Markov Chains and Stochastic Processes Questions and Answers

Q: Following are three statements. In a Markov chain, a persistent state i is a null persistent if :(I) \(\limsup p_{ii}^{(n)} = 0\)(II) mean recurrence time is \(\infty\)(III) \(\limsup p_{ii}^{(n)} \gt 0\)Which of the following is true ?

Both (I) and (II) are true.A persistent (recurrent) state is null persistent when its mean recurrence time \(\mu_{i}\) is infinite; that is the definition, so (II) is true. For such a state \(p_{ii}^{(n)} \to 0\), so \(\limsup p_{ii}^{(n)} = 0\) and (I) is true.A positive persistent state has finite mean recurrence time and, if aperiodic, \(p_{ii}^{(n)} \to 1/\mu_{i} \gt 0\). So \(\limsup p_{ii}^{(n)} \gt 0\) characterises positive persistence, and (III) is false for a null persistent state.

Q: Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3) :\[ P = \begin{bmatrix} 0.3 & \alpha & \beta \\ 0.4 & \gamma & 0.1 \\ \delta & 0.2 & \varepsilon \end{bmatrix} \]If P is a doubly stochastic matrix, the value of \(\beta\) :

\(\beta\) is 0.4.In a doubly stochastic matrix every row and every column sums to 1. Work through the entries that are forced:\[ \begin{aligned} \text{column 1: } 0.3 + 0.4 + \delta = 1 &\Rightarrow \delta = 0.3 \\ \text{row 3: } 0.3 + 0.2 + \varepsilon = 1 &\Rightarrow \varepsilon = 0.5 \\ \text{column 3: } \beta + 0.1 + 0.5 = 1 &\Rightarrow \beta = 0.4 \end{aligned} \]As a check, row 1 then gives \(\alpha = 0.3\) and column 2 gives \(\gamma = 0.5\), with row 2 summing to 1. So \(\beta\) can be computed and equals exactly 0.4.

Q: Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3) :\[ P = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 0 \\ \frac{3}{8} & \frac{5}{8} & 0 \\ \frac{1}{6} & \frac{1}{3} & \frac{1}{2} \end{bmatrix} \]Suppose X0 = 3. Then realized values of X1 and X2 corresponding to random numbers 0.36 and 0.64 respectively, from U(0, 1) distribution are :

The realized values are X1 = 2 and X2 = 2.To simulate a step, compare the uniform number with the cumulative probabilities of the current row.From \(X_{0} = 3\), row 3 is \((\frac{1}{6}, \frac{1}{3}, \frac{1}{2})\), with cumulative values 0.1667, 0.5, 1. Since \(0.1667 \lt 0.36 \le 0.5\), \(X_{1} = 2\).From state 2, row 2 is \((\frac{3}{8}, \frac{5}{8}, 0)\), with cumulative values 0.375, 1, 1. Since \(0.375 \lt 0.64 \le 1\), \(X_{2} = 2\).Reading 0.36 against row 1 or 0.64 against row 3 gives the other combinations, which use the wrong rows.

Q: Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3, 4} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3, 4) :\[ P = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{2}{3} \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{4} & 0 & \frac{3}{4} \end{bmatrix} \]Which of the following options is true ?

States 1 and 3 are essential states.From state 1 the chain moves only to 1 or 3, and from 3 only to 1 or 3. So {1, 3} is a closed class: once entered it is never left, and 1 and 3 communicate. A state is essential if every state it leads to leads back to it, which holds for 1 and 3.A finite closed communicating class is positive persistent, so 1 and 3 are not transient or inessential, and not null persistent (a finite chain has no null persistent states). The same holds for the separate closed class {2, 4}.

Q: Which of the following options is true ?Messages arrive on a mobile phone according to the Poisson process with rate 5 messages per hour. The probability that no message arrives during 10 : 00 am to 12 : 00 noon is :

The probability is e−10.From 10:00 am to 12:00 noon is 2 hours. In a Poisson process with rate 5 per hour, the number of messages in 2 hours is Poisson with mean \(5 \times 2 = 10\), so\[ P(N = 0) = e^{-10} \]\(1 - e^{-10}\) is the probability of at least one message, \(2e^{-5}\) uses a one-hour mean and doubles it incorrectly, and \(e^{-2/5}\) divides the time by the rate instead of multiplying.

Q: Suppose \(\{X(t), t \ge 0\}\) is a pure birth process with birth rate \(\lambda_{i} = \lambda\) for all \(i \in S\) and \(X(0) = 0\). The following are four statements.(I) \(\{X(t), t \ge 0\}\) is a process with stationary and independent increments.(II) \(E(X(t)) = e^{\lambda t}\)(III) X(t) follows Poisson P(\(\lambda t\)) distribution for every fixed t.(IV) X(0) cannot be 0 in a pure birth process.Which of the following is true ?

Only (I) and (III) are true.A pure birth process with constant rate \(\lambda\) starting at 0 is the Poisson process. It has stationary and independent increments (I), and X(t) is Poisson with mean \(\lambda t\) for each t (III).(II) is false: the mean is \(E(X(t)) = \lambda t\), not \(e^{\lambda t}\). The exponential mean belongs to the Yule process, where the birth rate is proportional to the population, \(\lambda_{i} = i\lambda\). (IV) is false: X(0) = 0 is the usual starting point of the Poisson process.

Q: Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space \(S = \{1, 2, 3, 4\}\) and the transition probability matrix P given by :\[ P = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{2}{3} \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{4} & 0 & \frac{3}{4} \end{bmatrix} \](rows and columns are indexed by the states 1, 2, 3, 4)Suppose \(f_{ij}\) denotes the probability of first visit to j from i. The following are four statements :(I) \(f_{12} = 0\), (II) \(f_{24} = 1\), (III) \(f_{13} = 1\), (IV) \(f_{33} \lt 1\).Which of the following is true ?

Only (I), (II) and (III) are true.From the matrix, state 1 moves only to 1 or 3, and state 3 only to 1 or 3; state 2 moves only to 2 or 4, and state 4 only to 2 or 4. So the chain splits into two closed classes, {1, 3} and {2, 4}.(I) Starting in {1, 3}, the chain can never reach 2, so \(f_{12} = 0\). (II) and (III) A finite closed irreducible class is recurrent, so from any state of the class every other state is reached with probability 1: \(f_{24} = 1\) and \(f_{13} = 1\).(IV) State 3 is recurrent, so \(f_{33} = 1\), not less than 1. That leaves exactly (I), (II) and (III).

Q: Suppose in a Markov chain, \(f_{22}\) denotes the probability of the first return to state 2 and \(\mu_{2}\) denotes the mean recurrence time to state 2. Suppose \(\sum_{n \ge 1} p_{22}^{(n)} = 2.45\). Then which of the following options is true ?

The true option is \(f_{22} \lt 1\).A state is recurrent exactly when \(\sum_{n} p_{22}^{(n)} = \infty\). Here the sum is the finite number 2.45, so state 2 is transient and \(f_{22} \lt 1\). In fact the sum equals \(\frac{f_{22}}{1 - f_{22}}\), so\[ f_{22} = \frac{2.45}{3.45} \approx 0.71 \]A transient state has \(f_{22} \lt 1\) (so option B fails) and an infinite mean recurrence time (option C fails), and since the series of \(p_{22}^{(n)}\) converges, its terms tend to 0 (option D fails).

Q: Suppose \(f_{ij}^{(n)}\) denotes the probability of the first visit to j from i in n steps. Suppose the Markov chain is periodic with period 2. The following are three statements :(I) \(f_{ii}^{(2)} = p_{ii}^{(2)} - p_{ii}^{(1)} f_{ii}^{(1)}\)(II) \(f_{ii}^{(4)} = p_{ii}^{(4)} - p_{ii}^{(2)} f_{ii}^{(2)}\)(III) \(f_{ii}^{(7)} = 0\)Which of the following is true ?

All three are true.The first-passage decomposition gives \(p_{ii}^{(n)} = \sum_{k=1}^{n} f_{ii}^{(k)} p_{ii}^{(n-k)}\), with \(p_{ii}^{(0)} = 1\).For \(n = 2\): \(p_{ii}^{(2)} = f_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(1)}\), which rearranges to (I). This holds in any chain.For \(n = 4\):\[ \begin{aligned} p_{ii}^{(4)} = &f_{ii}^{(4)} + f_{ii}^{(3)}p_{ii}^{(1)} \\ &+ f_{ii}^{(2)}p_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(3)} \end{aligned} \]With period 2, returns are possible only at even times, so \(p_{ii}^{(1)} = p_{ii}^{(3)} = 0\) and \(f_{ii}^{(1)} = f_{ii}^{(3)} = 0\), leaving (II).For the same reason a first return at the odd time 7 is impossible, so \(f_{ii}^{(7)} = 0\), which is (III).

31.

Following are three statements. In a Markov chain, a persistent state i is a null persistent if :

(I) \(\limsup p_{ii}^{(n)} = 0\)
(II) mean recurrence time is \(\infty\)
(III) \(\limsup p_{ii}^{(n)} \gt 0\)

Which of the following is true ?

A.Only (I) is true
B.Both (I) and (II) are true
C.Only (III) is true
D.Both (II) and (III) are true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

Both (I) and (II) are true.

A persistent (recurrent) state is null persistent when its mean recurrence time \(\mu_{i}\) is infinite; that is the definition, so (II) is true. For such a state \(p_{ii}^{(n)} \to 0\), so \(\limsup p_{ii}^{(n)} = 0\) and (I) is true.

A positive persistent state has finite mean recurrence time and, if aperiodic, \(p_{ii}^{(n)} \to 1/\mu_{i} \gt 0\). So \(\limsup p_{ii}^{(n)} \gt 0\) characterises positive persistence, and (III) is false for a null persistent state.

Discuss in Forum

32.

Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3) :

\[ P = \begin{bmatrix} 0.3 & \alpha & \beta \\ 0.4 & \gamma & 0.1 \\ \delta & 0.2 & \varepsilon \end{bmatrix} \]

If P is a doubly stochastic matrix, the value of \(\beta\) :

A.Cannot be computed from the given information
B.is \(\lt 0.4\)
C.is \(\gt 0.4\)
D.is 0.4

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

\(\beta\) is 0.4.

In a doubly stochastic matrix every row and every column sums to 1. Work through the entries that are forced:

\[ \begin{aligned} \text{column 1: } 0.3 + 0.4 + \delta = 1 &\Rightarrow \delta = 0.3 \\ \text{row 3: } 0.3 + 0.2 + \varepsilon = 1 &\Rightarrow \varepsilon = 0.5 \\ \text{column 3: } \beta + 0.1 + 0.5 = 1 &\Rightarrow \beta = 0.4 \end{aligned} \]

As a check, row 1 then gives \(\alpha = 0.3\) and column 2 gives \(\gamma = 0.5\), with row 2 summing to 1. So \(\beta\) can be computed and equals exactly 0.4.

Discuss in Forum

33.

Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3) :

\[ P = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 0 \\ \frac{3}{8} & \frac{5}{8} & 0 \\ \frac{1}{6} & \frac{1}{3} & \frac{1}{2} \end{bmatrix} \]

Suppose X₀ = 3. Then realized values of X₁ and X₂ corresponding to random numbers 0.36 and 0.64 respectively, from U(0, 1) distribution are :

A.\(X_{1} = 2\) and \(X_{2} = 2\)
B.\(X_{1} = 1\) and \(X_{2} = 1\)
C.\(X_{1} = 1\) and \(X_{2} = 2\)
D.\(X_{1} = 2\) and \(X_{2} = 1\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The realized values are X₁ = 2 and X₂ = 2.

To simulate a step, compare the uniform number with the cumulative probabilities of the current row.

From \(X_{0} = 3\), row 3 is \((\frac{1}{6}, \frac{1}{3}, \frac{1}{2})\), with cumulative values 0.1667, 0.5, 1. Since \(0.1667 \lt 0.36 \le 0.5\), \(X_{1} = 2\).

From state 2, row 2 is \((\frac{3}{8}, \frac{5}{8}, 0)\), with cumulative values 0.375, 1, 1. Since \(0.375 \lt 0.64 \le 1\), \(X_{2} = 2\).

Reading 0.36 against row 1 or 0.64 against row 3 gives the other combinations, which use the wrong rows.

Discuss in Forum

34.

Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space S = {1, 2, 3, 4} and the transition probability matrix P given by (rows and columns indexed by the states 1, 2, 3, 4) :

\[ P = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{2}{3} \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{4} & 0 & \frac{3}{4} \end{bmatrix} \]

Which of the following options is true ?

A.States 1 and 3 are essential states
B.States 1 and 3 are inessential states
C.States 1 and 3 are transient states
D.State 1 and 3 are null persistent states

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

States 1 and 3 are essential states.

From state 1 the chain moves only to 1 or 3, and from 3 only to 1 or 3. So {1, 3} is a closed class: once entered it is never left, and 1 and 3 communicate. A state is essential if every state it leads to leads back to it, which holds for 1 and 3.

A finite closed communicating class is positive persistent, so 1 and 3 are not transient or inessential, and not null persistent (a finite chain has no null persistent states). The same holds for the separate closed class {2, 4}.

Discuss in Forum

35.

Which of the following options is true ?

Messages arrive on a mobile phone according to the Poisson process with rate 5 messages per hour. The probability that no message arrives during 10 : 00 am to 12 : 00 noon is :

A.\(2e^{-5}\)
B.\(e^{-10}\)
C.\(1 - e^{-10}\)
D.\(e^{-\frac{2}{5}}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The probability is e⁻¹⁰.

From 10:00 am to 12:00 noon is 2 hours. In a Poisson process with rate 5 per hour, the number of messages in 2 hours is Poisson with mean \(5 \times 2 = 10\), so

\[ P(N = 0) = e^{-10} \]

\(1 - e^{-10}\) is the probability of at least one message, \(2e^{-5}\) uses a one-hour mean and doubles it incorrectly, and \(e^{-2/5}\) divides the time by the rate instead of multiplying.

Discuss in Forum

36.

Suppose \(\{X(t), t \ge 0\}\) is a pure birth process with birth rate \(\lambda_{i} = \lambda\) for all \(i \in S\) and \(X(0) = 0\). The following are four statements.

(I) \(\{X(t), t \ge 0\}\) is a process with stationary and independent increments.
(II) \(E(X(t)) = e^{\lambda t}\)
(III) X(t) follows Poisson P(\(\lambda t\)) distribution for every fixed t.
(IV) X(0) cannot be 0 in a pure birth process.

Which of the following is true ?

A.Only (I) is true
B.Only (I) and (II) are true
C.Only (I) and (III) are true
D.Only (IV) is true

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

Only (I) and (III) are true.

A pure birth process with constant rate \(\lambda\) starting at 0 is the Poisson process. It has stationary and independent increments (I), and X(t) is Poisson with mean \(\lambda t\) for each t (III).

(II) is false: the mean is \(E(X(t)) = \lambda t\), not \(e^{\lambda t}\). The exponential mean belongs to the Yule process, where the birth rate is proportional to the population, \(\lambda_{i} = i\lambda\). (IV) is false: X(0) = 0 is the usual starting point of the Poisson process.

Discuss in Forum

37.

The following are the transition probability matrices of the Markov chains with state space \(S = \{1, 2\}\) :

\[ P_{1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad P_{2} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \]

\[ P_{3} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{2}{3} & \frac{1}{3} \end{bmatrix}, \quad P_{4} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 0 & 1 \end{bmatrix} \]

(rows and columns are indexed by the states 1, 2)

Which Markov chains have a stationary distribution given by \(\left(\frac{1}{2}, \frac{1}{2}\right)\) ?

Markov chains with transition probability matrices.

A.\(P_{1}\), \(P_{3}\)
B.\(P_{1}\), \(P_{2}\)
C.\(P_{1}\), \(P_{4}\)
D.\(P_{2}\), \(P_{3}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The chains with stationary distribution \(\left(\frac{1}{2}, \frac{1}{2}\right)\) are \(P_{1}\) and \(P_{2}\).

Check \(\pi P = \pi\) with \(\pi = \left(\frac{1}{2}, \frac{1}{2}\right)\). For the identity \(P_{1}\), every distribution is stationary. \(P_{2}\) swaps the two states, so it maps \(\left(\frac{1}{2}, \frac{1}{2}\right)\) to itself; in general a doubly stochastic matrix has the uniform distribution stationary.

\[ \begin{aligned} \pi P_{3} &= \left(\tfrac{1}{4} + \tfrac{1}{3}, \tfrac{1}{4} + \tfrac{1}{6}\right) = \left(\tfrac{7}{12}, \tfrac{5}{12}\right) \\ \pi P_{4} &= \left(\tfrac{1}{4}, \tfrac{1}{4} + \tfrac{1}{2}\right) = \left(\tfrac{1}{4}, \tfrac{3}{4}\right) \end{aligned} \]

Neither is \(\left(\frac{1}{2}, \frac{1}{2}\right)\), so every option containing \(P_{3}\) or \(P_{4}\) is wrong. (\(P_{4}\)'s stationary distribution is (0, 1), since state 2 is absorbing.)

Discuss in Forum

38.

Suppose \(\{X_{n}, n \ge 0\}\) is a Markov chain with state space \(S = \{1, 2, 3, 4\}\) and the transition probability matrix P given by :

\[ P = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{2}{3} \\ \frac{2}{3} & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{4} & 0 & \frac{3}{4} \end{bmatrix} \]

(rows and columns are indexed by the states 1, 2, 3, 4)

Suppose \(f_{ij}\) denotes the probability of first visit to j from i. The following are four statements :

(I) \(f_{12} = 0\), (II) \(f_{24} = 1\), (III) \(f_{13} = 1\), (IV) \(f_{33} \lt 1\).

Which of the following is true ?

A.Only (I) and (II) are true
B.Only (I) and (III) are true
C.Only (I) and (IV) are true
D.Only (I), (II) and (III) are true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

Only (I), (II) and (III) are true.

From the matrix, state 1 moves only to 1 or 3, and state 3 only to 1 or 3; state 2 moves only to 2 or 4, and state 4 only to 2 or 4. So the chain splits into two closed classes, {1, 3} and {2, 4}.

(I) Starting in {1, 3}, the chain can never reach 2, so \(f_{12} = 0\). (II) and (III) A finite closed irreducible class is recurrent, so from any state of the class every other state is reached with probability 1: \(f_{24} = 1\) and \(f_{13} = 1\).

(IV) State 3 is recurrent, so \(f_{33} = 1\), not less than 1. That leaves exactly (I), (II) and (III).

Discuss in Forum

39.

Suppose in a Markov chain, \(f_{22}\) denotes the probability of the first return to state 2 and \(\mu_{2}\) denotes the mean recurrence time to state 2. Suppose \(\sum_{n \ge 1} p_{22}^{(n)} = 2.45\). Then which of the following options is true ?

A.\(f_{22} \lt 1\)
B.\(f_{22} = 1\)
C.\(\mu_{2} \lt \infty\)
D.\(\lim_{n \to \infty} p_{22}^{(n)} \gt 0\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The true option is \(f_{22} \lt 1\).

A state is recurrent exactly when \(\sum_{n} p_{22}^{(n)} = \infty\). Here the sum is the finite number 2.45, so state 2 is transient and \(f_{22} \lt 1\). In fact the sum equals \(\frac{f_{22}}{1 - f_{22}}\), so

\[ f_{22} = \frac{2.45}{3.45} \approx 0.71 \]

A transient state has \(f_{22} \lt 1\) (so option B fails) and an infinite mean recurrence time (option C fails), and since the series of \(p_{22}^{(n)}\) converges, its terms tend to 0 (option D fails).

Discuss in Forum

40.

Suppose \(f_{ij}^{(n)}\) denotes the probability of the first visit to j from i in n steps. Suppose the Markov chain is periodic with period 2. The following are three statements :

(I) \(f_{ii}^{(2)} = p_{ii}^{(2)} - p_{ii}^{(1)} f_{ii}^{(1)}\)
(II) \(f_{ii}^{(4)} = p_{ii}^{(4)} - p_{ii}^{(2)} f_{ii}^{(2)}\)
(III) \(f_{ii}^{(7)} = 0\)

Which of the following is true ?

A.Only (I) and (II) are true
B.Only (I) and (III) are true
C.Only (III) is true
D.All three are true

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

All three are true.

The first-passage decomposition gives \(p_{ii}^{(n)} = \sum_{k=1}^{n} f_{ii}^{(k)} p_{ii}^{(n-k)}\), with \(p_{ii}^{(0)} = 1\).

For \(n = 2\): \(p_{ii}^{(2)} = f_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(1)}\), which rearranges to (I). This holds in any chain.

For \(n = 4\):

\[ \begin{aligned} p_{ii}^{(4)} = &f_{ii}^{(4)} + f_{ii}^{(3)}p_{ii}^{(1)} \\ &+ f_{ii}^{(2)}p_{ii}^{(2)} + f_{ii}^{(1)}p_{ii}^{(3)} \end{aligned} \]

With period 2, returns are possible only at even times, so \(p_{ii}^{(1)} = p_{ii}^{(3)} = 0\) and \(f_{ii}^{(1)} = f_{ii}^{(3)} = 0\), leaving (II).

For the same reason a first return at the odd time 7 is impossible, so \(f_{ii}^{(7)} = 0\), which is (III).

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