Timeline for Validating ARIMA(1,0,0)(0,1,0)[12] with manual calculation
Current License: CC BY-SA 3.0
9 events
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| May 3, 2017 at 17:21 | comment | added | Richard Hardy | @ErvanSanjaya, see these questions, you should find quite a few relevant solutions. What I did above is separate differencing from the rest by changing variable names. So if the original variable $y_t$ is ARIMA(p,1,q), I consider $x_t$ instead that is ARIMA(p,0,q) with $x_t:=(y_t-y_{t-1})$, and similarly for seasonal integration (exactly what I did above). | |
| May 3, 2017 at 17:13 | comment | added | Ervan Sanjaya |
continued, I just tried to manually calculate an ARIMA(1,1,0)(0,1,0) following the previous count, the only difference is with the d=1 in this model, so I'm thinking of this equation : $$y_t = \varphi_1 (y_{t-1} - y_{t-2} - y_{t-13}) + y_t - y_{t-1}) + y_t - y_{t-12}) $$ But, as I calculated it, the function does not give the correct forecasted value from that model. Can you help me to understand the method? Because I'm kinda confused with these thing.
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| May 3, 2017 at 17:10 | comment | added | Ervan Sanjaya |
Richard, I have another question, actually I don't really get it how you can say "where xt=(yt−yt−12)xt = (yt−yt−12). Insert the latter ..." At this point, I'm kinda confused to fit each model to calculate manually. I have this following model to be manually calculated : ARIMA(1,0,0)(0,1,0) ARIMA(3,1,0)(0,1,0) ARIMA(1,1,0)(0,1,0) ARIMA(0,0,1)(0,1,0) ARIMA(2,0,0)(0,1,0) ARIMA(3,0,0)(0,1,0) ARIMA(2,1,0)(0,1,0) ARIMA(4,1,0)(0,1,0) ARIMA(2,1,0)(0,1,0) Can you give me solution on how to build the quotation to each model with a simple 'how-to'? Thanks.
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| May 3, 2017 at 16:31 | comment | added | Richard Hardy | @ErvanSanjaya, I am glad to have been able to help. | |
| May 3, 2017 at 16:31 | vote | accept | Ervan Sanjaya | ||
| May 3, 2017 at 16:30 | comment | added | Ervan Sanjaya |
Oh yea! Finally, it is actually a rounding problem on the coefficient, I just recheck by calling the function coef(fit) and found different result of my ar1, which is 0.7147584, not 0.7148 as displayed on the fit calls. Thank you so much Richard.
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| May 3, 2017 at 16:25 | comment | added | Richard Hardy |
Can it be a rounding problem or something? Maybe try using coef(fit)[1] instead of 0.7148 in your manual calculation.
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| May 3, 2017 at 16:23 | comment | added | Ervan Sanjaya |
Thank you for your brief explanation, I have tried your suggestion there, and the answer become more close to the auto.arima forecasting result. Here is the result I got : For my March 2016 forecast, I calculate with your suggested equation as follow : $$ \hat{y}_t(MARCH2016) = \{y}_t(MARCH2015) + 0.7148 * \{y}_t(FEB2016) - \{y}_t(FEB2015) $$ $$ = 415378 + 0.7148 * ( 565057 - 378703 ) $$ $$ = 548583.8392 $$ Expected : 548576.1 There was a little difference with the expected output in WA$mean variable, was I wrong in calculating the quotation? Thank you @richard
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| May 3, 2017 at 16:07 | history | answered | Richard Hardy | CC BY-SA 3.0 |