Rank Conditions and Model Solution

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Rank Conditions and Model Solution

Postby houwang » Thu Sep 25, 2014 6:54 pm

Hi, I have a very simple deterministic toy model of debt and government investment. In the model there is a parameter that controls how fast the tax rate changes to cover the fiscal gap (in the code, the symbol is lambda). If tax adjusts too slowly, the government has to borrow externally and debt could be unsustainable in come cases.

If I set the tax adjustment speed to fast, the rank condition is satisfied and everything goes back to the steady state eventually. If I set the speed to be low, the rank condition is not satisfied and debt explodes. But I could still get the paths of all the variables.

My questions are: 1. Why do I still get the solution even if the rank condition is not satisfied, and does the solution make sense? 2. How to interpret the jumps in the last period? I read some of the old posts and learned that what matters are the terminal values for those forward-looking variables, not those backward-looking ones. In my case, since I don't have the 'endval' block I assume that the forward-looking variables 'c' and 'tau' take their initial steady-state values. But still, I have a little problem understanding intuitively why debt explodes and drops to its initial level in the last period.

Thanks a lot!!!
Holly

Code: Select all
var c, b, tau, d, I, gap;
varexo u;

parameters beta r y lambda tau0 rho;

r=0.05;
beta=1/(1+r);
y=1;
lambda=1;
tau0=0.25;
rho=0.95;

model;
c(+1)/c=(1+tau)/(1+tau(+1))*beta*(1+r);
(1+tau)*c+(1+r)*b(-1)=y+b;
c*tau+d=I+(1+r)*d(-1);
gap=I-c*tau0+r*d(-1);
tau=lambda*tau(-1)+(1-lambda)*(tau0+gap/c);
//I=I(-1)*(1+u);
I-0.188233=rho*(I(-1)-0.188233)+(1-rho)*u;

end;

initval;
c =  0.792933;
b = 0.176685;
tau = 0.25;
d = 0.2;
I = 0.188233;
gap=0;
end;


//steady;
//check;

shocks;
var u;
periods 1:5;
values 0.10;
end;

options_.maxit_=100;
simul(periods=100);
houwang
 
Posts: 14
Joined: Wed Jul 30, 2014 9:56 pm
Location: Washington, DC

Re: Rank Conditions and Model Solution

Postby jpfeifer » Fri Sep 26, 2014 8:52 am

First of all, the BK condition is a local condition only valid around the steady state. It only provides an indication of the existence of a unique stable solution for a globally solved model.

In your case, as the BK condition correctly indicates, there exists no unique bounded solution. But as you can see, there exists a unique unbounded solution that Dynare finds. That is: debt explodes at the rate of interest growth.

As you correctly said, the endval values are only relevant for the computations for forward-looking variables. Nevertheless, to have vectors of the same length, the last "unused" periods for backward-looking variables are set to some value, which you now plot. In your case, the endval for debt is set to the initial steady state. It is not (!) an endogenously computed final values for debt, but rather what you set initially.

Thus, you should ignore this value and only plot the periods 2 to 101 which are endogenously computed.
------------
Johannes Pfeifer
University of Cologne
https://sites.google.com/site/pfeiferecon/
jpfeifer
 
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Location: Cologne, Germany

Re: Rank Conditions and Model Solution

Postby houwang » Fri Sep 26, 2014 8:16 pm

Thank you for your explanations, they are really helpful.

Let me rephrase it so that you can see if I really understand it or not. :) So there is always a 'solution' to the deterministic model provided that we feed into some terminal conditions for the forward-looking variables ('c' and 'tau'). The reason is that basically we are solving a system of equations, and there should be some solution to it as long as the equations are correctly specified (meaning no multicollinearity and things like that...) The real difference is whether the solution is bounded or not. If the BK conditions are satisfied, and the shocks are not too big, we should be able to get a solution where every endogenous variable is bounded. (Whether they go back to the steady state or not depends on whether we set the steady state to be the terminal condition.) But in other cases (like in my toy model where tax adjusts slowly), we get an unbounded solution where one of the endogenous variables (debt) is not bounded.

If the above is correct, my follow-up question is, in what case will we get not-unique solution?? And is it possible that we get no solution for this type of models?

Thanks a lot!
Holly

jpfeifer wrote:First of all, the BK condition is a local condition only valid around the steady state. It only provides an indication of the existence of a unique stable solution for a globally solved model.

In your case, as the BK condition correctly indicates, there exists no unique bounded solution. But as you can see, there exists a unique unbounded solution that Dynare finds. That is: debt explodes at the rate of interest growth.

As you correctly said, the endval values are only relevant for the computations for forward-looking variables. Nevertheless, to have vectors of the same length, the last "unused" periods for backward-looking variables are set to some value, which you now plot. In your case, the endval for debt is set to the initial steady state. It is not (!) an endogenously computed final values for debt, but rather what you set initially.

Thus, you should ignore this value and only plot the periods 2 to 101 which are endogenously computed.
houwang
 
Posts: 14
Joined: Wed Jul 30, 2014 9:56 pm
Location: Washington, DC

Re: Rank Conditions and Model Solution

Postby jpfeifer » Sat Sep 27, 2014 7:32 am

Close, but not entirely correct. Part of the confusion comes from the term "solution". In mathematical terms, a "solution" simply satisfies a system of equations. In economic terms, a "solution" typically involves that variables stay bounded, i.e. satisfy an additional transversality condition.

You are trying to solve a big non-linear equation system. Of course, there can be cases where the system has no solution at all and when there are multiple solutions. But in contrast to linear systems, no easy conditions exist for these cases.

The BK condition is a condition for linear systems and thus only valid close to the approximation point as it is based on a first-order Taylor approximation to the nonlinear problem. If the BK condition is satisfied, you will get a unique bounded solution close to the approximation point (the transversality condition will be satisfied, but there are still a bunch of pathological cases that can happen). In this case, if you have a small perturbation to the system, it will converge back to steady state. If the shock is too large, the local BK condition might not be representative of the overall dynamics and no convergence back will happen.

Now, in case of your model, the BK condition is not satisfied because the local dynamics imply explosive behavior. There is still a unique "solution" for the path of all endogenous variables, but this path violates the transversality condition for debt, which is explosive.
------------
Johannes Pfeifer
University of Cologne
https://sites.google.com/site/pfeiferecon/
jpfeifer
 
Posts: 6940
Joined: Sun Feb 21, 2010 4:02 pm
Location: Cologne, Germany

Re: Rank Conditions and Model Solution

Postby houwang » Sat Sep 27, 2014 6:55 pm

Thanks!!! Your explanation is very clear. I greatly appreciate it. 8)
houwang
 
Posts: 14
Joined: Wed Jul 30, 2014 9:56 pm
Location: Washington, DC


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