# Exchange Rate Dynamics

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

• Set $\dot p=0$ and $r=r^*$ and re-arrange $(1)$. Where are you struggling? Mar 30, 2016 at 5:34
• 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 $e=\bar{e}-\frac{1}{\lambda \theta}(p-\bar{p})$. Mar 30, 2016 at 9:50

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 $(2)$.