# How are error terms calculated for moving average model in R [duplicate]

For an ARIMA (0,0,1) model, I understand that R follows the equation: xt = mu + e(t) + theta*e(t-1) (Please correct me if I am wrong)

I assume e(t-1) is same as the residual of the last observation. But how is e(t) calculated?

For example, here are the first four observations in a sample data: 526 658 624 611

These are the parameters Arima(0,0,1) model gave: intercept = 246.1848 ma1 = 0.9893

And the first value that R fit using the model is: 327.0773

How do I get the second value? I used: 246.1848 + (0.9893*(526-327.0773)) = 442.979

But the 2nd fitted value given by R is : 434.7928

I assume the difference is because of the e(t) term. But I do not know how to calculate the e(t) term.

• I'm not sure I undestood your question. Do you want to know the calculations that yield the residuals from an ARIMA(0,0,1) model? If so, please post the last, say 5, values of the original series and residuals. – javlacalle Jul 28 '14 at 19:55
• Not the residuals.. I want to find the fitted values. What would be the 2nd value fitted by ARIMA (0,0,1)? – nancy Jul 28 '14 at 20:31
• Several of the questions under "Related" in the sidebar to the right -> are duplicates or near-duplicates. e.g here, or here – Glen_b Jul 29 '14 at 0:30

You could obtain the fitted values as one-step forecasts using the innovations algorithm. See for example proposition 5.5.2 in Brockwell and Davis; downloable from the internet I found these slides.

It is much easier to obtain the fitted values as the difference between the observed values and the residuals. In this case, your question boils down to obtaining the residuals.

Let's take this series generated as an MA(1) process:

set.seed(123)
x <- arima.sim(n = 150, model = list(ma = 0.4))
fit <- arima(x, order = c(0,0,1), include.mean = TRUE)
resid <- residuals(fit)


The residuals, $\hat{e}_t$, can be obtained as a recursive filter:

$$\hat{e}_t = x_t - \hat{\mu} - \hat{\theta} \hat{e}_{t-1}$$

For example, we can obtain the residual at time point $140$ as the observed value at $t=140$ minus the estimated mean minus $\hat{\theta}$ times the previous residual, $t=139$):

macoef <- coef(fit)
mu <- coef(fit)
as.vector(x - mu - macoef * resid)
#  0.7742192
# equal to
residuals(fit)
#  0.7742192


The function filter can be used to do these calculations:

resid.v2 <- filter(x = x - mu, filter = -macoef, method = "recursive")


You can see that the result are very close to the residuals returned by residuals. The difference in the first residuals is most likely due to some initialization that I may have omitted.

head(cbind(resid, resid.v2))
#            resid     resid.v2
# [1,] -0.39447063 -0.429102263
# [2,]  1.62425953  1.675613161
# [3,]  0.03344943  0.001881606
# [4,]  0.16839438  0.181951010
# [5,]  1.71983927  1.714147387
# [6,]  0.43595315  0.438334539
tail(cbind(resid, resid.v2))
#             resid   resid.v2
# [145,] -0.4803322 -0.4803322
# [146,] -1.4432094 -1.4432094
# [147,]  0.7463573  0.7463573
# [148,]  2.0810053  2.0810053
# [149,] -1.3126564 -1.3126564
# [150,]  0.8601761  0.8601761


The fitted values are just the observed values minus the residuals:

require(forecast)
# [1,]  -0.02526549 -0.05989712
# [2,]  -0.20897584 -0.15762221
# [3,]   0.69211011  0.66054229
# [4,]  -0.02445992 -0.01090329
# [5,]   0.05263269  0.04694081
# [6,]   0.70860766  0.71098905
tail(cbind(x - resid.v2, fitted(fit)))
# [145,]   -0.6911888  -0.6911888
# [146,]   -0.2309088  -0.2309088
# [147,]   -0.6431427  -0.6431427
# [148,]    0.2942704   0.2942704
# [149,]    0.8656695   0.8656695
# [150,]   -0.5872494  -0.5872494


In practice you should use the functions residuals and fitted but for pedagogical purpose you can try the recursive equation used above. You can start by doing some examples by hand as shown above. I recommend you to read also the documentation of function filter and compare some of your calculations with it. Once you understand the operations involved in the computation of the residuals and fitted values you will be able to make a knowledgeable use of the more practical functions residuals and fitted.

You may find some other information related to your question in this post.