Welfare analysis, logs-exp() and pruning

Stochastic simulations
1

Hi,

I have a model expressed in logs (using exp()) instead of levels and it is initially estimated in 1st order and it works properly.

The thing is that because i wan’t to make a welfare analysis, from what i have read, i need a 2nd order approximation to both the equilibrium condotion and the welfare criterion, so i try to proceed with 2nd order approximation.

Is it correct to apply it even in logs (leaving the exp()) in each variable or do i need to remove them and have the variables appearing in levels, recalculate the stady state and proceed with the 2nd order approximation?

Finally, in case i apply 2nd order approx. in log variables {exp()} and then because the IRFs are explosive, use Pruning would that be wrong for welfare analysis? My point is that whether Pruning is compatible with welfare analysis.

Kind regards

2

Doing the approximation in logs for welfare analysis is actually standard and not a problem. Pruning is a different issue. I am not aware of any analysis comparing welfare in a pruned and a non-pruned state space.There is indeed some tension when doing welfare analysis in a non-pruned state space and then turning on pruning for IRFs. But my reading of the literature is that this is quite standard.

3

Professor

Thank you very much for your help.

4

@jpfeifer Hello, I have a simple question, how to ensure that exp(welfare) is not negative in the steady state since log(ys) cannot be computed, for example in add. separable utility case

5

What do you mean? exp(welfare) can never be negative. The exponential function is never below 0.

6

Sorry Professor, here I meant that I want to use log linearization with exp(x) and log(y) at ss. I am putting in the mod file:
{exp(Wf)=\frac{((exp(C))^{1-\sigma})}{(1-\sigma)}-\frac{(exp(L)^{1+\phi})}{(1+\phi)}+\beta*exp(Wf(+1))}
and in the steady state file:
Wf=\frac{1}{(1-\beta)}*(\frac{C^{1-\sigma}}{(1-\sigma)}-\frac{L^{1+\phi}}{(1+\phi)})
where ys=\log(ys), and for \sigma>1 the steady state of {Wf} become negative. Now I realize that I may not be able to use log-linearization for welfare. Is it correct?

7
  1. Where is the problem with Wf being negative? Welfare is an ordinal concept: the larger the better. The absolute value has no meaning.
  2. Usually, you need to solve everything at second order, not just linearly.
8
  1. At the steady state, for {\sigma>1}, \frac{C^{1-\sigma}}{(1-\sigma)} is negative, so {Wf} becomes negative, log(Wf) at steady state is complex
  2. Yes, Professor, I am using second order.
9

Ok. In this case, don’t take the log of welfare. That won’t work for variables that can be negative.

10

I see. Thank you!