Genshiro Kitagawa. Introduction to Time Series Modeling (Введение в моделирование временных рядов)

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ANSWERS TO T H E PROBLEMS 269

Chapter 6

1.(1) Putting m = 1 in equation (6.21), the stationar ity condition is that

the root of 1 −a

B = 0 lies outside the unit circle. Therefore, from

|B| = |a

−1

|> 1, we have |a

|< 1.

(2) Putting m = 2 in equation (6.21), the roots are obtained by B =

+ 4a

)/2. Therefore, the stationarity condition s are that

+ 4a

≥ 0, a

< 1 −a

and a

< 1 + a

for a

+ 4a

< 0, a

−1. Therefore, the region satisfying the stationarity condition is

a triangle surrounded by three lines a

< 1 −a

, a

< 1 + a

and

> −1.

2.(1)

(2) From y

n+2

= ay

n+1

+ v

n+2

= a

+ av

n+1

+ v

n+2

, we have

n+2|n

= a

. Therefore, E (y

n+2

−y

n+2|n

)

= a

n+1

+Ev

n+2

(1 + a

)

(3) It can be expressed as y

n+k

= v

n+k

+ g

n+k−1

+ g

n+k−2

+ ···.

Here, from th e impu lse respon se function of AR(1), g

= a

. T hen,

since we can express y

n+k|n

= g

+ g

k+1

n+1

+ ···, we have

E (y

n+k

−y

n+k|n

)

= E (v

n+k

+ g

n+k−1

+ ···+ g

k−1

n+1

)

= (1 + g

+ ···+ g

k−1

)

= (1 + a

+ ···+ a

2(k−1)

)

1 −a

3.(1) From C

= −0.9C

+ 1, C

= −0.9C

, we have C

= 0.81C

+ 1,

= 1/0.19 = 5.26, C

= 5.26 ×(−0.9)

(2) C

= 25/7, C

= 75/28, C

= 15/14, C

= −9/28.

(3) C

= 1 + b

, C

= −b, C

= 0 (k > 1).

(4) Putting E[y

n−k

] = g

, we have g

= 1, g

= a −b. Therefore,

from C

= aC

+ g

−bg

, C

= aC

−bg

, we have

1 −2ab + b

1 −a

, C

(1 −ab)(a −b)

1 −a

k−1

4.(1) Substituting v

= y

−ay

n−1

into both sides of v

= bv

n−1

+ w

yields

= (a + b)y

n−1

−aby

n−2

+ w

and we obtain an AR model of order 2.

270 ANSWERS TO THE PROBLEMS

(2) The autocovariance function of y

is given by a

(1 −a

)

−1

, k =

0,1, 2,3. There fore, C

= (1 −a

)

−1

+ 0.01, C

= a

(1 −a

)

−1

5.(1) Substituting C

= 1 + b

, C

= −b, C

= 0 and k > 1 into p( f ) =

∞

∑

k=−∞

−2

ik f

, we have p( f ) = (1 + b

) −be

−2

i f

−be

i f

|1 −be

−2

i f

(2) From d p/d f = −4

a sin(2

f )/{1 −2a cos(2

f ) + a

}

= 0, we

have sin(2

f ) = 0. This means that the maximum (or minimum)

of the spectrum is attained e ither at f = 0 or 1/2. If a > 0, it has a

maximum at f = 0 and a minimum at f = 1/2. If a < 0, they a re

the other way around.

6. For y

= v

−bv

n−1

= (1−bB)v

, since (1−bB)

−1

= 1+bB+b

···, we ob ta in an AR mode l of inﬁnite order y

= −by

n−1

−b

n−1

−

···+ v

7.(1) From y

n+1|n

= −bv

and y

n+k|n

= 0, we have E

n+1|n

= 1,

n+k|n

) = 1 + b

(k ≥1).

(2) y

= v

n−1

+ a

n−2

+ ···, E(

n+k|n

) = 1 + a

+ ···+ a

2(k−1)

(3) From y

= v

+ v

n−1

+ v

n−2

+ ···, we have E(

n+k|n

) = k.

8.(1) From y

n+1|n

= −bv

and y

n+k|n

= 0, it follws that E(

n+1|n

) = 1

and E(

n+k|n

) = 1 + b

(k ≥1).

(2) Since y

n+k

= v

n+k

+ av

n+k−1

+ a

n+k−2

+ ··· and y

n+k|n

= a

n−1

+ ···, it follows that

n+k|n

= v

n+k

+ av

n+k−1

+ ··· +

k−1

n+1

. Therefo re, we have E(

n+k|n

) = 1 + a

+ ···+ a

2(k−1)

(3) Since y

n+k

= v

n+k

+ v

n+k−1

+ v

n+k−2

+ ··· and y

n+k|n

= v

n−1

+···, we have

n+k|n

= v

n+k

n+k−1

+···+ v

n+1

. Ther e fore,

n+k|n

) = k.

Chapter 7

1.(1) AIC

= N(log 2

n−1

+ log(1 −(a

)

) + 2(m + 1) = AIC

m−1

N lo g(1 −(a

)

) + 2. Therefore, if N log(1 −(a

)

) + 2 < 0, i.e. ,

if (a

)

> 1 −e

−2/N

, then we can conclude tha t A R(m) is better

than AR(m −1).

(2)

= (1 − 0.9

) × 1 = 0.19,

= 0.1216,

= 0.1107,

0.1095,

= 0.1071 (displayed up to the 4th decima l place)

(3) AIC

= 100(log2

×1+) + 2 ×1 = 285.78

AIC

= 100(log2

×0.19 + 1) + 2 ×2 = 121.71

ANSWERS TO T H E PROBLEMS 271

AIC

= 79.09, AIC

= 71.65, AIC

= 72.65, AIC

= 72.37.

Therefore, the third o rder model is considered best.

2. The Yule-Walker method has the advantage that it always yields a

stationary mode l. On the other hand, it may have large bias, espe-

cially for small sample sizes. These properties are due to the implicit

assumption that the time series takes a value 0 outside the a ctually

observed interval. The least squares method has the advantage that

it provides us with estimates having a small bias and th at it is easy

to pool additional data and perform various manipulations for ﬁtting

a more sophisticated model. On the other hand, it has problems in

that stationarity is not guaranteed, and with parameter estimation, the

ﬁrst several observations are used only for conditioning and cannot be

directly used fo r parameter estimation. In A R model e stimation, th e

PARCOR method is very efﬁcient in terms of computation and yields

very close estimates as maximu m likelihood estimates.

3. Fit a univariate AR model to each time series x

and y

, and obtain

AIC

and AIC

, respectively. On the other hand, ﬁt a 2-variate AR

model to (x

)

and compute AIC

. T hen, if AIC

+ AIC

< AIC

and y

are considere d independent.

Chapter 8

1. Co nsider an AR model y

+ a

n−1

+ ···+ a

n−m

+ v

, which

has an additional parameter corresponding to the mean of the process.

2. Co mpare the following two models; the switched mode l N(0,

) for

n = 1, ... ,k and N(0,

), and the stationary m odel y

∼ N(0,

) for

n = 1,... ,N. The AIC values of th e models are, respectively, g iven

AIC

= N log 2

+ k log

+ (N −k)log

+ N + 2 ×2

AIC

= N log 2

+ N log

+ N + 2.

If AIC

< A IC

, it is considered that the variance changed at n =

k + 1.

3. Assume that y

= a

+ a

+ ···+ a

for n = 1,. ..,k, and

that y

= b

+ b

+ ···+ a

for n = k + 1,...,N, where

∼

N(0,

∼ N(0,

)

AIC = N log 2

+ k log

+ (N −k)log

+ N + 2 ×(p + q + 4).

4. To ensure continuity of the trend, r equire the parameters to satisfy

+ a

k + ···+ a

= b

+ b

k + ···+ b

272 ANSWERS TO THE PROBLEMS

To ensure smoothness o f the trend, apply the further restriction

+ 2a

k + ···+ pa

= b

+ 2b

k + ···+ qb

Note that in computing AIC, the number of f ree parameters is de-

creased by 1 an d 2, respectively.

5. The amount of computation in ﬁtting an ordin ary AR mode l is

roughly Nm

/2. On the other hand, the two locally stationary AR

models need

−n

(p + 1)m

(N −n

−n

(p + 1)m

respectively. Therefore, the total amount of computation required is

evaluated as

−n

Even for p = 1, when the maximum of the sum of the second and

third terms is attained with a value of 3/2(n

−n

, the increase in

the amount o f computation is at most three times that of the original

AR model.

Chapter 9

1. Multiplying by T from the left on both sides of (9.1), w e obtain T x

T F

+ T G

= T F

−1

T x

+ T G

. T hen, putting z

= T x

and

using y

= H

−1

T x

, we c an verify that we obtain

an equivalent r e presentation.

2.(1)

Initial value : x

0|0

= 0,V

0|0

= 100

Prediction : x

n|n−1

= x

n−1|n−1

n|n−1

= V

n−1|n−1

Filter : G

= V

n|n−1

/(V

n|n−1

+ 1),

n|n

= x

n|n−1

+ G

−x

n|n−1

n|n

= (1 −G

n|n−1

= G

(2) V

n+1|n

= V

n|n−1

/(V

n|n−1

+ 1) +

(3) In (2), put V

n+1|n

= V

n|n−1

= V .

ANSWERS TO T H E PROBLEMS 273

(4) Denoting the solution of (3) by V , the variance of the ﬁlter is ob-

tained by V

′

= V /(V + 1), and

n+1|n

= x

n|n

= x

n|n−1

V + 1

−x

n|n−1

)

3. From V

= V , V

= G = V /(V + 1), A = V /(V + 1)/V = 1/(V + 1 )

and V

= V

+ A

−V

), we obtain V

= V /(V + 2).

1 0.1 0.01 0.001

1.1618 0.370 0.105 0.032

0.618 0 .270 0.095 0.0 31

0.447 0 .156 0.050 0.0 16

Chapter 10

1. From x

= (y

n−1

,.. ., y

n−m+1

)

, the (i, j )- element of Ex

n−i+1

n−j+1

= C

i−j

. Therefo re, compute the autocovariance fun c -

tions C

, k = 0, 1,. .. of the A R m odel of order m and c onstruct the

initial matrix by

0|0







··· C

m−1

··· C

m−2

m−1

m−2

··· C







2. The change in the log-likelihood for th e parameter transformed by

(10.9) becomes very small if the absolute value of the parameter is

large, and as a result, the optimization procedure may not work at all.

Therefore, it is suggested that the range of the transformed paramete r

be restricted to some appropriately deﬁned interval, say [−2 0,20].

Chapter 11

1.(1) For y

n+k

= y

n+k−1

+ v

n+k

= ··· = y

+ v

n+1

+ ··· + v

n+k

, since

n+i

= 0 a nd Ev

n+i

, we have y

n+k|n

= y

and E

n+k|n

= k

(2) From k

= 4 ×40,000 = 160, 000, we have

√

= 400. For the

normal distribution, th e probability of a value larger than 1

0.16.

(3) The variance

of the random walk ch anges over time. Therefore,

we may consider a model taking into acco unt changes in volatility

= y

n−1

log

= a + b log

n−1

+ v

(or log

= log

n−1

+ v

274 ANSWERS TO THE PROBLEMS

2. We may, for example, consider a trigonometric model.

3. Omitted.

Chapter 12

1. If a characteristic root of the AR model approa ches 1, discr imination

between the AR model and the trend model be comes difﬁcult. There-

fore, it is recommended to apply so me restriction so that the absolute

value of the characteristic roots of the estimated A R mod e l does not

approa c h 1.

2. A seasonal component model based on a trigo nometric function is

obtained by S

∑

j=1

cos

+ b

sin

3.(1) Assuming that

= ··· =

, we o btain a trading day

effect with 2 parameters. In this case, the trading day effect is ex-

pressed by td

′

+ d

′

) +

′

+ ···+ d

′

(2) Assuming that

= ··· =

, we obtain a trading day effect

model with 3 parameters.

4. The holiday pa ttern of Easter or “Golden week” in Japan c hanges

year by year. Therefore, it is necessary to take accou nt of these effects

carefully. The effect of a leap year may not be com pletely removed

by the standard trading day adjustment.

5. Omitted.

Chapter 13

1. (Example.) Stock-price data

2. In a time-varying coefﬁcient regression model y

···+

, consider, for example, a random walk model

j,n −1

+ v

for the coefﬁcients

,.. .,

. This model can be

expressed as a state-space model, so as in the case of time varying

coefﬁcient AR model, the parameter can be estimated by the Kalm a n

ﬁlter.

Chapter 14

1. Co mpared with the Gaussian distribution, they have characteristics

that the density is concentrated in the center and at the same time,

they also have heavy tails. Therefore, they ar e adequa te for handling

sudden structural changes or outliers that occur with small probabil-

ity.

2. For y ∼N(0,1), p ut z = y

, x = logz. Then the probability that z ta kes

ANSWERS TO T H E PROBLEMS 275

a smaller value than t is given by

F(t) = Prob(y

< t) =

√

−

dy.

Differentiating with respect to t yields

f (t) =

√

−

distribution with 1 degree of freedom.) From (4. 51), the distri-

bution of the log-transform x = h(z) is given by

g(x) =



−1



f (e

) = e

√

)

−

√

−

3. In the trend mod e l, deﬁne the distribution of the system noise by

(x) =



(0) w ith probability 1 −

N(0,

) with probability

where

(0) is the

function with mass at x = 0.

Chapter 15

1. Assuming that y

is a time series and t

is a trend, consider the model

= t

n−1

+ v

log

= log

n−1

+ u

= t

where u

∼ N(0,

), v

∼ N(0,

), w

∼ N(0, 1). Note that in this

case, the ob servation m odel is nonlinear.

2. In Monte Carlo simulation, in general, we need 100 times as many

particles to increase the accuracy b y one digit.

3. Without any re-sampling step, the weig hts of many par ticles approach

to zero and the acc uracy of approximatio n is rapidly reduced.

4. If the number of particles is m, the number of data values is N, the

dimension of the state is k and the lag is L, the ﬁxed-interval smoother

and th e ﬁxed-lag smoother need to store m ×k ×N and m ×k ×L data

elements, respectively. Note that N ≫ L.

Chapter 16

276 ANSWERS TO THE PROBLEMS

1. Assume that f (x) is the density of a uniform distribution, f (x) = 1

for 0 ≤ x ≤ 1, and 0 otherwise. Then since

g(x) =

∞

−∞

f (y) f (x−y)dy =

f (x−y)dy =







x 0 ≤ x ≤ 1

2 −x 1 ≤ x ≤ 2

0 otherwise

the density function of W is a triangle.

2. Assume that y

= 1, v

≡ 0; then y

= a

. There fore, if we take m

such that a

< c (for example, c = 10

−3

), then the effect of the initial

value can be ignored. For a = 0.9,0.5,0.1, m is given by 66, 10, 5,

respectively.

3. Usin g the observed time series, construct the state vector x

and

then by usin g ra ndom numbers, we can generate x

n+1

n+2

,.. .. To

simulate an AR model using the state given in (9.5), put x

n−1

,.. ., y

n−m+1

)

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