Difference equations steady states (Problem in the last step)

[sid9221]

http://dl.dropbox.com/u/33103477/model.png
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..

 

[Ray Vickson]

http://dl.dropbox.com/u/33103477/model.png
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..

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

 

[sid9221]

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