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Even though the title suggests just a simple question, it's a two-part.

First, can a repeated seasonal differencing filter also remove a polynomial trend? In which conditions? I would guess only when the degree of the trend is a multiple of the lag used in the filter.

Second, are both the seasonal differencing and the simple (repeated) differencing interchangeable? Is working with a time series by first taking out the trend and then seasonality the same as working with a t.s. by first taking out the seasonality and then the trend?

Any help would be appreciated.

Richard Hardy
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An old man in the sea.
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  • You asked "how do you distinguish between a deterministic linear trend, and a stochastic trend? Thanks ;) – An old man in the sea. Nov 26 at 22:07" . My answer is simple you approach model formulation in a universal way by evaluating/testing alternative/competing strategies and assess optimality via MAPEs form many origins for different lead times and general effectiveness in dealing/explaining/characterizing the observed history while being concerned with parsimony and sufficiency. – IrishStat Dec 13 '16 at 18:20

2 Answers2

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I do not have general answers, but here are some thoughts that will hopefully be helpful.

First, can a repeated seasonal differencing filter also remove a polynomial trend? In which conditions? I would guess only when the degree of the trend is a multiple of the lag used in the filter.

A simple example of seasonal integration with a seasonal period $m$ is a series made up of $m$ alternating independent random walks. That is, take random walks $x_{1,t},\dots,x_{m,t}$ and form a series $y_{\tau}$ of the form $$ x_{1,1},\dots,x_{m,1},x_{1,2},\dots,x_{m,2},x_{1,3},\dots,x_{m,3},\dots . $$ A seasonally differenced process then will be $$ \Delta x_{1,2},\dots,\Delta x_{m,2},\Delta x_{1,3},\dots,\Delta x_{m,3},\dots $$ where $\Delta x_{i,t}:=x_{i,t}-x_{i,t-1}$. So seasonal differencing applied on $y_{\tau}$ is equivalent to simple differencing applied on the alternating component series. The same is true for higher-order seasonal differencing of $y_{\tau}$ vs. higher-order simple differencing of the component series. So from this point on we can consider simple differencing of the component series instead of seasonal differencing of the original series.

If $x_{i,t}$ has a linear trend, $\Delta x_{i,t}$ will no longer have it.
If $x_{i,t}$ has a quadratic trend, $\Delta x_{i,t}$ will have a linear one, but $\Delta^2 x_{i,t}$ will have none.
And so on for higher-order polynomial trends vs. higher-order differencing.

Second, are both the seasonal differencing and the simple (repeated) differencing interchangeable?

No, they are not intechangeable.
Consider the setup presented above.
Seasonal differencing only involves simple differencing of the alternating component series. Therefore, $x_{i,s}$ never gets "mixed up" (i.e. added to or subtracted from) $x_{j,t}$, where $i\neq j$. The different component series remain separated, and a seasonal difference of $y_{\tau}$ of order $D$ can be written as an alternating sequence of simple differences of order $D$ of the component series.
Meanwhile, simple differencing of $y_{\tau}$ "mixes up" the component series: you get values like $x_{2,1}-x_{1,1},\dots,x_{m,1}-x_{m-1,1},x_{1,2}-x_{m,1},x_{2,2}-x_{1,2},\dots,x_{m,2}-x_{m-1,2},x_{1,3}-x_{m,2},x_{2,3}-x_{1,3},\dots,x_{m,3}-x_{m-1,3}$ etc. I do not think you could "unmix" the components by taking higher order simple differences, it would rather go the other way around -- you would be involving even more different series (up to a point where all series are involved, of course).

Is working with a time series by first taking out the trend and then seasonality the same as working with a t.s. by first taking out the seasonality and then the trend?

I guess it depends on how exactly you are doing this. Algebraically this is certainly possible; you can always write $(x_t+s_t)+t=(x_t+t)+s_t$ where $s_t$ stands for the seasonal component and $t$ for a linear time trend. But when you try to achieve this using a particular model being estimated on some data, the equivalence (or lack thereof) between the fitted values of the seasonal and the trend components will depend on the model.

Richard Hardy
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  • many thanks for your answer. After I posted this question, I continued reading the notes of Van der Vaart on Time Series. In section 1.2., exercise 1.24, he asks us to prove that filtering is commutative. But isn't differencing a linear filter? – An old man in the sea. Nov 26 '16 at 22:06
  • @Anoldmaninthesea, I missed this comment initially. I don't know how to prove that filtering is commutative or whether differencing is a linear filter (because I don't remember the definition of a linear filter). – Richard Hardy Nov 27 '16 at 16:11
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Seasonal differencing is applied once to remove a cyclical component. This would not remove a polynomial trend such as a linear or a quadratic trend. First differencing is used to remove a linear trend.

Michael R. Chernick
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    Could you please elaborate a bit more? Your answer seems imprecise... – An old man in the sea. Nov 26 '16 at 14:06
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    First differencing should not be used to eliminate a *linear* trend. It would get rid of the trend but at the same time introduce an integrated MA(1) component. Meanwhile, first differencing is to be used for removing a *stochastic* trend. – Richard Hardy Nov 26 '16 at 19:26
  • @RichardHardy how do you distinguish between a deterministic linear trend, and a stochastic trend? Thanks ;) – An old man in the sea. Nov 26 '16 at 22:07
  • @Anoldmaninthesea, a linear trend is literally linear, while a stochastic trend is some realization of a random walk (which you could simulate a number of times and see for yourself how it could look). They do not look alike. Or are you asking for some kind of formal test to be applied on a given time series to discover linear or stochastic trends? – Richard Hardy Nov 27 '16 at 07:12
  • @RichardHardy I was asking for a formula. In this case it would help to describe the concepts. – An old man in the sea. Nov 27 '16 at 09:29
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    A process with a stochastic trend: $y_t=x_t+\varepsilon_t$ where $\varepsilon_t$ is a stationary process and $x_t=\sum_{\tau=0}^t u_{\tau}$ where $u_{\tau}$ is an i.i.d. sequence; i.e., $x_t$ is a random walk. Meanwhile, a process with a linear trend: $z_t=\beta t+\varepsilon_t$. – Richard Hardy Nov 27 '16 at 14:12
  • @RichardHardy Thanks for your comment. By the way, did you read my other comment, but this time to your answer? – An old man in the sea. Nov 27 '16 at 15:57
  • Please merge your two new accounts http://stats.stackexchange.com/users/140029/michael-chernick and http://stats.stackexchange.com/users/140026/michael-chernick – Glen_b Nov 29 '16 at 01:14
  • Michael, it seems to me that seasonal differencing would remove linear trends in any series, because if we write the series in the form $x_t = \alpha+\beta t+r_t$ for a series $(r_t)$ without linear trends, then $$x_{t+s}-x_t=\beta s +(r_{t+s}-r_t)$$ exhibits the lag-$s$ differences (for a seasonality of lag $s$) as the sum of a constant $\beta s$ plus differences of a trendless series. Could you therefore provide more detailed justification for your assertions in this reply? – whuber Dec 13 '16 at 19:28
  • The merging was done a long time ago (i.e. last month). I think the questions are difficult to answer. I was thinking in the realm of ARIMA models where seasonal differencing removes period trends and first differencing removes linear trends. I think I was trying to show that these two things are not interchangeable. You introduced a second series in your example. I wouldn't have that in the model.. Also the trends are stochastic and not deterministic. – Michael R. Chernick Dec 18 '16 at 20:32
  • The only thing I think can be said with certainty is that my original deleted answer "The answer to both questions is no". – Michael R. Chernick Dec 18 '16 at 20:37