ITSDE2023_ExerciseBook (1)

1 k Exercises of Week One 1.1 Give the definition of: (a) Strictly stationary process (b) Weakly stationary process (c) iid sequence 1.2 Make use of the definitions given in the previous question to argue that • Every iid process is strictly stationary, but it may not be weakly stationary. • Not every strictly stationary process is weakly stationary and vice versa. 1.3 * The following Venn diagram specifies the relation between iid processes, weakly stationary processes and strictly stationary processes. (1) WS , SS , IID (4) WS , SS , IID (2) WS , SS , IID (5) WS , SS , IID (3) WS , SS , IID Give an example time series for every set (i.e. region 1 to 5) in the above Venn diagram. 1.4 Consider the following statement: There is no strictly stationary process which is iid and weakly stationary at the same time. Is this statement true or false. Justify your answer. 1.5 Let { X t } be weakly stationary. Show that E X 2 t = E X 2 t - 1 . 1.6 * Show that ρ X ( t + h, t ) = ρ X ( h ) if the time series is weakly stationary. 1.7 * Give the definition of white noise and random walk. 1.8 Give three examples of white noise processes. 1.9 * Complete the Venn diagram elaborated before with: (a) white noise process 3

2 k Exercises of Week Two 2.1 ** Suppose that { X t } is generated by the AR(1) below X t = 0 . 9 X t - 1 + ε t Use the sequence of innovations { ε t } below in order to complete the values of the time-series { X t } from time t = 4 to t = 9. Time t = 1 t = 2 t = 3 t = 4 t = 5 t = 6 t = 7 t = 8 t = 9 X t 0 . 31 1 . 25 0 . 79 ε t - 0 . 10 - 0 . 04 0 . 01 - 0 . 03 0 . 12 0 . 02 2.2 Repeat the exercise above assuming that: (a) X t = 0 . 7 X t - 1 + ε t (b) X t = X t - 1 + ε t (c) X t = - 0 . 9 X t - 1 + ε t 2.3 Suppose that { X t } is generated by the AR(1) below X t = 0 . 1 + 0 . 9 X t - 1 + ε t Use the sequence of innovations { ε t } below in order to complete the values of the time-series { X t } from time t = 4 to t = 9. Time t = 1 t = 2 t = 3 t = 4 t = 5 t = 6 t = 7 t = 8 t = 9 X t 0 . 31 1 . 25 0 . 79 ε t - 0 . 10 - 0 . 04 0 . 01 - 0 . 03 0 . 12 0 . 02 2.4 Repeat the exercise above assuming that (a) X t = 0 . 1 + 0 . 5 X t - 1 + ε t (b) X t = 0 . 5 + 0 . 5 X t - 1 + ε t (c) X t = 0 . 1 - 0 . 5 X t - 1 + ε t 2.5 ** Consider a time-series { X t } generated by a stable AR(1) with intercept. Verify that the variance, covariance and autocorrelation functions do not depend on the intercept parameter α . 2.6 Calculate the mean of the time-series { X t } generated by the following AR(1) models: X t = 0 . 1 + 0 . 5 X t - 1 + t X t = 0 . 1 + 0 . 9 X t - 1 + t X t = 0 . 5 + 0 . 9 X t - 1 + t X t = 0 . 5 - 0 . 9 X t - 1 + t 2.7 ** Derive the mean of a time series generated by a stable AR(1) model with intercept. 6

2.21 Let { X t } be generated according to the following AR(1) model: X t = 10 . 5 + 0 . 7 X t - 1 + t , { t } ∼ NID(0 , 2) . State the conditional distribution of X T + h for h = 1 , 2 , 3, given that X T = 16 . 4. In other words give the distribution of X T + h | X T for h = 1 , 2 , 3. 2.22 Let { X t } be generated according to the following MA(1) model: X t = ε t + 0 . 7 ε t - 1 , { ε t } ∼ NID(0 , 2) . State the conditional distribution of X T + h for h = 1 , 2 , 3, given that X T = - 0 . 5 and ε T - 1 = - 0 . 7. In other words give the distribution of X T + h | X T , ε T - 1 for h = 1 , 2 , 3. 2.23 ** Suppose that the dynamics of the IBM stock price are well approximated by the following AR(1) model X t = 1 . 42 + 0 . 99 X t - 1 + t , { t } ∼ NID(0 , 0 . 1 2 ) Suppose that you invested in a single IBM stock at time t - 1 for a price of x t - 1 = $112 . 60. If there are no transaction costs, (a) What is the probability of selling with a profit at time t ? (b) What is the probability of selling with a loss at time t ? 9

(b) φ 1 = 0 . 7, φ 2 = 0, φ 3 = - 0 . 7 and σ 2 = 0 . 1 (c) α = 0 . 5, φ 1 = 0 . 7, φ 2 = 0, φ 3 = - 0 . 3 and σ 2 = 0 . 1 3.18 * Derive the IRF with origin at x = E ( X t ) = 0 generated by an impulse of magnitude at time t = s for the AR(2) model with iid innovations. 3.19 Derive the IRF with origin x generated by an impulse of magnitude at time t = s for the AR(3) model with iid innovations. 3.20 * Suppose that the dynamics of the quarterly growth rate of the aggregate industrial production (AIP) in the Netherlands are well described by the following AR(4) model: X t = 0 . 008 + 0 . 92 X t - 1 - 0 . 14 X t - 4 + t where { t } ∼ NID(0 , 0 . 1). (a) Calculate the average growth rate E ( X t ) = μ X of the AIP in the Netherlands. (b) Calculate the IRF with origin at μ X generated by a negative shock of - 5% occurring at time t = s . 12

Suppose that we regress Y t on X t , Y t = α + βX t + u t The static regression model is misspecified and u t = φY t - 1 + ε t is correlated with its lag u t - 1 = φY t - 2 + ε t - 1 . (a) * Show that cov( u t , u t - 1 ) 6 = 0. (b) * Show that the same problem occurs when { Y t } is generated by an ADL(0 , 1). What about an ADL( p, q )? 4.13 Let { Y t } be generated by an ADL(1,1) with invertible autoregressive polynomial. (a) Calculate the mean of the process and relate your result to the long-run equilibrium of the process. (b) What is the contemporaneous effect on y t of a unit impulse on x t at time t ? (c) What is the long-run effect on y t of a unit impulse on x t at time t ? (d) Would you say that a change in X causes a change in Y ? 15

6 k Exercises of Week Six 6.1 * Show that any linear combination of independent I(0) sequences is I(0) 6.2 * Show that any linear combination of independent I(1) and I(0) sequences is not I(0). 6.3 * Show that any linear combination of independent I(1) sequences is I(1). 6.4 Consider the following statement: I(0) variables are always cointegrated because a linear combination of I(0) variables is I(0) Is this statement true or false. Justify your answer. 6.5 * Consider the following processes Z t = 1 . 83 Z t - 1 - 0 . 92 Z t - 2 + t , { t } ∼ WN (0 , σ 2 ) W t = 0 . 97 Z t - 1 + 0 . 03 W t - 2 + v t , { v t } ∼ WN (0 , σ 2 v ) R t = R t - 1 + u t , { u t } ∼ WN (0 , σ 2 u ) Y t = 0 . 2 Z t + 9 . 1 W t - 1 X t = - Y t + 2 R t H t = 3 Z t - v t + u t Q t = 3 Z t - v t u t Which of the above sequences is I(0)? Which time-series are cointegrated? 6.6 * Think about the properties of cointegrated time series. What would be the practical implications of the following statements: (a) Aggregate consumption and GDP are not cointegrated (b) Global temperatures and CO2 emissions are not cointegrated (c) The GDP of Netherlands and Belgium are not cointegrated (d) Oil prices and electricity prices are not cointegrated 6.7 * Consider the following regression results Y t = 0 . 68 (0 . 23) - 0 . 083 (0 . 035) Y t - 1 + 0 . 35 (0 . 012) X t + u t Y t = 0 . 68 (0 . 23) - 0 . 083 (0 . 001) Y t - 1 + 0 . 35 (0 . 492) X t - 1 + u t Y t = 0 . 68 (0 . 23) - 0 . 083 (0 . 035) Y t - 1 + 0 . 35 (0 . 17) X t - 1 + u t Y t = 0 . 68 (0 . 23) - 0 . 783 (0 . 981) X t - 1 + 0 . 35 (0 . 002) X t - 3 + u t In which cases does { X t } Granger cause { Y t } ? Are you sure? 18

1 9 2 3 4 5 6 7 8 Weakly stationary Strictly stationary IID White noise Random walk 1.9 (1) WS , SS, IID, WN, RW (2) WS , SS, IID, WN , RW (3) WS , SS , IID, WN , RW (4) WS , SS , IID , WN , RW (5) WS , SS , IID, WN, RW (6) WS , SS , IID , WN, RW (7) WS, SS , IID , WN, RW (8) WS, SS , IID, WN, RW (9) WS, SS, IID, WN, RW Legend: WS: weakly stationary SS: strictly stationary IID: identically independently distributed WN: white noise RW: random walk 1.10 (1) { X t } iid with X t ∼ N( μ, σ 2 ) for every t ≤ t * and X t ∼ t( μ, σ 2 ) for every t > t * , with μ 6 = 0. (2) { X t } iid with X t ∼ N( μ, σ 2 ) for every t ≤ t * and X t ∼ t( μ, σ 2 ) for every t > t * , with μ = 0. (3) { X t } is defined as X t = P t Y - (1 - P t ) Y , where { P t } is an iid sequence of Bernoulli random variables such that P t = 1 with probability 0.5 and P t = 0 with probability 0.5, and Y ∼ N (0 , 1). Note that Y does not have time index t . (4) { X t } iid with X t ∼ N(0 , σ 2 ) for every t (5) ( X t , X t - 1 ) ∼ N( μ , Σ ) for every t , where μ = [1 , 1] and Σ = " 1 0 . 5 0 . 5 1 # 21

(6) { X t } iid with X t ∼ N( μ, σ 2 ) for every t with μ 6 = 0 (7) { X t } iid with X t ∼ t( μ, σ 2 , λ ) for every t with λ < 2. (8) ( X t , X t - 1 ) ∼ t( μ , Σ , λ ) for every t , where λ < 2 and μ = [1 , 1] and Σ = " 1 0 . 5 0 . 5 1 # (9) { X t } is generated according to X t +1 = x t + t where { t } ∼ WN(0 , σ 2 ). 1.11 The random walk starting at time t = 1 is given by X t = ε 1 + ε 2 + . . . + ε t { ε t } ∼ WN (0 , σ 2 ε ) . Recall that V ar( X t ) = tσ 2 ε and C ov( X t , X t - h ) = ( t - h ) σ 2 ε . Therefore, ρ X ( t, t - h ) = γ X ( t, t - h ) p V ar( X t ) V ar( X t - h ) = ( t - h ) σ 2 ε p ( tσ 2 ε )(( t - h ) σ 2 ε ) = √ t - h √ t The correlations between elements of { X t } change over time! 1.12 The conditional mean of the random walk is given by E ( X t | X t - 1 ) = E ( X t - 1 + t | X t - 1 ) (using definition of random walk) = E ( X t - 1 | X t - 1 ) + E ( t | X t - 1 ) (linearity of conditional expectation) = X t - 1 + E ( t ) (definition of RW implies t ⊥ X t - 1 ) = X t - 1 . ( t is white noise so E ( t ) = 0) The conditional mean of X t is X t - 1 . This is very different from the unconditional mean of X t which is always zero. The conditional variance of the random walk is given by V ar( X t | X t t - 1 ) = V ar( X t - 1 + t | X t - 1 ) (using definition of random walk) = V ar( X t - 1 | X t - 1 ) + V ar( t | X t - 1 ) (definition of RW implies t ⊥ X t - 1 ) = V ar( X t - 1 | X t - 1 ) + V ar( t ) (definition of RW implies t ⊥ X t - 1 ) = 0 + V ar( t ) ( X t - 1 is a constant given the conditioning) = σ 2 ε . ( { t } is white noise with variance σ 2 ε ) The conditional variance of X t is fixed at σ 2 ε . This is very different from the unconditional variance of X t which grows ever time and is given by tσ 2 ε . 1.16 Since | X t | ≤ max { 1 , X 2 t } ≤ 1 + X 2 t we have E ( X 2 t ) < ∞ ⇒ E ( | X t | ) < ∞ . 22

Recall that strictly stationarity implies identically distributed. Since V ar( X t ) = E ( X 2 t ) - E ( X t ) 2 < ∞ , we can therefore conclude that expectation and variance are finite and constant. Finally, we use the Cauchy-Schwarz inequality with inner product h Z t , Z t - h i = E ( Z t Z t - h ) and Z t = X t - E X t to show that C ov( X t , X t - h ) = E ( Z t Z t - h ) ≤ q E Z 2 t E Z 2 t - h = p V ar( X t ) V ar( X t - h ) < ∞ , where the first and second equalities follow from the definition of the covariance and vari- ance, respectively. Lastly, by strict stationarity this covariance holds for all times t . 1.17 (Brockwell and Davies) Let { Z t } be a sequence of independent normal random variables, each with mean 0 and variance σ 2 , and let a , b and c be constants. Which of the following processes are weakly stationary? For each stationary process derive the mean and the autocovariance function: (a) X t = a + bZ t + cZ t - 2 E ( X t ) = E ( a + bZ t + cZ t - 2 ) = E ( a ) + E ( bZ t ) + E ( cZ t - 2 ) = a + b E ( Z t ) + c E ( Z t - 2 ) = a + b · 0 + c · 0 = a. V ar( X t ) = V ar( a + bZ t + cZ t - 2 ) = V ar( a ) + V ar( bZ t ) + V ar( cZ t - 2 ) = 0 + b 2 V ar( Z t ) + c 2 V ar( Z t - 2 ) = ( b 2 + c 2 ) σ 2 . C ov( X t , X t - h ) = C ov( a + bZ t + cZ t - 2 , a + bZ t - h + cZ t - h - 2 ) = b 2 C ov( Z t , Z t - h ) + bc C ov( Z t , Z t - h - 2 ) + cb C ov( Z t - 2 , Z t - h ) + c 2 C ov( Z t - 2 , Z t - h - 2 ) Now since C ov( Z t , Z t - h ) = 0 ∀ h 6 = 0, we have, γ (0) = ( b 2 + c 2 ) σ 2 γ (1) = 0 γ (2) = bcσ 2 γ ( h ) = 0 ∀ h > 2 . Since expectation, variance and covariance are all finite and constant over time we conclude that { X t } is weakly stationary. 23

(b) X t = a + bZ 0 E ( X t ) = E ( a + bZ 0 ) = a + b E ( Z 0 ) = a. V ar( X t ) = V ar( a + bZ 0 ) = b 2 σ 2 . C ov( X t , X t - h ) = C ov( a + bZ 0 , a + bZ 0 ) = b 2 σ 2 Since expectation, variance and covariance are all finite and constant over time we conclude that { X t } is weakly stationary. (c) X t = Z t Z t - 1 X t = Z t Z t - 1 E ( X t ) = E ( Z t Z t - 1 ) = 0 . V ar( Z t Z t - 1 ) = E ( Z 2 t Z 2 t - 1 ) - E ( Z t Z t - 1 ) 2 = V ar( Z t ) V ar( Z t - 1 ) = σ 4 C ov( X t , X t - h ) = C ov( Z t Z t - 1 , Z t - h Z t - 1 - h ) = 0 if h 6 = 0. Since expectation, variance and covariance are all finite and constant over time we conclude that { X t } is weakly stationary. 1.18 Let { W t } := { X t + Y t } . Then, E ( W t ) = E ( X t + Y t ) = E ( X t ) + E ( Y t ) = μ X + μ Y . V ar( W t ) = V ar( X t + Y t ) = V ar( X t ) + V ar( Y t ) = σ 2 X + σ 2 Y C ov( W t , W t - h ) = C ov( X t + Y t , X t - h + Y t - h ) = C ov( X t , X t - h ) + C ov( X t , Y t - h ) + C ov( Y t , X t - h ) + C ov( Y t , Y t - h ) = γ X ( h ) + 0 + 0 + γ Y ( h ) = γ X ( h ) + γ Y ( h ) . Since the mean, variance, and covariance are all finite and constant over time we conclude that { W t } is weakly stationary. 24

Solutions Week 2 2.1 Time t = 1 t = 2 t = 3 t = 4 t = 5 t = 6 t = 7 t = 8 t = 9 X t 0 . 31 1 . 25 0 . 79 0.611 0.509 0.468 0.392 0.472 0.445 ε t - 0 . 10 - 0 . 04 0 . 01 - 0 . 03 0 . 12 0 . 02 2.5 Since X t = α + φX t - 1 + ε t , we have: V ar( X t ) = V ar( α + φX t - 1 + ε t ) = φ 2 V ar( X t - 1 ) + σ 2 ⇒ V ar( X t ) = σ 2 1 - φ 2 C ov( X t , X t - h ) = C ov( α + φX t - 1 + ε t , X t - h ) = φ C ov( X t - 1 , X t - h ) ⇒ C ov( X t , X t - h ) = φ h σ 2 1 - φ 2 ρ ( h ) = γ ( h ) γ (0) = φ h . 2.7 We have X t = α + φX t - 1 + ε t (1 - φL ) X t = α + ε t X t = (1 - φL ) - 1 ( α + ε t ) = ∞ X j =0 ( φL ) j ( α + ε t ) = ∞ X j =0 φ j α + ∞ X j =0 φ j ε t - j E ( X t ) = E   ∞ X j =0 φ j ( α + ε t )   = ∞ X j =0 φ j α = α 1 - φ . 2.8 The autocovariance function is given by, γ (1) = C ov( X t , X t - 1 ) = C ov( φ 1 X t - 1 + φ 2 X t - 2 + ε t , X t - 1 ) = φ 1 γ (0) + φ 2 γ (1) and hence, γ (1) = φ 1 1 - φ 2 γ (0) γ (2) = C ov( X t , X t - 2 ) = C ov( φ 1 X t - 1 + φ 2 X t - 2 + ε t , X t - 2 ) = φ 1 γ (1) + φ 2 γ (0) and hence, γ (2) = φ 2 1 1 - φ 2 + φ 2 γ (0) . γ ( h ) = C ov( X t , X t - h ) = C ov( φ 1 X t - 1 + φ 2 X t - 2 + ε t , X t - h ) = φ 1 γ ( h - 1) + φ 2 γ ( h - 2) , h ≥ 2 Since we have that, γ (1) = φ 1 1 - φ 2 γ (0) γ (2) = φ 1 γ (1) + φ 2 γ (0) . . . 25

The autocorrelation function is given by, ρ (1) = γ (1) γ (0) = φ 1 1 - φ 2 ρ (2) = φ 1 γ (1) + φ 2 γ (0) γ (0) = φ 2 1 1 - φ 2 + φ 2 . . . 2.11 This statement is true. The solution to the exercise above reveals that the expressions for the variance and autocorrelation do not depend on α . 2.12 The autocovariance function is given by: γ (1) = C ov( X t , X t - 1 ) = C ov( φ 1 X t - 1 + ε t + θ 1 ε t - 1 , X t - 1 ) = C ov( φ 1 X t - 1 , X t - 1 ) + C ov( ε t , X t - 1 ) + C ov( θ 1 ε t - 1 , X t - 1 ) = φ 1 γ (0) + θ 1 σ 2 = φ 1 1 + θ 2 1 + 2 φ 1 θ 1 1 - φ 2 1 + θ 1 σ 2 γ (2) = C ov( X t , X t - 2 ) = C ov( φ 1 X t - 1 + ε t + θ 1 ε t - 1 , X t - 2 ) = C ov( φ 1 X t - 1 , X t - 2 ) + C ov( ε t , X t - 2 ) + C ov( θ 1 ε t - 1 , X t - 2 ) = φ 1 γ (1) . γ ( h ) = C ov( X t , X t - h ) = φ 1 γ ( h - 1) , h ≥ 2 . From the above it follows that γ (1) = φ 1 γ (0) + θ 1 σ 2 = φ 1 1 + θ 2 1 + 2 φ 1 θ 1 1 - φ 2 1 + θ 1 σ 2 γ ( h ) = φ 1 γ ( h - 1) , h ≥ 2 . The autocorrelation function is given by ρ (1) = φ 1 + θ 1 1 - φ 2 1 1 + θ 2 1 + 2 φ 1 θ 1 = ( φ 1 + θ 1 )(1 + φ 1 θ 1 ) 1 + θ 2 1 + 2 φ 1 θ 1 ρ ( h ) = φ 1 ρ ( h - 1) , h ≥ 2 . 2.16 (a) Since the sequence is iid, we know that all the random variables of the sequence { X t } are independent and have the same distribution. We also know that E ( X t ) = 0 ∀ t . Hence, the conditional expectation is given by E ( X t | x t - 1 , x t - 2 , ... ) = E ( X t ) = 0 . Note: The conditioning drops in the first equality because the X t ’s are all independent. (b) Our random walk satisfies X t = X t - 1 + t with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the innovations must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , x t - 2 , ... ) = E ( X t - 1 + t | x t - 1 ) = E ( X t - 1 | x t - 1 ) + E ( t | x t - 1 ) = x t - 1 + E ( t ) = x t - 1 . 26

Note: The first equality holds by the definition of the random walk. The second equality holds by linearity of the expectation. The conditioning in the third equality drops because the t is independent of past X t ’s in the random walk with iid innovations. The expectation of t is zero because the innovations are white noise. (c) An AR(1) process with intercept satisfies X t = α + φX t - 1 + t with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the inno- vations must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , x t - 2 , ... ) = E ( α + φX t - 1 + t | x t - 1 ) = α + φ E ( X t - 1 | x t - 1 ) + E ( t | x t - 1 ) = α + φx t - 1 + E ( t ) = α + φx t - 1 Note: The first equality holds by the definition of the AR(1). The second equality holds by linearity of the expectation. The conditioning in the third equality drops because the t is independent of past X t ’s in the AR(1) process with iid innovations. The expectation of t is zero because the innovations are white noise. An AR(2) process with intercept satisfies X t = α + φ 1 X t - 1 + φ 2 X t - 2 + t with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the innovations must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , x t - 2 , ... ) = E ( α + φ 1 X t - 1 + φ 2 X t - 2 + t | x t - 1 , x t - 2 , ... ) = α + φ 1 E ( X t - 1 | x t - 1 , x t - 2 ... ) + φ 2 E ( X t - 2 | x t - 1 , x t - 2 ... ) + E ( t | x t - 1 , x t - 2 , ... ) = α + φ 1 x t - 1 + φ 2 x t - 2 + E ( t ) = α + φ 1 x t - 1 + φ 2 x t - 2 Note: The first equality holds by the definition of the AR(2). The second equality holds by linearity of the expectation. The conditioning in the third equality drops because the t is independent of past X t ’s in the AR(2) process with iid innovations. The expectation of t is zero because the innovations are white noise. (d) Note that the mean of the innovations is zero and the variance is finite because the innovations must also be white noise. We therefore have E ( X t | x t - 1 , x t - 2 , . . . ) = E ( α + φ 1 X t - 1 + . . . + φ p X t - p + t | x t - 1 , x t - 2 , . . . ) = α + φ 1 E ( X t - 1 | x t - 1 , . . . , x t - p ) + . . . , + φ p E ( X t - p | x t - 1 , . . . , x t - p ) + E ( t ) = α + φ 1 x t - 1 + . . . + φ p x t - p (e) An MA(1) with intercept satisfies X t = α + t + θ 1 t - 1 with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the innovations 27

must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , ... ) = E ( α + t + θ 1 t - 1 | x t - 1 , ... ) = α + E ( t | x t - 1 , ... ) + θ 1 E ( t - 1 | x t - 1 , ... ) = α + E ( t ) + θ 1 t - 1 = α + θ 1 t - 1 Note: The first equality holds by the definition of the MA(1). The second equality holds by linearity of the expectation. In the third equality, the conditioning on the first expectation drops because the t is independent of past X t ’s in the MA(1) with iid innovations. Note also that t - 1 is known when we condition on past X t ’s. Finally, the expectation of t is zero because the innovations are white noise. An MA(2) with intercept satisfies X t = α + t + θ 1 t - 1 + θ 2 t - 2 with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the inno- vations must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , ... ) = E ( α + t + θ 1 t - 1 + θ 2 t - 2 | x t - 1 , ... ) = α + E ( t | x t - 1 , ... ) + θ 1 E ( t - 1 | x t - 1 , ... ) + θ 2 E ( t - 2 | x t - 1 , ... ) = α + E ( t ) + θ 1 t - 1 + θ 2 t - 2 = α + θ 1 t - 1 + θ 2 t - 2 . Note: The first equality holds by the definition of the MA(2). The second equality holds by linearity of the expectation. In the third equality, the conditioning on the first expectation drops because the t is independent of past X t ’s in the MA(2) with iid innovations. Note also that both t - 1 and t - 2 are known when we condition on past X t ’s. Finally, the expectation of t is zero because the innovations are white noise. An MA(q) with intercept satisfies X t = α + t + θ 1 t - 1 + ... + θ q t - q with { t } ∼ iid(0 , σ 2 ) ( note: the mean of the innovations is zero and the variance finite because the inno- vations must also be white noise). Hence, the conditional expectation is given by E ( X t | x t - 1 , ... ) = E ( α + t + θ 1 t - 1 + ... + θ q t - q | x t - 1 , ... ) = α + E ( t | x t - 1 , ... ) + θ 1 E ( t - 1 | x t - 1 , ... ) + ... + θ q E ( t - q | x t - 1 , ... ) = α + E ( t ) + θ 1 t - 1 + ... + θ q t - q = α + θ 1 t - 1 + ... + θ q t - q . Note: The first equality holds by the definition of the MA(q). The second equality holds by linearity of the expectation. In the third equality, the conditioning on the first expectation drops because the t is independent of past X t ’s in the MA(q) with iid innovations. Note also that the innovations t - 1 , ..., t - q are known when we condition on past X t ’s. Finally, the expectation of t is zero because the innovations are white noise. 28

(f) For the ARMA(1,1) model, we have E ( X t | x t - 1 , . . . ) = E ( α + φX t - 1 + t + θ 1 t - 1 | x t - 1 , . . . ) = α + φ E ( X t - 1 | x t - 1 , . . . ) + E ( t | x t - 1 , . . . ) + θ 1 E ( t - 1 | x t - 1 , . . . ) = α + φx t - 1 + θ 1 t - 1 . (g) An ARMA(p,q) with intercept satisfies X t = α + φ 1 X t - 1 + ... + φ p X t - p + t + θ 1 t - 1 + ... + θ q t - q with { t } ∼ WN (0 , σ 2 ). Hence, the conditional expectation is given by E ( X t | x t - 1 , ... ) = E ( α + φ 1 X t - 1 + ... + φ p X t - p + t + θ 1 t - 1 + ... + θ q t - q | x t - 1 , ... ) = α + φ 1 E ( X t - 1 | x t - 1 , ... ) + ... + φ p E ( X t - p | x t - 1 , ... ) + E ( t | x t - 1 , ... ) + θ 1 E ( t - 1 | x t - 1 , ... ) + ... + θ q E ( t - q | x t - 1 , ... ) = α + φ 1 x t - 1 + ... + φ p x t - 1 + E ( t ) + θ 1 t - 1 + ... + θ q t - q = α + φ 1 x t - 1 + ... + φ p x t - p + θ 1 t - 1 + ... + θ q t - q Note: The first equality holds by the definition of the ARMA(p,q). The second equality holds by linearity of the expectation. In the third equality, the conditioning on the expectation of t drops because t is independent of past X t ’s in the AR(p,q) with iid innovations.. Note also that the lagged data X t - 1 , ..., X t - p and the lagged innovations t - 1 , ..., t - q are known when we condition on past X t ’s. Finally, the expectation of t is zero because the innovations are white noise. 2.17 Since the process is given by X t = Z t + θZ t - 2 { Z t } ∼ WN (0 , 1), we have γ X (0) = V ar( X t ) = V ar( Z t + θZ t - 2 ) = V ar( Z t ) + θ 2 V ar( Z t - 2 ) = 1 + θ 2 Note: The second equality holds by the definition of the process. The third equality holds because the Z t ’s are white noise, hence they are all uncorrelated, which implies that the variance of the sum is equal to the sum of the variances. Finally, the last equality holds since the Z t ’s are white noise with unit variance and hence they all have the same variance of 1. γ X (1) = C ov( X t , X t - 1 ) = C ov( Z t + θZ t - 2 , Z t - 1 + θZ t - 3 ) = C ov( Z t , Z t - 1 ) + θ C ov( Z t , Z t - 3 ) + θ C ov( Z t - 2 , Z t - 1 ) + θ 2 C ov( Z t - 2 , Z t - 3 ) = 0 + 0 + 0 + 0 = 0 Note: The second equality holds by the definition of the process. The third equality holds by linearity of the covariance. Finally, the last equality holds since the Z t ’s are white noise, and hence, they are all uncorrelated. γ X (2) = C ov( X t , X t - 2 ) = C ov( Z t + θZ t - 2 , Z t - 2 + θZ t - 4 ) = C ov( Z t , Z t - 2 ) + θ C ov( Z t , Z t - 4 ) + θ C ov( Z t - 2 , Z t - 2 ) + θ 2 C ov( Z t - 2 , Z t - 4 ) = 0 + 0 + θ + 0 = θ = 0 . 8 29