00:01
Okay, so we're going to examine the stability of steady states in a dynamic system.
00:05
So this involves examining how the system responds to small perturbations around these equilibria.
00:11
So for 1, we have g of x equals βx plus θ, where β is in 0, 1, and θ greater than 0.
00:24
So, in the previous post, we have found that the steady state was θ over 1 -β, and to check the stability, we consider the small perturbation, εt, such that xt equals θ 1 -β plus θt.
00:47
Now, substituting this into the equation g of x, we obtain g of, or let's say, xt plus 1 equals beta times theta over 1 minus beta plus yt plus theta.
01:12
This is equal to beta theta over 1 minus beta plus theta plus beta times epsilon t beta times epsilon t so the evolution of the perturbation is theta of t plus 1 equals beta times epsilon t and since beta is within 0 and 1 epsilon t will 10 to 0 as t goes to infinity.
02:10
So if you want to solve this recurrence, let theta, or excuse me, epsilon 0 as the starting point, then epsilon t equals beta times epsilon 0.
02:30
So again, you can see easily here that if beta is in , which indeed is in this case, then the epsilon t would be 10 to 0.
02:44
So this indicates that the equilibrium x equals theta over 1 minus beta is stable.
02:56
Now in the case g equals x.
03:03
So here here we have found that the equilibrium is any x.
03:08
And since we have g of xt equals xt, any small perturbation around the steady state x equals x star will leave xt equals x star unchanged...