# What transfer function should I use to model a gradual permanent step positive intervention with increasing rate of growth?

Hello,

I want to ask what transfer function I should use to model a gradual permanent step intervention with increasing rate of growth. The relevant graph is presented at the attached image.

I want to model a step intervention that leads to an increase in the level of the series. I consider three cases, an abrupt (sudden) permanent step intervention, a gradual permanent positive intervention with decreasing rate of growth and a gradual permanent positive intervention with increasing rate of growth (see graph). In all the cases I create a binary dummy variable to model the intervention. 1) In the simple case of the abrupt (sudden) permanent step intervention I run the regression Yt = intercept + [omega] * Dummy. Example data are as follows:

… 20 20 33.5 33.5 33.5 33.5 33.5 33.5 33.5…

2) In the case of a gradual permanent step positive intervention with decreasing rate of growth I run the regression Yt = intercept + [omega/ (1-delta*B)] * Dummy. Example data are as follows:

… 20 25 28 30 32 33 33.5 33.5 33.5 …

3) In the case of a gradual permanent step positive intervention with increasing rate of growth what transfer function should I use to model it? Yt=intercept+[Transfer Function] * Dummy Example data are as follows:

… 20 20 20.25 21.5 23 25 29 33.5 33.5 …

Andreas

• Andreas, could you share some information about the purpose of this modeling? Absent that, you will only get generic answers that might or might not be appropriate for the question, which makes it difficult (if not impossible) for people to provide the great answers we look for here.
– whuber
Dec 9 '11 at 14:20
• It is not clear to me what this graph is showing. Is it the desired impulse or step response of the system for which you'd like a transfer function? Or, something else? It seems clear that if the graph is intended to be either an impulse or step response, then no rational transfer function should exist. Indeed, I could be mistaken, but I do not see how a linear system could produce such a response function at all. (But, I will give it some more thought. Please update your question in the meantime.) Dec 10 '11 at 3:26

I have edited this and tried to meet some of Cardinal's sage and continually positive and mature advice regarding postings/answers.

The question is how to efficiently convert/characterize an observed series in terms of a Transfer Function/ARMAX Model.There are four possible ways to include a deterministic/dummy 0/1 variable and the OP wishes to create a map between data and a possible representation. The 4 ways are Pulses, Level Shifts , Seasonal Pulses and Local Time Trends. A step/level (St) 0 before intervention and 1 after intervention) and a pulse Pt (1 at intervention and 0 elsewhere) interventions, the model then can be expressed as

Yt=β1St+β2Pt+ηt where nt is an ARIMA process

Also because there may different responses to the interventions, say graduate change in

level is by ωSt1−δB or decayed responses ωPt1−δB.

Your three examples reinforce the idea that all models are wrong but some are useful. It is possible to represent your three series with and wthout incorporating ARIMA structure. In

1) 20 20 33.5 33.5 33.5 33.5 33.5 33.5 33.5… might be handled by two pulses followed by a level shift

y(t)=b1*x1 + b2*x2 + b3*x3

No ARIMA structure needed

where x1 and x2 are pulse series and x3 is a step/level series

2) 20 25 28 30 32 33 33.5 33.5 33.5 … can be approximated by a combination of memory/ARIMA and dummy series.

[(1-B*1)]Y(T) = +[X1(T)][(1-B*1)][(+ .417)] :PULSE 5

              +[X2(T)][(1-B**1)][(-  .114)]        :PULSE               8

+[X3(T)][(1-B**1)][(+  .229)]        :PULSE               6

+     [(1-  .635B** 1)]**-1  [A(T)]


Alternatively if one wanted to supress the ARIMA component. Note that a step = [1-B]trend thus a trend = step/[1-B] . REstated a step is a difference of a trend and a pulse is a difference of a step

   +[X1(T)][(+ 2.6434)]   :TIME TREND          1

+[X2(T)][(- 2.5298)]   :TIME TREND          6

+[X3(T)][(-  .123)]    :TIME TREND         11

+[X4(T)][(- 2.4278)]   :PULSE               1

+[X5(T)][(- 1.0012)]   :PULSE               5

+       [A(T)]


OR in terms of step/level variables and pulses Y(T) = 19.784 gradual

   +[X1(T)][(+ 2.6434)]/[1-B]   : STEP/LEVEL   1

+[X2(T)][(- 2.5298)]/[1-B]   : STEP/LEVEL   6

+[X3(T)][(-  .123)]/[1-B]    : STEP/LEVEL  11

+[X4(T)][(- 2.4278)]   :PULSE               1

+[X5(T)][(- 1.0012)]   :PULSE               5

+       [A(T)]


3)… 20 20 20.25 21.5 23 25 29 33.5 33.5 . In terms of no ARIMA and no differencing operators

   +[X1(T)][(+  .932)]    :TIME TREND          1

+[X2(T)][(+ 1.2098)]   :TIME TREND          6

+[X3(T)][(- 2.1723)]   :TIME TREND         11

+[X4(T)][(+ 4.0747)]   :PULSE               8

+[X5(T)][(+ 1.9325)]   :PULSE               9

+       [A(T)]


If one wanted to represent this time series using a combination of ARIMA and dummies

     +[X1(T)][(1-B**1)][(1-B**1)][(+ 4.2000)]   :PULSE               8

+[X2(T)][(1-B**1)][(1-B**1)][(+ 1.8333)]   :PULSE               7

+[X3(T)][(1-B**1)][(1-B**1)][(+ 2.3000)]   :PULSE               9

+[X4(T)][(1-B**1)][(1-B**1)][(+  .833)]    :PULSE              10

+       [A(T)]


Andreas, these models were easily/automatically developed by commercially available software that I have had a role in developing (AUTOBOX). SAS and others attempt to form similar models but often fail, in my opinion. I am sure that some of this could be duplicated in other programming languages.

• I think some attention to the formatting of this answer would be very helpful. Right now it looks to me more or less like a bunch of screen dumps from a piece of software that I'm not familiar with, with little surrounding context. I don't mean that to sound too critical; I just feel that providing some context and words in between the software output would be helpful. Dec 10 '11 at 3:20