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Suppose I have an observed time series $y_t$, which I suspect has been smoothed out. It appears to be significant autocorrelation at lag 1 and 2, therefore I suppose that the observed series $y_t$ is in the form:

$$y_t = \theta_0 x_t + \theta_1 x_{t-1} + (1 - \theta_0 - \theta_1)x_{t-2}$$

where $x_t$ is the original series I am after.

How can I recover the "original" series $x_t$? Clearly I need a method to estimate $\theta_0$ and $\theta_1$ and then apply the relevant transformation. But how to do that? I don't see how to apply an arima process here.

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    $\begingroup$ It seems the OP will be able to figure out the MATLAB himself once he understands the ideas. Thus, I gather he is after a conceptual understanding here. I think this Q should be considered on topic. $\endgroup$ – gung - Reinstate Monica Jan 15 '16 at 12:59
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    $\begingroup$ @gung has persuaded me that this is on-topic, but putting particular software in a title and a last sentence is not the best signal if the main focus is statistical. $\endgroup$ – Nick Cox Jan 15 '16 at 13:06
  • $\begingroup$ @NickCox Yes, I see your point. I edited the question accoringly $\endgroup$ – Ant Jan 15 '16 at 13:16
  • $\begingroup$ question: is it true that $\Sigma_{i=t-2}^{i=t} \theta_{i} = 1$ ? $\endgroup$ – EngrStudent - Reinstate Monica Jan 15 '16 at 16:04
  • $\begingroup$ @EngrStudent, a great point. He should start with unconstrained fit, then test whether the relationship holds, then only constrain the fit. $\endgroup$ – Aksakal Jan 16 '16 at 3:42
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Your model can be written as arima model, or to be precise MA(2) model:

$$y_t= z_t+\alpha_1 z_{t-1}+\alpha_2 z_{t-2},$$

ARIMA models are usualy postulated with coefficient $1$ for $z_t$, because you can always move the multiplicative constant to the variance of the disturbances.

So your model

$$y_t= \theta_0x_t+\theta_1 x_{t-1}+(1-\theta_0-\theta_1) x_{t-2}$$

becomes MA(2) model

$$y_t= z_t+\frac{\theta_1}{\theta_0} z_{t-1}+\frac{1-\theta_0-\theta_1}{\theta_0} z_{t-2}$$

with $z_t=\theta_0x_t$.

So you can estimate MA(2) model and then recover $\theta_0$ and $\theta_1$ from $\alpha_1$ and $\alpha_2$:

$$\theta_0=\frac{1}{1+\alpha_1+\alpha_2},\quad \theta_1=\frac{\alpha_1}{1+\alpha_1+\alpha_2}$$

Here is the example in R:

set.seed(1001)
x<-rnorm(10000)
y<-filter(x,c(0.5,0.3,0.2),sides=1)
mod<-arima(y,order=c(0,0,2),include.mean=FALSE)
> mod
Call:
arima(x = y, order = c(0, 0, 2), include.mean = FALSE)

Coefficients:
         ma1     ma2
      0.6060  0.3874
s.e.  0.0093  0.0093

sigma^2 estimated as 0.2495:  log likelihood = -7246.14,  aic = 14498.28
> cf <- coef(mod) 
> 1/(1+sum(cf))
[1] 0.500497
> cf[1]/(1+sum(cf))
      ma1 
0.3007687 

You can recover $x_t$ as residuals of the arima model:

x_rec<- as.numeric(residuals(mod)*(1+sum(cf)))
plot(x-x_rec)

enter image description here

As you see the procedure recovered coefficients with the precision of 3 decimal places. The recovered $x_t$ is also recovered to similar precision. The difference is the initialisation. ARIMA model assumes that the process is infinite, but the data is never infinite, so each estimation procedure must assume some initialisation. As evidenced from the plot the first few elements of recovered $x_t$ have the biggest error, but then the error stabilizes.

Since ARIMA models are estimated via Kalman filter procedure, you could implement it yourself with the proper intialisation. Note that in this example I used quite a big sample of 10000 elements. Less data would result in worse precision, you should run some tests to see the extent of the impact of sample size to the precision of recovery.

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  • $\begingroup$ Ah, I see.. thank you, this helps. Initially I thought I could not apply directly an MA estimation because I thought the $x_t$ needed to be independent, but I take it that being uncorrelated is enough? :) $\endgroup$ – Ant Jan 15 '16 at 22:45
  • $\begingroup$ This assumes that original series are not correlated, which may not hold depending on the data. For instance, in economic data this almost never holds, series like sales or customer visits are correlated. $\endgroup$ – Aksakal Jan 15 '16 at 22:54
  • $\begingroup$ This approach can be generalised for arima type $x_t$, but then recovering the original $x_t$ might be difficult, without the additional restrictions. Furthemore even if $x_t$ are not independent, how would you test it, if they are not observed? $\endgroup$ – mpiktas Jan 15 '16 at 23:25
  • $\begingroup$ @mpiktas, it depends on the data. Some series are known to be dependent from prior experience or understanding the process. For instance, it's simply not reasonable to assume that deposit balances or area temperatures are independent or even not autocorrelated. I don't need to test this. Generally, you can't test for independence even if you observe the data. $\endgroup$ – Aksakal Jan 16 '16 at 3:38
  • $\begingroup$ I think a qqnorm plot might be more informative than a trendplot. $\endgroup$ – EngrStudent - Reinstate Monica Jan 16 '16 at 18:31
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You can start with assuming that your observed variable is obtained from the true value as $$y_t = \theta_0 x_t + \theta_1 x_{t-1} + e_t$$

It would help to know what is the process of the underlying variable, suppose it's $$x_t = \beta_0 + \beta_1 x_{t-1} + u_t$$

where $e_t,u_t$ are errors. If these equations make sense to you then, you can estimate them using Kalman filter, see example here.

Next, you test whether $\theta_0+\theta_1=1$, if it holds statistically, then maybe your specification holds, so you can proceed with a constrained fit.

You have to set the expectations though: smoothing leads to data loss, generally. So, you can't reproduce the original series exactly. That's why using Kalman filter we had to make an assumption about the observed and true processes, i.e. we needed to inject some outside data to compensate for lost data (from smoothing) in order to recover the true series.

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