Steady State Stability

patriks

Hey guys,

Just a quick question: With respect to steady state equilibria - in growth and developmental economics - is the stability of steady states determined and characterized by how and where the phase line intersects the 45° line? (i.e. If the phase line were to "cut" the 45° line from underneath, the steady state (equilibrium) would be unstable; whereas if the phase line were to "cut" the 45° line from above the 45° line, it would represent a stable steady state.)
If this is not the case, could anybody provide the basic concept of stability?

Thanks in advance!;)

Économiste Monétaire

If you have a difference equation of the form

[latex] \displaystyle y_{t+1} = f(y_{t})[/latex]

then setting

[latex] \displaystyle y_{t+1} = y_{t} = \bar{y}[/latex]

to find the steady state equilibria, you should find a stable steady state if the absolute value of the derivative ([latex] \displaystyle f'(\bar{y})[/latex]) is less than 1.

chave

patriks wrote: »

Hey guys,

Just a quick question: With respect to steady state equilibria - in growth and developmental economics - is the stability of steady states determined and characterized by how and where the phase line intersects the 45° line? (i.e. If the phase line were to "cut" the 45° line from underneath, the steady state (equilibrium) would be unstable; whereas if the phase line were to "cut" the 45° line from above the 45° line, it would represent a stable steady state.)
If this is not the case, could anybody provide the basic concept of stability?

Thanks in advance!;)

bascially what he said is the case. But using your diagram you can see that if your steady state is stable and you move from it in either direction and then use the function and 45 line you should end up converging from both sides where as the unstable SS usually diverges. But the derivative will tell you all.

patriks

Thanks a million, guys.
Utilizing the difference equations makes way more sense.

Économiste Monétaire

Just an example to help.

Take

[latex] \displaystyle y_{t+1} = y_{t}^{\alpha}[/latex]

Setting

[latex] \displaystyle y_{t+1} = y_{t} = \bar{y}[/latex]

we see that

[latex] \displaystyle \bar{y} = \bar{y}^{\alpha} \Rightarrow \bar{y}( \bar{y}^{\alpha-1} - 1) = 0[/latex]

0 and 1 are the steady state values. Setting [latex] \displaystyle \alpha = 0.5[/latex], you get this:



[latex] \displaystyle \frac{d y_{t+1}}{d y_{t}} = \alpha y_{t}^{\alpha-1} = 0.5 y_{t}^{-0.5}[/latex]

At our [latex] \displaystyle \bar{y} = 1[/latex], we get a slope of 0.5, which is less than 1. At [latex] \displaystyle \bar{y} = 0[/latex], we run into a problem of dividing by zero, which leaves us with an undefined number. So we say that our steady state value of 1 is locally stable. You can see this with a cobweb approach

patriks

Thanks a million. It's really appreciated.

I have one final question. It's with respect to the steady state again:
Supposing a model can be characterized by the following non-linear, first-order differential equation:
dx/dt = 0.2x2 – 0.51x

How could I determine - both graphically and analytically - the steady state value(s) of x?

I'm pretty sure in this instance that, with respect to stability, "delta k-overdot/delta k evaluated at k-overbar" has to be less than one to be locally stable. Do global transitional dynamics share the same criterion? ̇

Thanks!;)

Économiste Monétaire

The steady state values would be where

[latex] \displaystyle \frac{d x}{d t} = \dot{x} = 0[/latex],

so in the case of

[latex] \displaystyle 0.2x^{2} -0.51x = 0 \Rightarrow x(0.2x - 0.51) = 0[/latex]

the steady state values would be 2.55 and 0. The stability of a given steady state is determined by the sign of the first derivative at that value, so

[latex] \displaystyle \frac{d \dot{x}}{d x} = 0.4x - 0.51[/latex]

is positive for a steady state value of 2.55 and negative for 0. A steady state is stable for a negative derivative and unstable for a positive, so 2.55 is an unstable SS and 0 is stable.

Edit: Graphically it would look something like this:

patriks

Thanks again. That's great.
Yeah, I meant the stability condition is < 0 (negative) not < 1. Thanks for clearing that up though.
I know this is probably very basic, but how did you calculate the states 2.55 and 0? You see, I haven't had much opportunity to take quants classes, so I'm not very familiar with differential calculus; the vast majority of which is taken as read in the module I'm studying.

Économiste Monétaire

You're welcome. I calculated the steady state points here:

[latex] \displaystyle 0.2x^{2} -0.51x = 0 \Rightarrow x(0.2x - 0.51) = 0[/latex]

after I set dx/dt =0. From

[latex] \displaystyle x(0.2x - 0.51) = 0[/latex]

you just get the two roots of the equation, which are x = 0 and

[latex] \displaystyle 0.2x - 0.51 = 0 \Rightarrow x = 5 \times 0.51 = 2.55[/latex]