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I have a basic question about comparative statics. In Wikipedia, it is mentioned that:

As a study of statics it compares two different equilibrium states...

What exactly is an equilibrium? I have a fixed point in my model. This does not arise from a strategic utility maximisation. I start from an initial point, iterate repeatedly in a bounded space, and I can prove I converge to a given point. Now I would like to see how sensitive my final fixed point is to my initial starting point, which was exogenous.

Does this classify under comparative statics? Or is it just a sensitivity analysis? I am confused because the usual machinery of comparative statics, that is, the implicit function theorem, does not hold in my setting.

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    $\begingroup$ Others might disagree but to me, comparative statics is the right expression for your purpose, as for any exercise that consists in studying how the result of the model (in your case, the fixed point) varies with one of the parameters (in your case, the initial condition). $\endgroup$ – Oliv Feb 18 '16 at 18:24
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"Comparative statics" may refer to two closely related but not identical exercises:

a) Regarding a specific equilibrium point, we examine how the properties, associated values etc of it change as we vary the value of an exogenous parameter.

b) But also, we can put under "comparative statics" the case where the model has multiple equilibria. Inherently, when more than one equilibria exist, it means that they are dependent on the value of (or inter-relation between) some of the model parameters. Assessing this situation, i.e. contrasting the two solutions, the sets of parameter values for which we obtain the one equilibrium or the other ("values" do not necessarily mean numerical ones -they may be bounds expressed in terms of other parameters etc), understanding the economic essence behind the differences, is a comparison of two static situations.

So @Oliv comment below the question is valid, indeed, the initial condition structurally is just one more exogenous parameter of the model.

That said, in most cases, dependence (or not) on initial conditions (or "history-dependence" more generally) usually falls under stability analysis:

  • if a fixed point is globally asymptotically stable, it means that there is no dependence on the initial value, whatever it is (in the feasible space).

  • if a fixed point is asymptotically stable, it means that it has a basin of attraction, but which may be only a subset of the feasible space.

Proving global asymptotic stability is usually difficult (in most cases, one has to find an associated Liapunov function), and it becomes harder the more non-linear is the differential/difference equation/system under study. So computationally showing that it is indeed globally asymptotically stable, is useful.

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  • $\begingroup$ @Alecos_Papadopoulos That's a nice answer! May I ask you to elaborate on point a) or provide an example ? I always understood comparative statics as the analysis of the link between parameters and predictions (observables), whereas you seem to say that comparing predictions between themselves (in case the prediction is ambiguous) also falls into this category. $\endgroup$ – Oliv Feb 19 '16 at 6:49
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    $\begingroup$ @Oliv I had in mind models with multiple equilibria. Inherently, when more than one equilibria exist, it means that they are dependent on the value of (or inter-relation between) some of the model parameters. So assessing this situation, contrasting the two solutions, the sets of parameter values for which we obtain the one equilibrium or the other ("values" do not necessarily mean numerical ones -they may be bounds expressed in terms of other parameters etc), understanding the economic essence behind the differences, naturally falls in "comparative statics". $\endgroup$ – Alecos Papadopoulos Feb 19 '16 at 12:53
  • $\begingroup$ @Alecos_Papadopoulos Thanks! That makes perfect sense. I would suggest that you edit your answer to explain your point exactly as you did in the comment. $\endgroup$ – Oliv Feb 19 '16 at 13:58
  • $\begingroup$ @Oliv Good suggestion. Implemented. $\endgroup$ – Alecos Papadopoulos Feb 19 '16 at 14:58

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