Problem Set 1 - SOLUTIONS

240 C Time Series Econometrics

Oscar Jorda

Winter 2001

0.1 Theoretical Questions

1. Find the mean and the autocovariance function of the process

xt = 2+ 1.3xt−1 − 0.4xt−2 + zt + zt−1 {zt} ∼WN(0, σ2)

Solution:

• Start by checking stationarity, i.e. the roots of (1− 1.3x+0.4x2) = 0.These are x = 2, 1.25, clearly outside the unit circle so the process isstationary. Therefore. (1−1.3L+0.4L2) = (1−0.5L)(1−0.8L). Nowthat we know the process is stationary, we can calculate the mean,

E(xt) = 2 + 1.3E(xt)− 0.4E(xt−2)

stationarity means, E(xt) = E(xt−j), therefore, E(xt) = 2/(1 + 1.3−0.4) = 20.

• The autocovariance function can be easily calculated using the Yule-

Walker equations since xt is stationary

γ0 = 1.3γ1 − 0.4γ2 + 2σ2

γ1 = 1.3γ0 − 0.4γ1 + σ2

γ2 = 1.3γ1 − 0.4γ0

...

γr = 1.3γr−1 − 0.4γr−2 for r ≥ 3

1. (a) Is this process covariance-stationary? Yes

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(b) Is this process invertible? No since the MA polynomial (1 + L)is not invertible.

2. Find the coefficients ψj , j = 0, 1, 2, ... in the representation

xt =∞∑

j=0

ψjzt−j

of the ARMA(2,1) process

(1− 0.5L+ 0.04L2)xt = (1 + 0.25L)zt {zt} ∼WN(0, σ2)

Solution:

• First check stationarity.

(1− 0.5L+ 0.04L2) = (1− 0.1L)(1− 0.4L) so the process is covariance

stationary. Hence

xt =(1 + 0.25L)

(1− 0.1L)(1− 0.4L)zt

hence,

xt = (1 + 0.25L)(1 + 0.1L+ 0.12L2 + ...)(1 + 0.4L+ 0.42L2 + ...)zt

tedious algebra would deliver the ψ′

js requested.

1. (a) How would you find said coefficients computationally, say, usingEXCEL? Let cells A1 = A2 = B1 = B2 = 0 and let cell B3 = 1.

Then type the following formula in cell A3

= 0.5 ∗A2− 0.04 ∗A1 +B3 + 0.25 ∗B2

Copy cell A3 down the A column for as many ψ′

js you want to

compute. The first few ψj are, ψ1 = 0.75; ψ2 = 0.335; ψ3 =0.1375; ψ4 = 0.05535; and so on.

3. Let {yt} be a stationary, zero-mean, time series. Define

xt = (1− 0.4L)yt

wt = (1− 2.5L)yt

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(a) Express the autocovariance functions of {xt} and {wt} in termsof the autocorrelation function of {yt}.

γx0 = 1.16γy0 − 0.8γy1γx1 = 1.16γy1 − 0.4γy0 − 0.4γy2

...

γxr = 1.16γyr − 0.4γyr−1 − 0.4γyr+1 r ≥ 2

γw0 = 7.25γy0 − 5γy1γw1 = 7.25γy1 − 2.5γy0 − 2.5γy2

...

γwr = 7.25γyr − 2.5γyr−1 − 2.5γyr+1 r ≥ 2

(b) Show that {xt} and {wt} have the same autocorrelation function.

ρx1 =γx1γx0

=(1.16ρy1 − 0.4− 0.4ρy2) γ

y0

(1.16− 0.8ρy1)γy0

...

ρxr =(1.16ρyr − 0.4ρyr−1 − 0.4ρyr+1)

(1.16− 0.8ρy1)r ≥ 2

ρw1 =(7.25ρy1 − 2.5− 2.5ρy2)

(7.25− 5ρy1)=

6.25 (1.16ρy1 − 0.4− 0.4ρy2)

6.25 (1.16− 0.8ρy1)= ρx1

similarly for r ≥ 2.

(c) Show that the process ut = −∑∞

j=1(0.4)jxt+j satisfies the differ-

ence equation ut − 2.5ut−1 = xt

ut = −∞∑

j=1

(0.4)jxt+j = xt −∞∑

j=0

(0.4L−1)jxt;

ut = xt −1

1− 0.4L−1xt;

(1− 0.4L−1)ut = −0.4L−1xt;

ut = xt − 2.5ut−1 Q.E.D.

4. The series {xt} is generated by the process xt + αxt−1 = εt + βεt−1where −1 < α, β < 1.

(a) Show that stationarity holds under the appropriate initial condi-tions. Stationarity holds since |α| < 1 and as long as |x0| <∞,and ε0 ∼WN(0, σ2) with σ2 <∞.

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(b) Let V (xt) denote the variance of xt, establish that under station-arity

V (xt) = σ2(1− 2αβ + β2)

(1− α2)

and

ρ1 =(β − α)(1− αβ)

(1− 2αβ + β2)

with ρr = αρr−1 for r = 2, 3, ... where ρr = E(xtxt−r)/E(x2t ).

Solution:

xt =1

1 + αLεt +

β

1 + αLεt−1

hence

E(x2t) = E

({[εt − αεt−1 + α2εt−2 − ...

]+ β

[εt−1 − αεt−2 + α2εt−3 − ...

]}2)=

1

1− α2σ2 +

β2

1− α2σ2 −

2α

1− α2σ2

=1+ β2 − 2αβ

1− α2σ2 Q.E.D.

Similarly, for ρ1 we can now use the Yule-Walker equations toobtain

E(xtxt−1) = −αE(x2

t) + βσ2

so that

ρ1 = −α+β

E(x2t)σ2 =

β(1− α2)σ2

(1− 2αβ + β2)σ2− α

=(β − α)(1− αβ)

(1− 2αβ + β2)Q.E.D.

and ρr= −αρ

r−1, for r ≥ 2.

(c) Derive the values of α and β from knowledge of ρr, r = 2, 3, ...

Solution:

α = −ρr/ρr−1. Note β is the MA parameter so it can only be

identified from ρ1 once we obtain the value for α.

5. Let {εt} be a white noise. Verify that the process defined as xt = εtand yt = (−1)tεt are stationary. Show that their sum, zt = xt + yt isnot stationary.

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Solution:

E(xt) = 0, V (xt) = σ2, therefore, x is covariance-stationary. Sim-

ilarly, E(yt) = 0, V (yt) = σ2, also covariance-stationary. However,

E(zt) = 0 but V (zt) = 0 if t odd, 2σ2 if t even, so that the variance

depends on t and thus the process is not stationary.

6. Consider an AR(1) process following

yt = c+ φyt−1 + εt

where the process started at t = 0 with the initial value y0. Suppose|φ| < 1 and that y0 is uncorrelated with the subsequent white noiseprocess (ε1, ε2, ...).

(a) Write the variance of yt as a function of σ2 (≡ V (εt)); V (y0), and

φ. Hint: By successive substitution, show

yt = (1 + φ+ φ2 + ...+ φt−1)c+ εt + φεt−1 + ...+ φt−1ε1 + φty0

Solution:

Using the above formula,

V (yt) = (1 + φ2 + φ4 + ...+ φ2(t−1))σ2 + φ2tV (y0)

=1− φ2t

1− φ2 σ2 + φ2tV (y0)

(b) Let µ and {γj} be the mean and the autocovariances of thecovariance-stationary AR(1) process, so µ = c/(1− φ) and γj =

φjσ2/(1− φ2). Show that:

limt→∞

E(yt) = µ

limt→∞

V (yt) = γ0

limt→∞

COV (yt, yt−j) = γj

Solution:

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limt→∞

E(yt) = limt→∞

E

{1− φt

1− φc+

1− φt

1− φLεt + φty0

}=

1

1− φc

limt→∞

V (yt) = limt→∞

{1− φ2t

1− φ2σ2 + φ2tV (y0)

}=

1

1− φ2σ2

limt→∞

COV (ytyt−j) = limt→∞

{φj (1− φ2t)

1− φ2σ2

}=

φj

1− φ2σ2

0.2 Applied Questions

1. The models for question 3 are

xt = 2 + 0.5xt−1 + εt + 0.7εt−1

yt = 1 + 0.5yt−1 − 0.2yt−2 + εt

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