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I am trying to understand the way MA(q) models work. For this purpose I have created a simple data set with only three values. I then adapted a MA(1) model to it. The results are shown below:

x<-c(2,5,3)
m<-arima(x,order=c(0,0,1))

Series: x 
ARIMA(0,0,1) with non-zero mean 

Coefficients:
          ma1  intercept
      -1.0000     3.5000
s.e.   0.8165     0.3163

sigma^2 estimated as 0.5:  log likelihood=-3.91
AIC=13.82   AICc=-10.18   BIC=11.12

While the MA(1) model looks like this: $$X_t = c +a_t - \theta*a_{t-1}$$

and $a_t$ is White Noise.

What I cant figure out is how to get the fitted values:

library(forecast)
fitted(m)
Time Series:
Start = 1 
End = 3 
Frequency = 1 
[1] 3.060660 4.387627 3.000000

I tried different ways, but I cant find out how the fitted values (3.060660, 4.387627 and 3.000000) are calculated.

I would be very thankful for an answer!

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I found it out, but I can't tell you the exact reason for it:

The problem is, that the initialization is unkown/strange, if you do an example with more values, you will see, that a simple MA(1) forecasting according to the following rule will work (notation in R of the MA is slightly different to yours, the sign of the theta is different): \begin{align}\hat{X}_{T|T-1}=E(c+a_T+\theta∗a_{T−1})=c+\theta*a_{T-1}\end{align}

You can calculate these values manually, consider the following example:

z<-c(2,5,3,4,3,4,5,4.3,4.3,4.5,4.3,4.5,3.4,5.3,4.2,3.4,2.3,2.3,4.5,3.4,5,5.4,5.4,3.4,5.43,5.64,5.6,3.4,5.3,5,6.3,4.5)
m<-arima(z,order=c(0,0,1))

This gives

Series: z 
ARIMA(0,0,1) with non-zero mean 

Coefficients:
         ma1  intercept
      0.1162     4.2748
s.e.  0.1500     0.2076

sigma^2 estimated as 1.11:  log likelihood=-47.09
AIC=100.18   AICc=101.03   BIC=104.57

The values of the output can be used via m$coef[1] and m$coef[1]

So you now compare the values with the following code:

m$coef[2]+ m$coef[1]*(2-4.259580)
fitted(m)[2]
m$coef[2]+ m$coef[1]*(5-4.013978)
fitted(m)[3]
m$coef[2]+ m$coef[1]*(3-4.389402)
fitted(m)[4]
m$coef[2]+ m$coef[1]*(4-4.113285)
fitted(m)[5]
m$coef[2]+ m$coef[1]*(3-4.261626)
fitted(m)[6]

You will notice, that at the beginning, there is a small difference, at the end, the values are the same! So the simple forecasting rule of a MA(1) does hold, but R seems to do some specific initialization. I know that STATA uses a certain Kalman filter setting, maybe R is doing the same. I hope this helps.

If it did help you, you can accept my answer by clicking on the hook to the left of my post.

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  • $\begingroup$ Where did $\theta$ come from, given this is meant to be $\hat X_{T|T-1}$? $\endgroup$ – conjectures May 22 '13 at 16:54
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    $\begingroup$ @conjectures $\theta$ is the estimated coefficient. The calcualted values with these coefficient are all in-sample fits. This is what the fitted method does. As already stated, this is not about model checking or something like this, it is just to understand how to forecast a MA(1). Normally, at this time point you would not know this value (you would get a different estimate, since you do not have the following values if you do the estimation). $\endgroup$ – Jen Bohold May 22 '13 at 17:03
  • $\begingroup$ Thank you for your effort, it looks pretty good! The question actually was about the initial value, or rather the one-step-ahead prediction for the first period. And arima somehow estimates fitted(m)[1]=4.25958, and in order to understand it, I should understand the way Kalman filter works, did i get it right? $\endgroup$ – DatamineR May 22 '13 at 17:13
  • $\begingroup$ @JohnnyB yes, but the Kalman filter is normally not necessary to forecast a MA(1). So this could be done without the Kalmna filter and it is pretty simple: Just use the given formula and calculate it the R code does. But this does not work, since most statistical packages use the Kalman filter for the initialization period. The first "shock", or call it error is not known, since you have no estimate for this. Normally, people set the observation equal to the fitted values, so the shock would be equal to zero. $\endgroup$ – Jen Bohold May 22 '13 at 17:17
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In terms of where those particular numbers come from, this seems to do the trick:

> x - m$residuals
Time Series:
Start = 1 
End = 3 
Frequency = 1 
[1] 3.060660 4.387627 3.000000

Or

> x + (m$residuals * ma1)
Time Series:
Start = 1 
End = 3 
Frequency = 1 
[1] 3.060660 4.387627 3.000000

I don't know in what sense these are meant to be the 'fitted' values (maybe someone else can help?). For an ARIMA model our explanatory data are also our fitted values.

Our estimator for $Y_k$ given the history up to time $T$, $H_T$, where $T>k$, is $\hat y_k = \mathbb E [Y_k | H_T ] = y_k$, i.e. we already have the answer in the process history. So we have to ask, our fitted values anchored at what time? For a length $T$ series I guess there would be $T$ possible fitted values for each point.

If anyone asks, based on having the data $y_{1:T}$ what's the best estimate of $y_k$ - seems there's only one pragmatic answer, $y_k$. The residuals are interesting in the sense that if they're massive relative to the data, that's saying something about the magnitude of the process variance.

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    $\begingroup$ seriously, that is not an answer, because still the question is, how those values are generated? So where do the residuals come from, how can one calculate them manually? $\endgroup$ – Jen Bohold May 22 '13 at 15:43
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    $\begingroup$ @conjectures downvote, because: Asking for the fitted values is indeed the same problem as asking for how to compute the residuals. Normally, the way is to 1. Set an initial value for the a_0, normally this is zero. 2. Compute the next fitted value and so on. So the fitted values imply, that you know the data up to the previous data point. So at T you know T-1. Normally, forecasting MA(1) isn't that hard anymore, just X_{t+1}=constant+theta*a_{t}. But this does not work here, even without the problem of not knowing the inital value you could check this by using the second value: $\endgroup$ – Stat Tistician May 22 '13 at 15:47
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    $\begingroup$ @all 4.387627 = 3.5 + (-1)*(5-3.060660) or 4.387627 = 3.5 - (-1)*(5-3.060660) but this does not work. $\endgroup$ – Stat Tistician May 22 '13 at 15:48
  • $\begingroup$ @Jen The origin of the residuals is explained in the arima docs (Kalman filter). So I've replaced an unkown with something that can be tracked down. The fitted function is from a different library - I think we need to wonder why. $\endgroup$ – conjectures May 22 '13 at 16:05
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    $\begingroup$ @conjectures that is right, normally this in-sample approach would not be good, but that is not the point. JohnnyB just wants to understand how the fitted values are generated, as far as I have understand him correctly. The way of Stat Tistician is indeed correct - I implemented this, but the values differ slightly, because of the Kalman initialization. $\endgroup$ – Jen Bohold May 22 '13 at 16:35

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