). We are given that wS=2wCw_S = 2w_C. Since wS+wC=1w_S + w_C = 1, we can substitute to get 2wC+wC=12w_C + w_C = 1, which means 3wC=13w_C = 1, so wC=1/3w_C = 1/3 and wS=2/3w_S = 2/3. The steady-state vector is W=(2/3,1/3)W = (2/3, 1/3). This vector must satisfy WP=WWP = W. Using the first component of the resulting equation: (2/3)(1−p)+(1/3)(q)=2/3(2/3)(1-p) + (1/3)(q) = 2/3. Multiplying the entire equation by 3 gives 2(1−p)+q=22(1-p) + q = 2. This simplifies to 2−2p+q=22 - 2p + q = 2, which implies q=2pq = 2p.

Question 11

Voters in a district are registered with either Party A or Party B. Each election cycle, 85% of Party A voters remain with their party, while 15% switch to Party B. Concurrently, 95% of Party B voters remain with their party, while 5% switch to Party A. In the long run, what proportion of the electorate is expected to be aligned with Party A?

  1. 20%
  2. 25% (correct answer)
  3. 50%
  4. 75%
Explanation: Let the states be Party A and Party B. The transition matrix is P=(0.850.150.050.95)P = \begin{pmatrix} 0.85 & 0.15 \\ 0.05 & 0.95 \end{pmatrix}. Let the steady-state vector be W=(wA,wB)W = (w_A, w_B). We need to solve WP=WWP = W with wA+wB=1w_A + w_B = 1. The first component of WP=WWP=W is 0.85wA+0.05wB=wA0.85w_A + 0.05w_B = w_A. This simplifies to 0.05wB=0.15wA0.05w_B = 0.15w_A, or wB=3wAw_B = 3w_A. Using the constraint wA+wB=1w_A + w_B = 1, we can substitute to get wA+3wA=1w_A + 3w_A = 1, which gives 4wA=14w_A = 1, so wA=0.25w_A = 0.25. Therefore, in the long run, 25% of the electorate is expected to be aligned with Party A.

Question 12

Two companies, Apex and Bedrock, compete in a market. Each year, Apex keeps 70% of its customers while 30% switch to Bedrock. Bedrock keeps 80% of its customers while 20% switch to Apex. Assuming these trends continue and the total number of customers in the market is stable, what is the long-run market share for Apex?

  1. 30%
  2. 40% (correct answer)
  3. 50%
  4. 60%
Explanation: Let the states be Apex (A) and Bedrock (B). The transition matrix is P=(0.70.30.20.8)P = \begin{pmatrix} 0.7 & 0.3 \\ 0.2 & 0.8 \end{pmatrix}, where PijP_{ij} is the probability of moving from state ii to state jj. Let the steady-state vector be W=(wA,wB)W = (w_A, w_B). We need to solve the system of equations given by WP=WWP = W and wA+wB=1w_A + w_B = 1. The equation WP=WWP = W gives 0.7wA+0.2wB=wA0.7w_A + 0.2w_B = w_A and 0.3wA+0.8wB=wB0.3w_A + 0.8w_B = w_B. Both simplify to 0.2wB=0.3wA0.2w_B = 0.3w_A, or 2wB=3wA2w_B = 3w_A. Using the constraint wB=1−wAw_B = 1 - w_A, we substitute to get 2(1−wA)=3wA2(1 - w_A) = 3w_A. This simplifies to 2−2wA=3wA2 - 2w_A = 3w_A, which gives 2=5wA2 = 5w_A, so wA=2/5=0.4w_A = 2/5 = 0.4. Thus, Apex's long-run market share is 40%.

Question 13

The economy of a region is modeled as being in one of three states: Boom (B), Stagnation (S), or Recession (R). The year-to-year transition probabilities are given by the matrix P=(0.60.30.10.20.60.20.10.30.6)P = \begin{pmatrix} 0.6 & 0.3 & 0.1 \\ 0.2 & 0.6 & 0.2 \\ 0.1 & 0.3 & 0.6 \end{pmatrix}, where the states are in the order (B, S, R).

If the economy is currently in a Boom, what is the long-run probability that it will be in a Recession?

  1. 1/10
  2. 9/50
  3. 2/7 (correct answer)
  4. 3/7
Explanation: For a regular Markov chain, the long-run probabilities are given by the steady-state vector WW, and these probabilities are independent of the initial state. We need to find W=(wB,wS,wR)W = (w_B, w_S, w_R) by solving WP=WWP=W. Due to the symmetry of the matrix (P12=P32P_{12}=P_{32}, P21=P23P_{21}=P_{23}, etc.), we can deduce that wB=wRw_B = w_R. The first component equation is 0.6wB+0.2wS+0.1wR=wB0.6w_B + 0.2w_S + 0.1w_R = w_B. Substituting wR=wBw_R=w_B, we get 0.7wB+0.2wS=wB0.7w_B + 0.2w_S = w_B, which simplifies to 0.2wS=0.3wB0.2w_S = 0.3w_B, or wS=32wBw_S = \frac{3}{2}w_B. Using the condition wB+wS+wR=1w_B + w_S + w_R = 1, we have wB+32wB+wB=1w_B + \frac{3}{2}w_B + w_B = 1. This becomes 3.5wB=13.5w_B = 1, so wB=1/3.5=2/7w_B = 1/3.5 = 2/7. Then wR=2/7w_R = 2/7 and wS=3/7w_S=3/7. The long-run probability of a Recession is wR=2/7w_R = 2/7. The fact that the economy starts in a Boom is irrelevant to the long-run probability. Distractor 1/10 is the one-step probability. Distractor 9/50 is the two-step probability.

Question 14

A transition matrix PP is called regular if some power PkP^k contains only positive entries. The existence of a regular power guarantees that the Markov chain has a unique steady-state vector that is independent of the initial state. Which of the following transition matrices is regular?

  1. A=(0110)A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
  2. B=(100.50.5)B = \begin{pmatrix} 1 & 0 \\ 0.5 & 0.5 \end{pmatrix}
  3. C=(0.50.5000.50.50.500.5)C = \begin{pmatrix} 0.5 & 0.5 & 0 \\ 0 & 0.5 & 0.5 \\ 0.5 & 0 & 0.5 \end{pmatrix} (correct answer)
  4. D=(0.50.500.50.50001)D = \begin{pmatrix} 0.5 & 0.5 & 0 \\ 0.5 & 0.5 & 0 \\ 0 & 0 & 1 \end{pmatrix}
Explanation: We need to check if any power of the matrices has all positive entries. A) A2=IA^2 = I, A3=AA^3 = A, so powers of AA will always contain zeros. B) State 1 is an absorbing state. The entry B12B_{12} will always be 0 in any power BkB^k. C) Let's compute C2C^2. C2=(0.250.50.250.250.250.50.50.250.25)C^2 = \begin{pmatrix} 0.25 & 0.5 & 0.25 \\ 0.25 & 0.25 & 0.5 \\ 0.5 & 0.25 & 0.25 \end{pmatrix}. Since all entries of C2C^2 are positive, matrix CC is regular. D) States 1 and 2 form a sub-chain that cannot reach state 3. The entries D13D_{13} and D23D_{23} will always be 0 in any power DkD^k. Therefore, only matrix C is regular.

Question 15

Let W=(w1,w2,...,wn)W = (w_1, w_2, ..., w_n) be the unique steady-state vector for a regular transition matrix PP. Which of the following statements about WW is NOT always true?

  1. The sum of the components of WW must be equal to 1.
  2. WW is an eigenvector of the matrix PP corresponding to an eigenvalue of 1.
  3. For any initial probability vector V0V_0, the distribution V0PnV_0P^n approaches WW as n→∞n \to \infty.
  4. The component wiw_i represents the one-step probability of transitioning to state ii from any other state. (correct answer)
Explanation: The steady-state vector WW represents the long-run probability distribution of the states. Choice A is true by definition of a probability vector. Choice B is the mathematical definition of a steady-state vector (WP=WWP=W is equivalent to W(P−I)=0W(P-I)=0, which is a left eigenvector problem for eigenvalue 1). Choice C is the fundamental theorem for regular Markov chains. Choice D is incorrect. The one-step probability of transitioning to state ii from state jj is given by the matrix entry PjiP_{ji}, which is not generally equal to wiw_i and depends on the starting state jj. The component wiw_i is the long-run average probability of being in state ii.