Forecasting - Rob J Hyndman · 2014. 9. 16. · Forecasting: Principles and Practice Seasonal or cyclic? 21. Outline 1Time series graphics 2Seasonal or cyclic? 3Autocorrelation 4Forecast

2. The forecaster’s toolbox

OTexts.com/fpp/2/Forecasting: Principles and Practice 1

Rob J Hyndman

Forecasting:

Principles and Practice

Outline

1 Time series graphics

2 Seasonal or cyclic?

3 Autocorrelation

4 Forecast residuals

5 White noise

6 Evaluating forecast accuracy

Forecasting: Principles and Practice Time series graphics 2

Time series graphics


Economy class passengers: Melbourne−Sydney

Year

Tho

usan

ds

1988 1989 1990 1991 1992 1993

05

1015

2025

30

plot(melsyd[,"Economy.Class"])



Antidiabetic drug sales

Year

$ m

illio

n

1995 2000 2005

510

1520

2530

> plot(a10)



510

1520

2530

Seasonal plot: antidiabetic drug sales

Year

$ m

illio

n

●● ●

● ●

●

●

● ● ● ● ●● ● ● ●

●

●●

●● ● ● ●

●● ● ●

●

●●

●●

●● ● ●

● ● ●●

●●

●

● ●● ● ●

●●

● ●

●

●

● ●●

●●

● ● ●● ●

●●

●●

● ●●

●●

●●

●

●

●

●

● ●●

●●

●

● ●●

●●

●

●

●

● ●●

● ● ●

●

●

●

●

● ●

● ●

● ●

●

●●

●

●

●

●●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

● ●

●●

●●

●

●

●

●

●

●

● ●

● ●

● ●

●●

●

●

●

●

●

●●

● ●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●●

●

●

●●

●●

●

●

●

● ●

●

1991

1992

19931994

1995

1996

19971998199920002001

2002

2003

2004

2005

2006

2007

1992199319941995

199619971998

1999

2000

2001

20022003

2004

2005

2006

2007

2008

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Seasonal plots

Data plotted against the individual “seasons” inwhich the data were observed. (In this case a“season” is a month.)

Something like a time plot except that the datafrom each season are overlapped.

Enables the underlying seasonal pattern to beseen more clearly, and also allows anysubstantial departures from the seasonalpattern to be easily identified.

In R: seasonplot


Seasonal plots




In R: seasonplot


Seasonal plots




In R: seasonplot


Seasonal plots




In R: seasonplot


Seasonal subseries plots


Seasonal subseries plot: antidiabetic drug sales

Month

$ m

illio

n

510

1520

2530

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

> monthplot(a10)


Data for each season collected together in timeplot as separate time series.

Enables the underlying seasonal pattern to beseen clearly, and changes in seasonality overtime to be visualized.

In R: monthplot





In R: monthplot





In R: monthplot


Quarterly Australian Beer Production

beer <- window(ausbeer,start=1992)

plot(beer)

seasonplot(beer,year.labels=TRUE)

monthplot(beer)




Australian quarterly beer production

meg

alite

rs

1995 2000 2005

400

450

500



400

450

500

Seasonal plot: quarterly beer production

Quarter

meg

alitr

es

Q1 Q2 Q3 Q4

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

19921993

1994

1995

1996

199719981999

2000

2001

200220032004

2005

2006

20072008

1992

1993

1994

1995

1996

199719981999

20002001

2002

2003

2004

20052006

2007



Seasonal subseries plot: quarterly beer production

Quarter

Meg

alitr

es

400

450

500

Jan Apr Jul Oct


Time plotsR command: plot or plot.ts

Seasonal plotsR command: seasonplot

Seasonal subseries plotsR command: monthplot

Lag plotsR command: lag.plot

ACF plotsR command: Acf


Outline



3 Autocorrelation


5 White noise


Forecasting: Principles and Practice Seasonal or cyclic? 14

Time series patterns

Trend pattern exists when there is a long-termincrease or decrease in the data.

Seasonal pattern exists when a series isinfluenced by seasonal factors (e.g., thequarter of the year, the month, or day ofthe week).

Cyclic pattern exists when data exhibit rises andfalls that are not of fixed period (durationusually of at least 2 years).




Australian electricity production

Year

GW

h

1980 1985 1990 1995

8000

1000

012

000

1400

0



Australian clay brick production

Year

mill

ion

units

1960 1970 1980 1990

200

300

400

500

600



Sales of new one−family houses, USA

Tota

l sal

es

1975 1980 1985 1990 1995

3040

5060

7080

90



US Treasury bill contracts

Day

pric

e

0 20 40 60 80 100

8586

8788

8990

91



Annual Canadian Lynx trappings

Time

Num

ber

trap

ped

1820 1840 1860 1880 1900 1920

010

0020

0030

0040

0050

0060

0070

00

Seasonal or cyclic?

Differences between seasonal and cyclicpatterns:

seasonal pattern constant length; cyclic patternvariable length

average length of cycle longer than length ofseasonal pattern

magnitude of cycle more variable thanmagnitude of seasonal pattern

The timing of peaks and troughs is predictable withseasonal data, but unpredictable in the long termwith cyclic data.


Seasonal or cyclic?







Seasonal or cyclic?







Seasonal or cyclic?







Seasonal or cyclic?







Outline



3 Autocorrelation


5 White noise


Forecasting: Principles and Practice Autocorrelation 22

Autocorrelation

Covariance and correlation: measure extent oflinear relationship between two variables (y andX).

Autocovariance and autocorrelation: measurelinear relationship between lagged values of atime series y.

We measure the relationship between: yt and yt−1

yt and yt−2

yt and yt−3

etc.


Autocorrelation




yt and yt−2

yt and yt−3

etc.


Autocorrelation




yt and yt−2

yt and yt−3

etc.


Example: Beer production


lag 1

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18 19

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42 43

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

62

63

64

65

66400

450

500

400 450 500

lag 2

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

1819

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

4243

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

62

63

64

65

lag 3

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

1819

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

4243

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

62

63

64

400 450 500

lag 4

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18 19

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42 43

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

62

63

lag 5

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18 19

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42 43

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

62

lag 6

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

1819

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

4243

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

61

400

450

500

lag 7

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

1819

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

4243

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

60

400

450

500

lag 8

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

1819

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42 43

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

59

400 450 500 lag 9

beer

1

2

3

4

5

6

7

8

9

10

11

12

13

1415

16

17

18 19

20

21

2223

24

25

2627

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42 43

44

45

46

47

48

49

50

51

52

53

5455

56

57

58

> lag.plot(beer,lags=9)

Example: Beer production


●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

lag 1

beer

400

450

500

400 450 500

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

lag 2

beer

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

lag 3

beer

400 450 500

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

lag 4

beer

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

lag 5

beer

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

lag 6

beer

400

450

500

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

lag 7

beer

400

450

500

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

lag 8

beer

400 450 500

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

● ●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

lag 9

beer

> lag.plot(beer,lags=9,do.lines=FALSE)

Lagged scatterplots

Each graph shows yt plotted against yt−k fordifferent values of k.

The autocorrelations are the correlationsassociated with these scatterplots.


Lagged scatterplots

Each graph shows yt plotted against yt−k fordifferent values of k.

The autocorrelations are the correlationsassociated with these scatterplots.


AutocorrelationWe denote the sample autocovariance at lag k by ck and thesample autocorrelation at lag k by rk. Then define

ck =1

T

T∑t=k+1

(yt − y)(yt−k − y)

and rk = ck/c0

r1 indicates how successive values of y relate to eachother

r2 indicates how y values two periods apart relate toeach other

rk is almost the same as the sample correlation betweenyt and yt−k.



ck =1

T

T∑t=k+1

(yt − y)(yt−k − y)

and rk = ck/c0






ck =1

T

T∑t=k+1

(yt − y)(yt−k − y)

and rk = ck/c0






ck =1

T

T∑t=k+1

(yt − y)(yt−k − y)

and rk = ck/c0





AutocorrelationResults for first 9 lags for beer data:

r1 r2 r3 r4 r5 r6 r7 r8 r9

−0.126 −0.650 −0.094 0.863 −0.099 −0.642 −0.098 0.834 −0.116


AutocorrelationResults for first 9 lags for beer data:

r1 r2 r3 r4 r5 r6 r7 r8 r9

−0.126 −0.650 −0.094 0.863 −0.099 −0.642 −0.098 0.834 −0.116−

0.5

0.0

0.5

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1711 13 17


Autocorrelation

r4 higher than for the other lags. This is due tothe seasonal pattern in the data: the peakstend to be 4 quarters apart and the troughstend to be 2 quarters apart.

r2 is more negative than for the other lagsbecause troughs tend to be 2 quarters behindpeaks.

Together, the autocorrelations at lags 1, 2, . . . ,make up the autocorrelation or ACF.

The plot is known as a correlogram


Autocorrelation






Autocorrelation






Autocorrelation






ACF

Acf(beer)


−0.

50.

00.

5

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1711 13 17

ACF

Acf(beer)


−0.

50.

00.

5

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1711 13 17

Recognizing seasonality in a time series

If there is seasonality, the ACF at the seasonal lag(e.g., 12 for monthly data) will be large andpositive.

For seasonal monthly data, a large ACF valuewill be seen at lag 12 and possibly also at lags24, 36, . . .

For seasonal quarterly data, a large ACF valuewill be seen at lag 4 and possibly also at lags 8,12, . . .


Recognizing seasonality in a time series

If there is seasonality, the ACF at the seasonal lag(e.g., 12 for monthly data) will be large andpositive.

For seasonal monthly data, a large ACF valuewill be seen at lag 12 and possibly also at lags24, 36, . . .

For seasonal quarterly data, a large ACF valuewill be seen at lag 4 and possibly also at lags 8,12, . . .


Australian monthly electricity production


Australian electricity production

Year

GW

h

1980 1985 1990 1995

8000

1000

012

000

1400

0



−0.

20.

00.

20.

40.

60.

8

Lag

AC

F

0 10 20 30 40


Time plot shows clear trend and seasonality.The same features are reflected in the ACF.

The slowly decaying ACF indicates trend.

The ACF peaks at lags 12, 24, 36, . . . , indicateseasonality of length 12.



Time plot shows clear trend and seasonality.The same features are reflected in the ACF.

The slowly decaying ACF indicates trend.

The ACF peaks at lags 12, 24, 36, . . . , indicateseasonality of length 12.


Which is which?

chirp

s pe

r m

inut

e0 20 40 60

4060

80

1. Daily morning temperature of a cow

thou

sand

s

1973 1975 1977 1979

78

910

2. Accidental deaths in USA (monthly)

thou

sand

s

1950 1952 1954 1956

100

200

300

400

3. International airline passengers

thou

sand

s

1850 1870 1890 1910

2060

100

4. Annual mink trappings (Canada)

A

AC

F

5 10 15 20

-0.4

0.2

0.6

1.0

B

AC

F

5 10 15 20

-0.4

0.2

0.6

1.0

C

AC

F

5 10 15 20

-0.4

0.2

0.6

1.0

D

AC

F

5 10 15 20

-0.4

0.2

0.6

1.0

Outline



3 Autocorrelation


5 White noise


Forecasting: Principles and Practice Forecast residuals 36

Forecasting residuals

Residuals in forecasting: difference betweenobserved value and its forecast based on allprevious observations: et = yt − yt|t−1.

Assumptions1 {et} uncorrelated. If they aren’t, then

information left in residuals that should be usedin computing forecasts.

2 {et} have mean zero. If they don’t, thenforecasts are biased.

Useful properties (for prediction intervals)3 {et} have constant variance.4 {et} are normally distributed.
















Forecasting Dow-Jones index


Day

Dow

−Jo

nes

inde

x

0 50 100 150 200 250 300

3600

3700

3800

3900


Naïve forecast:

yt|t−1 = yt−1

et = yt − yt−1

Note: et are one-step-forecast residuals



Naïve forecast:

yt|t−1 = yt−1

et = yt − yt−1




Naïve forecast:

yt|t−1 = yt−1

et = yt − yt−1





Day

Dow

−Jo

nes

inde

x

0 50 100 150 200 250 300

3600

3700

3800

3900



Day

Cha

nge

in D

ow−

Jone

s in

dex

0 50 100 150 200 250 300

−10

0−

500

50



Histogram of residuals

Change in Dow−Jones index

Fre

quen

cy

−100 −50 0 50

010

2030

4050

60

Normal?



−0.

15−

0.05

0.05

0.10

0.15

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 12 14 16 18 20 22


fc <- rwf(dj)

res <- residuals(fc)

plot(res)

hist(res,breaks="FD")

Acf(res,main="")


Outline



3 Autocorrelation


5 White noise


Forecasting: Principles and Practice White noise 45

Example: White noise


White noise

Time

x

0 10 20 30 40 50

−3

−2

−1

01

2



White noise

Time

x

0 10 20 30 40 50

−3

−2

−1

01

2

White noise data is uncorrelated acrosstime with zero mean and constant variance.(Technically, we require independence aswell.)



White noise

Time

x

0 10 20 30 40 50

−3

−2

−1

01

2

White noise data is uncorrelated acrosstime with zero mean and constant variance.(Technically, we require independence aswell.)

Think of white noise as completelyuninteresting with no predictable patterns.


r1 = 0.013r2 = −0.163r3 = 0.163r4 = −0.259r5 = −0.198r6 = 0.064r7 = −0.139r8 = −0.032r9 = 0.199

r10 = −0.240

Sample autocorrelations for white noise series.For uncorrelated data, we would expect eachautocorrelation to be close to zero.


−0.

4−

0.2

0.0

0.2

0.4

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 1511 13

Sampling distribution of autocorrelations

Sampling distribution of rk for white noise data isasymptotically N(0,1/T).

95% of all rk for white noise must lie within±1.96/

√T.

If this is not the case, the series is probably notWN.

Common to plot lines at ±1.96/√

T whenplotting ACF. These are the critical values.





√T.








√T.








√T.





Autocorrelation


−0.

4−

0.2

0.0

0.2

0.4

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 11 12 13 14 1511 13

Example:T = 50 and socritical values at±1.96/

√50 =

±0.28.All autocorrelationcoefficients lie withinthese limits,confirming that thedata are white noise.(More precisely, the data cannot bedistinguished from white noise.)

Example: Pigs slaughtered


Number of pigs slaughtered in Victoria

thou

sand

s

1990 1991 1992 1993 1994 1995

8090

100

110



−0.

20.

00.

2

Lag

AC

F

0 10 20 30 40


Monthly total number of pigs slaughtered in thestate of Victoria, Australia, from January 1990through August 1995. (Source: Australian Bureau ofStatistics.)

Difficult to detect pattern in time plot.

ACF shows some significant autocorrelation atlags 1, 2, and 3.

r12 relatively large although not significant.This may indicate some slight seasonality.

These show the series is not a white noise series.





































ACF of residuals

We assume that the residuals are white noise(uncorrelated, mean zero, constant variance). Ifthey aren’t, then there is information left in theresiduals that should be used in computingforecasts.So a standard residual diagnostic is to checkthe ACF of the residuals of a forecastingmethod.We expect these to look like white noise.

Dow-Jones naive forecasts revisited

yt|t−1 = yt−1

et = yt − yt−1


ACF of residuals



yt|t−1 = yt−1

et = yt − yt−1


ACF of residuals



yt|t−1 = yt−1

et = yt − yt−1


ACF of residuals



yt|t−1 = yt−1

et = yt − yt−1


ACF of residuals



yt|t−1 = yt−1

et = yt − yt−1




Day

Cha

nge

in D

ow−

Jone

s in

dex

0 50 100 150 200 250 300

−10

0−

500

50



−0.

15−

0.05

0.05

0.10

0.15

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 12 14 16 18 20 2211 13 15 17 19 21 23

Example: Dow-Jones residuals

−0.

15−

0.05

0.05

0.10

0.15

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 12 14 16 18 20 2211 13 15 17 19 21 23

These look like white noise.

But the ACF is a multiple testing problem.


Example: Dow-Jones residuals

−0.

15−

0.05

0.05

0.10

0.15

Lag

AC

F

1 2 3 4 5 6 7 8 9 10 12 14 16 18 20 2211 13 15 17 19 21 23

These look like white noise.

But the ACF is a multiple testing problem.


Portmanteau testsConsider a whole set of rk values, and develop atest to see whether the set is significantly differentfrom a zero set.



Box-Pierce test

Q = Th∑

k=1

r2k

where h is max lag being considered and T isnumber of observations.

My preferences: h = 10 for non-seasonal data, h = 2mfor seasonal data.If each rk close to zero, Q will be small.If some rk values large (positive or negative), Q will belarge.



Box-Pierce test

Q = Th∑

k=1

r2k





Box-Pierce test

Q = Th∑

k=1

r2k





Box-Pierce test

Q = Th∑

k=1

r2k





Ljung-Box test

Q∗ = T(T + 2)h∑

k=1

(T − k)−1r2k


My preferences: h = 10 for non-seasonal data, h = 2mfor seasonal data.Better performance, especially in small samples.


Portmanteau testsIf data are WN, Q∗ has χ2 distribution with(h− K) degrees of freedom where K = no.parameters in model.

When applied to raw data, set K = 0.For the Dow-Jones example,

res <- residuals(naive(dj))

# lag=h and fitdf=K> Box.test(res, lag=10, fitdf=0)Box-Pierce test

X-squared = 14.0451, df = 10, p-value = 0.1709> Box.test(res, lag=10, fitdf=0, type="Lj")Box-Ljung test

X-squared = 14.4615, df = 10, p-value = 0.153
















Exercise

1 Calculate the residuals from a seasonal

naive forecast applied to the quarterly

Australian beer production data from

1992.2 Test if the residuals are white noise.


Exercise

1 Calculate the residuals from a seasonal

naive forecast applied to the quarterly

Australian beer production data from

1992.2 Test if the residuals are white noise.

beer <- window(ausbeer,start=1992)fc <- snaive(beer)res <- residuals(fc)Acf(res)Box.test(res, lag=8, fitdf=0, type="Lj")


Outline



3 Autocorrelation


5 White noise


Forecasting: Principles and Practice Evaluating forecast accuracy 61

Measures of forecast accuracy

Let yt denote the tth observation and yt|t−1 denote its forecastbased on all previous data, where t = 1, . . . , T. Then thefollowing measures are useful.

MAE = T−1T∑

t=1

|yt − yt|t−1|

MSE = T−1T∑

t=1

(yt − yt|t−1)2 RMSE =

√√√√T−1

T∑t=1

(yt − yt|t−1)2

MAPE = 100T−1T∑

t=1

|yt − yt|t−1|/|yt|

MAE, MSE, RMSE are all scale dependent.

MAPE is scale independent but is only sensible if yt � 0for all t, and y has a natural zero.




MAE = T−1T∑

t=1

|yt − yt|t−1|

MSE = T−1T∑

t=1


√√√√T−1

T∑t=1

(yt − yt|t−1)2

MAPE = 100T−1T∑

t=1

|yt − yt|t−1|/|yt|






MAE = T−1T∑

t=1

|yt − yt|t−1|

MSE = T−1T∑

t=1


√√√√T−1

T∑t=1

(yt − yt|t−1)2

MAPE = 100T−1T∑

t=1

|yt − yt|t−1|/|yt|





Mean Absolute Scaled Error

MASE = T−1T∑

t=1

|yt − yt|t−1|/Q

where Q is a stable measure of the scale of the timeseries {yt}.




MASE = T−1T∑

t=1

|yt − yt|t−1|/Q


Proposed by Hyndman and Koehler (IJF, 2006)




MASE = T−1T∑

t=1

|yt − yt|t−1|/Q


For non-seasonal time series,

Q = (T − 1)−1T∑

t=2

|yt − yt−1|

works well. Then MASE is equivalent to MAE relativeto a naive method.




MASE = T−1T∑

t=1

|yt − yt|t−1|/Q


For seasonal time series,

Q = (T −m)−1T∑

t=m+1

|yt − yt−m|

works well. Then MASE is equivalent to MAE relativeto a seasonal naive method.




Forecasts for quarterly beer production

1995 2000 2005

400

450

500

Mean methodNaive methodSeasonal naive method



Forecasts for quarterly beer production

1995 2000 2005

400

450

500

Mean methodNaive methodSeasonal naive method


Mean method

RMSE MAE MAPE MASE38.0145 33.7776 8.1700 2.2990

Naïve method


Seasonal naïve method





Dow Jones Index (daily ending 15 Jul 94)

Day

0 50 100 150 200 250 300

3600

3700

3800

3900

Mean methodNaive methodDrift model



Dow Jones Index (daily ending 15 Jul 94)

Day

0 50 100 150 200 250 300

3600

3700

3800

3900

Mean methodNaive methodDrift model


Mean method


Naïve method


Drift model



Training and test sets

Available data

Training set Test set(e.g., 80%) (e.g., 20%)

The test set must not be used for any aspect ofmodel development or calculation of forecasts.

Forecast accuracy is based only on the test set.



Available data

Training set Test set(e.g., 80%) (e.g., 20%)

The test set must not be used for any aspect ofmodel development or calculation of forecasts.

Forecast accuracy is based only on the test set.



beer3 <- window(ausbeer,start=1992,end=2005.99)beer4 <- window(ausbeer,start=2006)

fit1 <- meanf(beer3,h=20)fit2 <- rwf(beer3,h=20)

accuracy(fit1,beer4)accuracy(fit2,beer4)

In-sample accuracy (one-step forecasts)accuracy(fit1)accuracy(fit2)



beer3 <- window(ausbeer,start=1992,end=2005.99)beer4 <- window(ausbeer,start=2006)

fit1 <- meanf(beer3,h=20)fit2 <- rwf(beer3,h=20)

accuracy(fit1,beer4)accuracy(fit2,beer4)

In-sample accuracy (one-step forecasts)accuracy(fit1)accuracy(fit2)


Beware of over-fitting

A model which fits the data well does notnecessarily forecast well.A perfect fit can always be obtained by using amodel with enough parameters. (Compare R2)Over-fitting a model to data is as bad as failingto identify the systematic pattern in the data.Problems can be overcome by measuring trueout-of-sample forecast accuracy. That is, totaldata divided into “training” set and “test” set.Training set used to estimate parameters.Forecasts are made for test set.Accuracy measures computed for errors in testset only.














Poll: true or false?

1 Good forecast methods should have normallydistributed residuals.

2 A model with small residuals will give goodforecasts.

3 The best measure of forecast accuracy is MAPE.

4 If your model doesn’t forecast well, you shouldmake it more complicated.

5 Always choose the model with the best forecastaccuracy as measured on the test set.


Documents

Forecasting - Rob J Hyndman · 2014. 9. 16. · Forecasting: Principles and Practice Seasonal or cyclic? 21. Outline 1Time series graphics 2Seasonal or cyclic? 3Autocorrelation 4Forecast