2
$\begingroup$

I have calculated an ARDL(24,36) model with 1 independent variable. The data is monthly, hence the inclusion of so many lags.

I am trying to calculate the interim multiplier (the cumulative effect at a given point in time) and the total long-run multiplier.

I have seen the latter formula given as (B0+B1)/(1-A1)

Where B0 is the coefficient on the independent variable at time t, B1 is the coefficient on the independent variable at t-1 and A1 is the coefficient on the dependent variable at t-1. However, this formula seems to be limited to ARDL(1,1) models.

In models with more than 1 lag, what is the interim multiplier at lag k and what is the total long-run multiplier?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

For an ARDL(n,m) model using your notation the multiplier is given by:

$\frac{\sum_{i=0}^mb_i}{1-\sum_{j=1}^na_j}$

Improve this answer
$\endgroup$
6
  • $\begingroup$ Hi Ryan, thanks a lot. Does this hold if there is more than 1 independent variable as well? $\endgroup$
    – ts_highbury
    Commented Sep 16, 2016 at 10:09
  • 1
    $\begingroup$ Yes, as long as two things hold. 1, You analyze each variable separately, i.e. the b coefficients are on lags of the same variable. And 2, there are no interaction terms. $\endgroup$
    – Ryan
    Commented Sep 17, 2016 at 13:36
  • $\begingroup$ Hi, any idea how to do the above long-run multiplier calculation (with also the standard error obtained) in Stata? $\endgroup$
    – Jonas
    Commented Oct 1, 2019 at 15:21
  • $\begingroup$ I'm not aware of any way to do what you are asking in STATA. It has been over 5 years since I worked with STATA daily though. My suggestion for an unknown way to calculate the standard error for that unknown value would be to just bootstrap the standard error by sampling randomly from your data with replacement, run the regression on the sampled data, doing the calculation that I describe on using the method I describe above, record the outcome, and then rinse and repeat like 1000X. Take the 97.5% percentile and the 2.5% percentile of the results to get your confidence interval. $\endgroup$
    – Ryan
    Commented Oct 3, 2019 at 23:49
  • $\begingroup$ Or take the standard deviation of your results and claim that as your bootstrapped standard error. $\endgroup$
    – Ryan
    Commented Oct 3, 2019 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.