# Validating ARIMA(1,0,0)(0,1,0) with manual calculation

I am using the forecast package in R to do ARIMA forecasting with auto.arima() function by Professor Hyndman.

I have 27 months of sales data (June 2014 to August 2016), I'm using the first 21 months as training set, and the rest 6 months as test set.

dataset <- c(380075, 367137, 320548, 290192, 256514, 365335, 356847, 287760, 378703, 415378, 331637, 228744, 407655, 496475, 340695, 410041, 725069, 594173, 614646, 530843, 565057, 487604, 565945, 410603, 405337, 283004, 211718)
datasets = ts(dataset, start=c(2014,6), frequency=12)
trainsets <- window(datasets, end=time(datasets)[length(dataset)-6])
testsets <- window(datasets, start=time(datasets)[length(dataset)-5])
fit <- auto.arima(trainsets, ic='aic', D=1, seasonal=TRUE, approximation=FALSE,trace=TRUE, stepwise=FALSE, allowdrift=FALSE)
WA <- forecast(fit, 6)


And I got the return

> fit
Series: trainsets
ARIMA(1,0,0)(0,1,0)

Coefficients:
ar1
0.7148
s.e.  0.1989

sigma^2 estimated as 2.489e+10:  log likelihood=-120.32
AIC=244.63   AICc=246.63   BIC=245.03

> WA

Point          Forecast     Lo 80    Hi 80      Lo 95    Hi 95
Mar 2016       548576.1 346405.21 750747.0  239382.42 857769.8
Apr 2016       426841.5 178337.38 675345.5   46787.27 806895.6
May 2016       296792.2  27678.71 565905.7 -114781.35 708365.7
Jun 2016       456293.0 177237.44 735348.6   29514.34 883071.7
Jul 2016       531239.4 247238.93 815239.9   96898.14 965580.7
Aug 2016       365543.2  79049.34 652037.0  -72611.33 803697.7


I tried to do the manual calculation to understand the output, so because I have ARIMA(1,0,0)(0,1,0)

So I expect the calculation to be

$$\hat{Y_t}(1) = \mu + \phi*(Y_{t-1} - Y_{t-2}) + Y_{t-12}$$

I think I can leave the $\mu$ = 0

So, for the March 2016 with the forecast of 548576.1, I calculate

$$\hat{Y_t}(MARCH2016) = 0 + 0.7148*(FEB2016 - JAN2016) + MARCH2015$$ $$\hat{Y_t}(MARCH2016) = 0 + 0.7148*(565057 - 530843) + 415378$$ $$\hat{Y_t}(MARCH2016) = 589513.1672$$

Question : What makes my calculation doesn't meet the correct forecasting? Can you please suggest me the right way to calculate the function?

Thank you.

• I don't think you got the model equation right. The model is (omitting the intercept) $x_t=\varphi_1 x_{t-1} + \varepsilon_t$ where $x_t=(y_t-y_{t-12})$. Insert the latter expression of $x$ into the first equation, and you will get something different than you have now. – Richard Hardy May 3 '17 at 14:10
• Hello, thank you for your reply. I tried to calculate the way you suggest me, but I think I got lost and needs to be clear. In your formula, what does Xt, Yt, and Et actually means? What I got is, Et for error term (but I don't know if there's exist in my ARIMA, so I leave it 0), Xt forecasted value?, and Yt is my actual sales? Or I'm getting wrong here? Thanks. @RichardHardy – Ervan Sanjaya May 3 '17 at 14:30

The model is $$x_t=\varphi_1 x_{t−1}+\varepsilon_t$$ where $x_t=(y_t−y_{t−12})$. Insert the latter expression into the equation above to get $$(y_t - y_{t-12}) = \varphi_1 (y_{t-1} - y_{t-13}) + \varepsilon_t,$$ thus $$y_t = y_{t-12} + \varphi_1 (y_{t-1} - y_{t-13}) + \varepsilon_t.$$ When you forecast one step ahead, you get \begin{aligned} \hat y_{t+1|t} &= y_{t-12+1} + \varphi_1(y_{t-1+1} - y_{t-13+1}) + \hat\varepsilon_{t+1|t} \\ &= y_{t-11} + \varphi_1(y_{t} - y_{t-12}) + 0 \\ &= y_{t-11} + \varphi_1(y_{t} - y_{t-12}). \\ \end{aligned}
• 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 – Ervan Sanjaya May 3 '17 at 16:23
• Can it be a rounding problem or something? Maybe try using coef(fit) instead of 0.7148 in your manual calculation. – Richard Hardy May 3 '17 at 16:25
• 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. – Ervan Sanjaya May 3 '17 at 16:30
• 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. – Ervan Sanjaya May 3 '17 at 17:10