# Difference equations steady states !(Problem in the last step)

1. Apr 24, 2012

### sid9221

http://dl.dropbox.com/u/33103477/model.png [Broken]
Form of the equation

$$\gamma -\delta p_t = \frac{p_{t-1}-\alpha}{\beta}$$

or

$$p_t = \frac{-1}{\delta \beta}p_{t-1} + \frac{\alpha + \gamma \beta}{\delta \beta}$$

So the steady state is $$p^* = \frac{\alpha + \gamma \beta}{\delta \beta + 1}$$

For the steady state to be stable it should tend to zero hence

$$\alpha + \gamma \beta < \delta \beta + 1$$

or $$\alpha + \beta \gamma - \beta \delta < 1$$

Now the general solution from the def of a steady state can be given by:

$$P_n = (\frac{-1}{\delta \beta})^n [1-\frac{\alpha + \gamma \beta}{\delta \beta + 1}] + \frac{\alpha + \gamma \beta}{\delta \beta + 1}$$

Subbing in values:

$$P_n = (\frac{-5}{6})^n [\frac{-9}{11}] + \frac{20}{11}$$

So say I put $$P_n = (\frac{20}{11} - \frac{1}{100})$$ ie within a penny(approaching downwards)

than $$[\frac{-5}{6}]^n = \frac{11}{900}$$

Which can't be solved as log of negative is the "end of the universe". So where am I going wrong ?

The condition for the steady state to be stable might be wrong but that's a secondry issue, is there anyway I can get the minus out cause without it I get an reasonable answer of around 24 days..

Last edited by a moderator: May 5, 2017
2. Apr 24, 2012

### Ray Vickson

You might not be able to get to exactly one penny away from equilibrium in an integer number of periods. However, you do want to get *within* one penny of equilibrium. Your $P_n$ fluctuates above and below equilibrium, so you need
$$\left|P_n - \frac{11}{20} \right| \leq \frac{1}{100},$$ hence
$$\frac{9}{11} \left| \left(\frac{-5}{6} \right)^n \right| \leq \frac{1}{100},$$
or $$\left(\frac{5}{6} \right)^n \leq \frac{11}{900}.$$

RGV

Last edited by a moderator: May 5, 2017
3. Apr 24, 2012

### sid9221

Thank you, can you tell me if my condition for stability of the steady state is ok ?