Markov Chains and Stochastic Processes Questions and Answers

Q: Consider a Markov chain on S = {0, 1, 2, ......}. The transition probabilities are given by :\(p_{00} = \dfrac{1}{2}\) \(p_{01} = \dfrac{1}{2}\)and for \(i \ge 1\),\(p_{ii - 1} = \dfrac{1}{2}\) \(p_{ii + 1} = \dfrac{1}{2}\)Then :

All states are transient or persistent null (in fact null persistent).This is a symmetric random walk on \(\{0, 1, 2, \ldots\}\), with the step down from 0 replaced by staying put. It behaves like the distance from the origin of a simple symmetric walk, which returns with probability 1, so every state is persistent.A stationary distribution would satisfy \(\pi_{0} = \tfrac{1}{2}\pi_{0} + \tfrac{1}{2}\pi_{1}\) and \(\pi_{j} = \tfrac{1}{2}\pi_{j-1} + \tfrac{1}{2}\pi_{j+1}\), giving \(\pi_{j} = \pi_{0}\) for all \(j\), which cannot sum to 1. So there is no stationary distribution, and the persistent states are null.Hence no state is non-null persistent, and no stationary distribution exists, ruling out A, B and C.

Q: Consider a branching process \(\{X_{n}, n \ge 0\}\). Let \(X_{0} = 1\) and assume that \(E\left(\sum_{\theta}^{\infty} X_{n}\right) \lt \infty\). Then, the extinction probability :

The extinction probability equals 1.The sum is printed from \(\theta\), meaning from generation 0. With mean offspring \(m\), \(E(X_{n}) = m^{n}\) when \(X_{0} = 1\), so\[ E\left(\sum_{n} X_{n}\right) = \sum_{n} m^{n} \]which is finite only if \(m \lt 1\). A subcritical branching process dies out with probability 1. (Equivalently, a finite expected total progeny forces the total progeny to be finite almost surely, so the population cannot survive forever.)An extinction probability in (0, 1) arises only for \(m \gt 1\), when the expected total is infinite; 0 is impossible once there is any chance of having no offspring, and the extinction probability always exists as a limit of \(P(X_{n} = 0)\).

Q: Let \(Y_{1}, Y_{2} \ldots\) be a sequence of independent random variables with mean zero and let \(\mathbf{F}_{n} = \sigma(Y_{1}, \ldots, Y_{n})\).Define \(X_{n} = \sum_{k=1}^{n} Y_{k}\); \(W_{1} = Y_{1}\) and \(W_{n+1} = W_{n} + Y_{n+1}W_{n}^{2}\) and \(T_{n} = \sum_{k=1}^{n} (Y_{k} + 1)\). Then :

The answer is both \(\{X_{n}\}\) and \(\{W_{n}\}\) are \(\mathbf{F}_{n}\)-martingales.\(X_{n+1} = X_{n} + Y_{n+1}\), and \(Y_{n+1}\) is independent of \(\mathbf{F}_{n}\) with mean 0, so \(E[X_{n+1} \mid \mathbf{F}_{n}] = X_{n}\).\(W_{n}\) is \(\mathbf{F}_{n}\)-measurable, so\[ \begin{aligned} E[W_{n+1} \mid \mathbf{F}_{n}] &= W_{n} + W_{n}^{2}E[Y_{n+1}] \\ &= W_{n} \end{aligned} \](assuming the variables are integrable). But \(T_{n} = X_{n} + n\), so \(E[T_{n+1} \mid \mathbf{F}_{n}] = T_{n} + 1\): it is a submartingale, not a martingale. That rules out the options including \(T_{n}\), and 'only \(X_{n}\)' misses \(W_{n}\).

Q: Let \(\{X_{t}\}\) be a stationary time series model given by :\(X_{t} = \mu + \phi X_{t-1} + Z_{t}\), \(Z_{t} \sim\) iid normal \((0, \sigma^{2})\). Then, which of the following statements is not true if \(P_{X_{t}}(X_{t+k})\) is the forecast of \(X_{t+k}\). Given \(X_{1}, X_{2} \ldots X_{t}\) ?

The statement that is not true is the forecast mean square error approaches \(\mu^{2} + \sigma^{2}/(1 - \phi^{2})\).For a stationary AR(1) with constant, the \(k\)-step forecast is\[ \begin{aligned} P(X_{t+k}) &= m + \phi^{k}(X_{t} - m), \\ m &= \frac{\mu}{1 - \phi} \end{aligned} \]a weighted average of the mean \(E(X_{t}) = m\) and the latest value \(X_{t}\) (D), which tends to \(m\) as \(k \to \infty\) (C). Its error variance is \(\sigma^{2}(1 - \phi^{2k})/(1 - \phi^{2})\), which tends to the process variance \(\sigma^{2}/(1 - \phi^{2})\) (B).A forecast error has mean 0, so the constant \(\mu\) never adds \(\mu^{2}\) to the mean square error; (A) is the false statement.

Q: Let \(\{X_{n}\}_{0}^{\infty}\) be a Markov chain with \(P[X_{0} = 0] = P[X_{0} = 1] = \dfrac{1}{4}\) and \(P[X_{0} = 2] = \dfrac{1}{2}\). The t.p.m. is given by :01201/21/31/6101/32/321/201/2Then \(P[X_{1} = 1]\) is given by :

\(P[X_{1} = 1] = \) 1/6.Multiply the initial distribution by the column of the transition matrix for state 1:\[ \begin{aligned} P[X_{1} = 1] &= \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{2}\cdot 0 \\ &= \tfrac{1}{12} + \tfrac{1}{12} = \tfrac{1}{6} \end{aligned} \]1/3 is just the transition probability from 0 (or 1) to 1, ignoring the starting distribution, and 1/4 is the initial probability of state 1.

Q: Let \(\{X(t), t \ge 0\}\) be a time homogeneous Poisson process with rate \(\lambda\). Let \(X(T) = n \gt 0\) and let \(T_{1} \lt \cdots \lt T_{n} \lt T\) be times at which these \(n\) events occur. Then, \(E(T_{1} \mid X(T) = n)\) is given by :

\(E(T_{1} \mid X(T) = n) = \) \(\dfrac{T}{n + 1}\).For a (homogeneous) Poisson process, given that \(n\) events occurred in [0, T], the event times are distributed as the order statistics of \(n\) independent \(U(0, T)\) variables. The smallest of \(n\) uniforms on [0, T] has mean\[ E(U_{(1)}) = \frac{T}{n + 1} \](the \(n\) points split [0, T] into \(n + 1\) gaps of equal expected length). The answer does not depend on \(\lambda\), which rules out \(T\lambda^{n}\), and \(n/(n + 1) - T\) can be negative.

Q: Consider a Branching process with \(\mu\) as the mean of the off-spring distribution. Let \(\mu = \dfrac{1}{2}\) and \(X_{0} = 1\). Then \(E\left(\sum_{0}^{\infty} X_{n}\right)\) equals :

\(E\left(\sum X_{n}\right) = \) 2.In a branching process with \(X_{0} = 1\), \(E(X_{n}) = \mu^{n}\). With \(\mu = 1/2\), the expected total progeny is\[ \begin{aligned} \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n} &= \frac{1}{1 - 1/2} \\ &= 2 \end{aligned} \](by monotone convergence the expectation and the sum can be exchanged). The process is subcritical and dies out, so the total is finite, not \(\infty\). 1 counts only the ancestor and \(1\frac{1}{2}\) stops after the first generation.

Q: Let \(\{X_{t}\}\) be a stationary time series with autocovariance function \(r(h)\). Then, which of the following statements is not true about \(r(h)\) ?

The statement that is not true is \(|\gamma(k)| \ge \gamma(0)\).By the Cauchy-Schwarz inequality, \(|\gamma(k)| = |\mathrm{Cov}(X_{t}, X_{t+k})|\), which is at most \(\sqrt{\mathrm{Var}X_{t}\,\mathrm{Var}X_{t+k}} = \gamma(0)\). So the true inequality is the reverse, \(|\gamma(k)| \le \gamma(0)\).The others are standard properties of an autocovariance function: \(\gamma(0)\) is the variance (A), \(\gamma\) is even (B), and \(\gamma\) is non-negative definite (D), because \(\sum\sum \alpha_{i}\alpha_{j}\gamma(|t_{i} - t_{j}|)\) equals \(\mathrm{Var}(\sum \alpha_{i}X_{t_{i}}) \ge 0\).

Q: Given two time series,\(X_{t} = 0.3\,X_{t-1} - 0.2\,X_{t-2} + Z_{t}\)\(Y_{t} - 0.4\,Y_{t-1} = Z_{t} + 1.2\,Z_{t-1}\)where \(\{Z_{t}\}\) iid WN(0, \(\sigma^{2}\)), which of the following statements is true ?

The true statement is that \(\{Y_{t}\}\) is causal, but not invertible.\(Y_{t}\) is ARMA(1, 1) with AR polynomial \(1 - 0.4z\), whose root \(z = 2.5\) lies outside the unit circle, so it is causal. Its MA polynomial \(1 + 1.2z\) has root \(z = -1/1.2 \approx -0.83\), inside the unit circle, so it is not invertible.\(X_{t}\) is AR(2) with polynomial \(1 - 0.3z + 0.2z^{2}\). Its roots are complex with \(|z|^{2} = 1/0.2 = 5\), so \(|z| = \sqrt{5} \gt 1\) and \(X_{t}\) is causal; a pure AR process is always invertible. So B and D are false, and A has Y's properties reversed.

Q: Let \(\{X_{t}\}\) be a stationary AR(1) model with \(X_{t} = \mu + \phi X_{t-1} + Z_{t}\), where \(Z_{t} \sim \mathrm{WN}(0, \sigma^{2})\). Then, which of the following statements is true ?(i) \(E(X_{t}) = \mu\)(ii) \(V(X_{t}) = \sigma^{2}/1 - \phi^{2}\)(iii) \(\mathrm{Corr}(X_{t}, X_{t+s}) = \phi^{s}\)(iv) \(V(X_{t}) = 1 - \phi^{2}\)

The true statements are (ii) and (iii).Taking expectations of \(X_{t} = \mu + \phi X_{t-1} + Z_{t}\) under stationarity gives \(E(X_{t}) = \mu/(1 - \phi)\), not \(\mu\), so (i) is false. Taking variances gives \(\gamma(0) = \phi^{2}\gamma(0) + \sigma^{2}\), so \(V(X_{t}) = \sigma^{2}/(1 - \phi^{2})\), statement (ii) as printed without brackets. The autocorrelation of an AR(1) at lag \(s\) is \(\phi^{s}\) (iii).(iv) leaves out \(\sigma^{2}\). So the options including (i) or (iv) are wrong.

11.

Consider a Markov chain on S = {0, 1, 2, ......}. The transition probabilities are given by :

\(p_{00} = \dfrac{1}{2}\) \(p_{01} = \dfrac{1}{2}\)

and for \(i \ge 1\),

\(p_{ii - 1} = \dfrac{1}{2}\) \(p_{ii + 1} = \dfrac{1}{2}\)

Then :

A.all states are persistent non-null
B.there exists a persistent non-null state
C.there exists a unique stationary distribution
D.all states are transient or persistent null

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

All states are transient or persistent null (in fact null persistent).

This is a symmetric random walk on \(\{0, 1, 2, \ldots\}\), with the step down from 0 replaced by staying put. It behaves like the distance from the origin of a simple symmetric walk, which returns with probability 1, so every state is persistent.

A stationary distribution would satisfy \(\pi_{0} = \tfrac{1}{2}\pi_{0} + \tfrac{1}{2}\pi_{1}\) and \(\pi_{j} = \tfrac{1}{2}\pi_{j-1} + \tfrac{1}{2}\pi_{j+1}\), giving \(\pi_{j} = \pi_{0}\) for all \(j\), which cannot sum to 1. So there is no stationary distribution, and the persistent states are null.

Hence no state is non-null persistent, and no stationary distribution exists, ruling out A, B and C.

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12.

Consider a branching process \(\{X_{n}, n \ge 0\}\). Let \(X_{0} = 1\) and assume that \(E\left(\sum_{\theta}^{\infty} X_{n}\right) \lt \infty\). Then, the extinction probability :

A.equals 0
B.lies in (0, 1)
C.does not exist
D.1

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The extinction probability equals 1.

The sum is printed from \(\theta\), meaning from generation 0. With mean offspring \(m\), \(E(X_{n}) = m^{n}\) when \(X_{0} = 1\), so

\[ E\left(\sum_{n} X_{n}\right) = \sum_{n} m^{n} \]

which is finite only if \(m \lt 1\). A subcritical branching process dies out with probability 1. (Equivalently, a finite expected total progeny forces the total progeny to be finite almost surely, so the population cannot survive forever.)

An extinction probability in (0, 1) arises only for \(m \gt 1\), when the expected total is infinite; 0 is impossible once there is any chance of having no offspring, and the extinction probability always exists as a limit of \(P(X_{n} = 0)\).

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13.

Let \(Y_{1}, Y_{2} \ldots\) be a sequence of independent random variables with mean zero and let \(\mathbf{F}_{n} = \sigma(Y_{1}, \ldots, Y_{n})\).

Define \(X_{n} = \sum_{k=1}^{n} Y_{k}\); \(W_{1} = Y_{1}\) and \(W_{n+1} = W_{n} + Y_{n+1}W_{n}^{2}\) and \(T_{n} = \sum_{k=1}^{n} (Y_{k} + 1)\). Then :

A.Both \(\{X_{n}\}\) and \(\{W_{n}\}\) are \(\mathbf{F}_{n}\)-martingales
B.Only \(\{X_{n}\}\) is a \(\mathbf{F}_{n}\)-martingale
C.Both \(\{X_{n}\}\) and \(\{T_{n}\}\) are \(\mathbf{F}_{n}\)-martingales
D.All the three sequences are \(\mathbf{F}_{n}\)-martingales

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The answer is both \(\{X_{n}\}\) and \(\{W_{n}\}\) are \(\mathbf{F}_{n}\)-martingales.

\(X_{n+1} = X_{n} + Y_{n+1}\), and \(Y_{n+1}\) is independent of \(\mathbf{F}_{n}\) with mean 0, so \(E[X_{n+1} \mid \mathbf{F}_{n}] = X_{n}\).

\(W_{n}\) is \(\mathbf{F}_{n}\)-measurable, so

\[ \begin{aligned} E[W_{n+1} \mid \mathbf{F}_{n}] &= W_{n} + W_{n}^{2}E[Y_{n+1}] \\ &= W_{n} \end{aligned} \]

(assuming the variables are integrable). But \(T_{n} = X_{n} + n\), so \(E[T_{n+1} \mid \mathbf{F}_{n}] = T_{n} + 1\): it is a submartingale, not a martingale. That rules out the options including \(T_{n}\), and 'only \(X_{n}\)' misses \(W_{n}\).

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14.

Let \(\{X_{t}\}\) be a stationary time series model given by :

\(X_{t} = \mu + \phi X_{t-1} + Z_{t}\), \(Z_{t} \sim\) iid normal \((0, \sigma^{2})\). Then, which of the following statements is not true if \(P_{X_{t}}(X_{t+k})\) is the forecast of \(X_{t+k}\). Given \(X_{1}, X_{2} \ldots X_{t}\) ?

A.The forecast mean square error of \(P_{X_{t}}(X_{t+k})\) approaches to \(\mu^{2} + \sigma^{2}/(1 - \phi^{2})\)
B.The forecast mean square error of \(P_{X_{t}}(X_{t+k})\) approaches to \(\sigma^{2}/(1 - \phi^{2})\)
C.\(P_{X_{t}}(X_{t+k}) \to E(X_{t})\) as \(k \to \infty\)
D.The forecast \(P_{X_{t}}(X_{t+k})\) is a weighted average of \(E(X_{t})\) and the latest observation \(X_{t}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The statement that is not true is the forecast mean square error approaches \(\mu^{2} + \sigma^{2}/(1 - \phi^{2})\).

For a stationary AR(1) with constant, the \(k\)-step forecast is

\[ \begin{aligned} P(X_{t+k}) &= m + \phi^{k}(X_{t} - m), \\ m &= \frac{\mu}{1 - \phi} \end{aligned} \]

a weighted average of the mean \(E(X_{t}) = m\) and the latest value \(X_{t}\) (D), which tends to \(m\) as \(k \to \infty\) (C). Its error variance is \(\sigma^{2}(1 - \phi^{2k})/(1 - \phi^{2})\), which tends to the process variance \(\sigma^{2}/(1 - \phi^{2})\) (B).

A forecast error has mean 0, so the constant \(\mu\) never adds \(\mu^{2}\) to the mean square error; (A) is the false statement.

Discuss in Forum

15.

Let \(\{X_{n}\}_{0}^{\infty}\) be a Markov chain with \(P[X_{0} = 0] = P[X_{0} = 1] = \dfrac{1}{4}\) and \(P[X_{0} = 2] = \dfrac{1}{2}\). The t.p.m. is given by :

	0	1	2
0	1/2	1/3	1/6
1	0	1/3	2/3
2	1/2	0	1/2

Then \(P[X_{1} = 1]\) is given by :

A.\(\dfrac{1}{3}\)
B.\(\dfrac{1}{4}\)
C.\(\dfrac{1}{5}\)
D.\(\dfrac{1}{6}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

\(P[X_{1} = 1] = \) 1/6.

Multiply the initial distribution by the column of the transition matrix for state 1:

\[ \begin{aligned} P[X_{1} = 1] &= \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{4}\cdot\tfrac{1}{3} + \tfrac{1}{2}\cdot 0 \\ &= \tfrac{1}{12} + \tfrac{1}{12} = \tfrac{1}{6} \end{aligned} \]

1/3 is just the transition probability from 0 (or 1) to 1, ignoring the starting distribution, and 1/4 is the initial probability of state 1.

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16.

Let \(\{X(t), t \ge 0\}\) be a time homogeneous Poisson process with rate \(\lambda\). Let \(X(T) = n \gt 0\) and let \(T_{1} \lt \cdots \lt T_{n} \lt T\) be times at which these \(n\) events occur. Then, \(E(T_{1} \mid X(T) = n)\) is given by :

A.\(\dfrac{T}{n + 1}\)
B.\(T.\lambda^{n}\)
C.\(\dfrac{n}{n + 1} - T\)
D.\(\dfrac{1}{2T}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

\(E(T_{1} \mid X(T) = n) = \) \(\dfrac{T}{n + 1}\).

For a (homogeneous) Poisson process, given that \(n\) events occurred in [0, T], the event times are distributed as the order statistics of \(n\) independent \(U(0, T)\) variables. The smallest of \(n\) uniforms on [0, T] has mean

\[ E(U_{(1)}) = \frac{T}{n + 1} \]

(the \(n\) points split [0, T] into \(n + 1\) gaps of equal expected length). The answer does not depend on \(\lambda\), which rules out \(T\lambda^{n}\), and \(n/(n + 1) - T\) can be negative.

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17.

Consider a Branching process with \(\mu\) as the mean of the off-spring distribution. Let \(\mu = \dfrac{1}{2}\) and \(X_{0} = 1\). Then \(E\left(\sum_{0}^{\infty} X_{n}\right)\) equals :

A.1
B.\(1\frac{1}{2}\)
C.2
D.\(\infty\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

\(E\left(\sum X_{n}\right) = \) 2.

In a branching process with \(X_{0} = 1\), \(E(X_{n}) = \mu^{n}\). With \(\mu = 1/2\), the expected total progeny is

\[ \begin{aligned} \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n} &= \frac{1}{1 - 1/2} \\ &= 2 \end{aligned} \]

(by monotone convergence the expectation and the sum can be exchanged). The process is subcritical and dies out, so the total is finite, not \(\infty\). 1 counts only the ancestor and \(1\frac{1}{2}\) stops after the first generation.

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18.

Let \(\{X_{t}\}\) be a stationary time series with autocovariance function \(r(h)\). Then, which of the following statements is not true about \(r(h)\) ?

A.\(\gamma(0) = \mathrm{Var}(X_{t})\)
B.\(\gamma(k) = \gamma(-k)\)
C.\(|\gamma(k)| \ge \gamma(0)\)
D.\(\sum_{i}\sum_{j} \alpha_{i}\alpha_{j}\, \gamma(|t_{i} - t_{j}|) \ge 0\) for every set of time points \(\{t_{1}, t_{2}, \ldots, t_{n}\}\) and real members \(\{\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}\}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The statement that is not true is \(|\gamma(k)| \ge \gamma(0)\).

By the Cauchy-Schwarz inequality, \(|\gamma(k)| = |\mathrm{Cov}(X_{t}, X_{t+k})|\), which is at most \(\sqrt{\mathrm{Var}X_{t}\,\mathrm{Var}X_{t+k}} = \gamma(0)\). So the true inequality is the reverse, \(|\gamma(k)| \le \gamma(0)\).

The others are standard properties of an autocovariance function: \(\gamma(0)\) is the variance (A), \(\gamma\) is even (B), and \(\gamma\) is non-negative definite (D), because \(\sum\sum \alpha_{i}\alpha_{j}\gamma(|t_{i} - t_{j}|)\) equals \(\mathrm{Var}(\sum \alpha_{i}X_{t_{i}}) \ge 0\).

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19.

Given two time series,

\(X_{t} = 0.3\,X_{t-1} - 0.2\,X_{t-2} + Z_{t}\)
\(Y_{t} - 0.4\,Y_{t-1} = Z_{t} + 1.2\,Z_{t-1}\)

where \(\{Z_{t}\}\) iid WN(0, \(\sigma^{2}\)), which of the following statements is true ?

A.\(\{Y_{t}\}\) is not causal, but invertible
B.\(\{X_{t}\}\) is not causal
C.\(\{Y_{t}\}\) is causal, but not invertible
D.\(\{X_{t}\}\) is not invertible, but causal

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The true statement is that \(\{Y_{t}\}\) is causal, but not invertible.

\(Y_{t}\) is ARMA(1, 1) with AR polynomial \(1 - 0.4z\), whose root \(z = 2.5\) lies outside the unit circle, so it is causal. Its MA polynomial \(1 + 1.2z\) has root \(z = -1/1.2 \approx -0.83\), inside the unit circle, so it is not invertible.

\(X_{t}\) is AR(2) with polynomial \(1 - 0.3z + 0.2z^{2}\). Its roots are complex with \(|z|^{2} = 1/0.2 = 5\), so \(|z| = \sqrt{5} \gt 1\) and \(X_{t}\) is causal; a pure AR process is always invertible. So B and D are false, and A has Y's properties reversed.

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20.

Let \(\{X_{t}\}\) be a stationary AR(1) model with \(X_{t} = \mu + \phi X_{t-1} + Z_{t}\), where \(Z_{t} \sim \mathrm{WN}(0, \sigma^{2})\). Then, which of the following statements is true ?

(i) \(E(X_{t}) = \mu\)
(ii) \(V(X_{t}) = \sigma^{2}/1 - \phi^{2}\)
(iii) \(\mathrm{Corr}(X_{t}, X_{t+s}) = \phi^{s}\)
(iv) \(V(X_{t}) = 1 - \phi^{2}\)

A.(ii) and (iii)
B.(i) and (iii)
C.(i) and (ii)
D.(iii) and (iv)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The true statements are (ii) and (iii).

Taking expectations of \(X_{t} = \mu + \phi X_{t-1} + Z_{t}\) under stationarity gives \(E(X_{t}) = \mu/(1 - \phi)\), not \(\mu\), so (i) is false. Taking variances gives \(\gamma(0) = \phi^{2}\gamma(0) + \sigma^{2}\), so \(V(X_{t}) = \sigma^{2}/(1 - \phi^{2})\), statement (ii) as printed without brackets. The autocorrelation of an AR(1) at lag \(s\) is \(\phi^{s}\) (iii).

(iv) leaves out \(\sigma^{2}\). So the options including (i) or (iv) are wrong.

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