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I am a beginner in Time Series Analysis and I am reading something about the most simple processes like AR and MA. So far I understood that one wants to have the following result:

The MA($\infty$)-process is weakly stable if the absolute sum of the coefficients is convergent.

The definition for the MA($\infty$) process is

$$X_t = p(L)Z_t$$ where $p(z) = \sum_{i=0}^\infty a_i z^i$ is a (formal) power series in $L$ such that $\sum_{i=0}^\infty |a_i| < \infty$ and $L : \{\text{random variables}\} \mapsto \{\text{random variables}\}$ is the lag operator $LX_t = X_{t-1}$ and $(Z_t)_t$ is a process of independent random variables with finite mean $E[Z_t]=0$ independent of time and finite variance $\text{Var}(Z_t)=\sigma$ independent of time as well.

Now people make a great mystery out of the fact whether or not their time series starts at $t=0$ (i.e. the index set is $\mathbb{N}_0$) or whether the index set is $\mathbb{Z}$. Let us assume that the index set is $\mathbb{Z}$ then I understand (up to questions of convergence but this seems ok as $\sum|a_i| < \infty$ implies that the $a_i$ have to tend to zero so $a_i^2 \leq |a_i|$ and so forth) why the variance does not depend on time:

$$\text{Var}(X_t) = \sum_{i=0}^\infty a_i^2 \text{Var}(Z_{t-i}) = \sigma \sum_{i=0}^\infty a_i^2$$

However, if the process starts at $t=0$ (i.e. $Z_t = 0$ for $t < 0$) then actually we have $X_t = \sum_{i=0}^t a_i Z_{t-i}$ and thus

$$\text{Var}(X_t) = \sum_{i=0}^t a_i^2 \text{Var}(Z_{t-i}) = \sigma \sum_{i=0}^t a_i^2$$

depends strongly on $t$!

So the question is: Is it true that for all time series we assume the index set to be $\mathbb{Z}$?

This is hardly true for any real data we observe because we start to observe the data at some point (so we do not have $x_{-1}, x_{-2}, ...$). We could overcome this by saying that we could have collected data before $t=0$ and the process that produced the data would produce it by the same rules. Then one could argue that the mere process itself that produces the data does start at some point (so we could definitely not collect more data). Again, we can overcome this by stating something like 'well, if the process was active then it would have behaved in the same way'... Is that somewhat reasonable?

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  • $\begingroup$ It is a bit difficult to read. Where did you get that definition? What do you mean by all time series in your bold question. The time series defined by an infinite moving average process or also the others? $\endgroup$ – Sextus Empiricus Jun 22 '18 at 12:56
  • $\begingroup$ You might find it useful to distinguish a time series qua dataset from a time series qua statistical model. The former obviously cannot be bi-infinite (or even infinite), but the latter may be. See stats.stackexchange.com/a/126830/919 for more about this distinction. $\endgroup$ – whuber Jun 22 '18 at 13:50
  • $\begingroup$ @Martijn: It was a “too broad”, general question. I guess I am asking about the mere definition of time series. Is it a collection of random variables like $(X_t)_{t \in \mathbb{N}_0}$ or is it $(X_t)_{t \in \mathbb{Z}}$? $\endgroup$ – Fabian Werner Jun 22 '18 at 15:41
  • $\begingroup$ @FabianWerner I believe you make a false dichotomy by proposing that the definition for a time series needs to include a range either N or Z. Another possibility is that the definition for a time-series does not include any statement about the range of the index. (and the range will just be a specification of a time series, not something that is necessary to make a set of points be defined as a time-series) $\endgroup$ – Sextus Empiricus Jun 22 '18 at 15:47
  • $\begingroup$ @Martijn: I am confused by your words: let’s do it your way: the index set is part of the time series. So when we say AR or MA then we must always include N or Z, i.e. Z-AR or N-AR. This seems to be rather clumsy and then one has these things like that the Z-MA is stationary but the N-MA is not for a very „stupid“ reason of the finite beginning. By the way: what is a N-MA? How is X_1 defines if the order is 4 for example? $\endgroup$ – Fabian Werner Jun 22 '18 at 18:13

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