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

  1. Apr 24, 2012 #1
    http://dl.dropbox.com/u/33103477/model.png [Broken]
    Form of the equation

    [tex] \gamma -\delta p_t = \frac{p_{t-1}-\alpha}{\beta} [/tex]

    or

    [tex] p_t = \frac{-1}{\delta \beta}p_{t-1} + \frac{\alpha + \gamma \beta}{\delta \beta} [/tex]

    So the steady state is [tex] p^* = \frac{\alpha + \gamma \beta}{\delta \beta + 1} [/tex]

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

    [tex] \alpha + \gamma \beta < \delta \beta + 1 [/tex]

    or [tex] \alpha + \beta \gamma - \beta \delta < 1 [/tex]

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

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

    Subbing in values:

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

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

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

    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. jcsd
  3. Apr 24, 2012 #2

    Ray Vickson

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    Homework Helper

    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 [itex] P_n [/itex] fluctuates above and below equilibrium, so you need
    [tex] \left|P_n - \frac{11}{20} \right| \leq \frac{1}{100},[/tex] hence
    [tex] \frac{9}{11} \left| \left(\frac{-5}{6} \right)^n \right| \leq \frac{1}{100}, [/tex]
    or [tex] \left(\frac{5}{6} \right)^n \leq \frac{11}{900}.[/tex]

    RGV
     
    Last edited by a moderator: May 5, 2017
  4. Apr 24, 2012 #3
    Thank you, can you tell me if my condition for stability of the steady state is ok ?
     
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