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@sbxkoenk Thank You.

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@sbxkoenk Yes It is.

Golf

Hi everyone, I'm looking for help with the correct code to run the following model using "proc ARIMA": dy = a0 + a1*trend + a2*coint(-1) + b1*dy(-1) + b2*dy(-2) +....+b5*dy(-5) + c1*dx(-1) + c2*dx(-2) + .... +c5*dx(-5) +error In this model, dy is the first difference of y, dy(-1) represents the 1 period lag of dy, and dy(-5) represents the 5 period lag of dy. I generated the dy and dx variables from dif(y) and dif(x), respectively. Additionally, I'm wondering if the backward selection can be applied to the lag of dy(-1) through dy(-5), dx(-1) through dx(-5). Thank you for your help.

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@FreelanceReinh Thank You.

Golf

Hello everyone, I have been using proc reg with backward selection and noticed some discrepancies in my results. I ran the same independent variable in the last step of backward selection and obtained different output. Below, I have included the code and the results. Can someone please explain which output is correct? Thank you. Compared to the below procedure.

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Thank You. @Rick_SAS

Golf

Hello, I have a model; Y = b0+ b1X1+b2X2 + u I want to use proc IML to create a column vector: [beta] = [1 -b0 b1 -b2]'. [Dimension: 4x1] I also want to create a matrix of X [X ] = [Y ones X1 X2]. [Dimension: nx4] Then I calculate the vector of coint: coint = [X]*[beta]. [Dimension: nx1] I started with the following codes; proc reg data = a outest = coefficients; model y = x1 x2; run; How can I use Proc IML to obtain the desired results? Thank You

Golf · ‎10-25-2023

@Ksharp Thank You so much for your suggestion.

Golf · ‎10-25-2023

@Kurt_Bremser Thank you for your help. I will try your suggestion.

Golf · ‎10-25-2023

Here it is. data time_series; input year inflation; datalines; 1948 . 1949 -0.988480068 1950 1.053828352 1951 7.366171193 1952 2.227075801 1953 0.761468753 1954 0.351633428 1955 -0.262872574 1956 1.462744884 1957 3.277625189 1958 2.662459493 1959 0.909736704 1960 1.481626961 1961 1.058582189 1962 1.158617264 1963 1.236272118 1964 1.305567525 ; run; Thanks

Golf · ‎10-25-2023

@Kurt_Bremser Here is the codes: PROC IMPORT DATAFILE="/home/u45083887/my_content/SM518/SM518 Ch5/time_series_ex1.xlsx" OUT=Ch6DOLS.time_series DBMS=XLSX REPLACE; RUN; data ch6dols.time_series; set ch6dols.time_series; dinf = dif(inflation); proc expand data= ch6dols.time_series out=ch6dols.time_series; convert dinf = dinf_lead1 / transform=(lead 1); convert dinf = dinf_lead2 / transform=(lead 2); convert dinf = dinf_lead3 / transform=(lead 3); convert dinf = dinf_lead4 / transform=(lead 4); convert dinf = dinf_lead5 / transform=(lead 5); convert dinf; run; proc print data = ch6dols.time_series; var dinf dinf_lead1 dinf_lead2 dinf_lead3 dinf_lead4 dinf_lead5; run; Thank You.

Golf · ‎10-25-2023

@Kurt_Bremser Here is my desired data for lead variables. I use the simple codes as follows: proc expand data= ch6dols.time_series out=ch6dols.time_series; convert dinf = dinf_lead1 / transform=(lead 1); convert dinf = dinf_lead2 / transform=(lead 2); convert dinf = dinf_lead3 / transform=(lead 3); convert dinf = dinf_lead4 / transform=(lead 4); convert dinf = dinf_lead5 / transform=(lead 5); convert dinf; run; I would like to know how to generate this dataset in a loop. It would be helpful to generate longer leads and lags. Thank You.

Golf · ‎10-25-2023

Hi Kurt_Bremser, Thank you for helping me with this problem. Can you provide a code to generate the lead variable dinf(t+1), dinf(t+2), ..., dinf(t+5) using a loop? @Kurt_Bremser

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Different output in proc reg

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proc ARIMA

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Different output in proc reg

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Regression in proc IML

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