forecast::ets, auto.arima offset by one
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I'm not sure if this is intended behaviour. Consider the following snippet of code -
library(forecast)
x <- c(
0, 0, 0, 0, 0.00217764964493354, 0.00339032724317772, 0.00357374918778428,
0.00282328811130057, 0.00272679331678393, 0.0030360769697858,
0.00316665914235777, 0.00163300219677676, 0.00249817841157489,
0.00207838479809976, 0.00192104504850639, 0.00209700948212983,
0.00216356555603635, 0.00250983016815862, 0.0017474879860201
)
tsData <- ts(data = x, start = 2000, frequency = 1)
df <- data.frame(
x = x,
fittedets = fitted(forecast(ets(tsData), h = 7)),
fittedarima = fitted(forecast(auto.arima(tsData), h = 7))
)
df
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000
3 0.000000000 -7.133162e-15 0.000000000
4 0.000000000 -7.201966e-19 0.000000000
5 0.002177650 0.000000e+00 0.000000000
6 0.003390327 2.177430e-03 0.002007587
7 0.003573749 3.390205e-03 0.003125561
8 0.002823288 3.573731e-03 0.003294659
9 0.002726793 2.823364e-03 0.002602805
10 0.003036077 2.726803e-03 0.002513846
11 0.003166659 3.036046e-03 0.002798976
12 0.001633002 3.166646e-03 0.002919360
13 0.002498178 1.633157e-03 0.001505474
14 0.002078385 2.498091e-03 0.002303084
15 0.001921045 2.078427e-03 0.001916074
16 0.002097009 1.921061e-03 0.001771022
17 0.002163566 2.096992e-03 0.001933245
18 0.002509830 2.163559e-03 0.001994603
19 0.001747488 2.509795e-03 0.002313826
The actual values are 0 until the fifth value, while in case of both models, the fitted values are about 0 until the sixth value.
I would assume them to be approximately 0 for the first five values, like the x column. Am I missing something basic?
r forecasting forecast
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
add a comment |
I'm not sure if this is intended behaviour. Consider the following snippet of code -
library(forecast)
x <- c(
0, 0, 0, 0, 0.00217764964493354, 0.00339032724317772, 0.00357374918778428,
0.00282328811130057, 0.00272679331678393, 0.0030360769697858,
0.00316665914235777, 0.00163300219677676, 0.00249817841157489,
0.00207838479809976, 0.00192104504850639, 0.00209700948212983,
0.00216356555603635, 0.00250983016815862, 0.0017474879860201
)
tsData <- ts(data = x, start = 2000, frequency = 1)
df <- data.frame(
x = x,
fittedets = fitted(forecast(ets(tsData), h = 7)),
fittedarima = fitted(forecast(auto.arima(tsData), h = 7))
)
df
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000
3 0.000000000 -7.133162e-15 0.000000000
4 0.000000000 -7.201966e-19 0.000000000
5 0.002177650 0.000000e+00 0.000000000
6 0.003390327 2.177430e-03 0.002007587
7 0.003573749 3.390205e-03 0.003125561
8 0.002823288 3.573731e-03 0.003294659
9 0.002726793 2.823364e-03 0.002602805
10 0.003036077 2.726803e-03 0.002513846
11 0.003166659 3.036046e-03 0.002798976
12 0.001633002 3.166646e-03 0.002919360
13 0.002498178 1.633157e-03 0.001505474
14 0.002078385 2.498091e-03 0.002303084
15 0.001921045 2.078427e-03 0.001916074
16 0.002097009 1.921061e-03 0.001771022
17 0.002163566 2.096992e-03 0.001933245
18 0.002509830 2.163559e-03 0.001994603
19 0.001747488 2.509795e-03 0.002313826
The actual values are 0 until the fifth value, while in case of both models, the fitted values are about 0 until the sixth value.
I would assume them to be approximately 0 for the first five values, like the x column. Am I missing something basic?
r forecasting forecast
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
1
The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37
add a comment |
I'm not sure if this is intended behaviour. Consider the following snippet of code -
library(forecast)
x <- c(
0, 0, 0, 0, 0.00217764964493354, 0.00339032724317772, 0.00357374918778428,
0.00282328811130057, 0.00272679331678393, 0.0030360769697858,
0.00316665914235777, 0.00163300219677676, 0.00249817841157489,
0.00207838479809976, 0.00192104504850639, 0.00209700948212983,
0.00216356555603635, 0.00250983016815862, 0.0017474879860201
)
tsData <- ts(data = x, start = 2000, frequency = 1)
df <- data.frame(
x = x,
fittedets = fitted(forecast(ets(tsData), h = 7)),
fittedarima = fitted(forecast(auto.arima(tsData), h = 7))
)
df
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000
3 0.000000000 -7.133162e-15 0.000000000
4 0.000000000 -7.201966e-19 0.000000000
5 0.002177650 0.000000e+00 0.000000000
6 0.003390327 2.177430e-03 0.002007587
7 0.003573749 3.390205e-03 0.003125561
8 0.002823288 3.573731e-03 0.003294659
9 0.002726793 2.823364e-03 0.002602805
10 0.003036077 2.726803e-03 0.002513846
11 0.003166659 3.036046e-03 0.002798976
12 0.001633002 3.166646e-03 0.002919360
13 0.002498178 1.633157e-03 0.001505474
14 0.002078385 2.498091e-03 0.002303084
15 0.001921045 2.078427e-03 0.001916074
16 0.002097009 1.921061e-03 0.001771022
17 0.002163566 2.096992e-03 0.001933245
18 0.002509830 2.163559e-03 0.001994603
19 0.001747488 2.509795e-03 0.002313826
The actual values are 0 until the fifth value, while in case of both models, the fitted values are about 0 until the sixth value.
I would assume them to be approximately 0 for the first five values, like the x column. Am I missing something basic?
r forecasting forecast
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
I'm not sure if this is intended behaviour. Consider the following snippet of code -
library(forecast)
x <- c(
0, 0, 0, 0, 0.00217764964493354, 0.00339032724317772, 0.00357374918778428,
0.00282328811130057, 0.00272679331678393, 0.0030360769697858,
0.00316665914235777, 0.00163300219677676, 0.00249817841157489,
0.00207838479809976, 0.00192104504850639, 0.00209700948212983,
0.00216356555603635, 0.00250983016815862, 0.0017474879860201
)
tsData <- ts(data = x, start = 2000, frequency = 1)
df <- data.frame(
x = x,
fittedets = fitted(forecast(ets(tsData), h = 7)),
fittedarima = fitted(forecast(auto.arima(tsData), h = 7))
)
df
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000
3 0.000000000 -7.133162e-15 0.000000000
4 0.000000000 -7.201966e-19 0.000000000
5 0.002177650 0.000000e+00 0.000000000
6 0.003390327 2.177430e-03 0.002007587
7 0.003573749 3.390205e-03 0.003125561
8 0.002823288 3.573731e-03 0.003294659
9 0.002726793 2.823364e-03 0.002602805
10 0.003036077 2.726803e-03 0.002513846
11 0.003166659 3.036046e-03 0.002798976
12 0.001633002 3.166646e-03 0.002919360
13 0.002498178 1.633157e-03 0.001505474
14 0.002078385 2.498091e-03 0.002303084
15 0.001921045 2.078427e-03 0.001916074
16 0.002097009 1.921061e-03 0.001771022
17 0.002163566 2.096992e-03 0.001933245
18 0.002509830 2.163559e-03 0.001994603
19 0.001747488 2.509795e-03 0.002313826
The actual values are 0 until the fifth value, while in case of both models, the fitted values are about 0 until the sixth value.
I would assume them to be approximately 0 for the first five values, like the x column. Am I missing something basic?
r forecasting forecast
r forecasting forecast
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
asked Nov 22 '18 at 1:48
AmeyaAmeya
1,0711920
1,0711920
1
The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37
add a comment |
1
The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37
1
1
The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37
add a comment |
1 Answer
1
active
oldest
votes
It also has to do with the ARIMA model that auto.arima is fitting to your data. If you look at the model that it's being fitted:
Series: tsData
ARIMA(1,0,0) with zero mean
Coefficients:
ar1
0.9219
s.e. 0.0638
sigma^2 estimated as 6.076e-07: log likelihood=108.59
AIC=-213.17 AICc=-212.42 BIC=-211.28
Remember that ARIMA stands for Autoregressive Integrated Moving Average, and the output tells us that only the AR part of the model was fitted, which makes it an AR(1) model:
y[t] = c + p1 * y[t-1]
With this equation you can get a sense of what happened here:
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000 # .9219 * 0 = 0
3 0.000000000 -7.133162e-15 0.000000000 # .9219 * 0 = 0
4 0.000000000 -7.201966e-19 0.000000000 # .9219 * 0 = 0
5 0.002177650 0.000000e+00 0.000000000 # .9219 * 0 = 0
6 0.003390327 2.177430e-03 0.002007587 # .9219 * .00217 = .002007
7 0.003573749 3.390205e-03 0.003125561 # .9219 * .00339 = .003125
You can also observe this behavior with a plot:
library(ggplot2)
fcast <- forecast(auto.arima(tsData), h = 7)
autoplot(fcast) +
autolayer(fitted(fcast))
For the ets model a similar thing happens, but I hope this made it clear why auto.arima had such result. Next time you could explore more forecasting models that are included in the forecast package.
Hope this helped!
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
add a comment |
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It also has to do with the ARIMA model that auto.arima is fitting to your data. If you look at the model that it's being fitted:
Series: tsData
ARIMA(1,0,0) with zero mean
Coefficients:
ar1
0.9219
s.e. 0.0638
sigma^2 estimated as 6.076e-07: log likelihood=108.59
AIC=-213.17 AICc=-212.42 BIC=-211.28
Remember that ARIMA stands for Autoregressive Integrated Moving Average, and the output tells us that only the AR part of the model was fitted, which makes it an AR(1) model:
y[t] = c + p1 * y[t-1]
With this equation you can get a sense of what happened here:
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000 # .9219 * 0 = 0
3 0.000000000 -7.133162e-15 0.000000000 # .9219 * 0 = 0
4 0.000000000 -7.201966e-19 0.000000000 # .9219 * 0 = 0
5 0.002177650 0.000000e+00 0.000000000 # .9219 * 0 = 0
6 0.003390327 2.177430e-03 0.002007587 # .9219 * .00217 = .002007
7 0.003573749 3.390205e-03 0.003125561 # .9219 * .00339 = .003125
You can also observe this behavior with a plot:
library(ggplot2)
fcast <- forecast(auto.arima(tsData), h = 7)
autoplot(fcast) +
autolayer(fitted(fcast))
For the ets model a similar thing happens, but I hope this made it clear why auto.arima had such result. Next time you could explore more forecasting models that are included in the forecast package.
Hope this helped!
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
add a comment |
It also has to do with the ARIMA model that auto.arima is fitting to your data. If you look at the model that it's being fitted:
Series: tsData
ARIMA(1,0,0) with zero mean
Coefficients:
ar1
0.9219
s.e. 0.0638
sigma^2 estimated as 6.076e-07: log likelihood=108.59
AIC=-213.17 AICc=-212.42 BIC=-211.28
Remember that ARIMA stands for Autoregressive Integrated Moving Average, and the output tells us that only the AR part of the model was fitted, which makes it an AR(1) model:
y[t] = c + p1 * y[t-1]
With this equation you can get a sense of what happened here:
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000 # .9219 * 0 = 0
3 0.000000000 -7.133162e-15 0.000000000 # .9219 * 0 = 0
4 0.000000000 -7.201966e-19 0.000000000 # .9219 * 0 = 0
5 0.002177650 0.000000e+00 0.000000000 # .9219 * 0 = 0
6 0.003390327 2.177430e-03 0.002007587 # .9219 * .00217 = .002007
7 0.003573749 3.390205e-03 0.003125561 # .9219 * .00339 = .003125
You can also observe this behavior with a plot:
library(ggplot2)
fcast <- forecast(auto.arima(tsData), h = 7)
autoplot(fcast) +
autolayer(fitted(fcast))
For the ets model a similar thing happens, but I hope this made it clear why auto.arima had such result. Next time you could explore more forecasting models that are included in the forecast package.
Hope this helped!
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
add a comment |
It also has to do with the ARIMA model that auto.arima is fitting to your data. If you look at the model that it's being fitted:
Series: tsData
ARIMA(1,0,0) with zero mean
Coefficients:
ar1
0.9219
s.e. 0.0638
sigma^2 estimated as 6.076e-07: log likelihood=108.59
AIC=-213.17 AICc=-212.42 BIC=-211.28
Remember that ARIMA stands for Autoregressive Integrated Moving Average, and the output tells us that only the AR part of the model was fitted, which makes it an AR(1) model:
y[t] = c + p1 * y[t-1]
With this equation you can get a sense of what happened here:
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000 # .9219 * 0 = 0
3 0.000000000 -7.133162e-15 0.000000000 # .9219 * 0 = 0
4 0.000000000 -7.201966e-19 0.000000000 # .9219 * 0 = 0
5 0.002177650 0.000000e+00 0.000000000 # .9219 * 0 = 0
6 0.003390327 2.177430e-03 0.002007587 # .9219 * .00217 = .002007
7 0.003573749 3.390205e-03 0.003125561 # .9219 * .00339 = .003125
You can also observe this behavior with a plot:
library(ggplot2)
fcast <- forecast(auto.arima(tsData), h = 7)
autoplot(fcast) +
autolayer(fitted(fcast))
For the ets model a similar thing happens, but I hope this made it clear why auto.arima had such result. Next time you could explore more forecasting models that are included in the forecast package.
Hope this helped!
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
It also has to do with the ARIMA model that auto.arima is fitting to your data. If you look at the model that it's being fitted:
Series: tsData
ARIMA(1,0,0) with zero mean
Coefficients:
ar1
0.9219
s.e. 0.0638
sigma^2 estimated as 6.076e-07: log likelihood=108.59
AIC=-213.17 AICc=-212.42 BIC=-211.28
Remember that ARIMA stands for Autoregressive Integrated Moving Average, and the output tells us that only the AR part of the model was fitted, which makes it an AR(1) model:
y[t] = c + p1 * y[t-1]
With this equation you can get a sense of what happened here:
x fittedets fittedarima
1 0.000000000 -6.997521e-07 0.000000000
2 0.000000000 -7.065016e-11 0.000000000 # .9219 * 0 = 0
3 0.000000000 -7.133162e-15 0.000000000 # .9219 * 0 = 0
4 0.000000000 -7.201966e-19 0.000000000 # .9219 * 0 = 0
5 0.002177650 0.000000e+00 0.000000000 # .9219 * 0 = 0
6 0.003390327 2.177430e-03 0.002007587 # .9219 * .00217 = .002007
7 0.003573749 3.390205e-03 0.003125561 # .9219 * .00339 = .003125
You can also observe this behavior with a plot:
library(ggplot2)
fcast <- forecast(auto.arima(tsData), h = 7)
autoplot(fcast) +
autolayer(fitted(fcast))
For the ets model a similar thing happens, but I hope this made it clear why auto.arima had such result. Next time you could explore more forecasting models that are included in the forecast package.
Hope this helped!
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
answered Nov 22 '18 at 5:23
Kean HerreraKean Herrera
112
112
add a comment |
add a comment |
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The forecast is made on the known information up to that point, meaning that the prediction for the fifth value are made using the first four observations (well, depending on the chosen model ofc). Since the historical data is 0 until then, there will be no trend. If there is also no intercept, the prediction is 0 as well.
– CIAndrews
Nov 22 '18 at 2:17
Makes sense, thanks!
– Ameya
Nov 22 '18 at 2:25
have a look at this post this will help - stackoverflow.com/questions/35448610/…
– Hunaidkhan
Nov 22 '18 at 4:37