# How to interpret the constant for an ARMA model

I'm trying to fit an ARMA(1,0) model for a timeseries that start at $$10$$ and drops slowly to $$4$$ in around $$180$$ steps. For this, I've tried to fit an ARMA model in python using the following:

# contrived dataset
data1 = data['beta_0'].tolist()

# fit model
model = ARMA(data1, order = (1,0))
model_fit = model.fit()

The results are as follows:

I'm having trouble understanding why the constant would be so big. Wouldn't this mean that if you say for instance that $$y_0 = 10$$ (as it is in my time series), then $$y_1 = 6.8840 + 0.9916y_0 + \epsilon$$. I would assume that then the values of this time series would blow up right?

Then how can it be that this is the output for a series that drops slowly from 10 to 4?

Any help would be much appreciated.

$$[y(t)-6.8840][1-.9916B]= +ϵ(t)$$

or

$$y(t)= .0084\times6.8840 + .9916\cdot y(t-1)$$

$$y(t)= .0578 + .9916\cdot y(t-1)$$

What has you confused is for your stationary model the constant that is estimated is a Left-Hand side constant not the right-hand side that you were (normally !) expecting.

Your model in my opinion should be changed to a first difference model as your ar coefficient is not different from 1.0 .

EDITED AFTER COMMENTS BY @whuber:

I was intrigued by the interesting discussion and decided to roll out my trusty time series simulator and generated a realization of the following model . The nature of the model suggests that other realizations could be expected to have different patterns.

The series has its up's and down's much like what might be expected from a process where the expected value is the last value more or less.

The ACF and PACF are here

The estimated model is here and here (with a non-significant BUT necessary constant )

Expressed as a pure right-hand side equation here

The ACTUAL/FIT/FORECAST graph is here

The history shows trends up and down for different segments .

Treated objectively the model is simply a random walk

• Your formula doesn't match the conventions used by Statsmodels: see stats.stackexchange.com/questions/280507. I agree that the output indicates the advisability of first-differencing, but the short answer is that this looks like a random walk and, since the innovation variance is estimated as $0.294^2=0.0684,$ over the course of $180$ independent steps the variance of the total change is $15.56,$ indicating a typical change of $\sqrt{15.56}\approx 4.$ This value and the estimated constant of $6.88$ are perfectly consistent with data ranging from $10$ to $4.$
– whuber
Nov 18 '19 at 21:15
• a random walk is a particular case of a first difference model allowing for a possible drift parameter or even arma structure. In my parlance a random walk is a first difference model without drift and without arma structure ... this data suggests drift which may not be significant. I pursued your suggested thread but as usual I am (slightly) confused . Are you sure your url is correct ? Nov 18 '19 at 21:47
• I apologize: the final digit of the question number got cut off during the copy-and-paste operation. I have restored it. It should link to a thread titled "Generating equation from python ARMA model summary."
– whuber
Nov 18 '19 at 21:48
• it would appear that their general expression is inconsistent with the reported/estimated constant C . The OP's reflection about the series blowing up is correct while if one uses .0578 (which may not be statistically significant) the series will blow up slower. I surmize from his "dropping slowly" reflection that there may be outliers in the series that are falsely imputing a positive but not significant positive drift from 10. to 4. . His model's forecast converge to an assymptote as ultimately the ar effect dies out. If the last value is near 4.the limiting value will be close to 4 .... . Nov 18 '19 at 22:32