Deriving approx. Uncovered Interest Parity

We derive approximated version of Uncoverd Interest Parity (UIP) from original UIP with/without using logarism and Taylor expansion.

In other words, we convert

$$1+i = (1+i^*)E^e/E$$

to

$$i \approx i^* + \frac{E^e-E}{E}$$

simple algebra

$$ 1 + i = (1 + i^*) \frac{E^e}{E}
$$

$$ i = (1 + i^*) \frac{E^e}{E} - 1 $$

$$ i = \frac{E^e}{E} + \frac{E^e}{E}i^* - 1 $$

$$ i = \frac{E^e - E}{E} + \frac{E^e}{E}i^* $$

$$ i = \frac{E^e - E}{E} + \frac{E^e}{E}i^* + \left( - \frac{E}{E}i^* + \frac{E}{E}i^* \right) $$

$$ i = \frac{E^e - E}{E} + \frac{E^e - E}{E}i^* + \frac{E}{E}i^* $$

$$ \therefore i \approx i^* + \frac{E^e - E}{E} $$

The term $\displaystyle \frac{E^e - E}{E}i^*$ is very small and we can ignore it up to first order.

Using Taylor approximation

Note that $\log(1+x) \approx x$ for small $x$. (See below)

$$ 1 + i = (1 + i^*) \frac{E^e}{E}
$$

$$ \log(1 + i) = \log(1 + i^*) + \log\left(\frac{E^e}{E}\right)
$$

$$ i \approx i^* + \log\left(\frac{E^e-E}{E} + \frac{E}{E}\right)
$$

$$ i \approx i^* + \log\left(1 + \frac{E^e-E}{E} \right)
$$

$$ i \approx i^* + \frac{E^e-E}{E} $$

Taylor expansion around $x=x^*$

$$ f(x) \approx f(x^*) + f'(x^*)(x-x^*) $$

Taylor expansion of $f(x)=\log(1+x)$ around $x^*=0$

$$ \log(1+x) \approx \log(1+x^*) + \frac{1}{1+x^*}(x-x^*) = x$$ $$ \therefore \log(1+x) \approx x$$

See Also International Finance: Theory and Policy.