I’ve put the model by Liu and Pappa (JEDC, 2008) and solved it under Nash for both discretion and commitment (Ramsey). It’s meant to be a simple two-country model but as is typical in the literature, under Nash it resembles a closed economy.
Strangely, the loss is higher under commitment than under discretion.
I am at a loss in trying to figure out how I got this result but the model is extremely simple so it should not have occurred. I also get something similar under coordination.
Does anyone know what’s going on?PappaLiuNashComm.mod (1.1 KB)
What do I need to do to see the problem?
Thank you for your reply. If you run the mod file to obtain the loss (the uploaded file is under discretion) and then re-run it by replacing discretionary_policy with ramsey I get that under discretion the loss is 283.09046901 while under ramsey it’s 283.0906.
Having read the description of the two policies that use Dennis’ algorithm these are losses and the only difference between the two models is the policy, as I am changing it within the same file with identical parameter values. I’m certain there’s a simple explanation for this, but I just can’t see it.
This seems to be a problem with numerical accuracy. If I use
options_.discretionary_tol=1e-10;
discretionary_policy(instruments=(r),irf=20,planner_discount=beta);
i.e. decrease the tolerance in the algorithm, the losses are virtually identical.
Thank you very much for looking into this. I now see that the losses are the same up to the seventh decimal and that, as expected, Ramsey is superior to discretion.
I wasn’t aware that the tolerance level would be so important in this setting so I’ve certainly learnt something new!
Usually, it is not that important as welfare differences tend to be larger. But if you are comparing the third or fourth digit behind the decimal separator, it may matter.
Hi Mr. Pfeifer, can you tell me what does discretionary_tol=1e-10 term mean ?
The code uses a numerical optimizer. That option determines the convergence criterion for the optimizer.
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