1
$\begingroup$

Higher interest rates offer lenders in an economy a higher return relative to other countries. Therefore, higher interest rates attract foreign capital and cause the exchange rate to rise.

Does to rise means the value of currency decreasing?

$\endgroup$
3
$\begingroup$

This is a constant source of confusion in international macreconomics. There are two ways to define the exchange rate.

$$E = \frac {\text {home}}{\text{foreign}}$$

Here, we express "units of home currency per unit of foreign currency" (or basket of foreign currencies). In this definition $$E \uparrow \Rightarrow \text {home currency depreciates}= \text {value} \downarrow$$ so it can be confusing at times, since usually the object of study is a specific "home" currency.

The other way is the reciprocal

$$S = 1/E = \frac {\text{foreign}}{\text {home}}$$

Here we measure "foreign currency units per unit of home currency", and so $$S \uparrow \Rightarrow \text {home currency appreciates} = \text {value} \uparrow$$

Some argue that this second expression is more "natural" since it expresses the "price" of home currency in terms of foreign currency, i.e. "what is the price of 1 euro ? 1.22 USD", "what is the price of 1 USD? 0.82 euro".

But not even the notation is standard (for example, the wikipedia article uses the symbol $S$ to denote the exchange rate that we denoted by $E$).

One way to avoid confusion is to write the various relations with subscripts, and use just one symbol for the exchange rate. Using $S$ as this symbol, the subscript $f$ for "foreign" and the subscript $h$ for "home" we have

$$S_{h/f} = \frac {\text {home}}{\text{foreign}},\;\; S_{f/h} = \frac {\text{foreign}}{\text {home}}$$

For the specific question, since interest rates are in focus, it appears that the underlying law is the Uncovered Interest Rate Parity (UIRP), where

$$1+i_h = (1+i_f)\left(\frac {S_{h/f, t+1}}{S_{h/f,t}}\right)$$

$$\Rightarrow_{\text{approx}}\;\; i_h = i_f + \Delta s_{h/f,t+1}$$

or

$$i_h= i_f - \Delta s_{f/h,t+1}$$

where $s$ denotes the natural logarithm, and $\Delta$ is the first-difference operator. By looking at the UIRP, we see that if the home interest rate is higher than the foreign interest rate, $i_h > i_f$, it must be the case that $\Delta s_{h/f,t+1} >0$ to satisfy the relation. $\Delta s_{h/f,t+1} >0$ is expected percentage change in the "home per foreign unit" exchange rate. So we say that the home currency is expected to depreciate...

But hey, intuition appears to say otherwise: Assume $i_h$ rises above $i_f$. Then foreign investors have a motive to deposit their funds in the home economy, since the interest is higher. But in order to do so, they increase the demand for the home currency, since they need to convert their funds in home currency units. Assuming constant home money supply, this leads to an increase in the value of home currency compared to other currencies. But it does that now so mathematically it decreases the current exchange rate, $S_{h/f,t}$:

$$i_h > i_f \Rightarrow S_{h/f,t} < S_{h/f,t+1} \Rightarrow \Delta s_{h/f,t+1} > 0$$

In other words a higher home interest rate increases the value of the home currency now, creating expectations for home currency depreciation in the future.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Is it really a constant source of conduction, or isn't it rather rising? I mean it decreases. $\endgroup$ – FooBar Dec 19 '14 at 15:20
  • $\begingroup$ @FooBar Sorry, I have difficulty understanding your comment/question. Can you re-phrase please? $\endgroup$ – Alecos Papadopoulos Dec 19 '14 at 19:59
  • $\begingroup$ Meh, it was supposed to be a joke (and the typo didn't help). I guess the following formulation would have made that more clearly: "Does to rise means the value of confusion decreasing?" $\endgroup$ – FooBar Dec 20 '14 at 0:48
  • 1
    $\begingroup$ @FooBar Well, not being a native English speaker, I guess I don't dare to be certain about when a word that I don't recognize is a humorous typo, or a word I just don't know! $\endgroup$ – Alecos Papadopoulos Dec 20 '14 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy