Steady State with Endogenous Growth
Posted: Fri May 15, 2015 12:28 pm.
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Steady State with Endogenous Growth
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Posted: Fri May 15, 2015 12:28 pmRe: Steady State with Endogenous Growth
Posted: Tue May 19, 2015 6:33 pmUPDATE: I fixed Problem 1) and scaled leisure by the education level. Nevertheless the other problems remain. Any ideas are greatly appreciated.
Re: Steady State with Endogenous Growth
Posted: Sat May 23, 2015 1:33 pmI don't know your model, so it is hard to give specific answers. My general answers are:
2. In steady state, the SDF is almost always equal to beta (the discount factor in the non-stationary economy) times an adjustment for the growth rate and risk aversion. If that is not the case, I guess you made a mistake in your detrending or in the FOCs. Equation 15 looks a bit weird as it implies that the SDF is the inverse of Delta_e.
3. In general, there are as many equations as variables (things with a time index) in your model. In steady state, the time indices are dropped. Given the parameters, you still have as many variables in steady state as you have equations. Given the parameters, should thus be able to solve for the steady states. However, people have found it to often be convenient to instead of fixing a parameter and compute the "corresponding" variable, to fix a variable (e.g. labor in steady state) and then solve for the "corresponding" parameter. But the general lesson remains: you have as many unknowns as equations. If there is a unique mapping from A_e to the steady state of learning time, the latter is uniquely determined given the former. That is, if you fix the learning time h at a particular steady state value, there will be one value for A_e that is consistent with this value. But you can also fix A_e to a value consistent with empirically observed growth rates and then compute the h consistent with this. What you cannot do is fix both A_e and h simultaneously or try to compute them both endogenously.
4. As long as you are consistent, i.e. you do not use e.g. c to denote consumption in one equation and then c to denote log consumption in another, this is not a problem. Put differently, you need to be aware that if c is the log of consumption, you must use exp(c) if the model equation calls for the unlogged level of consumption.
I hope this helps.
2. In steady state, the SDF is almost always equal to beta (the discount factor in the non-stationary economy) times an adjustment for the growth rate and risk aversion. If that is not the case, I guess you made a mistake in your detrending or in the FOCs. Equation 15 looks a bit weird as it implies that the SDF is the inverse of Delta_e.
3. In general, there are as many equations as variables (things with a time index) in your model. In steady state, the time indices are dropped. Given the parameters, you still have as many variables in steady state as you have equations. Given the parameters, should thus be able to solve for the steady states. However, people have found it to often be convenient to instead of fixing a parameter and compute the "corresponding" variable, to fix a variable (e.g. labor in steady state) and then solve for the "corresponding" parameter. But the general lesson remains: you have as many unknowns as equations. If there is a unique mapping from A_e to the steady state of learning time, the latter is uniquely determined given the former. That is, if you fix the learning time h at a particular steady state value, there will be one value for A_e that is consistent with this value. But you can also fix A_e to a value consistent with empirically observed growth rates and then compute the h consistent with this. What you cannot do is fix both A_e and h simultaneously or try to compute them both endogenously.
4. As long as you are consistent, i.e. you do not use e.g. c to denote consumption in one equation and then c to denote log consumption in another, this is not a problem. Put differently, you need to be aware that if c is the log of consumption, you must use exp(c) if the model equation calls for the unlogged level of consumption.
I hope this helps.