ARIMA Model Coefficients

Introduction

The process for finding the best values for the coefficients of an ARIMA(p, d, q) model for given values of p, q and d is identical to that described in Calculating ARMA Model Coefficients using Solver, except that we need to take differencing into account.

We show how this is done using the Real Statistics ARIMA data analysis tool, introduced in Real Statistics Tool for ARMA Models.

Example

Example 1: Repeat Example 1 of Real Statistics Tool for ARMA Models using an ARIMA(2,1,1) model without a constant.

Start by pressing Ctr-m and choosing the Time Series option. Select the ARIMA Model and Forecast option on the dialog box that appears and click on the OK button. Now, fill in the dialog box that appears as shown in Figure 1 of Real Statistics Tool for ARMA Models except that you need to insert a 1 in the MA order field and a 1 in the Differences field. You also choose to uncheck the Constant included in the model field.

The output is shown in Figures 1 and 2 (the forecast data will be shown later in Figure 1 of ARIMA Forecasts). This output is similar to that shown in Figures 2 of Real Statistics Tool for ARMA Models. The main difference is that the data being analyzed (as shown in column E) are the differences between the input data values (as shown in column B). For example, cell E4 contains the formula =B5-B4.

Real Statistics ARIMA model

Figure 1 – ARIMA(2,1,1) model: part 1

From Figure 1, we see that the model takes the form

Δyi = .456 Δyi-1 + .559 Δyi-2 + εi – 1.112 εi-1

where Δyi = yi – yi-1. Thus

yi – yi-1 = .456 (yi-1 – yi-2) + .559 (yi-2 – yi-3) + εi – 1.112 εi-1

which can be re-expressed as

yi  = 1.456 yi-1 + .103 yi-2 – .559 yi-3 + εi – 1.112 εi-1

ARIMA model output 2

Figure 2 – ARIMA(2,1,1) model: part 2

Examples Workbook

Click here to download the Excel workbook with the examples described on this webpage.

References

Pelgrin, F. (2011) Estimation of ARIMA models
https://math.univ-cotedazur.fr/~frapetti/CorsoP/Chapitre_4_IMEA_1.pdf

Triacca, U. (2014) Estimation of the parameters of an ARMA model
https://www.lem.sssup.it/phd/documents/Lesson12.pdf

Hamilton, J. D. (1994) Time series analysis. Princeton University Press
https://press.princeton.edu/books/hardcover/9780691042893/time-series-analysis

Hitchcock, D. (2016) Parameter estimation in time series models
https://people.stat.sc.edu/Hitchcock/stat520ch7slides.pdf

33 thoughts on “ARIMA Model Coefficients”

  1. Hi Charles,

    Hope you are keeping well.

    Should the equation be re-expressed as yi = 0.456 yi-1 + .103 yi-2 – .559 yi-3 + εi – 1.112 εi-1
    instead of yi = 1.456 yi-1 + .103 yi-2 – .559 yi-3 + εi – 1.112 εi-1

    if it’s supposed to stay as 1.456 yi-1, could you perhaps explain why?

    thanks

    kind regards
    Declan

    Reply

  2. Dear Charles, Thank You very much for the content of this site. It has been very helpful.
    The question i have is that coefficients (Phi and Theta) could be more than 1 or less than -1… ? I was using solver always restricting this values between -1 and 1…
    thanks for your answer.

    Reply

    • I tend to use lower-case letters for variables and upper-case letters for matrices or vectors. I am not always consistent about this though. It is also common to use an upper-case letter such as U to represent the statistic used in the Mann-Whitney test.
      Charles

      Reply

  6. Hi Charles,
    first of all thanks for your amazing work.
    I’m having problems with the ARIMA model, the s.e. in the model paramaters always return a VALUE error.
    Did I miss something, I have tried with the ARIMA_SE formula and gives me the same result.
    Thank you very much,

    Reply

    • Hi Anna,
      If you use the Solver approach, then it is very likely that you will get a VALUE error for the s.e. since the algorithm used to calculate the s.e. often doesn’t converge to a solution. This should not be the case if you use Newton’s method. You should not get an error in this case.
      If you send me an Excel file with your data by email, I will try to figure out why you are getting error values for the s.e.
      Charles

      Reply

  7. Hi Charles,

    Sorry, i thought that you know that approcah when MA is also included for Reformat linear regression.

    So to solve this problem, either i need to depend on excel solver(which takes more time, especially when q increases i.e more than 4 minutes) or LM procedure(which i dont know, but i will learn for sure).

    Could you please help me preparing the data columns which you are solving by LM(Levenberg-Marquart procedure).

    I am not asking the algorithm to explain, just i need data columns for ARIMA(p,q,d) where p = 1,2… q = 1,2,… and d = 1(fixed)

    I will use another method to solve the system of linear equations to get the coefficents values.

    Waiting for your reply

    Thanks,
    Venkat

    Reply

      • Hi Charles,

        Thanks for the link. As this is a itertive procedure, i dont think coeffiecent values will get with in 1-minute. If you have any sample example then please forward to me. Although this may not be usefull for me.

        I expected that you will provide data columns for ARIMA(p,q,d) where p = 1,2… q = 1,2,… and d = 1(fixed).

        Please provide data columns or matrix in excel for ARIMA(p=2,q=2,d=1) for the coeffiecnts values. Later i can extend this when q increases upto 40.

        Waiting for your reply.

        Thanks,
        Venkat

        Reply

  8. Hi Charles,

    While referring to your example sheet “AR4”, here you have solved both with ARIMA and Reformat Linear regression Analysis.

    Could you please provide an another example with AR = 2, Diff = 1, MA = 2.

    Thanks,
    Venkat

    Reply

  9. Hi Charles,

    Sorry to hear that you had flu. Wish you a speedy recovery. Please check the issue when are healthy or message me so that i can explain it again.

    Sending you well wishes for your quick recovery and good health.

    Thanks,
    Venkat

    Reply

      • Hi Charles,

        After using that formula, results are not matching to tool vs regression in excel. I have sent a mail to you.

        Could you please solve and update your excel file with A(2,1,1) using Regression Analysis for reformatted one. I am not getting same results. I need your support in helping data columns X1 X2 X3 …. Y.

        In Arima(p,q,d=1), p>1(2,3,4,5), q > 1 (2,3,4,5,….50). I need to know how we can prepare data columns for regression.

        Just like you did for AR(3) Model under sheet “AR 4” from excel file “Real-Statistics-Time-Series-Examples”. Here SSE = 5.055049434 is same when solved by your tool and excel regression. Both are exactly matching.

        Kindly help on this, got stuck with this work.

        Thanks,
        Venkat

        Reply

        • Hi Charles,

          Could you please provide the data columns X1, X2,X3,X4,…Y when MA(=3,4,5) is included for Reformatted regression analysis.

          Eagerly waiting for your reply from past 1 week.

          Thanks,
          Venkat

          Reply

  11. Hi Charles,

    Query is regarding ARIMA(2,1,1) Model Coefficients for the above example

    As you mentioned that finding ARIMA Model Coefficients is same as that of Calculating ARMA Model Coefficients using Solver, except that we need to take differencing into account.

    ARIMA(2,1,1)

    1. Solving by Excel solver by minimising SSE, it took around 4 minutes to get the coefficent values for phi1,phi2,theta1. Initial values are set to zero. Results as

    Model statistics

    index phi theta
    2 0.548889167 0
    1 0.465595967 -1.105339913
    const 0 0

    SSE 123.2225837

    2. Solving by your Real Statistics Tool, it took less than 1 minute to get the coefficent values for phi1,phi2,theta1 as

    Model statistics

    index phi theta
    2 0.559115427 0
    1 0.456347574 -1.111806918
    const 0 0

    SSE 122.6052706

    3. Does the Real statistis tool is using excel solver to find the coefficient values? I think it is not using excel solver inside your tool to get the values as it taking longer time. Both SSE or not equal. SSE 123.2225837 SSE 122.6052706

    Please suggest which technique it is using to get the values with less time and improved results in accuracy?.

    Let me know if i did any mistake or forgot to use any values.

    Waiting for your reply.

    Thanks,
    Venkat

    Reply

    • Venkat,
      By default Real Statistics doesn’t use Solver (although if you check the Solver option then Solver will be used). I have not yet described the algorithm used, although the approach uses the Levenberg-Marquart procedure.
      Charles

      Reply

      • Hi Charles,

        Is it possible to describe Levenberg-Marquart procedure for computation of coefficient values? As it is used to solve non-linear least squares problems.

        Can you send some links which can be solved in excel with this procedure that was used in your tool.

        Thanks,
        Venkat

        Reply

          • Hi Charles,

            Thanks. I donot know the Levenberg-Marquart procedure for computation of coefficient values. I will study it later.

            It seems like the values are similar to that of “Reformat for Linear Regression”.

            Could you please explain how we can reformat data in excel when difference d = 1, with constant and without constant.

            For the same example, ARIMA(2,1,1). I tried as below, but unable to get the required results to that of same coefficent values.

            I hope, i am missing something here. Below is the excel difference values which i considered

            Please suggest on this

            Y delta yi delta yi-1 delta yi-2 y
            1 3.117675648
            2 5.808251086 2.690575438
            3 4.906768598 -0.901482487 -3.592057926
            4 6.119474482 1.212705884 2.114188371 5.706246297 2.01585183
            5 8.135326312 2.01585183 0.803145946 -1.311042426 1.209541115
            6 9.344867427 1.209541115 -0.806310715 -1.60945666 -0.197216743
            7 9.147650684 -0.197216743 -1.406757858 -0.600447143 -1.046157403
            8 8.101493281 -1.046157403 -0.84894066 0.557817198 2.044925748
            9 10.14641903 2.044925748 3.09108315 3.94002381 0.89877732
            10 11.04519635 0.89877732 -1.146148428 -4.237231578 -0.922618269
            11 10.12257808 -0.922618269 -1.821395589 -0.675247161 0.051024164
            12 10.17360224 0.051024164 0.973642433 2.795038022 0.647707556
            13 10.8213098 0.647707556 0.596683392 -0.376959041 -0.737246551
            14 10.08406325 -0.737246551 -1.384954108 -1.9816375 0.261990806
            .
            .
            .

            Thanks,
            Venkat

          • Venkat,
            It is difficult for me to answer these sort of questions without seeing the calculations. If you send me an Excel file with your data and calculations, I will try to answer your question.
            Charles

          • Hi Charles,

            I have sent excel file for ARIMA(2,1,1). Kindly look at it so that i can use data for ““Reformat for Linear Regression”.

            Thanks,
            Venkat

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