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Is a mean-zero time series simply minusing the average value of the series.

Say I have data $Y=1,1,2,2,3,3$ would a mean zero series simply be, $$y=(1-2),(1-2),(2-2),(2-2),(3-2),(3-2)$$

as the mean of $Y$ is 2?

user22485
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This operation is called de-meaning or more informally making the series/signal zero-mean. The "de-mean"ed version of $Y$ is surely $y=\{-1 -1,0,0,1,1\}$ as you wrote. But, this operation shouldn't be confused with the definition of zero-mean time series, i.e. a random process / signal is zero mean when $E[X(t)]=0$.

You can make the signal zero-mean by subtracting different values from each entry. For example, $y=\{1,1,2,2,3,-9\}$ is also a zero-mean signal. I just subtracted the sum of your series from the last element.

gunes
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  • A good answer but perhaps I should include more details so that I am clear. I am reading a paper and if the author says "a mean-zero time series y1 and y2 is generated from a VAR(1) model" are they simulating the de-meaned version. – user22485 Mar 05 '19 at 13:41
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    VAR is vector version of AR, and AR(1) has the following form: $y_t=c+\alpha y_{t-1}+\epsilon_t$. The mean is, if assumed constant, $c/(1-\alpha)$ and if the generated signals are zero-mean, then $c=0$. – gunes Mar 05 '19 at 13:46
  • Thanks very much, your time has helped a lot. One last thing, is it, therefore, ok to leave out c for the regression when I am estimating it? – user22485 Mar 05 '19 at 14:01
  • afais, if you want to simulate zero-mean processes from the zero-mean noise signal, yes. – gunes Mar 05 '19 at 14:04
  • perfect youve been a big help – user22485 Mar 05 '19 at 14:06
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This is not generally true. While the empirical mean of the series is 2, the second moment of the process isn't necessarily equal to 2. This number is just a reasonable estimate of the mean if and only if the process is stationary and has a finite second moment. By subtracting 2 from the observations you would not necessarily get a mean zero series.

You would get zero mean series if and only if

  • $\mathbb{E}[X_t]\equiv \mu = 2, \forall t \in \mathbb{Z}$ where you observations $x_t$ are a realization of the family of random variables $\{X_t\}$ defined on a probability space $(\Omega, \mathcal{F},P)$

Note that subtracting an estimate of the mean from the observations is called demeaning. But as the series might not be stationary, the most reasonable estimate of the mean might be time-varying (since $\mathbb{E}[X_t]=\mu_t$ - where the subscript $t$ implies time dependency). Finally, if the condition $\mathbb{E}[X_t] < \infty$ (e.g. Cauchy- distributed time series) does not hold, subtracting any set of finite numbers from the process won't get the process zero mean.

Stats
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  • A good answer but perhaps I should include more details so that I am clear. I am reading a paper and if the author says "a mean-zero time series y1 and y2 is generated from a VAR(1) model" are they simulating the de-meaned version, without a constant term? – user22485 Mar 06 '19 at 08:45
  • !!!!!!!!!!!!!!!!!!!!!!!!!!!! – user22485 Mar 07 '19 at 07:19
  • It just means that the expected value of both $y_1$ and $y_2$ equals zero. De-meaning is the procedure we follow when our aim is to get zero mean series. When series already zero mean, you don't need to follow such procedure. Note that simulation is something else - you assume/know what the generating model is and you generate random paths based on the model you've assumed. The sentence you quote seems to just mean that the generating process of a given series is a zero mean VAR(1) model. – Stats Mar 07 '19 at 15:32