程序代写案例-STATS 786

STATS 786
SEMESTER 1, 2021
STATISTICS
Special Topic in Statistical Computing
(Time Series Forecasting for Data Science)
NOTE: For constant
s and , ( + )2 = 2 + 2 + 2.
Page 1 of 15
STATS 786
1 Select FOUR of the following scenarios. State whether the underlined statements
are true or false. You MUST provide reasoning for your answer.
a Consider a time series generated from the following model:
= 0 + 12 + −12 + ,
where is a white noise series, 0 and 1 are constants. A seasonal differenc-
ing and a first differencing are sufficient to make the time series stationary.
b The following sample autocorrelations are computed using a time series of
length 500:
lag () 1 2 3 4 5
k 0.208 -0.44 -0.166 -0.036 -0.021
Assume that the sample autocorrelations are approximately normally dis-
tributed. Only the first two autocorrelations are statistically significant at
the 5% level.
c The MA(2) model given below is stationary and invertible:
= 0.3 + + 1.2−1 + 0.8−2,
where the white noise series ∼ (0, 4).
d Suppose the error sum of squares has decreased after including an additional
term in the model. As a result, the value of the Akaike Information Criterion
(AIC) will also decrease.
e The ETS(A,N,N) model has a flat forecast function, and the width of the
pointwise prediction intervals is fixed.
f The ACF plot shown to the right of Figure 1 does not match with the time
plot given.
0.5
0.7
0.9
1.1
2002 Jan 2004 Jan 2006 Jan 2008 Jan
Month
Sa
le
s
(in
m
illio
ns
)
−0.4
0.0
0.4
0.8
5 10 15
lag
a
cf
Figure 1: Time and ACF plots.
[Total: 20 marks]
Page 2 of 15
STATS 786
2 Figures 2 and 3 show the time, seasonal, and subseries plots for quarterly sales (in
millions of dollars) from food and beverage services in New Zealand over the period
1995 Q3–2020 Q4.
1000
1500
2000
2500
3000
2000 Q1 2010 Q1 2020 Q1
Quarter [1Q]
Sa
le
s
(in
m
illio
ns
of
do
lla
rs)
1995199678
19992000
20012002
2003
2004
20052006
20072008
200910
20112
2013
2014
2015
2016
2017
2018
201920
1000
1500
2000
2500
3000
Q1 Q2 Q3 Q4
Quarter
Sa
le
s
(in
m
illio
ns
of
do
lla
rs)
Figure 2: Time and seasonal plots for sales in food and beverage services.
Page 3 of 15
STATS 786
Q1 Q2 Q3 Q4
20
00
20
10
20
20
20
00
20
10
20
20
20
00
20
10
20
20
20
00
20
10
20
20
1000
1500
2000
2500
3000
Quarter
Sa
le
s
(in
m
illio
ns
of
do
lla
rs)
Figure 3: Subseries plot for sales in food and beverage services.
a Using Figures 2 and 3, describe the sales data for food and beverage services
in New Zealand. Your answer must refer to information obtained from all
three plots.
[9 marks]
b The sales time series is decomposed into its components using two different
settings. The estimates of the decomposition are shown in Figure 4.
Comment on
• what is plotted in all eight panels of Figure 4;
• the behaviour of each component over time;
• the effect of using robust = TRUE.
Which setting would you consider appropriate for this time series?
[16 marks]
[Total: 25 marks]
Page 4 of 15
STATS 786
log(Sales)
trend
se
a
so
n
_ye
a
r
re
m
ainder
2000 Q1 2010 Q1 2020 Q1
7.0
7.4
7.8
6.75
7.00
7.25
7.50
7.75
8.00
−0.05
0.00
0.05
−0.4
−0.3
−0.2
−0.1
0.0
0.1
Quarter
STL(log(Sales) ~ trend(window = 17))
Decomposition setting 1
log(Sales)
trend
se
a
so
n
_ye
a
r
re
m
ainder
2000 Q1 2010 Q1 2020 Q1
7.0
7.4
7.8
6.8
7.2
7.6
8.0
−0.025
0.000
0.025
0.050
−0.4
−0.2
0.0
Quarter
STL(log(Sales) ~ trend(window = 17), robust = TRUE)
Decomposition setting 2
Figure 4: Decomposition settings 1 and 2.
Page 5 of 15
STATS 786
3 Figure 5 shows the number of employees’ in food and beverage stores in the US
over the period January 1990–March 2021.
2.7
2.8
2.9
3.0
3.1
3.2
1990 Jan 2000 Jan 2010 Jan 2020 Jan
Month [1M]
N
um
be
r o
f e
m
pl
oy
e
e
s'
(in
m
illio
ns
)
Figure 5: Time plot for the number of employees’ in food and beverage stores in the US.
a Briefly comment on the main features that you can observe in this time series?
Can you identify any unusual observations?
[4 marks]
b The R code below is used to fit two models to the employees’ data shown in
Figure 5 and to extract summary output from each model.
The estimated components for the two models are shown in Figure 6.
Use this information to answer questions 3(b)i–3(b)vi.
fit <- employees_food %>%
model(
additive = ETS(Persons ~ trend("A")),
damped = ETS(Persons ~ trend("Ad"))
)
Page 6 of 15
STATS 786
fit %>% select(additive) %>% report()
## Series: Persons
## Model: ETS(A,A,A)
## Smoothing parameters:
## alpha = 0.253
## beta = 0.0545
## gamma = 0.000104
##
## Initial states:
## l b s1 s2 s3 s4 s5 s6
## 2.77 0.000889 0.0364 0.0254 0.00395 -0.00354 0.0128 0.0207
## s7 s8 s9 s10 s11 s12
## 0.0205 -0.00885 -0.0308 -0.0352 -0.0273 -0.014
##
## sigma^2: 1e-04
##
## AIC AICc BIC
## -1156 -1154 -1089
fit %>% select(damped) %>% report()
## Series: Persons
## Model: ETS(M,Ad,A)
## Smoothing parameters:
## alpha = 0.754
## beta = 0.217
## gamma = 0.24
## phi = 0.879
##
## Initial states:
## l b s1 s2 s3 s4 s5 s6
## 2.79 -0.00323 0.0548 0.0333 0.0119 -0.0112 0.00827 0.0197
## s7 s8 s9 s10 s11 s12
## 0.0106 -0.0194 -0.0363 -0.0435 -0.0308 0.00272
##
## sigma^2: 0
##
## AIC AICc BIC
## -1286 -1284 -1215
Note: The ̂2 for the damped model appears in the summary output as zero
due to the rounding.
Page 7 of 15
STATS 786
P
e
rso
n
s
le
vel
slope
se
a
so
n
re
m
ainder
1990 Jan 2000 Jan 2010 Jan 2020 Jan
2.7
2.8
2.9
3.0
3.1
3.2
2.8
2.9
3.0
3.1
−0.005
0.000
0.005
−0.02
0.00
0.02
−0.04
−0.02
0.00
0.02
0.04
Month
Estimated components
ETS(A,A,A) decomposition
P
e
rso
n
s
le
vel
slope
se
a
so
n
re
m
ainder
1990 Jan 2000 Jan 2010 Jan 2020 Jan
2.7
2.8
2.9
3.0
3.1
3.2
2.8
2.9
3.0
3.1
−0.010
−0.005
0.000
0.005
0.010
−0.025
0.000
0.025
0.050
−0.01
0.00
0.01
0.02
Month
Estimated components
ETS(M,Ad,A) decomposition
Figure 6: Estimated components from the two models.
Page 8 of 15
STATS 786
i Describe the differences between the two model specifications.
[5 marks]
ii Describe the estimated components shown in Figure 6 for the ETS(A,A,A)
model. Explain how they are related to the estimated parameters.
[4 marks]
iii Considering the names of the R objects created above for this question,
write R code to assess the fit of the additive model.
[6 marks]
iv What modifications would you make to the R code written in 3(b)iii to
assess the fit of the damped model?
[3 marks]
v Figure 7 shows forecasts from the two fitted models. Based on these
forecasts, which model would you choose for the given data. Give reasons
for your selection.
[6 marks]
vi Write down the equations for the model you have chosen in 3(b)v.
[5 marks]
[Total: 33 marks]
Page 9 of 15
STATS 786
2.7
2.9
3.1
3.3
3.5
1990 Jan 2000 Jan 2010 Jan 2020 Jan
Month
N
um
be
r o
f e
m
pl
oy
e
e
s'
(in
m
illio
ns
)
level
80
95
Forecasts from ETS(A,A,A) model
2.8
3.0
3.2
3.4
1990 Jan 2000 Jan 2010 Jan 2020 Jan
Month
N
um
be
r o
f e
m
pl
oy
e
e
s'
(in
m
illio
ns
)
level
80
95
Forecasts from ETS(M,Ad,A) model
Figure 7: Forecasts from the two models fitted to the employees’ data.
Page 10 of 15
STATS 786
4 Consider the employees’ time series data used in Question 3. The following R code
creates three new variables.
employees_food %>%
mutate(diff_persons = difference(Persons),
sdiff_persons = difference(Persons, 12),
diff_sdiff_persons = difference(difference(Persons, 12)))
Figures 8 and 9 show time, ACF, and PACF plots for the original employees’ time
series and the new variables constructed in the R code above.
P
e
rso
n
s
diff_persons
sdiff_persons
diff_sdiff_persons
1990 Jan 2000 Jan 2010 Jan 2020 Jan
2.7
2.8
2.9
3.0
3.1
3.2
−0.04
0.00
0.04
−0.05
0.00
0.05
0.10
−0.02
0.00
0.02
0.04
0.06
Month
Figure 8: Time plots related to the employees’ time series.
Page 11 of 15
STATS 786
0.00
0.25
0.50
0.75
1.00
6 12 18 24
lag [1M]
a
cf
ACF of Persons
0.0
0.5
6 12 18 24
lag [1M]
a
cf
ACF of diff_persons
0.00
0.25
0.50
0.75
1.00
6 12 18 24
lag [1M]
a
cf
ACF of sdiff_persons
−0.2
−0.1
0.0
0.1
6 12 18 24
lag [1M]
a
cf
ACF of diff_sdiff_persons
−0.5
0.0
0.5
1.0
6 12 18 24
lag [1M]
pa
cf
PACF of Persons
−0.2
0.0
0.2
0.4
0.6
0.8
6 12 18 24
lag [1M]
pa
cf
PACF of diff_persons
0.00
0.25
0.50
0.75
1.00
6 12 18 24
lag [1M]
pa
cf
PACF of sdiff_persons
−0.3
−0.2
−0.1
0.0
0.1
6 12 18 24
lag [1M]
pa
cf
PACF of diff_sdiff_persons
Figure 9: ACF and PACF plots related to the employees’ time series.
Page 12 of 15
STATS 786
a Use Figures 8 and 9 to find an appropriate differencing to obtain a stationary
time series for employees’ data. Give reasons for your selection.
[6 marks]
b The R code below is used to fit three models to the employees’ data and extract
summary output from each model.
Consider the model for your choice of differencing in 4a to answer questions
4(b)i–4(b)iii.
fit <- employees_food %>%
model(arima1 = ARIMA(Persons ~ pdq(d = 1) + PDQ(D = 0),
stepwise = FALSE),
arima2 = ARIMA(Persons ~ pdq(d = 0) + PDQ(D = 1),
stepwise = FALSE),
arima3 = ARIMA(Persons ~ pdq(d = 1) + PDQ(D = 1),
stepwise = FALSE))
fit %>% select(arima1) %>% report()
## Series: Persons
## Model: ARIMA(4,1,0)(0,0,2)[12]
##
## Coefficients:
## ar1 ar2 ar3 ar4 sma1 sma2
## 0.0810 -0.181 -0.1915 -0.1115 0.7616 0.5468
## s.e. 0.0547 0.051 0.0511 0.0539 0.0602 0.0572
##
## sigma^2 estimated as 0.0001507: log likelihood=1112
## AIC=-2210 AICc=-2210 BIC=-2182
fit %>% select(arima2) %>% report()
## Series: Persons
## Model: ARIMA(2,0,1)(1,1,2)[12]
##
## Coefficients:
## ar1 ar2 ma1 sar1 sma1 sma2
## 1.9639 -0.965 -0.9190 -0.737 0.121 -0.6135
## s.e. 0.0221 0.022 0.0319 0.168 0.155 0.0936
##
## sigma^2 estimated as 6.55e-05: log likelihood=1231
## AIC=-2449 AICc=-2448 BIC=-2421
Page 13 of 15
STATS 786
fit %>% select(arima3) %>% report()
## Series: Persons
## Model: ARIMA(0,1,0)(0,1,2)[12]
##
## Coefficients:
## sma1 sma2
## -0.5482 -0.1959
## s.e. 0.0595 0.0602
##
## sigma^2 estimated as 6.721e-05: log likelihood=1223
## AIC=-2440 AICc=-2440 BIC=-2428
i Describe the relationship between the relevant ACF and PACF plots given
in Figure 9 to the orders estimated in the ARIMA model.
[5 marks]
ii Write down the estimated model using the backward shift operator.
[3 marks]
iii Use the information given below to compute a 1-step-ahead forecast and
its 95% prediction interval for the model written in 4(b)ii.
[8 marks]
[Total: 22 marks]
Information about arima1 model
## # A tsibble: 15 x 3 [1M]
## Month Persons .resid
##
## 1 2020 Jan 3.06 -0.0124
## 2 2020 Feb 3.05 -0.00915
## 3 2020 Mar 3.03 -0.0222
## 4 2020 Apr 3.01 -0.0302
## 5 2020 May 3.08 0.0590
## 6 2020 Jun 3.14 0.0306
## 7 2020 Jul 3.13 -0.0202
## 8 2020 Aug 3.13 0.0261
## 9 2020 Sep 3.11 0.00132
## 10 2020 Oct 3.13 0.0111
## 11 2020 Nov 3.16 0.0165
## 12 2020 Dec 3.18 0.0164
## 13 2021 Jan 3.14 -0.0125
## 14 2021 Feb 3.14 0.0212
## 15 2021 Mar 3.13 0.00427
Page 14 of 15
STATS 786
Information about arima2 model
## # A tsibble: 15 x 3 [1M]
## Month Persons .resid
##
## 1 2020 Jan 3.06 0.00101
## 2 2020 Feb 3.05 0.00147
## 3 2020 Mar 3.03 -0.0190
## 4 2020 Apr 3.01 -0.0229
## 5 2020 May 3.08 0.0578
## 6 2020 Jun 3.14 0.0334
## 7 2020 Jul 3.13 -0.0170
## 8 2020 Aug 3.13 0.0105
## 9 2020 Sep 3.11 0.00434
## 10 2020 Oct 3.13 0.00494
## 11 2020 Nov 3.16 0.00419
## 12 2020 Dec 3.18 0.0107
## 13 2021 Jan 3.14 0.00339
## 14 2021 Feb 3.14 0.0122
## 15 2021 Mar 3.13 -0.00289
Information about arima3 model
## # A tsibble: 15 x 3 [1M]
## Month Persons .resid
##
## 1 2020 Jan 3.06 0.000272
## 2 2020 Feb 3.05 -0.000909
## 3 2020 Mar 3.03 -0.0207
## 4 2020 Apr 3.01 -0.0259
## 5 2020 May 3.08 0.0551
## 6 2020 Jun 3.14 0.0335
## 7 2020 Jul 3.13 -0.0174
## 8 2020 Aug 3.13 0.0123
## 9 2020 Sep 3.11 0.00191
## 10 2020 Oct 3.13 0.00453
## 11 2020 Nov 3.16 0.00465
## 12 2020 Dec 3.18 0.0119
## 13 2021 Jan 3.14 0.00414
## 14 2021 Feb 3.14 0.0153
## 15 2021 Mar 3.13 -0.0000886
Page 15 of 15

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