<p>Suppose you have a dynamic system 
$$ x_{t+1} = Ax_{t} $$
with a stationary point (or steady state as used in growth or RBC literature), say, $x^*$, i.e. $x^{*} = Ax^{*}$.</p>

<p>Now, consider the following question. Starting from an initial value $x_0$, how many paths are there leading to the stationary point $x^*$? If there is an unique path going from $x_{0}$ to $x^{*}$, then your model is well behaved in the sense that you can track the vector of variables, $x_t$, along the transition without worrying which path you are actually on. This is the saddle-path stability case that you are referring to. On the other hand, if the answer is not affirmative, it means that you have at least two routes from $x_0$ to $x^{*}$. There is another case: no matter where you start from you eventually end up at the stationary point $x^{*}$, in which case your model is said to be indeterminate. </p>

<p>So you can think of saddle-path stability as a desirable feature that you would like that your model to manifest in order to analyze the problem at hand. For example, standard RBC models all posses this property. </p>

<p>There are some mathematical conditions that ensure the saddle-path stability. </p>

<p>For the details and more, check out the section 7.8 (The q-Theory Investment and Saddle-Path Stability) in Introduction to Modern Economic Growth (Acemoglu, 2009).</p>

<p><a href="https://i.sstatic.net/d04hZ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/d04hZ.png" alt="enter image description here"></a></p>

<p>Ok, I'll try to explain intuitively for the Ramsey model (but it's not rigorous). Suppose that from the first order conditions, you can conclude that the equilibrium is unique and furthermore you can express consumption as a function of capital stock, that is $c_{t} = g(k_t)$, for some (smooth) function $g$. If that is the case, for a given initial capital stock $k_0$, you know how the economy is going to evolve. That is, if we denote the vector of variables by $x_t = (c_t, k_t)$, you know $\{ x_{t}\}_{t=0}^{\infty}$; because </p>

<p>$t=0: \quad$ $k_0$ is given, and $c_0 = g(k_0)$, </p>

<p>$t=1: \quad$ $k_1 = (1-\delta)k_0 + f(k_0) - c_0$, and $c_1 = g(k_1)$, 
(here $f$ is the production function and $\delta$ is depreciation)</p>

<p>$\vdots $</p>

<p>$t=\tau: \quad$ $k_{\tau} = (1-\delta)k_{\tau -1} + f(k_{\tau -1}) - c_{\tau -1}$, and $c_{\tau} = g(k_{\tau})$, </p>

<p>so on .... </p>

<p>Now, suppose the economy loses some of its capital stock for some reason while it was on $x^*=(k^*, c^*)$ and the new capital stock is, say, $k_{\tau}$. </p>

<p>Where will the economy be with this new capital stock? </p>

<p>Well, we have just claimed that there is a corresponding consumption $c_{\tau}=g(k_{\tau})$, and therefore the economy will jump to the point $x_{\tau}=(k_{\tau}, c_{\tau})$ on impact the shock and then move toward back to steady state $x^*$. You can think of $x_{\tau}$ as the red point in the figure. </p>

<p>Finally, a few remarks on the intuition behind the mathematical conditions ensuring saddle-path stability. </p>

<p>1) Note that, in the model we have actually two  variables $c_t$ and $k_t$, but we have assumed  that one of them can be  expressed in terms of the other, so the number of free variables is reduced to one, which is $k_t$ here. Basically, in order to be able to do this, you want to have a smooth curve mapping capital stock to consumption, i.e. $g: \mathcal{R} \to \mathcal{R}$ is a smooth curve (1-dimensional manifold). </p>

<p>2) Once the economy is at $x_{1}$ after the shock, you want it go back to $x^*$ and this is related to the number of eigenvalues that are less than one in absolute value.  </p>

<p>For precise statements (in a general environment) see Acemoglu as I mentioned before.  </p>

<p>Regarding rational expectations(RE): it is a solution concept and RE alone does not imply saddle-path stability. </p>

CC BY-SA 3.0
2017-02-20T18:48:21.567Z
Thank you for the answer. Am I right to understand that on the path you are mentioning, the economy is in equilibrium, although this equilibrium is not stationary (until its end point x* has been reached) ?
user11629
CC BY-SA 3.0
2017-02-21T03:39:34.923Z
Yes, you are right; the economy is in equilibrium throughout the transition. However, instead of saying "this equilibrium is not stationary", you can say "the economy (which is already in equilibrium) hasn't reached the stationary(steady state) point".
Piazzolla
CC BY-SA 3.0
2017-02-21T09:53:42.813Z
Thank you. What I still don't understand is this : imagine that your model has saddle path stability. This means that from each point $x$ there is one single path $p(x)$ leading to $x^*$. Now imagine that you are in $x_0$ but deviate on a point $x_1$ that is not on path $p(x_0)$. Now are not there two paths to go to $x^*$ : either follow $p(x_1)$ or go back on $p(x_0)$ ? Is it impossible to go back ?
user11629
CC BY-SA 3.0
2017-02-21T18:57:01.440Z
I have included a figure (corresponding to a standard Ramsey problem) to clarify. Suppose the economy is hit by an exogenous shock (as in RBC models) and no longer is at $x^*$. The point of saddle-path stability is that, _*right after the shock the economy falls somewhere on the saddle-path*_  (point $x_1$ in the figure) and then will move back towards $x^*$. In your notation, it will still be on $p(x_0)$ (assuming that the shock doesn't change the steady state point).
Piazzolla
CC BY-SA 3.0
2017-02-21T22:14:29.807Z
"_right after the shock the economy falls somewhere on the saddle-path_" Is this because of rational expectations and some transversality condition ? Tobin seems to say that this does not need to happen. (Sorry for asking so many questions but I still don't understand.)
user11629
CC BY-SA 3.0
2017-02-22T02:51:10.593Z
Please check out the updated answer above.
Piazzolla
CC BY-SA 3.0
2017-02-19T02:40:21.607Z
2017-02-22T02:28:58.600Z
2017-02-22T02:28:58.600Z
Piazzolla
<p>In fact, saddle path equilibrium is one of the most common equilibrium in canonical growth models, mainly in dynamic optimization problems. It is possible that he makes a reference to a decentralized equilibrium without cycles around steady state. </p>

<p>Otherwise, if there are cycles around a steady-state equilibrium. It means that there are oscillations (see indeterminacy and bifurcation issues for technical details), which means that your equilibrium is not a saddle path. (I skip the mathematical details for those technical terms, which is beyond the scoop of your question.)</p>

<p>In which situation, you don't have a saddle path equilibrium and some cycles around steady state ? </p>

<p>A decentralized market economy with externalities which are not taken into account by agents in the market can show indeterminate equilibrium. I think he makes reference to this kind of economy.</p>

<p>A very good and illustrative paper to respond to your question is <a href="http://128.97.206.178/NewWeb/PdfFiles/indeterminacy%20and%20ir.pdf" rel="nofollow noreferrer">Benhabib and Farmer (1994)</a></p>

<p>I quote one of their sentences from introduction. Then, you can look at references cited ;</p>

<p>"...Previous work on indeterminate equilibria has shown that stable steady states are associated with rational expectations models in which there exists a continuum of self-fulfilling belief-driven equilibria each of which is stationnary..."</p>

<p>Hope it answers to your question.</p>

CC BY-SA 3.0
2017-01-07T00:49:09.957Z
Thank you for the answer. But do you think he is referring to a growth model, or a model describing a whole economy ? The text suggests more a model dedicated to a single (financial) market. And according to the quote (which I expanded in my question above), he does not seem to be referring to multiple equilibria neither, but rather to out-of-equilibrium dynamics, don't you think ?
user11629
CC BY-SA 3.0
2017-01-07T13:20:54.623Z
He probably refers to a growth model. Because at these years, concepts like saddle path equilibrium are primarily treated in growth models to my knowledge. I guess that he refers to cycles. (note that I did not read entirely the paper.) because I think he focuses on uncommon situations on the market that are not described through mathematical models, where the transitional dynamics represents saddle path equilibrium.
optimal control
CC BY-SA 3.0
2017-01-07T16:15:48.357Z
So you think he is referring to some general equilibrium result ?
Transitional dynamics is disequilibrium dynamics ?
When you talked about "indeterminate equilibrium" you meant multiple possible equilibria ?
user11629
CC BY-SA 3.0
2017-01-07T20:28:46.990Z
Transitional dynamics show the path that makes you reach to a stable equilbrium (if the stable equilibrium exists.) Multiple equilibria is a different concept. You can have a multiple equilibria but your equilibrium points can have saddle path equilibrium. An easy example to read could be https://gargantua.polytechnique.fr/siatel-web/linkto/mICYYYWjYS
optimal control
CC BY-SA 3.0
2017-01-09T23:52:48.580Z
So there is no saddle path in partial equilibrium ?
user11629
CC BY-SA 3.0
2017-01-06T19:10:24.487Z
2017-01-06T19:10:24.487Z
optimal control
<p>In his <a href="http://college.holycross.edu/eej/Volume4/V4N3_4P153_159.pdf" rel="nofollow noreferrer">1978 paper</a> introducing the Tobin tax Tobin states that :</p>

<blockquote>
  <p>As a technical matter, we know that a rational expectations
  equilibrium on markets of this kind is a saddle point. That is, there
  is only a singular path that leads from disequilibrium to equilibrium.
  If the markets are not on that path, or if they don't jump to it from
  wherever they are, they can follow any of a number of paths that lead
  away from equilibrium - paths along which, nonetheless, expectations
  are on the average fulfilled.</p>
</blockquote>

<p>He seems to consider this result as common knowledge among the economists and gives no reference. To which result or paper is he referring ? Do you have an idea of what he means with "markets of this kind" ?</p>

<reference-request><financial-markets><dynamic-optimization><rational-expectations>
Saddle path equilibrium on financial market with rational expectations
user11629
# Saddle path equilibrium on financial market with rational expectations

(Asked by: username_0 on 2016-12-31)

In his [1978 paper](http://college.holycross.edu/eej/Volume4/V4N3_4P153_159.pdf) introducing the Tobin tax Tobin states that :

> As a technical matter, we know that a rational expectations
> equilibrium on markets of this kind is a saddle point. That is, there
> is only a singular path that leads from disequilibrium to equilibrium.
> If the markets are not on that path, or if they don't jump to it from
> wherever they are, they can follow any of a number of paths that lead
> away from equilibrium - paths along which, nonetheless, expectations
> are on the average fulfilled.

He seems to consider this result as common knowledge among the economists and gives no reference. To which result or paper is he referring ? Do you have an idea of what he means with "markets of this kind" ?

**Tags:** reference-request, financial-markets, dynamic-optimization, rational-expectations

---

## 2 answers

### Answer by username_1 on 2017-02-19. Score: 2 | Accepted answer

Suppose you have a dynamic system
$$ x\_{t+1} = Ax\_{t} $$
with a stationary point (or steady state as used in growth or RBC literature), say, $x^\*$, i.e. $x^{\*} = Ax^{\*}$.

Now, consider the following question. Starting from an initial value $x\_0$, how many paths are there leading to the stationary point $x^\*$? If there is an unique path going from $x\_{0}$ to $x^{\*}$, then your model is well behaved in the sense that you can track the vector of variables, $x\_t$, along the transition without worrying which path you are actually on. This is the saddle-path stability case that you are referring to. On the other hand, if the answer is not affirmative, it means that you have at least two routes from $x\_0$ to $x^{\*}$. There is another case: no matter where you start from you eventually end up at the stationary point $x^{\*}$, in which case your model is said to be indeterminate.

So you can think of saddle-path stability as a desirable feature that you would like that your model to manifest in order to analyze the problem at hand. For example, standard RBC models all posses this property.

There are some mathematical conditions that ensure the saddle-path stability.

For the details and more, check out the section 7.8 (The q-Theory Investment and Saddle-Path Stability) in Introduction to Modern Economic Growth (Acemoglu, 2009).

[![enter image description here](https://i.sstatic.net/d04hZ.png)](https://i.sstatic.net/d04hZ.png)

Ok, I'll try to explain intuitively for the Ramsey model (but it's not rigorous). Suppose that from the first order conditions, you can conclude that the equilibrium is unique and furthermore you can express consumption as a function of capital stock, that is $c\_{t} = g(k\_t)$, for some (smooth) function $g$. If that is the case, for a given initial capital stock $k\_0$, you know how the economy is going to evolve. That is, if we denote the vector of variables by $x\_t = (c\_t, k\_t)$, you know $\{ x\_{t}\}\_{t=0}^{\infty}$; because

$t=0: \quad$ $k\_0$ is given, and $c\_0 = g(k\_0)$,

$t=1: \quad$ $k\_1 = (1-\delta)k\_0 + f(k\_0) - c\_0$, and $c\_1 = g(k\_1)$,
(here $f$ is the production function and $\delta$ is depreciation)

$\vdots $

$t=\tau: \quad$ $k\_{\tau} = (1-\delta)k\_{\tau -1} + f(k\_{\tau -1}) - c\_{\tau -1}$, and $c\_{\tau} = g(k\_{\tau})$,

so on ....

Now, suppose the economy loses some of its capital stock for some reason while it was on $x^\*=(k^\*, c^\*)$ and the new capital stock is, say, $k\_{\tau}$.

Where will the economy be with this new capital stock?

Well, we have just claimed that there is a corresponding consumption $c\_{\tau}=g(k\_{\tau})$, and therefore the economy will jump to the point $x\_{\tau}=(k\_{\tau}, c\_{\tau})$ on impact the shock and then move toward back to steady state $x^\*$. You can think of $x\_{\tau}$ as the red point in the figure.

Finally, a few remarks on the intuition behind the mathematical conditions ensuring saddle-path stability.

1) Note that, in the model we have actually two variables $c\_t$ and $k\_t$, but we have assumed that one of them can be expressed in terms of the other, so the number of free variables is reduced to one, which is $k\_t$ here. Basically, in order to be able to do this, you want to have a smooth curve mapping capital stock to consumption, i.e. $g: \mathcal{R} \to \mathcal{R}$ is a smooth curve (1-dimensional manifold).

2) Once the economy is at $x\_{1}$ after the shock, you want it go back to $x^\*$ and this is related to the number of eigenvalues that are less than one in absolute value.

For precise statements (in a general environment) see Acemoglu as I mentioned before.

Regarding rational expectations(RE): it is a solution concept and RE alone does not imply saddle-path stability.

> **Comments:**
> **username_0**: Thank you for the answer. Am I right to understand that on the path you are mentioning, the economy is in equilibrium, although this equilibrium is not stationary (until its end point x* has been reached) ?  
> **username_1**: Yes, you are right; the economy is in equilibrium throughout the transition. However, instead of saying "this equilibrium is not stationary", you can say "the economy (which is already in equilibrium) hasn't reached the stationary(steady state) point".  
> **username_0**: Thank you. What I still don't understand is this : imagine that your model has saddle path stability. This means that from each point $x$ there is one single path $p(x)$ leading to $x^*$. Now imagine that you are in $x_0$ but deviate on a point $x_1$ that is not on path $p(x_0)$. Now are not there two paths to go to $x^*$ : either follow $p(x_1)$ or go back on $p(x_0)$ ? Is it impossible to go back ?  
> **username_1**: I have included a figure (corresponding to a standard Ramsey problem) to clarify. Suppose the economy is hit by an exogenous shock (as in RBC models) and no longer is at $x^*$. The point of saddle-path stability is that, _*right after the shock the economy falls somewhere on the saddle-path*_  (point $x_1$ in the figure) and then will move back towards $x^*$. In your notation, it will still be on $p(x_0)$ (assuming that the shock doesn't change the steady state point).  
> **username_0**: "_right after the shock the economy falls somewhere on the saddle-path_" Is this because of rational expectations and some transversality condition ? Tobin seems to say that this does not need to happen. (Sorry for asking so many questions but I still don't understand.)  
> **username_1**: Please check out the updated answer above.  

---

### Answer by username_2 on 2017-01-06. Score: 1

In fact, saddle path equilibrium is one of the most common equilibrium in canonical growth models, mainly in dynamic optimization problems. It is possible that he makes a reference to a decentralized equilibrium without cycles around steady state.

Otherwise, if there are cycles around a steady-state equilibrium. It means that there are oscillations (see indeterminacy and bifurcation issues for technical details), which means that your equilibrium is not a saddle path. (I skip the mathematical details for those technical terms, which is beyond the scoop of your question.)

In which situation, you don't have a saddle path equilibrium and some cycles around steady state ?

A decentralized market economy with externalities which are not taken into account by agents in the market can show indeterminate equilibrium. I think he makes reference to this kind of economy.

A very good and illustrative paper to respond to your question is [Benhabib and Farmer (1994)](http://128.97.206.178/NewWeb/PdfFiles/indeterminacy%20and%20ir.pdf)

I quote one of their sentences from introduction. Then, you can look at references cited ;

"...Previous work on indeterminate equilibria has shown that stable steady states are associated with rational expectations models in which there exists a continuum of self-fulfilling belief-driven equilibria each of which is stationnary..."

Hope it answers to your question.

> **Comments:**
> **username_0**: Thank you for the answer. But do you think he is referring to a growth model, or a model describing a whole economy ? The text suggests more a model dedicated to a single (financial) market. And according to the quote (which I expanded in my question above), he does not seem to be referring to multiple equilibria neither, but rather to out-of-equilibrium dynamics, don't you think ?  
> **username_2**: He probably refers to a growth model. Because at these years, concepts like saddle path equilibrium are primarily treated in growth models to my knowledge. I guess that he refers to cycles. (note that I did not read entirely the paper.) because I think he focuses on uncommon situations on the market that are not described through mathematical models, where the transitional dynamics represents saddle path equilibrium.  
> **username_0**: So you think he is referring to some general equilibrium result ? Transitional dynamics is disequilibrium dynamics ? When you talked about "indeterminate equilibrium" you meant multiple possible equilibria ?  
> **username_2**: Transitional dynamics show the path that makes you reach to a stable equilbrium (if the stable equilibrium exists.) Multiple equilibria is a different concept. You can have a multiple equilibria but your equilibrium points can have saddle path equilibrium. An easy example to read could be https://gargantua.polytechnique.fr/siatel-web/linkto/mICYYYWjYS  
> **username_0**: So there is no saddle path in partial equilibrium ?  

---

