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Due to my project in mathematics I am trying to understand the dynamics of exchange rate.

Consider the following:

$$\dot{p}= \pi \ln (D/Y)= \pi[u+\delta(e-p)+(\gamma -1) y-\sigma r] \ \ \ \ \ \ \ \ \ \ (1)$$

where $e=\bar{e}-\frac{1}{\lambda \theta}(p-\bar{p})$

This describes the rate of increase in the price of domestic goods.

Then we have

$$\bar{e}=\bar{p} + \frac{1}{\delta}[\sigma r^*+(1-\gamma)y-u] \ \ \ \ \ \ \ \ \ \ (2)$$

This is the long-run equilibrium exchange rate implied by $(1)$. Then the paper says that equation $(1)$ is obtained by setting $\dot{p}=0$ and $r=r^*$.

But I want to show step by step how $(1)$ implies $(2)$. Can anyone help or give a hint.

BTW

$ r =$ Domestic Interest Rate

$ r^{*} =$ Foreign Interest Rate

$ e =$ Current Exchange Rate

$ \bar{e} =$ Long-Run Exchange Rate

$ p =$ Current Price Level

$ \bar{p} =$ Long-Run Price Level

$ \delta =$ Constant

$ \sigma =$ Constant

$ \gamma =$ Constant

$ y =$ Logarithm of Real Income

$ u =$ Constant

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  • $\begingroup$ Set $\dot p=0$ and $r=r^*$ and re-arrange $(1)$. Where are you struggling? $\endgroup$
    – David
    Commented Mar 30, 2016 at 5:34
  • $\begingroup$ If I do so, then I get $e=p+\frac{1}{\delta}[\sigma r^*+(1-\gamma)y-u]$ but $e$ and $p$ have to be long-run. Hereafter I realized that I have to substitute $e$ first that is $e=\bar{e}-\frac{1}{\lambda \theta}(p-\bar{p})$. $\endgroup$
    – Alim Teacher
    Commented Mar 30, 2016 at 9:50

1 Answer 1

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In the long run we have $p=\bar{p}$, i.e the current price level in the long run is exactly the long run price. Therefore $e=\bar{e}$. As in the long run we have $\dot{p}=0$ we may write $$u+\delta(\bar{e}-\bar{p})+(\gamma -1) y-\sigma r^*=0$$ which upon rearragning gives equation $(2)$.

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