<p>The condition that unstable eigenvalues equal in number the control/decision/non-predetermined variables is equivalent to the requirement that a model possess the "saddle point property".  </p>

<p>Of more interest is <em>why economists generally want their models to possess such a property</em>, something that holds for the DSGE models.</p>

<p>In Economics we call models with the saddle point property as "saddle path <em>stable</em>"; a mathematician would disagree. From a mathematical point of view these models are unstable, because there is a <em>unique path</em> to the fixed point/equilibrium. It should then be evident why such models are characterized as unstable: the slightest perturbation will push the dynamics off the unique path to equilibrium, never to return... enter human will and purposeful action by economic agents: we <em>keep</em> the model on the unique path on purpose, in order to fulfill the optimization conditions that describe our will and desires. In other words, going to the fixed point is the optimal thing to do from a utility-maximization point of view. And if a shock happens we forcefully "jump" back on the saddle path (that may have changed or stayed the same due to the shock), correcting the course of the economic system by will and decision.</p>

<p>Consider now a model that is mathematically stable proper. It means that it does not matter where we start, and it does not matter what we do, we will end up in the long-term equilibrium. Hardly an appropriate model to reflect the observed human affairs, or decision-making. </p>

<p><em>And decision-making is structurally an unstable phenomenon, mathematically speaking</em>. This may help glean a bit more intuition: to possess "as much decision-making as needed" in order to have a unique way to go to the fixed point, we need the model to allow us to actually decide optimally on the number of variables the model premises say that we can decide upon, no less and no more: so, as many <em>unstable</em> roots as "unstable"(decision) variables.  </p>

<p>If the unstable roots are <em>less</em> than the unstable/decision variables, the model does not give us the "degrees of freedom" we need - it is "too stable", hence the many solutions: it is mathematically stable. Boring.</p>

<p>If the unstable roots are <em>more</em> than the "unstable" variables, we cannot control the model through our decision making, and it will explode/implode.</p>

CC BY-SA 3.0
2017-12-21T22:58:59.427Z
2017-12-21T22:58:59.427Z
Alecos Papadopoulos
<p>In order for a DSGE model to have a unique solution, it is required to satisfy the Blanchard Kahn conditions. However, these conditions seem very abstract to me. </p>

<p><strong>Is there an intuition behind the blanchard kahn conditions? Why does a DSGE model have a unique solution iff the number of unstable eigenvalues at the linearized steady state is equal to the number of control variables?</strong></p>

<mathematical-economics><dsge>
intuition behind the blanchard kahn conditions?
user56834
# intuition behind the blanchard kahn conditions?

(Asked by: username_0 on 2017-12-21)

In order for a DSGE model to have a unique solution, it is required to satisfy the Blanchard Kahn conditions. However, these conditions seem very abstract to me.

**Is there an intuition behind the blanchard kahn conditions? Why does a DSGE model have a unique solution iff the number of unstable eigenvalues at the linearized steady state is equal to the number of control variables?**

**Tags:** mathematical-economics, dsge

---

## 1 answers

### Answer by username_1 on 2017-12-21. Score: 3 | Accepted answer

The condition that unstable eigenvalues equal in number the control/decision/non-predetermined variables is equivalent to the requirement that a model possess the "saddle point property".

Of more interest is *why economists generally want their models to possess such a property*, something that holds for the DSGE models.

In Economics we call models with the saddle point property as "saddle path *stable*"; a mathematician would disagree. From a mathematical point of view these models are unstable, because there is a *unique path* to the fixed point/equilibrium. It should then be evident why such models are characterized as unstable: the slightest perturbation will push the dynamics off the unique path to equilibrium, never to return... enter human will and purposeful action by economic agents: we *keep* the model on the unique path on purpose, in order to fulfill the optimization conditions that describe our will and desires. In other words, going to the fixed point is the optimal thing to do from a utility-maximization point of view. And if a shock happens we forcefully "jump" back on the saddle path (that may have changed or stayed the same due to the shock), correcting the course of the economic system by will and decision.

Consider now a model that is mathematically stable proper. It means that it does not matter where we start, and it does not matter what we do, we will end up in the long-term equilibrium. Hardly an appropriate model to reflect the observed human affairs, or decision-making.

*And decision-making is structurally an unstable phenomenon, mathematically speaking*. This may help glean a bit more intuition: to possess "as much decision-making as needed" in order to have a unique way to go to the fixed point, we need the model to allow us to actually decide optimally on the number of variables the model premises say that we can decide upon, no less and no more: so, as many *unstable* roots as "unstable"(decision) variables.

If the unstable roots are *less* than the unstable/decision variables, the model does not give us the "degrees of freedom" we need - it is "too stable", hence the many solutions: it is mathematically stable. Boring.

If the unstable roots are *more* than the "unstable" variables, we cannot control the model through our decision making, and it will explode/implode.


---

