<p>From the basic equation you get
\begin{eqnarray*}
\frac{e^E_{t+1}}{e_{t}} &amp; = &amp; \frac{1+i_t}{1+i^*_t} \\
\\
(1+i^*_t) \cdot e^E_{t+1} &amp; = &amp; (1+i_t) \cdot e_t \\
\\
e^E_{t+1} - e_t &amp; = &amp; i_t \cdot e_t - i^*_t \cdot e^E_{t+1}.
\end{eqnarray*}
Now comes the approximation. You divide either by $e_t$ or by $e_{t+1}^E$.</p>

<p>First method
$$
\frac{e^E_{t+1} - e_t}{e_t} = i_t - i^*_t \cdot \frac{e^E_{t+1}}{e_t} \approx
i_t - i^*_t.
$$
Second method
$$
\frac{e^E_{t+1} - e_t}{e^E_{t+1}} = i_t \cdot \frac{e_t}{e^E_{t+1}} - i^*_t  \approx
i_t - i^*_t.
$$
I would say there is no crucial difference, both of these just assume that the proportional change in the exchange rate multiplied by the interest rate is an order of magnitude lower than either the interest rate or the change in the exchange rate alone.</p>

CC BY-SA 3.0
2015-12-20T07:53:34.423Z
2015-12-22T09:54:39.300Z
2015-12-22T09:54:39.300Z
Giskard
<p>Until otherwise $e=\frac{\text{Domestic currency}}{\text{Foreign currency}}$.</p>

<p>The UIP condition can be written in the following form: Investing now in a domestic bond, must give the same return as investing in a foreign bond, which means, $1+i_t=e^{-1}_t\left(1+i^*_t\right)e^E_{t+1}$, implying</p>

<p>$$\frac{1+i_t}{1+i^*_t}=\frac{e^E_{t+1}}{e_{t}}=\frac{1}{1+\frac{e_{t}-e^E_{t+1}}{e^E_{t+1}}}$$ where $i_t$ is the domestic nominal interest rate at time t, $i^*_t$ is the foreign nominal interest rate at time t, $e^E_{t+1}$ is the expected exchange rate at time $t+1$.</p>

<p>Using the derivations presented <a href="https://en.wikipedia.org/wiki/Fisher_equation#Derivation" rel="nofollow">here</a> I get:
$$i_t-i^*_t=-\frac{e_{t}-e^E_{t+1}}{e^E_{t+1}}$$</p>

<p>However, if we define the $e=\frac{\text{Foreign currency}}{\text{Domestic currency}}$, we get $1+i_t=e_t\left(1+i^*_t\right)\left(e^E_{t+1}\right)^{-1}$, and using an analogous reasoning to the above:
$$i_t-i^*_t=-\frac{e^E_{t+1}-e_{t}}{e_{t}}$$</p>

<p>However, this <a href="https://en.wikipedia.org/wiki/Interest_rate_parity#Approximation" rel="nofollow">wiki link</a> gives(notice the missing minus sign):</p>

<p>$$i_t-i^*_t=\frac{e^E_{t+1}-e_{t}}{e_{t}}$$</p>

<p>It's not the first time I see this last(wiki) approximation being used.</p>

<p>What's the reasoning for this last approximation? are my approximations wrong? I would really like to know the reason for the minus signs to be missing.</p>

<p>Any help would be appreciated. </p>

CC BY-SA 3.0
2015-12-19T22:08:37.400Z
@denesp the problem is the extra minus sign
An old man in the sea.
<macroeconomics><interest-rate><exchange-rates>
Uncovered Interest Parity (UIP) condition approximation
An old man in the sea.
# Uncovered Interest Parity (UIP) condition approximation

(Asked by: username_0 on 2015-12-19)

Until otherwise $e=\frac{\text{Domestic currency}}{\text{Foreign currency}}$.

The UIP condition can be written in the following form: Investing now in a domestic bond, must give the same return as investing in a foreign bond, which means, $1+i\_t=e^{-1}\_t\left(1+i^\*\_t\right)e^E\_{t+1}$, implying

$$\frac{1+i\_t}{1+i^\*\_t}=\frac{e^E\_{t+1}}{e\_{t}}=\frac{1}{1+\frac{e\_{t}-e^E\_{t+1}}{e^E\_{t+1}}}$$ where $i\_t$ is the domestic nominal interest rate at time t, $i^\*\_t$ is the foreign nominal interest rate at time t, $e^E\_{t+1}$ is the expected exchange rate at time $t+1$.

Using the derivations presented [here](https://en.wikipedia.org/wiki/Fisher_equation#Derivation) I get:
$$i\_t-i^\*\_t=-\frac{e\_{t}-e^E\_{t+1}}{e^E\_{t+1}}$$

However, if we define the $e=\frac{\text{Foreign currency}}{\text{Domestic currency}}$, we get $1+i\_t=e\_t\left(1+i^\*\_t\right)\left(e^E\_{t+1}\right)^{-1}$, and using an analogous reasoning to the above:
$$i\_t-i^\*\_t=-\frac{e^E\_{t+1}-e\_{t}}{e\_{t}}$$

However, this [wiki link](https://en.wikipedia.org/wiki/Interest_rate_parity#Approximation) gives(notice the missing minus sign):

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

It's not the first time I see this last(wiki) approximation being used.

What's the reasoning for this last approximation? are my approximations wrong? I would really like to know the reason for the minus signs to be missing.

Any help would be appreciated.

**Tags:** macroeconomics, interest-rate, exchange-rates
> **Comments:**
> **username_0**: @denesp the problem is the extra minus sign  

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## 1 answers

### Answer by username_1 on 2015-12-20. Score: 1 | Accepted answer

From the basic equation you get
\begin{eqnarray\*}
\frac{e^E\_{t+1}}{e\_{t}} & = & \frac{1+i\_t}{1+i^\*\_t} \\
\\
(1+i^\*\_t) \cdot e^E\_{t+1} & = & (1+i\_t) \cdot e\_t \\
\\
e^E\_{t+1} - e\_t & = & i\_t \cdot e\_t - i^\*\_t \cdot e^E\_{t+1}.
\end{eqnarray\*}
Now comes the approximation. You divide either by $e\_t$ or by $e\_{t+1}^E$.

First method
$$
\frac{e^E\_{t+1} - e\_t}{e\_t} = i\_t - i^\*\_t \cdot \frac{e^E\_{t+1}}{e\_t} \approx
i\_t - i^\*\_t.
$$
Second method
$$
\frac{e^E\_{t+1} - e\_t}{e^E\_{t+1}} = i\_t \cdot \frac{e\_t}{e^E\_{t+1}} - i^\*\_t \approx
i\_t - i^\*\_t.
$$
I would say there is no crucial difference, both of these just assume that the proportional change in the exchange rate multiplied by the interest rate is an order of magnitude lower than either the interest rate or the change in the exchange rate alone.


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