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Stuggling with this one: In a simple economic model, time t is set up as a discrete parameter (possibly referring to days or months), while demand $D_{t}$ and supply $S_t$ are described in terms of the price $p_{t-1}$ and $p_{t}$ as follows: since prices influence demand "instantaneously", one puts $D_{t}=\gamma-\delta p_{t}$, but one sets $S_{t}=(p_{t-1}-\alpha)/\beta$ because the production of new commodities consumes time. Assume that all parameters $\alpha, \beta, \gamma, \delta$ are positive, and set $p_{0}=1.$

1) Dervive and solve the difference equation describing economic equilibrium, i.e the situation when $D_{t}=S_{t}$.

$\Rightarrow$ $\gamma-\delta p_{t}= \frac {p_{t-1}-\alpha}{\beta}$ $\Rightarrow$ $\gamma(\beta)-\delta(\beta)\ p_{t}=p_{t-1}-\alpha$ $\Rightarrow$ $p_{t}=\frac {-p_{t-1}+\alpha+\gamma\beta}{-\delta\beta}$

Not sure if this is on the correct lines?

2)Compute the steady states $p^{*}$ and determine a condition on the parameters $\alpha, \beta, \gamma, \delta$ that ensures stabilty of $p^{*}$.

Steady states $p^{*}$:

$p^{*}=\frac {-p^{*}=\alpha+\gamma\beta}{-\delta\beta}$ $\Rightarrow$ $p^{*}-p^{*}(\delta\beta)=\alpha+\gamma\beta$ $\Rightarrow$ $p^{*}=\frac{\alpha+\gamma\beta}{1-\delta\beta}$

Then we want $\alpha, \beta, \gamma, \delta$ that ensures stabilty:

So we call $f(p)= \frac {-p+\alpha+\gamma\beta}{-\delta\beta}$ $\Rightarrow$ $f^\prime= \frac{1}{\delta\beta}$ as being stable $\frac{1}{\delta\beta}<1$ $\Rightarrow$ $1<\delta\beta$.

3) Assume t is measured in days and $p_{t}$ in Pounds sterling. Setting $\alpha=1, \beta=\frac{3}{2}, \gamma=2, \delta=\frac {4}{5}$, how long does it take until $p_{t}$ stays within a penny of $p^{*}$?

Not sure if I'm going down the right lines here, any help will be much appreciated, many thanks in advance.

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1 Answer 1

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In 1), you have derived the difference equation $$p_t={-p_{t-1}+\alpha+\gamma\beta\over\delta\beta}$$ but you haven't solved it (you also had a minus sign in there where it didn't belong). It is a first-order, linear, constant-coefficient difference equation. Have you learned how to solve that kind of equation?

You're going to need that for part 3), because you're going to need a formula for $p_t$ and then you'll want the value of $t$ that makes $p_t$ differ from $p^*$ by less than a penny.

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We have kearnt breifly for but in the cases I have we have used $p_{n+1}$ and $p_{n}$. Then say for example using our steady states to obtain the general $p_{n}=\alpha^{n}(p_{0}-p*)+p*$. so struggling how to relate this question to what I have seen, since it asks about the steady states after? –  user24930 Mar 1 '12 at 11:34
    
I don't know what "kearnt breifly for" means. If you are happier with $p_{n+1}$ and $p_n$, can you solve the equation $p_{n+1}=(-p_n+\alpha+\gamma\beta)/\delta\beta$? I don't know what you mean by "it asks about the steady states after." After what? –  Gerry Myerson Mar 1 '12 at 11:49
    
Sorry that was a typo, should of been learnt. So are we using $p_{n}=\alpha^{n}p_{0}+\frac {\alpha ^{n} -1}{\alpha-1} \beta$? –  user24930 Mar 1 '12 at 13:19
    
I don't know what "learnt breifly for" means. I suspect what you have seen is that the general solution to $p_{n+1}=\alpha p_n+\beta$ is $p_n=\alpha^np_0+{\alpha^n-1\over\alpha-1}\beta$. That's what you have to use, but remember that the $\alpha$ and $\beta$ in this last equation are not the $\alpha$ and $\beta$ in your question, a question which also involves $\gamma$ and $\delta$. –  Gerry Myerson Mar 1 '12 at 23:55
    
Think im there now, so in this case our $ \alpha= \frac {-1}{\delta \beta}$ and $\beta= \frac {\beta \gamma +\alpha}{\delta \beta}$. Then consequently from this we can just plug into the general form equation. Really appreciate your help. –  user24930 Mar 3 '12 at 14:27

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